1. NUMBER THEORY 1.0 INTRODUCTION Since vedic period Indians used numbers and fractions. They also used irrational numbers. Greeks did not regard irrational numbers as numbers at all. But Indians treated all numbers alike and this served them well in the invention of zero, creation of negative numbers and development of the concept of infinity. Empty space was created and was granted a symbol 0 and was given a name ‘Sunya’. The most significant achievement of Indians was the creation of decimal system. The numbers 1 to 9 had been used by Indians even before the time of the Emperor Asoka. Brahmagupta (598-665 AD) was the first to introduce negative numbers. He applied negative numbers to represent debts. He gave the rules for 4 basic operations of +, –, ×, ÷ . In 766 AD Indian numerals 0, 1, 2, ..., 9 were carried by Arab mathematicians to Baghdad. Indian numerical system which was much superior to the complicated Roman numerical system was readily adopted by the European traders ignoring the orders of the Roman Emperor. Srinivasa Ramanujan the most celebrated Indian Mathematical genius made a significant contribution to man’s knowledge of Mathematics, specially in the field of number theory which has been unique and unparalleled in the world. His famous note books contain mathematical results and theorems that can fascinate and stimulate not only research mathematicians but also school students. Ramanujan’s jottings in his note books cover Bernoulli numbers, continued fractions, infinite series including divergent series, analytical theory of numbers, expression for numbers and so on. These jottings began with magic squares, his first passion begun in his school days.
1.1 SEQUENCES In earlier classes, you might have come across various patterns of numbers like 1, 2, 3, 4, 5, ... 1, 3, 5, 7, 9, ... 1, 8, 27, 64, 125, ... These patterns are generally known as sequences. An arrangement of numbers of which one number is designated as the first, another as second, another as third and so on is known as a sequence. Consider the set of numbers 2,3,5,8, ... Here we find the numbers arranged according to some specific rule and this helps us to find out other numbers that follow. Such an arrangement is called a sequence. Thus we may define a sequence formally as follows: If for every positive integer n there is associated only one number an, according to some rule, then the ordered set of numbers a1, a2, a3, … an is said to define a sequence. The various numbers occurring in a sequence are called its terms. an the nth term is also called the general 1
term of the sequence. A sequence can be described either by listing its first few terms till the rule for writing down the other terms becomes clear or, by writing the algebraic formula for the nth term of the sequence. For example, the sequence of odd natural numbers 1, 3, 5, 7, 9, ... can be described as an = 2n – 1 where n = 1, 2, 3, 4,... The sequence 1, 8, 27, 64, 125,... can be described as an = n3 where n = 1, 2, 3, 4, ... (−1) n find the sequence 2n (−1)1 1 (−1) 2 1 Solution: a1 = 1 = − , a 2 = = , 2 2 4 2 2 1 1 1 1 1 ∴ sequence is − , , − , , − ,... 2 4 8 16 32
Example 1: If
an =
a3 =
(−1)3 1 =− 3 8 2
Example 2: Let a sequence be defined by a1 = 1, a2 = 1, an = an–1 + an–2 for n > 2. Find the sequence. Solution: a1 = 1, a2 = 1 an = an–1 + an–2 for n > 2 a3 = a2 + a1 = 1 + 1 = 2 a4 = a3 + a2 = 2 + 1 = 3 a5 = a4 + a3 = 3 + 2 = 5 ∴ The sequence is : 1, 1, 2, 3, 5,... Do it yourself
I.
Write the first four terms of the sequences whose general terms are given below: 1 + (−1) n 3n − 1 2) 3) 4) 2n2 – 3n+1 5) (–1)n 2n 1) n3 – 1 n 5
II. Find the indicated term in each of the following sequences. n +2 2) a7 if an 1) a12, a15 if an = 5n – 4, 2n + 3 n(n + 1) 3) a3 if an = 4) a10 if an = 5 + 2 (n – 1) 2
5) a5 if an = (–1)n n
1.1.1 Arithmetic Progression (A.P) In this section, we shall discuss a particular type of sequences in which each term, except the first, progresses in a definite manner. For example in the sequence 2, 5, 8, 11, 14... every term except the first is obtained by adding 3 to the preceding term. Such sequences are called Arithmetic progression. An Arithmetic progression is a sequence of numbers in which each term except the first is obtained by adding a fixed number to the immediately preceding term. This fixed number is called the common difference.
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For example: 1, 2, 3, 4,... is an A.P with C.D. = 1 5, 7, 9, 11,... is an A.P with C.D. = 2 1 1 3 1 , , , 1 ,... is an A.P with C.D.= 4 2 4 4 102, 97, 92, 87,... is an A.P with C.D = –5
General form of an A.P. is a, a + d, a + 2d,... with first term a, and C.D. = d The general term or the nth term of an A.P. is t n = a+ (n -1)d
Properties of an A.P. 1.
An A.P. remains an A.P if a constant quantity is added to or subtracted from each term of the A.P. For example: 9, 13, 17, 21, 25,... is an A.P with C.D = 4. Add 3 to each term of the given A.P. The resulting sequence 12, 16, 20, 24, 28, ...is also an A.P with C.D = 4. Subtract 2 from each term of the given A.P. The resulting sequence 7, 11, 15, 19, 23,... is also an A.P with C.D = 4.
2.
An A.P remains an A.P. if each term of the A.P is multiplied or divided by a nonzero constant quantity. For example : 2, 4, 6, 8,... is an A.P with C.D = 2 Multiply the given A.P by 5 The resulting sequence 10, 20, 30, 40, ... is also an A.P. with C.D. 10 Divide the given A.P by 2 The resulting sequence 1, 2, 3, 4,… is also an A.P with C.D = 1
Example 3: Is the sequence 10, 4, –2, –8, … an A.P.? Solution: In the given sequence we find 4 – 10 = –2 – 4 = –8 – (–2) = – 6 The common difference is –6. Hence the given sequence is an A.P. Example 4: Is the sequence described by an = 2n2 + 1 an A.P.? Solution: an = 2n2 + 1 a2 = 2(2)2 + 1 = 9 a1 = 2(1)2 + 1 = 3, a3 = 2(3)2 + 1 = 19, a4 = 2(4)2 + 1 = 33
The sequence is 3, 9, 19, 33, ... Here,
9– 3=6 19 – 9 = 10 33 – 19 = 14 The difference is not the same. ∴The given sequence is not an A.P.
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Example 5: Find the common difference and the next three terms of the A.P. 1, 4, 7, ... Solution: The common difference = 4 – 1 = 3 The next three terms are 7 + 3 = 10, 10 + 3 = 13 13 + 3 = 16 Example 6: Find the 12th term of an A.P. 6, 1, –4... Solution: Consider the A.P in the form a, a + d, a + 2d, ... Here, a = 6, d = 1 – 6 = –5, n = 12 = a + (n–1) d tn t12 = 6 + (12 – 1) (–5) = 6 + 11 x (–5) = 6 – 55 = – 49 ∴ The 12th term is –49 Example 7: The 7th term of an A.P is –15 and the 16th term is 30. Find the A.P. Solution: Consider the A.P in the form a, a + d, a + 2d,... = a + 6d = –15 t7 t16 = a + 15d = 30 t16–t7 ⇒ 9d = 45, d = 5
Substituting d = 5 in t7 we get a + 30 = –15, a = –45 ∴ The A.P is –45, –40, –35... Example 8: Write down the A.P. and its general term if a = 3, d = 7. Solution: Consider the A.P in the form a, a + d, a + 2d. ∴ The A.P is 3, 3 + 7, 3 + 14, … or 3, 10, 17… General term tn = a (n – 1)d = 3 + (n – 1) 7 = 7n – 4 Example 9: The nth term of a sequence is 7n – 3. Show that it is an A.P and find the first term and the common difference. Solution: tn = 7n – 3 t1 = 7–3=4 = 14 – 3 = 11 t2 t3 = 21 – 3 = 18 = 28 – 3 = 25 t4 ∴ The sequence is 4, 11, 18, 25, ... First term is 4, Common difference = 11 – 4 = 18 – 11 = 25 – 18 = 7 ∴ The given sequence is an A.P with C.D = 7 Example 10: If an office clerk is fixed in the pay scale 3200 – 85 – 4900, when will he reach his maximum? Solution: Pay Scale : 3200 – 85 – 4900 Starting salary = Rs.3200 = a, Annual increment = Rs. 85 = d
Maximum salary = Rs.4900 = tn tn = a + (n – 1) d ⇒ 4900 = 3200 + (n–1) 85 1700 = 20, n = 20 + 1= 21 n −1 = 85 The clerk will reach his maximum in his 21st year of service. 4
Example 11: Find 4 numbers between 3 and 38 which are in an A.P. Solution: Consider the A.P in the form a, a + d, a + 2d, ... Here a = 3, and a + 5d = 38 ⇒ 5d = 35, ⇒ d = 7 ∴ The A.P. is 3, 10, 17, 24, 31, 38... ∴ The 4 numbers between 3 and 38 are 10, 17, 24, 31 Example 12: If five times the fifth term of an A.P is equal to 8 times its eighth term, show that its 13th term is zero. Solution: Given: 5t5 = 8t8 5(a + 4d) = 8 (a + 7d) 3a + 36d = 0 ⇒ a + 12d = 0 ∴ t13 = a + 12d = 0 Example 13: Divide 20 into 4 parts which are in A.P such that the product of the first and fourth is to the product of the second and third in the ratio 2: 3 Solution: Let 20 be divided into 4 parts a – 3d, a – d, a + d, a + 3d which are in A.P. ⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20 ⇒ 4a = 20 ⇒ a =5 Product of the first and fourth part (a − 3d) (a + 3d) 2 = = Product of the second and third part (a − d) (a + d) 3
a 2 − 9d 2 2 = 3 a 2 − d2 2 ⇒ 3(25 – 9d ) = 2(25 – d2) ⇒ 25d2 = 25 ⇒ d = +1 When a = 5, d = 1, the 4 parts are 2, 4, 6, 8 When a = 5, d = –1, the 4 parts are 8, 6, 4, 2
⇒
Example 14: The sum of 3 numbers in an A.P is 21 and their product is 280. Find the numbers. Solution: Assume that the 3 numbers in A.P are in the form a – d, a, a + d Sum of the numbers = a – d + a + a + d = 21 ⇒ 3a = 21 ⇒ a = 7 Product of the numbers = (a – d) a (a + d) = 280 ⇒ (a2 – d2) a = 280 2 2 2 (49 –d ) 7 = 280, 49 – d = 40, d = 9 ⇒ d = +3 ∴ The numbers are a – d = 7 – 3 = 4, a = 7, a + d = 10 ∴ The required numbers are 4, 7, 10 Note : Taking d = –3, we get the same set of numbers. Example 15 : The angles of a triangle are in A.P. If its greatest angle equals the sum of the other two, find the angles. Solution: Let the angles of a triangle be a – d, a, a + d
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⇒ a – d + a + a + d = 180o ⇒ 3a = 180o ⇒ a = 60o Again, a + d = a – d + a ⇒ 2d = a, 2d = 60o ∴ d = 30o a – d = 60 – 30 = 30o, a = 60o, a + d = 60 + 30 = 90o ∴ The angles of the triangle are 30o, 60o, 90o Example 16: Find the number of integers between 60 and 600 which are divisible by 9. Solution: The first number divisible by 9 between 60 and 600 is 63. The last number divisible by 9 which is less than 600 is 594. The sequence 63, 72, 81, ... 594 is an A.P. Here, a = 63, d = 72 – 63 = 9 tn = 594 ⇒ a + (n –1)d = 594 ⇒ 63 + (n–1) 9 = 594 ⇒ (n–1) 9 = 594 – 63 = 531 ⇒ n – 1 = 59 ⇒ n = 60 ∴ There are 60 integers between 60 and 600 which are divisible by 9. Example 17: If a, b, c are in A.P then prove that (a–c)2 = 4(b2 – ac) Solution: a, b, c are in A.P ⇒ b – a = c – b = common difference ⇒ 2b = a + c ⇒ 4b2 = a2 + 2ac + c2 ⇒ 4b2 – 4ac = a2 – 2ac + c2 ⇒ 4(b2 – ac) = (a – c)2 Example 18: If a2, b2, c2 are in A.P show that
1 1 1 , , are also in A.P. b+c c +a a +b
Solution: a2, b2, c2 are in A.P
⇒ b2 – a2 = c2 – b2
= common difference
⇒ (b – a) (b + a)
= (c – b) (c + b)
⇒ (b + c – c – a) (b + a)
= (c + a – a – b) (c + b)
⇒ [(b + c) – (c + a)] (b + a) = [(c + a) – (a + b)] (c + b) dividing both sides by (b +c) (c + a) (a + b), we get 1 1 1 1 1 1 1 are in A.P − = − ⇒ , , c+a b+c a +b c+a b+c c+a a +b Example 19: If ax = by = cz and b2 = ac show that
1 1 1 , , are in A.P x y z
Solution: Let ax = by = cz = k (say) 1
1
1
⇒ a = kx , b = ky , c =kz 2
1 1 ⎛ 1y ⎞ x Given b = ac ⇒ ⎜ k ⎟ = k .k z ⇒ ⎜ ⎟ ⎝ ⎠
2
2 y
k =k
6
1 1 + x z
⇒
2 1 1 1 1 1 = + ⇒ , , are in A.P. y x z x y z
Exercise 1.1.1.
1.
2.
3.
4.
5.
6.
7.
8. 9. 10. 11. 12. 13.
Which of the following sequences are A.P.? a) 11/3, 13/3, 15/3, 17/3, ... b) 0, –3, –6, –9, ... c) 12, 32, 52, 72, … d) 5, 5, 5, 5, ... e) a, a + 2, a + 4, a + 6, ... Find the common difference of the A.P. : a) 3, 1, –1, –3, ... b) 1.0, 1.7, 2.4, 3.1, ... c) 0, 1/4, 1/2, 3/4, ... d) 7, 4, 1, –2, ... e) tn = 4n + 5 Write the next three terms of the following sequences: a) 14, 11, 8, ... b) 6, 4.5, 3, ... c) 22, 29, 36, ... d) 1, 1½, 2, ... e) –1, –5/6, –2/3, ... Write down the A.P if a) a = –5, d = 6 b) a = 3½, d = 1½ c) a = p, d = q d) a = 0.7, d = 0.02 e) tn = 3n – 2 Find the required term in the following a) 15th term of 40, 43, 46, ... b) 10th term of 10, 10.5, 11, ... c) 9th term if a = 3/5, d = 2/5 d) 20th term if a = 18, d = –4 e) 7th term if tn = 4n + 5 a) In an A.P., t7 = 45, t9 = 57. Find the first three terms and the common difference. b) Find the 30th term of an A.P whose third term is 14 and the 9th term is –52. c) Find the middle term of an A.P with 21 terms if a = –3, d = 3 d) In an A.P., 24th term is twice the 10th term and the sixth term is 10. Write the first three terms of the A.P. e) If 10 times the 10th term of an A.P is equal to 15 times the 15th term show that 25th term of the A.P is zero. a) Which term of the sequence 21, 42, 63, ... is 420? b) If a = 5, d = 3 which term of the A.P is 320? c) How many terms are there in the A.P –1, –5/6, –2/3, ... 10/3? d) Is 68 a term of the A.P. 7, 10, 13, ...? e) How many numbers of two digits are divisible by 6? The sum of three numbers in an A.P is 9 and their product is –48. Find the numbers. Find four numbers in A.P whose sum is 50 and in which the greatest number is 4 times the least. The angles of a quadrilateral are in A.P whose common difference is 10o. Find the angles. The sum of three numbers in an A.P is 12 and the sum of their squares is 56. Find the numbers. b+c−a c +a −b a +b − c 1 1 1 , , If are in A.P show that , , are in A.P. a b c a b c –1 –1 –1 If a, b, c are in A.P show that (ab) , (bc) , (ca) are also in A.P
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1.1.2 Geometric progression
Observe the following sequences a) 2, 4, 8, 16, ... b) 100, 20, 4, 4/5
c) 3, 32, 33, 34, ...
d) 4, –2, 1, –1/2, ...
These are not A.P. But they have definite pattern. t t t t t t 1 In a) We find 2 = 3 = 4 = 2 , In b) We find 2 = 3 = 4 = t1 t2 t3 t1 t2 t3 5 t t t2 t 1 In d ) We find 2 = 3 = In c) We find = 3 =3, t1 t2 t1 t2 2 It means each term of the sequence except the first is obtained by multiplying the preceding term by a constant factor. Such a sequence is called Geometric progression. The constant factor is called common ratio (C.R) The general form of a G.P is a, ar, ar2, ar3, ... with a ≠ 0 th
C.R = r ≠ 0
n–1
The n term of the G.P is tn = ar Note:
If each term of a G.P be multiplied or divided by the same non zero number, the resulting series is also a G.P.
Example 20: Find the 5th term of the G.P 64, 16, 4... 16 1 Solution: a = 64, r = = , n=5 64 4 tn = arn–1, t5 = ar5–1 = ar4 4
64 1 ⎛1⎞ = t5 = 64 ⎜ ⎟ = 256 4 ⎝4⎠ 1 ∴ 5th term of the given G.P. is 4 Example 21: The sixth and the tenth term of a G.P are 63 and 5103 respectively. Find the G.P. Solution: t6 = 63, t10 = 5103 t10 ar 9 5103 = 5 = ⇒ r 4 = 81 ∴ r = ± 3 t6 63 ar
∴
Substituting r = 3 in t6, we get 63 7 a(3)5 = 63 ⇒ a = = 243 27 −7 If r = –3, then we get a 27 7 21 63 −7 21 −63 , , , .... (or) , , ,... The G.P is 27 27 27 27 27 27
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Example 22: Find three numbers in G.P whose sum is 14 and product is 64. Solution: Let the numbers be a/r, a, ar Product of the numbers = a/r × a × ar = 64 ⇒ a3 = 64, ∴a = 4 ⎛1 ⎞ Sum of the numbers = a/r + a + ar = 14, a ⎜ + 1 + r ⎟ = 14 ⎝r ⎠ 2 ⎛1+ r + r ⎞ 2 4⎜ ⎟ = 14 ⇒ 2(1 + r + r ) = 7r r ⎝ ⎠
⇒ 2r2 – 5r + 2 = 0, ∴ r = ½ or 2 If r = 2, the numbers are 2, 4, 8. If r = 1/2 the numbers are 8, 4, 2 Example 23: Find three numbers in G.P such that their sum is 7 and the sum of the reciprocals is 7/4. Solution: Let the three numbers in G.P be a, ar, ar2 Sum of the numbers = a + ar + ar2 = 7, a (1 + r + r2) = 7 (1)
1 1 1 7 1 + r + r2 + + 2 = = (2) a ar ar 4 ar 2 Dividing (1) and (2) we get (ar)2 = 4, ar = + 2 ⇒ a = + 2/r Substituting a = 2/r in (1) we get 2/r (1 + r + r2) = 7 ⇒ 2 (1 + r + r2) = 7r ⇒ 2r2 – 5r + 2 = 0 ⇒ r = 1/2 or 2 If r = 1/2 then a = 4. ∴ The numbers are 4, 2, 1,... If r = 2 then a = 1 ∴ The numbers are 1, 2, 4 22n −1 Example 24: The nth term of a G.P is for all values of n. Write down the first three 3 terms and also the 10th term. 22n −1 22(1) −1 2 , t1 = Solution: t n = = 3 3 3 4 −1 6 −1 2 8 2 32 t2 = = , t3 = = 3 3 3 3 8/3 2 a= , r= =4 ∴ 3 2/3 2 8 32 The first three terms are , , 3 3 3 2 2 18 219 t10 = ar 9 = (4)9 = (2) = 3 3 3 Sum of the reciprocals
=
Example 25: The sum of the first two terms of a G.P is 2 and the sum of the first four terms is 20. Find the G.P. Solution: Consider the GP a, ar, ar2 Given : a + ar = 2, a(1 + r) = 2 (1) a + ar + ar2 + ar3 = 20 ⇒ a (1 + r) (1 + r2) = 20 (2)
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By virtue of (1), (2) becomes 2(1 + r2) = 20 ⇒ 1 + r2 = 10 ⇒ r2 = 9 ⇒ r = + 3 Substituting r = + 3 in (1) If r = 3 then a = 1/2 If r = – 3 then a = –1 1 3 9 27 ∴ The G.P is , , , ,...(or) –1, 3, –9, 27, ... 2 2 2 2 Exercise 1.1.2 1. Find which of the following are not a G.P. c) 1, –1, 1, –1, ... d) 5, 55, 555, 5555,... a) 2, 4, 8, 16,... b) 12, 22, 32, 42, ... 2. Find the common ratio of the following G.P. a) 25, –5, 1, –1/5, ... b) 1, –5/2, 25/4, –125/8, … c) 0.2, 0.06, 0.018, 0.0054 d) 2/5, 6/25, 18/125, 54/625, … b) Find the 8th term of the G.P. 3,6, 12,… 3. a) In a G.P 1, ½, ¼, … find t7 c) Find the 7th term of a G.P 9, 3, 1, … d) Find the 9th term of a G.P. 2/5, 8/25, 32/125,… 4. a The third and fifth term of a G.P are 4/3 and 16/27 respectively. Find the G.P. b) In a G.P t3 = 16, t7 = 1 find the G.P c) The 5th term of a G.P is 4/9 and the seventh term is 16/81. Find the 10th term. d) The fourth term is 27 and the 7th term is 729. Find the first term and the common ratio. 5.
6. 7. 8. 9. 10. 11. 12. 13. 14.
a) Find the three terms in G.P whose sum is 6
1 and the product is 8. 3
b) The sum of three numbers of a G.P is 26. Their product is 216. Find the numbers. c) The product of 3 numbers in G.P is 216. The sum of their product taken in pairs is 156. Find the numbers. d) The sum of the first two terms of a G.P is –1 and the sum of the first four terms is –5. Find the G.P. Find three numbers in G.P such that their sum is 19/3 and the sum of their reciprocals is 19/12. The sum of the first three terms of a G.P is 7 and the sum of their squares is 21. Find the first five terms of the G.P. The first term of a G.P is 64 and the common ratio is r. Find the value of r if average of the first and the fourth term is 140. The second term of G.P is b and the common ratio is r. Write down the value of b if the product of the first three terms is 64. Find the number of terms of the G.P 1, 4, 16, … 4096. Find three numbers a, b, c between 2 and 18 such that their sum is 25, the numbers 2, a, b are in A.P and the numbers b, c, 18 are in G.P. In a set of 4 numbers the first three are in G.P and the last three are in A.P with common difference 6. If the first number is the same as the fourth, find the four numbers. If a, b, c, d are in G.P., prove that a + b, b + c, c + d are in G.P. If a1, a2, a3, … are in G.P (ai > 0) then show that log a1, log a2, log a3, …. are in A.P.
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1.2 SERIES When the terms of a sequence are connected by the sign +, it is called a series. Thus a1 + a2 + a3 + … an + … is an infinite series. The symbol Σ an is used to denote a series. 1.2.1 Sum to n terms of an A.P
Let Sn denote the sum of the terms of the A.P a, a + d, a + 2d, …, a+(n–1) d Sn = a + (a + d) + (a + 2d) + … + [a + (n–1)d] Writing this series in the reverse order Sn = [a + (n – 1)d] + [a + (n – 2)d] + …. + a Adding (1) and (2) 2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + …. + [2a + (n – 1)d] = n [2a + (n – 1) d] n ∴ Sn = [2a + (n-1) d] 2 n n Sn = [a + a+(n-1) d] , Sn = 2 [a + l] 2
(1) (2)
where l = tn = a + (n – 1) d = last term Example 26: Find the sum of the first 11 terms of the A.P 3, 8, 13… Solution: Given A.P is 3, 8, 13, …. n Here a = 3, d = 8 – 3 = 5, n = 11, Sn = [2a + (n − 1)d] 2 11 11 11 = [(2 × 3) + (11 − 1)5] = [6 + 50] = × 56 = 308 2 2 2 ∴ The sum of the first 11 terms of the given A.P is 308. Example 27: Find the sum: 3 + 11 + 19 + … + 787. Solution: The given series is an A.P. Here a = 3, d = 8, tn = 787 = l tn = a + (n – 1) d = 787 3 + (n – 1) 8 = 787 787 − 3 + 1 = 99 ∴ n = 8 n 99 99 Sn = 2 [a + l] = [3 + 787] = x 790 = 39105 2 2 Hence the sum of the given series is 39105. Example 28: Find the sum of all the numbers between 300 and 500 divisible by 11. Solution: The first number greater than 300 and divisible by 11 is 308. The last number less than 500 and divisible by 11 is 495. ∴ Series is 308 + 319 + … + 495 a = 308, d = 11, l = 495, tn = a + (n – 1)d = 495
11
308 + (n – 1) 11
= 495
∴n
=
495 − 308 + 1 = 18 11
n ∴ Sn = 2 [a + l] 18 S18 = [308 + 495] = 7227 2 Example 29: Find the sum to n terms of an A.P whose nth term is an = 5 – 6 n Solution: General term of the given A.P is an = 5 – 6 n; a1 = 5 – 6(1) = –1 a2 = 5 – 6(2) = –7; a3 = 5 – 6(3) = –13 ∴ The A.P is –1, –7, – 13 … with C.D. = –6 a = –1, l = 5 – 6 n n n Sn = 2 [a + l] = [4 − 6n] = n[2–3n] = 2n – 3n2 2 Example 30: In an A.P the sum of the first 7 terms is 10 and that of the next 7 terms is 17. Find the A.P. Solution: Given: S7 = 10, S14 = 10 + 17 = 27 7 10 = 10 ⇒ [2a + 6d] = 10 ⇒ a + 3d = S7 (1) 2 7 14 27 [2a + 13d] = 27 ⇒ 2a + 13d = S14 = 27 ⇒ (2) 2 7 Solving (1) and (2) we get a = 1, d = 1/7 ∴The A.P is 1, 8/7, 9/7, 10/7, …. Example 31: How many terms of the A.P 3, 7, 11, … are needed to yield the sum 1275? n [2a + (n − 1)d] = 1275 Solution: a = 3, d = 4, Sn = 1275, 2 n [6 + (n − 1)4] = 1275 , n [3 + (n–1)2] = 1275 2 ⇒ 2n2 + n – 1275 = 0 ⇒ (2n + 51) (n – 25) = 0 ⇒ n = 25 ∴ 25 terms of the A.P will yield the sum 1275. Example 32: i) If a clock strikes appropriate number of times at each hour how many times will it strike in a day? (ii) If it strikes the half hour also, how many times does it strike in a day? Solution: i) Number of times the clock strikes at each hour form on A.P. The A.P is 1,2,3 … 12 12 n Sn = 2 [a + l] = [1 +12] = 78 2
If the clock strikes appropriate number of times at each hour, total number of times the clock strikes in a day = 2 × 78 = 156 ii) If it strikes half hour also then the total number of times it strikes in a day = 156 + 24 = 180. 12
Example 33: A machine costs Rs.5,00,000/–. If the value depreciates 15% the first year, 13½% the next year 12% the third year and so on. What will be its value at the end of 10 years, all percentages applying to the original cost. Solution: The percentage of depreciation in value in consecutive years form an A.P. ∴ Total depreciation (in %) = 15 + 13½ + 12 + … to 10 terms, Here a = 15, d = – 1.5 n Sn = [2a + (n − 1)d] 2 10 S10 = [30 − 13.5] in % = 82.5% 2 Value of the machine after 10 years = 100 – 82.5 = 17.5% Original cost of the machine = Rs.5,00,000
∴Value of the machine after 10 years = Rs.5,00,000 ×
17.5 = Rs.87,500 100
Example 34: In a school sports day, picking balls kept in a line was one of the games. 20 balls were placed in a straight line on the ground at intervals of 3 meters. The starting point was at a distance 3 meters from the first ball in line with the balls. How far a boy would have to run to bring the balls one by one to a basket kept at his starting point? Solution: The distances run by the boy to pick each ball form an A.P. Each distance is run twice. The A.P is 3, 6, 9, … Total distance run by the boy = 2(3 + 6 + 9 … to 20 terms) n 20 = 2. Sn = 2 × [2a + (n − 1)d] = 2 × [2(3) + (20 − 1) 3] = 20 × 63 = 1260 m. 2 2 The boy has to run 1260 m to bring all balls to the starting point. Example 35: Show that the sum of an A.P whose first term is a, second term is b and the last
term is c is equal to
(a + c) (b + c − 2a) 2(b − a)
Solution: t1 = a, t2 = b ⇒ C.D = b – a, Last term l = c
⇒ a + (n – 1)d = c ⇒ a + (n – 1) (b – a) = c ⇒ n – 1 =
c−a l−a
c−a b + c − 2a +1 = b−a b−a n (b + c − 2a) n ∴ Sn = 2 [a + l] = [a + c] = (a + c) 2 2(b − a)
⇒ n =
Exercise 1.2.1
1.
Find the sum of the following:
1 1 3 +5 +7 +...+ to 23 terms 4 2 4
a) 17 + 19 + 21 + … to 30 terms
b) 3
c) 7 + 3 + (–1) + (–5) + … to 15 terms e) 2 + 3.5 + 5 + 6.5 + … + 38
d) 4 + 9 + 14 + … + 199
13
2.
a) Find the sum of all numbers between 200 and 400 divisible by 13. b) Find the sum of all multiples of 9 between 400 and 600 c) Find the sum of all numbers between 100 and 300 not divisible by 5 d) Find the sum of all multiples of 6 between 500 and 700. e) Find the sum of all odd numbers between 0 and 1000. 3. The sum of the first 25 terms of a series in A.P is 1175. The last term is 83. Find the first term and the common difference. 4. The first term of an A.P is 12. The sum of the first 15 terms is 390. Find the middle term. 5. The sum of the first 11 terms of an A.P is 44 and that of the next 11 terms is 55. Find the A.P. 6. The 24th term of an A.P is 47 and the sum of 24 terms is 576. Find the common difference and the sum of the first 12 terms. 7. How many terms of the sequence 18, 16, 14, … should be taken so that their sum is zero. 8. If the sum of certain number of terms of an A.P 25, 22, 19, … is 116. Find the last term. 9. Find the sum of 15 terms of an A.P if the nth term is 6–n. 10. 30 consecutive terms of the A.P 100 + 99 + 88 … are taken to get a sum 1155. At which term will you begin to get the given total. 11. A man saved Rs.16,500/– in ten years. Each year, after the first, he saved Rs.100 more than in the preceding year. Find his savings in the first year. 12. In boring a well 50 m deep, the cost is Rs.4/– for the first meter. The cost of each successive meter increases by Rs.2/–. What is the cost of boring the entire well?
1.2.2
Sum to n terms of a G.P
Let Sn denote the sum of n terms of the G.P a, ar, ar2, …. Sn = a + ar + ar2 + …. + arn–1 Multiplying both sides by r r. Sn = ar + ar2 + ar3 + … + arn (2) – (1) ⇒ r. Sn – Sn = arn – a, Sn (r – 1) = a(rn – 1)
Sn =
a (r n −1) if r > 1 or r −1
Sn =
(1) (2)
a(1 − r n ) if r < 1 or Sn = na if r = 1 1− r
If r < 1 say r = ½ then r2 = ¼, r3 = 1/8, …, rn = 1/2n is very small and rn Æ 0 when n is very large or , r < 1 ⇒ rn Æ 0 as n Æ ∞. So the sum of infinite geometric series is S∞ = Example 36: Sum the series to n terms Solution:
a=
1 1 ,r= 1 3
a(r n − 1) ∴ Sn = r −1
a . 1− r
1 + 1 + 3 + 3 + ... 3
= 3 >1 3 =
1 3
(
n
3 −1 3−1 14
)
n
3 −1 = 3− 3
Example 37: Find the sum to 8 terms of the G.P 2, 4, 8, … Solution: a = 2, r = 4/2 = 2 > 1, n = 8
∴ Sn =
a(r n − 1) 2(28 − 1) , S8 = = 2 (256 – 1) = 2 × 255 = 510 2 −1 r −1
Example 38: Find the sum to infinity of the series 54, 18, 6, 2, …. Solution: a = 54, r = 18/54 = 1/3 < 1
∴ S∞ =
a 54 3 = = 54 × = 81 1− r 1 − 1/ 3 2
Example 39: Find the sum to n terms of the series 3 + 33 + 333 + …. Solution: Sn = 3 + 33 + 333 + … n terms 3 = 3 (1 + 11 + 111 + … n terms) = (9 + 99 + 999 + ... to n terms) 9 3 = [(10 − 1) + (100 − 1) + (1000 − 1) + ... to n terms] 9 ⎤ 3 ⎡10(10n − 1) 30 3n 10 n = (10n − 1) − = (10n −1) − − n⎥ = ⎢ 9 ⎣ 9 81 9 27 3 ⎦
0.9 = 1.0
Example 40: Prove that Solution:
0.9 = 0.999.... = 0.9 + 0.09 + 0.009 + …. = 9 (0.1 + 0.01 + 0.001 + …) = 9[10–1 + 10–2 + ….] = 9 ×
10−1 1 = 9× =1 −1 9 1 − 10
Example 41: Express 0.241 as a fraction Solution: 0.241 = 0.2414141…. = 0.2 + 0.041 + 0.00041 + … 2 ⎡ 41 41 41 2 41/103 ⎤ = + ⎢ 3 + 5 + 7 + ...⎥ = + 10 ⎣10 10 1 − 1/102 10 10 ⎦ 2 41 198 + 41 + = = 239/990 = 10 990 990 Example 42: The first term of an infinite G.P is 6 and its sum is 8. Find the G.P. Solution: Consider the G.P a, ar, ar2, … a 6 a = 6, S∞ = 8 ⇒ = 8 ⇒ 1 – r = 6/8 = 3/4 ⇒ r = 1/4 =8⇒ 1− r 1− r ∴ The G.P is 6, 3/2, 3/8, 3/32,…. Example 43: Find the value of Solution:
1 2
1 4
3 3 3... 1 8
⎛1 1 1 ⎞ ⎜ + + + ... ⎟ ⎝2 4 8 ⎠
3 3 3... = 3 × 3 × 3 × .... = 3
15
Now
1 1 1 + + + ... is an infinite G.P. 2 4 8 ⎛1
S∞ = 1
1
a 1− r
⎞
⎜ + + + ... ⎟ 1 1 1 1/ 2 ∴ + + + ... = = 1 ⇒ 3⎝ 2 4 8 ⎠ = 31 = 3 ∴ The value of 2 4 8 1 − 1/ 2
3 3 3... = 3
Example 44: The sum of an infinite G.P is 4 and the sum of the cubes of the terms is 192. Find the common ratio. a Solution: a + ar + ar2 + … = 4 ⇒ =4 1− r a3 / (1–r)3 = 64 (1) 3 a (2) a3 + a3r3 + a3 r6 + … = 192; ⇒ = 129 1− r3
(2) (1 − r)3 192 (1 − r) 2 ⇒ = = 3 ⇒ = 3 ⇒ 2r 2 + 5r + 2 = 0 3 2 (1) 64 1− r 1+ r + r ⇒ (2r + 1) (r + 2) = 0 ⇒ r = − 1/ 2 or − 2 ∴ The common ratio of the G.P is –1/2 or –2. Example 45: A rubber ball dropped from a height of 50 m rebounds at every impact from the floor to a height half of that from which it has fallen. Find the total distance described by the time it comes to rest. Solution: Distance described in the first impact = 50 m = 2[1/2× 50] = 2 × 25 m Distance described in the 2nd impact = 2 × 25/2 m Distance described in the 3rd impact 25 25 ⎛ ⎞ ∴ Distance described by the time it comes to rest = 50 + 2 ⎜ 25 + + + ... ⎟ 2 4 ⎝ ⎠ ⎛ 25 ⎞ = 50 + 2 ⎜ ⎟ = 50 + 2 × 25 × 2 = 50 + 100 = 150 m ⎝ 1 − 1/ 2 ⎠ ∴ Distance travelled by the ball by the time it comes to rest is 150 m. Exercise 1.2.2
1.
2.
3.
Find the sum of the following G.P. a) 5 + 25 + 125 + … to 8 terms c) 4 + 3 + 2¼ + … to 7 terms
b) 1 + 0.1 + 0.01 + 0.001 + … to 10 terms d) 1 + 3 + 32 + … to 8 terms 2 2n-1 e) Sum of the first 10 terms if the nth terms of the G.P is 3 Find the sum to infinity of the following a) 36, 12, 4, … b) ¼, –3/16, 9/64, … c) (–2) + 1 + (–1/2), … d) 2/3 + 2/27 + 2/243 + … e) 8/5 + 1 + 5/8 + … f) 1 + (1 + x) + (1 + x + x2) + … Find the sum to n terms of the G.P a) 3 + (–6) + 12 + … b) 9 + 99 + 999 + … c) 0.7 + 0.97 + 0.977 + … e) 1 + 12 + 104 + 1006 + … d) 1 + 11 + 111 + …
16
4.
Find the least number of terms of the series 1 + 3 + 32 + … that must be taken to give a sum exceeding 1500.
5.
A ball is dropped from a height of 1 m. At every bounce it travels half the height it travelled with the previous flight. Find the total distance travelled by the ball.
6.
Express 0.736 into a fraction.
7.
Find the value of
8.
The sum of an infinite number of terms of a G.P is 15 and the sum of their squares is 45. Find the sequence.
3
9 3 9 3 9...
1.2.3. Summation of some special series Sum of the first n natural numbers :
1 + 2 + 3 + ... + n =
n
∑n 1
This is an A.P where a = 1, d = 1, l = n n
n n n(n + 1) ∴ Sn = 2 [a + l] = 2 [1 + n] ∴ Σ n = 2 1
Example 46: Find the sum of 1 + 2 + 3 + … + 30 n(n + 1) Solution: ∑ n = 2 30 30 (30 + 1) ∑1 n = 2 = 15 × 31 = 465
Example 47: Find the sum of 11 + 12 + 13 + … + 31 31 × 32 Solution: 1 + 2 + 3 + … + 31 = = 496 2 10 × 11 1 + 2 + … + 10 = = 55 2 ∴ 11 + 12 + 13 + … 31 = (1 + 2 + … + 31) – (1 + 2 + 3 … + 10) = 496 – 55 = 441 Sum of the first n odd numbers n
1 + 3 + 5 + … + (2n – 1) =
∑ (2n − 1) 1
This is an A.P with a = 1, d = 2, l = 2n − 1 n Sn = 2 [a + l],
n
Σ 1
n n (2n − 1) = 2 [1 + 2n − 1] = 2 × 2n = n2
17
∴ Sum of the first n odd natural numbers is n2.
2
l+1 ⎡l + 1⎤ Note: If l is the last odd number of the series then Sn = ⎢ 2 ⎥ since n = 2 ⎦ ⎣ Example 48: Find the sum of 11 + 13 + … +35 2 ⎛ 35 + 1 ⎞ 2 Solution: 1 + 3 + … + 35 = ⎜ ⎟ = 18 = 324 2 ⎝ ⎠
⎛ 9 +1 ⎞ = 52 = 25 1+3+…+ 9= ⎜ ⎟ ⎝ 2 ⎠ 2
∴ 11 + 13 + … + 35 = 324 – 25 = 299 Sum of the squares of the first n natural numbers n
∑n
2
= 12 + 22 + 32 + ... + n 2
1
This is neither an A.P nor a G.P. Look at the figure below : 3 +
1
+ 3
3
12
+
32
2
22
1
32
3
2
+
22
2 1 1
Total number of squares in the Fig. = (1 + 2 + 3) (2 × 3 + 1) = 3(12 + 22 + 32) 3(12 + 22 + 32) = (1 + 2 + 3) [(2 × 3) + 1] ∴ 12 + 22 + 32 = 1/3 (1 + 2 + 3) [(2 × 3) + 1] Extending this pattern of n terms 12 + 22 + 32 + … + n2 = 1/3 (1 + 2 + 3 + …+ n) [(2 × n) + 1] n(n + 1) n(n + 1)(2n + 1) = 1/3 (2n + 1) = 2 6
∴
n
∑ 1
n2 =
n(n + 1) (2n + 1) 6
Verify! The total number of squares in a chess board 8 8 × 9 × 17 ∑1 n 2 = 12 + 22 + 32 + ... + 82 = 6 = 204 18
Example 49: Find the sum of 12 + 22 + … + 202
n(n + 1)(2n + 1) 20 2 20(20 + 1) (2 × 20 + 1) 20 × 21× 41 ; ∑n = = = 2870 6 6 6 1 1 Sum of the cubes of the first n natural numbers n
Solution:
∑n
n
∑n
3
2
=
= 13 + 23 + 33 + ... + n 3
1
Observe the pattern 13 13 + 23 13 + 23 + 33 13 + 23 + 33 + 43
= = = =
1 9 36 100
= 12 = 32 = (1 + 2)2 = 62 = (1 + 2 + 3)2 = 102 = (1 + 2 + 3 + 4)2
Extending this to n terms we get 13 + 23 + 33 + .. + n3 = (1 + 2 + 3+ + … + n)2 ⎡ n(n + 1) ⎤ = ⎢ ⎣ 2 ⎥⎦ ∴
n
∑n 1
3
2
⎡ n(n + 1) ⎤ = ⎢ ⎣ 2 ⎥⎦
2
⎡ 20 × 21 ⎤ = 2102 = 4410 Example 50: 13 + 23 + 33 + … + 203 = ⎢ ⎥ ⎣ 2 ⎦ Verify! The total number of rectangles (including squares) in a chess board. Solution: 2
8
∑ 1
⎡8 × 9 ⎤ n =⎢ = 362 = 1296 ⎥ ⎣ 2 ⎦ 2
3
Exercise 1.2.3
1.
2. 3. 4.
Find the sum of the following a) 1 + 2 + 3 + … + 70 b) 1 + 2 + 3 + … + 112 c) 50 + 51 + 52 + … + 98 d) 15 + 17 + … + 65 e) 1 + 3 + 5 + … to 100 terms f) 12 + 14 + 16 + … + 88 i) 400 + 441 + … + 1600 g) 5 + 10 + 15 + … + 200 h) 162 + 172 + … + 302 j) 1 + 8 + 2 + … + 8000 k) 213 + 223 + … + 413 Find the sum of the volumes of the 15 cubes, whose sides are 1 cm, 2 cm, 3 cm, …15 cm respectively. Find the total area of the 10 squares whose sides are 20 cm, 21 cm, … 29 cm respectively. On each birthday Mr.Kumar gave his son square the amount of his age. Find the total amount Mr.Kumar gave his son by the time he was 17 years old.
19
ANSWERS Exercise 1.1.1 1) a, b, d, e are in A.P. (2) (a) –2 (b) 0.7 (c) 1/4 (d) –3 (e) 4 3) (a) 5, 2, –1 (b) 1.5, 0, –1.5 (c) 43, 50, 57 (d) 2½, 3½ (e) –½ , –1/3, –1/6 4) (a) –5, 1, 7, … (b) 3½, 5, 6½ , … (c) p, p + q, p + 2p, … (d) 7, 0.72, 0.74, … (e) 1, 4, 7, … 5) (a) 82 (b) 14.5 (c) 19/5 (d) –58 (e) 33 6) (a) 9, 6 (b) –283 (c) 27 (d) 5, 6, 7 th th 7) (a) 20 (b) 106 (c) 27 (d) No (e) 15 8) –2, 3, 8 (9) 5, 10, 15, 20 (10) 75, 85, 95, 105 (11) 2, 4, 6 Exercise 1.1.2 217 (1) (b),(d) (2) (a) –1/5 (b) –5/2 (c) 0.3 (d) 3/5 (3) (a) 1/64 (b) 384 (c) 1/81 (d) 9 5
(4) (a) 3,2, 4/3, 3, −2, 4/3 … (b) 64, 32, 16, … (c) 128/2187 (d) 1, 3 (5) (a) 3, 2, 4/3 or 4/3, 2, 3 (b) 18, 6, 2 or 2, 6, 18 (c) 2, 6, 18 (d) 1,–2, 4, –8 or –1/3, –2/3, –4/3, –8/3 (6) 3, 2, 4/3 (7) 1, 2, 4, 8, 16 (8) 3/2 (9) 4 (10) 7 (11) 5, 8, 12 (12) 8, –4, 2, 8 Exercise 1.2.1 1) (a) 1380 (b) 644 (c) –315 (d) 4060 (e) 500 2) (a) 4485 (b) 10989 (c) 32000 (d) 19800 (e) 250000 3) 11,3 (4) 26 (5) 39/11, 40/11, 41/11, … (6) 144 (7) 19 (8) 4 9) –30 (10) 53 (11) Rs.1,200/(12) Rs.2,650/Exercise 1.2.2 1111111111 14197 (c) (d) 3280 (e) 2/9 (410 – 1) 1) (a) 488280 (b) 1000000000 1024 2) (a) 54 (b) 1/7 (c) –4/3 (d) 3/4 (e) 64/15 3) (a) 1 –( –2)n
(c) n −
(b) 10/9 (10n –1)–n
1 3
1 ⎞ ⎛ ⎜1 − n ⎟ 10 ⎠ ⎝
(d)
10 n (10n − 1) − 81 9
n
x(1 − x) n 10n − 1 + n(n − 1) (f) (e) − 2 1 − x 9 (1 − x) 4) 8
(5) 3
10 20 81 , ,... (6) 110 (7) 3 (8) 5, 3 9
Exercise 1.2.3 (1) (a) 2485 (b) 6328 (c) 3626 (g) 4100 (h) 8215 (i) 19670 2 (4) Rs.1785 (3) 6085 cm
(d) 1040 (j) 44100
20
(e) 10000 (f) 1950 (k) 697221 (2) 14400 cm3
2. MENSURATION 2.0
INTRODUCTION
We do measurement in our daily life in many situations. For example, we measure the length of cloth for stitching, the area of a wall for painting, the perimeter of a plot for fencing, the volume of a water tank for filling. Similar to these activities we do measurements for our further needs. This branch of mathematics is called mensuration. In earlier classes, we have learnt about perimeters and areas of plane figures. We have also learnt about figures other than plane figures, which occupy space and have more than two dimensions called solids. In this chapter we shall study first the volume and surface area of combined solids and finally the invariant volumes. Although we have studied about volume and surface area of various solids in earlier classes let us now recall the formulae. Right circular cylinder A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides. Consider a rectangle ABCD which revolves about its side AB and completes one full round to arrive at its initial position. The revolution generates a right circular cylinder as shown in Fig.2.1. A right circular cylinder has two plane ends. Each plane end is circular in shape and the plane ends are parallel. Each of these ends is called the base of the cylinder. The line segment joining the centres of the plane ends is the axis of the cylinder. The radius of the base is the radius of the cylinder. The curved surface joining the two plane ends is the lateral surface.
C
A
D
B
Fig.2.1 For a right circular cylinder of radius r and height h, (a) Base Area = πr2 sq. units; (b) Curved surface area = 2πrh sq. units (c) Total surface area = 2πr (h + r) sq. units; (d) Volume = πr2h cubic units. Hollow cylinder
R
A solid bounded by two coaxial cylinders of same height and different radii is a hollow cylinder. If R and r are the external and internal radii of a hollow cylinder of height h then, a) Area of the base = π(R2– r2) sq. units = π(R+r)(R– r) sq. units b) Curved surface area = 2πh (R+r) sq. units c) Total surface area = 2πr (R+r) (h+R– r) sq.units d) Volume of the material = πh (R2– r2) = πh (R+r) (R – r) cubic units. 21
h r
Fig.2.2
Right circular cone A right circular cone is a solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with a fixed line. In Fig.2.3, V is a fixed point. VO is a fixed line and VA is a revolving line which makes constant angle with VO. The point A will describe a circle with centre O such that the line segment VO is perpendicular to the base. VO is the height `h' of the cone and OA is the base radius ‘r' of the cone. VA is the slant height `l' of the cone. Clearly l2 = r2 + h2. For a right circular cone of radius r, height h and slant height l, a) Area of the base = πr2 sq. units b) Curved surface area = πrl sq. units c) Total surface area = πr (l + r) sq. units d) Volume
V
l
h B
O r
A
Fig.2.3
A
1 2 = πr h cubic units. 3
r O
B
r
Fig.2.4 Sphere A sphere is generated by revolving a semi-circle about its diameter. A plane through the centre of a sphere divides the sphere into two equal parts. Each part is a hemisphere. For a sphere of radius r (a) Surface area = 4πr2 square units
(b) Volume =
4 3 πr cubic units 3
For a hemisphere of radius r (a) Curved surface area = 2πr2 square units
(b)
Volume =
(c) Total surface area of a solid hemisphere = 3πr2 square units
2 3 πr cubic units 3
Review Exercise 1. 2. 3. 4. 5.
A cylindrical pillar is 3.5 m in diameter and 20 m high. Find the volume and the cost of painting its curved surface at the rate of Rs.20 per square metre. A rectangular sheet of aluminium foil 44 cm x 20 cm is rolled along to form a cylinder of height 20 cm. Find the volume of the cylinder formed. The radius and height of a cone are 7 cm and 24 cm respectively. Find the volume and curved surface area of the cone. A cone of height 24 cm has a curved surface area of 550 cm2. Find the volume of the cone. The diameter of hemispherical tank is 14 m. It contains 50 m3 of water. Find the volume of water needed to fill the tank. 22
2.1 VOLUMES AND SURFACE AREAS In our day-to-day life we come across objects, toys etc. which are combinations of more than one solid. Some objects are shown in the Figs. below:
Fig.2.5
Fig.2.6
Fig.2.7
Name and draw such objects you see around. We will now work out a few examples involving volume and surface area of combination of solids. Example 1 : A circus tent is cylindrical to a height of 3 m and conical above it. If the base radius is 52.5 m and slant height of the cone is 53 m, find the area of canvas required to make the tent. Solution: Radius of the cylindrical part r = 52.5m Height of the cylindrical part h = 3m ∴Curved surface area of the cylindrical part = 2πrh square units. =
2×π×
53
m
52.5 m
3m
52.5 m
105 × 3 m2 = 315 πm2 2
Radius of the conical part
= 52.5m;
CSA of the conical part
= πrl sq.units = π ×
Fig.2.8
Slant height of the cone = 53m
105 × 53 m2 = 2782.5 πm2 2
Area of the canvas required = CSA of the cylindrical part + CSA of the conical part = π (315 + 2782.5) = (3097.5) πm2 Example 2 : A rocket is in the form of a cylinder surmounted by a cone. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 6.5m. Calculate the surface area and volume of the rocket. Solution : Radius of the cylinder = radius of the cone ; r = 2.5m ; Height of the cylinder h1 = 21m ; Height of the cone h2
=
l2 − r2 23
(l = slant height of the cone).
21 m
1 2 πr h2 3
= π × 2.5 × 2.5 × 21 +
1 π × 2.5 × 2.5 × 6 3
2.5m
= πr2 h1 +
6 .5 m
h 2 = 6.52 − 2.52 = (6.5 + 2.5) (6.5 − 2.5) = 9 × 4 = 36 ⇒ h2 = 6m Surface area of the rocket = CSA of the cylinder + CSA of the cone = 2πrh1 + πrl = 2 × π × 2.5 × 21 + π × 2.5 × 6.5 = 105 π + 16.25π = (121.25)π square metre Volume of the rocket = Volume of cylinder + Volume of the cone
= (131.25) π + (12.5) π = (143.75) π cubic metre.
Fig.2.9 Example 3 : A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5cm. Find the volume of the toy. Solution : Radius of the cone = Radius of the hemisphere = 3.5 cm = 7/2 cm 15.5 Total height of the toy = 15.5cm cm Total height of the toy = Radius of the hemisphere + height of the cone. B A 3.5 Height of the cone = 15.5 – 3.5 = 12 cm. Volume of the toy = Volume of the cone + volume of the hemisphere
=
1 2 2 πr h + πr3 cubic units 3 3
Fig.2.10
7 7 7 7 7⎞ 343π ⎛1 ⎞ ⎛2 = ⎜ π × × × 12 ⎟ + ⎜ π × × × ⎟ = 49π + 2 2 2 2 2⎠ 12 ⎝3 ⎠ ⎝3 = 49π + 28.58π = (77.58) π cm3
10 cm
12 cm
3.5cm
Example 4 : A solid is in the form of right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the heights of the cylindrical and conical portions are 10 cm and 12 cm respectively. Find the total surface area of the solid.
Fig.2.11 Solution: Radius of the hemisphere = 3.5 cm Curved surface area of the hemisphere = 2πr2 sq.units = 2 × π × (3.5)2 = (24.5)π sq.units. Radius of the cylinder r = 3.5 cm ; Height of the cylinder h = 10 cm Curved surface area of the cylinder = 2πrh square units = 2 × π × 3.5 × 10 = 70 π cm2 Radius of the cone r = 3.5 cm ; Height of the cone = 12 cm
24
625 25 49 + 144 = = 12.5 cm = 4 4 2 Curved Surface area of the cone = πrl square units = π × 3.5 × 12.5 = (43.75) π cm2 Total surface area of the solid = Curved Surface area of the hemisphere + curved surface area of the cylinder + curved surface area of the cone = 24.5π + 70π + 43.75 π = (138.25) π cm2 Total Surface area of the solid = (138.25) π cm2. r 2 + h 2 = 3.5 2 + 12 2 =
= =
7 cm
7 cm
Fig.2.12
2 3 πr + πr2h cubic units 3
2 π×7×7×7 + π×7×7×6 cm3 3 ⎡⎛ 2 ⎤ ⎞ = π × 7 × 7 ⎢⎜ × 7 ⎟ + 6 ⎥ cm3 ⎠ ⎣⎝ 3 ⎦
32 1568 cm3 = π cm3 = (522.67) π cm3 3 3
Example 6 : A capsule is in the form of a cylinder with hemispherical ends. The total height of the capsule is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the capsule. Solution: Diameter of the cylinder = 7cm Radius of the cylinder = 3.5 cm = 7/2 cm Radius of the cylinder = Radius of the hemisphere Total height of the capsule = 19 cm Height of the cylinder = 19 – (2 × 3.5) = 12 cm Volume of the capsule = Volume of the cylinder + volume of two hemisphere ⎡ ⎛2 ⎞⎤ = ⎢πr 2 h + 2⎜ πr 3 ⎟⎥ cubic units ⎝3 ⎠⎦ ⎣
25
3.5 cm
12 cm
= π × 49 ×
Fig.2.13
13 cm
Example 5 : A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the sphere is 14 cm and the total height of the vessel is 13 cm. Find its capacity. Solution: Diameter of the hemisphere = 14 cm Radius of the hemisphere = 7 cm Radius of the cylinder = Radius of the hemisphere Height of the cylinder = 13 – 7 = 6 cm Capacity of the vessel = volume of the hemisphere + volume of the cylinder.
6 cm
Slant height of the cone =
7 7 2 7 7 7⎞ ⎛ = ⎜ π × × ×12 + 2 × × π × × × ⎟ cm3 2 2 3 2 2 2⎠ ⎝ 25 1225π = 49π × = = (204.17) π cm3 6 6 Total surface area of the capsule = CSA of the cylinder + surface area of two hemispheres. = 2πrh + 2 × 2 πr2 sq. units = 2 × π × 3.5 (12 + 2 × 3.5) cm2 =2×π×
7 × 19 = 133 π cm2. 2
Example 7 : In a cylindrical wooden block of radius 7 cm and height 14 cm hemispherical blocks of radius 7 cm are carved out from both ends. Find the 7 cm volume of the resulting solid. Solution Radius of the cylinder = 7cm; Height of the cylinder = 14cm ; Volume of the cylinder = πr2h cubic units = π×7×7×14 = 686 π cm3 Radius of the hemisphere = 7cm
2 × π r3 3 2 1372 =2× ×π×7×7×7= πcm3 3 3 1372 ⎛ 2⎞ π = 686 π ⎜ 1 - ⎟ Volume of the resulting solid = 686π – 3 ⎝ 3⎠ Volume of the 2 hemispheres = 2 ×
686 π 3
= 228.67 πcm3.
Example 8 : The radius of the top of a bucket is 18 cm and that of the bottom is 6 cm. Its depth is 24 cm. Find the capacity of the bucket. Solution: The bucket is obtained by cutting a smaller cone of radius 6 cm from a bigger cone of radius 18 cm. Let h be the height of the smaller cone 24 + h 18 3 ΔABC | | | ΔDEC; ∴ = = h 6 1 3h = 24 + h or 2h = 24 or h = 12 cm. ∴Height of the smaller cone = 12 cm ∴Height of the bigger cone = 12 + 24 = 36 cm Capacity of the bucket = Volume of the bigger cone – Volume of the smaller cone 1 ⎛1 ⎞ = ⎜ π×18 ×18 × 36 − π 6 × 6 ×12 ⎟ cm3 3 ⎝3 ⎠ 26
A 18 cm
B
24 cm
=
Fig.2.14
D 6 cm
E
h
C
Fig.2.15
= 144 π(27–1) cm3 = 144 π × 26 cm3 = 3744 π cm3
Note : Solid obtained by cutting a right circular cone by a plane parallel to the base is called frustum of a cone. Try a formula for the volume of a frustum. Example 9 : Find the volume of the hay stock as shown in the Fig.2.16 Solution: Upper part of the hay stock is a cone of radius 3 m and height 7m.
Volume of the lower part
1 1 π×3×3×10.5 – × π×2×2×7 3 3 1 π × (94.5 – 28) m3 = 3 1 π × 66.5 = 22.17 π m3. = 3
3 cm
10.5 cm
The lower part is a frustum. Radius of the upper end = 3m; Radius of the lower end = 2 m Height of the frustum = 10.5 – 7 = 3.5 m Let H be the height of the bigger cone; Let h be the height of the smaller cone. 3 H 3.5 + h Using similar triangles, = ; = 2 h h 3h = 7 + 2 h or h = 7 ⇒ H = 7 + 3.5 = 10.5 m
7 cm
1 2 1 πr h = π × 3 × 3 × 7 = 21 π m3. 3 3
10.5 cm
∴Volume of the cone =
=
2 cm
Fig.2.16
Volume of the hay stock = 21π + 22.17 m3 = 43.17 π m3.
Example 10 : A godown building is in the form as shown in the figure. Find its volume (π ~ 3.14). Solution : Area of the vertical cross section of the godown = Area of the of the rectangle + Area of the 3m semicircle 1 7 7⎞ ⎛1 = l × b + πr2 = (7 x 3) + ⎜2 × 3.14 × 2 × 2 ⎟ m2 ⎠ ⎝ 2 2 2 10 m = 21 + 19.23m = 40.23 m . 7m Length of the building = 10m. Volume of the godown =Area of cross section × length Fig.2.17 3 3 = 40.23 × 10 m = 402.3 m . Aliter : Volume of the godown = Volume of the cuboid + Volume of the cuboid
= l×b×h
Volume of the cylinder
= πr2h
∴ Volume of the godown
= 210 +
1 (Volume of the cylinder) 2
= 7 × 10 × 3 = 210 m3. 7 7 = 3.14 × × × 10 = 384.65 m3. 2 2
1 (384.65)m3 = 402.325 m3 = 402.3 m3. 2 27
Exercises 2.1 1.
An iron pillar is in the form of a cylinder of height 2.8 m and 20 cm in diameter surmounted by a cone of height 42 cm. Find the weight of the pillar if 1 cu.cm of iron weighs 7.5 gm (π ~ 3.14).
2.
A circus tent is cylindrical upto a height of 3 m and conical above it. If the diameter of the base is 105 m and the slant height of the cone is 53 m, find the length of the canvas to be bought if the width of the canvas is 5 m (π=3.14).
3.
The interior of a building is in the form of a cylinder of diameter 4.2 m and height 3.8m surmounted by a cone whose vertical angle is a right angle. Find the surface area and volume of the building.
4.
The height of a solid cylinder is 15 cm and the diameter of the base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm are cut off. Find the volume of the remaining solid.
5.
A solid is composed of a cylinder with hemispherical ends. If the length of the solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 ps/cm2 (π = 3.14).
6.
A toy is in the shape of a right circular cylinder with a hemisphere on one end a cone on another end. The radius and height of the cylindrical part are 5 cm and 30 cm respectively. The radius of the hemisphere and conical parts are the same as that of the cylinder. The height of the conical part is 12 cm. Find the surface area of the toy.
7.
A toy is in the form of a cone mounted on a hemisphere of radius 4.5 cm. The total height of the toy is 24.5 cm. Find the total surface area of the toy.
8.
A vessel in the form of a hemispherical bowl is mounted by a hollow cylinder. The diameter of the bowl is 14 cm and the total height of the vessel is 13 cm. Find the capacity of the vessel.
E
13 F A
0 10.5
In the adjoining figure, find the total surface area if AC = 13 cm; CE = 3m and FC = OE = 10.5m.
3
10.5
9.
C
B
D
Fig.2.18 10.
An icecream cone has a hemispherical top. If the height of the cone is 9 cm and base radius is 2.5 cm. Find the volume of the icecream in the icecream cone.
11.
A cup is in the form of a hemisphere surmounted by a cylinder. The radius of the hemispherical part is 12.5 cm and the total height of the cup is 25 cm, find the capacity of the cup.
12.
The cross section of a hay shed is a rectangle of height 21 m and width 18 m surmounted by a semicircle on the breadth wise side. If the length of the shed is 24 m, find the volume of the hay that can be stored in it (π = 3.14).
28
13.
If the radii of the circular ends of a bucket of height 45 cm are 28 cm and 7cm. Find the capacity of the bucket.
14.
The perimeters of the ends of the Fig.2.19 are 14π and 8.4π cm. If the h eight is 12 cm, find its volume.
15. 16.
The radii of the ends of a bucket 32cm high are 21 cm and 7 cm. Find its capacity in litres (π = 3.14).
12 cm
Fig.2.19
Find the volume of the solid given by the Fig.2.20.
2 cm
3 cm
12.5 cm
7 cm
Fig.2.20
2.2 INVARIANT VOLUMES At home you would have seen your mother making a number of `laddus' spherical in shape with `sweet boondhis' kept in vessels of any shape. In industries too metal sheets or rods are melted and cast into solids of different shapes and sizes. In this unit we will learn to compute the number of new shapes made out of given ones when total volume remains unchanged.
Example 11 : A conical piece of lead has a radius of 5.25 cm and is 8 cm in height. If it is melted and made into smaller cones 2 cm high and 1.75 cm in diameter, find how many such cones can be made. Solution: Radius of the cone = 5.25 cm; Height of the cone = 8cm. Volume of the cone =
1 2 1 πr h cubic units = π × 5.25 × 5.25 × 8 cm3 3 3
Diameter of the smaller cone = 1.75 cm ⇒ Radius of the smaller cone = 1 1.75 1.75 π× × ×2 3 2 2 Volume of the bigger cone = Volume of the smaller cone
Volume of the smaller cone = Number of smaller cones
29
1.75 cm 2
1 π × 5.25 × 5.25 × 8 3 = 3 × 3 × 8 × 2 = 144. = 1 1.75 1.75 ×π × × ×2 3 2 2 ∴ 144 smaller cones can be made.
Example 12 : A cylindrical bucket is internally 12 cm in diameter and 16 cm deep. It is full of sand and when emptied the sand stands in the shape of a cone 18 cm in diameter. Find the height of the heap. Solution : Diameter of the cylinder bucket = 12 cm. 12 = 6 cm; Height of the cylinder = 16 cm. Radius of the cylinder = 2 ∴Volume of the sand in the bucket = πr2h = π × 6 × 6 × 16 cm3. 18 Diameter of the sand heap = 18 cm; ∴Radius of the heap = = 9 cm. 2 Let the height of the heap be `h' cm Volume of the conical heap = volume of the sand in the bucket =
1 × π × 9 × 9 × h = π × 6 × 6 × 16 3
π × 6 × 6 × 16 64 cm = 21.33 cm. = 1 3 ×π×9×9 3 ∴ The height of the heap = 21.33 cm.
∴ h =
Example 13 : The curved surface area of a cone made of lead is 204.1 cm2 and its slant height is 13 cm. It is melted and cast in the form of a cylinder. If the area of the base of the cylinder is 16π cm2, find the height (Take π = 3.14). Solution: Slant height of the cone = 13 cm ; Curved surface area of the cone = 204.1cm2. 204.1 πrl = 3.14 × r × 13 = 204.1 cm2 or r = 3.14 × 13
Radius of the cone
= 5 cm ;
l2 − r2
h =
=
13 2 − 5 2
∴Height ‘h’ of the cone = 12 cm Volume of the cone
=
1 2 πr h 3
=
1 π × 5 × 5 × 12 cm3 3
Base area of the cylinder = 16 π cm2 ; Let the height of the cylinder be H cm. Volume of the cylinder made = Volume of the cone melted ⇒ πr2 H =
1 1 π × 5 × 5 × 12 (or) 16π × H = π × 5 × 5 × 12 (or) 3 3 1 1 (or) H = 6.25 cm. H = π × 5 × 5 × 12 × 3 16π
Base area × H =
30
1 2 πr h 3
Example 14 : A hemispherical bowl of radius 30 cm is filled with soap paste. If this paste is made into cylindrical soap cakes each of radius 5 cm and height 2 cm, how many cakes do we get? Solution: Radius of the hemisphere = 30 cm ; Volume of the hemisphere =
∴Volume of the soap paste =
2 3 πr cubic units 3
2 π × 30 × 30 × 30 cm3. 3
Radius of the soap cake Volume of the soap cake
= 5 cm; Height of the soap cake = 2 cm; = π × 5 × 5 × 2 cm3. 2 π × 30 × 30 × 30 Volume of the soap paste 3 = 360 Number of soap cakes = = Volume of the soap cake π × 5 × 5× 2
∴We get 360 soap cakes out of the given paste. Example 15 : A solid metal cylinder is 20 cm in height and has a radius of 1.5 cm. This is melted down and cast into spheres each of radius 1.5 cm. How many spheres can be cast from the cylinder? Solution : Height of the cylinder = 20 cm; Radius of the cylinder = 1.5 cm Volume of the cylinder = π × 1.5 × 1.5 × 20 cm3 Radius of the sphere = 1.5 cm 4 Volume of the sphere cast = × π × 1.5 × 1.5 × 1.5 cm3. 3 Volume of the cylinder melted π × 1.5 × 1.5 × 20 × 3 = = 10 Number of spheres cast = Volume of a sphere cast 4 × π × 1.5 × 1.5 × 1.5
⇒ 10 spheres can be cast from the cylinder. Example 16 : A cylindrical jar of diameter 14 cm and depth 20 cm is half full of water. 300 lead shots of the same size are dropped into the jar and the level of water rises by 2.8 cm. Find the diameter of each lead shot. 14 Solution: Diameter of the jar = 14 cm ; ∴Radius of the jar = = 7 cm 2 20 = 10 cm; Rise in the level of water = 2.8 cm. Depth of water = 2 Volume of water raised in the jar = π × 7 × 7 × 2.8 cm3. Let r denote the radius of a lead shot. Volume of 300 lead shots = Volume of water raised.
300 ×
4 × π × r3 3
=
π × 7 × 7 × 2.8
π × 7 × 7 × 2.8 × 3 = r = 300 × 4 × π
7×7×7 7 ⎛7⎞ = 0.7 cm. = ⎜ ⎟ (or) r = 10 1000 ⎝ 10 ⎠ Radius of each lead shot = 0.7 cm ∴Diameter of each lead shot = 1.4 cm. 3
3
31
Example 17 : A cylindrical vessel of diameter 14 cm contains water. A metal sphere of diameter 7 cm is lowered into the water until it is completely immersed. By how much does the level of water rise? 7 Solution: Diameter of the sphere = 7 cm ⇒ Radius of the sphere = cm 2 4 7 7 7 Volume of the sphere = π × × × cm3. 3 2 2 2 Let h denote the level of water raised in the cylindrical vessel. Volume of water raised in the cylindrical vessel = Volume of the sphere immersed. 4 7 7 7 π×7×7× h= π× × × 3 2 2 2 4 7 7 7 × π× × × 2 2 2 = 7 cm ~ 1.17 cm. h= 3 6 π×7×7
The level of water in the vessel is rises by 1.17 cm. Example 18 : A slab of iron whose dimensions are 60 cm × 20 cm × 28.26 cm is used to caste an iron pipe. The outer diameter of the pipe is 10 cm. If the wall of the pipe is 1 cm thick. Calculate the length of the pipe that can be cast from the slab (π = 3.14). Solution: Volume of the iron slab melted = 60 × 20 × 28.26 cm3. Volume of the hollow cylindrical pipe = πh (R + r) (R – r) Outer radius of the pipe R = 5 cm. Inner radius of the pipe r = 5 – 1 = 4 cm Volume of the metal in the pipe = Volume of the melted slab. π (5 + 4) (5 – 4)h = 60 × 20 × 28.26 60 × 20 × 28.26 h = = 1200 cm = 12 m. 3.14 × 9 × 1
Length of the pipe that can be cast from the slab is 12 m. Example 19 : An over head tank has been constructed to supply water to a village with a population of 3140 at the rate of 25 litres per head per day. Water is pumped into it through a pipe of 10 cm diameter, the rate of flow being 4 m per sec. How long will it take to fill the tank every morning? (π = 3.14) Solution: Volume of water supplied for 1 person = 25 litres = 25000 cm3. Volume of water in the tank = 3140 × 25000 cm3 Volume of water pumped in 1 sec = 3.14 × 5 × 5 × 400 cm3. (π = 3.14) Volume of the tank Time taken to fill the tank = Volume of water pumped in 1 sec
=
3140 × 25000 3.14 × 5 × 5 × 400
= 2500 seconds = 41 m 40 seconds. Pipe will take 41 minutes and 40 seconds to fill the tank. 32
Exercise 2.2 1. 2. 3.
4.
5. 6.
7. 8. 9. 10.
11. 12.
13.
14. 15. 16.
17. 18.
A cubic centimeter of iron is drawn into a wire of diameter 3.5 mm. Find the length of the wire (π = 3.14). A solid rectangular block of iron of dimensions 4.396 m × 2.5 m × 1.6 m is recast into a cylinder of radius 0.4 m. Find the height of the recast cylinder (π=3.14). A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, find the rise of the water level in the vessel. A cylindrical bowl of diameter 15 cm contains some liquid to a height of 6 cm. This liquid is to be filled in conical bottles of radius 1.25 cm and height 6 cm. Find the number of bottles that can be filled with the liquid in the bowl. A rod 4 m in length and 1.2 cm in radius is drawn into another rod of length 72 cm. Find the radius of the new rod. Rain water falls in a tub of dimensions 6 m x 4 m x 2.75 m and is transferred to a cylindrical tub of diameter 20 meters. Find the height of water in the cylindrical tub (π = 3.14). A sphere of radius 12.6 cm is melted and recast into a right circular cone of height 12.6 cm. Find the diameter of the base of the cone. A hemisphere of lead of radius 8 cm is cast into a right circular cone of base 6 cm. Find the height of the cone correct to two decimal places. The surface area of a sphere of radius 5 cm is five times the curved surface area of a cone of radius 4 cm. Find the height and volume of the cone. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball is dropped into the tub and the water level is raised by 6.75 cm. What is the radius of the ball? Water flowing through a pipe of diameter 7 cm irrigates a field of area 4.9 hectares to a depth of 10 cm in 70 hours. Find the speed of water. A hollow cylindrical metal pipe is 40 cm long. Its internal and external radii are 4 cm and 12 cm respectively. It is melted and cast into a solid cylinder 20 cm. long. What is the diameter of the solid cylinder? A cone and a cylinder have the same base area and the same curved surface area. If the height of the cylinder is 2.5 m and the radius of the cone is 3 m find the curved surface area and volume of the cone. A hollow sphere of internal and external diameter 4 cm and 8 cm respectively is melted to form a cone of diameter 8 cm. Find the height of the cone. Three solid spheres of radii 6cm, 8 cm and 10 cm respectively are melted to form a single solid sphere. Find the radius of the resulting sphere. A solid right circular cylinder has a base radius of 12 cm and height of 16 cm. It is melted and made into 8 spherical balls of equal size. Calculate the radius of each spherical ball. A well of diameter 3 m is dug to a depth of 20m. The earth taken out is heaped into a cone of height 15 m. Find the radius and slant height of the cone. For the noon meal, food is kept ready in 3 cylindrical vessels of diameter 1.4 m and height 56 cm. How many students can be served if the food is served twice with a hemispherical bowl of radius 14 cm?
33
ANSWERS Exercise 2.1
(1) 692.37 kg (4) (159.75) πcm3. (7) (132.75) πcm2 (10) (29.17) πcm3 (13) 15335 πcm3 (15) (21.35) h
(3) (22.2) πm2, (19.745) πm3 (6) 415 πcm2 (9) 420 πm2 (12) 12124.08 cm3
(2) 1945.23 m (5) Rs.854.58 (8) (522.67) πcm3 (11) (3255.21) πcm3 (14) (384.16) πcm2 (16) (48.5) π cm3.
Exercise 2.2
(1) 10.4 cm (6) 5.25 cm (11) 18.19 km/hr (16) 6cm
(2) 35m (7) 50.4 cm (12) 32 cm (17) 3m, 15.3 m
(3) 1 cm (4) 108 (8) 28.44 cm (9) 50.3 cm3 (13) 15πcm2, πm3 (14) 14 cm (18) 225.
34
(5) 2√2 cm (10) 9 cm (15) 12cm
3. THEORY OF SETS 3.0
INTRODUCTION
Cantor (1845-1918 A.D), father of modern theory of sets published a paper in 1874 A.D that the set of real numbers could not be put into one-to-one correspondence with the integers. From 1879 onwards, he published many research papers on properties of abstract sets. Famous mathematicians Dedekind (1831-1916 A.D) and Kronecker (1810-1893 A.D) have continued research on theory of sets. German mathematician Gottlob presented the set theory as principles of logic. Bertrand Russell (1872-1970 A.D) and Paul Halmos proposed paradoxes in set theory. The axiomatization of set theory was published by Ernest Zermelo (1908), Abraham Fraenekel 1922 and John Von Neumann (1925). Paul Bernays (1937) and Kurt Godel (1940) gave a set of more satisfactory axiomatization. But Cantor’s set theory is used in the present day mathematics. Descartes (1596-1650 A.D) introduced the word ‘function’ in 1637 A.D to mean xn. An explicit definition of function was given by James Gregory (1638-1675 A.D). Leibnitz (1673 A.D) used the word ‘function’ to mean any quantity varying from point to point on a curve, such as the co-ordinates of a point on the curve, the slope of the curve, the tangent and the normal at a point on the curve. Later Leibnitz (1714 A.D) used the word function to mean quantities that depend on a variable. He was the first to use the phrase ‘function of x’. The notation f(x) was introduced by Euler (1734 A.D). Fourier (1768-1830 A.D) while investigating heat conduction problem gave a broad definition of a function. Drichlet (18051859 A.D) gave the definition of function which is in use today. Cantor gave set theoretic definition of the function.
3.1
SETS
We have learnt about sets in the previous classes. All branches of Mathematics can be brought into the frame work of set theory. Let us recall what we have learnt about sets by answering the following questions: Define a set, empty set, disjoint sets, subsets, power set, universal set, complement of a set, operations on sets and give some examples. In this chapter we are going to learn in detail about the laws of set operations, Relations and Functions. Union of two sets is commutative
A B
A
B A
A
B
Fig.3.1
B
Fig.3.2 35
From the diagrams we see that A ∪ B = B ∪ A. For example, given A = {–2, 3, 5, 7} and B = {3, 9, 11} we see that A ∪ B = {–2, 3, 5, 7, 9, 11}, B ∪ A = {–2, 3, 5, 7, 9, 11} ∴ A ∪ B = B ∪ A Intersection of two sets is commutative
A B
B A
A
A
B
B
Fig.3.3
Fig.3.4
From the above diagrams we see that A ∩ B = B ∩ A. For example, given A = {–7, 5, 2, 3,6} and B = {3, 6, 7, 12}. We see that A ∩ B = {3, 6}; B ∩ A = {3, 6} ∴ A ∩ B = B ∩ A. Idempotency of Union and Intersection of two sets A ∪ A = A and A ∩ A = A If A = {1, 3, 5, 7}, then A ∪ A = {1, 3, 5,7} = A and A ∩ A = {1, 3, 5, 7} = A. Venn diagrams to show the idempotency of union and intersection of sets:
A A
A A
A
A
B
B
Fig.3.5 Associative law for union sets
Fig.3.6
A ∪ (B ∪ C) = (A ∪ B) ∪ C Verification of the associative law for the union of sets using Venn diagrams: A A (B C) B C
B
A
C Fig.3.7
B
A
C Fig.3.8 36
B
A
C Fig.3.9
(A B) C
C
A B
B
A
A
B
B
A
C
C
C
Fig.3.10
Fig.3.11
Fig.3.12
From Fig.3.9 and Fig.3.12 it is clear that A ∪ (B ∪ C) = (A ∪ B) ∪ C Associative law for intersection of sets A ∩ (B ∩ C) = (A ∩ B) ∩ C Verification of the associative law for intersection of sets using Venn diagrams” B C A
B
C
Fig.3.13
A B
A (B C) A
B
A
(A B) C
B
C
C
Fig.3.14
Fig.3.15
A
B
C
Fig.3.16
From Fig.3.14 and Fig.3.16 it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C Example 1: Verify the associative law of union and intersection of the following sets. A = {4, 5, 6}, B = {6, 7, 8}, C = {7, 8, 9} Solution: (a) B ∪ C = {6, 7, 8, 9}; A ∪ (B ∪ C) = {4, 5, 6, 7, 8, 9} A ∪ B = {4, 5, 6, 7, 8}; (A ∪ B) ∪ C = {4, 5, 6, 7, 8, 9} From (1) and (2) it is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C. (b) B ∩ C = {7, 8}; A ∩ (B ∩ C) = { } A ∩ B = {6}; (A ∩ B) ∩ C = { } From (3) and (4) it is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C Distributive laws (a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Consider the sets A = {1, 3, 5, 7}, B = {1, 2, 4, 6, 8} and C = {1, 3, 6,8} (b) B ∩ C = {1, 6, 8}; A ∪ (B ∩ C) = {1, 3, 5, 7, 6, 8} A ∪ B = {1, 3, 5, 7, 2, 4, 6, 8} A ∪ C = {1, 3, 5, 6, 7, 8} (A ∪ B) ∩ (A ∪ C) = {1, 3, 5, 7, 6, 8} From (1) and (2) we see that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (b) B ∪ C = {1,2,4,6,8,3}; A ∩ (B ∪ C) = {1, 3} A ∩ B = {1} ; A ∩ C = {1, 3} (A ∩ B) ∪ (A ∩ C) = {1, 3} From (3) and (4) we see that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 37
(1) (2) (3) (4)
(1) (2) (3) (4)
Verification of the law: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using Venn diagrams:
B C
A (B C)
A
B
A
C Fig.3.17 A B A
B
C Fig.3.18 (A B) (A C)
A C B A
B A
C Fig.3.19
C Fig.3.20
B
C Fig.3.21
From Fig.3.18 and Fig.3.21 we find A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Verification of the law: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) using Venn diagrams: A (B C) B C A
A
B
C Fig.3.23
C Fig.3.22
(A B) (A
A C
A B A
B
B
A
B
C C C Fig.3.24 Fig.3.25 Fig.3.26 From Fig.3.23 and Fig.3.26 we observe that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 38
C) B
De Morgan’s Laws Let A, B, C be any three sets then (i) (A ∪ B)′ = A′ ∩ B′; (ii) (A ∩ B)′ = A′ ∪ B′ (iii) A – (B ∪ C) = (A – B) ∩ A – C) (iv) A – (B ∩ C) = (A – B) ∪ (A – C) are called De Morgan’s laws. Example 2: ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {2, 4, 6}, B = {1, 2, 3, 4, 5} verify that (i) (A ∪ B)′ = A′ ∩ B′ and (ii) (A ∩ B)′ = A′ ∪ B′ Solution: ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A ∪ B = {1, 2, 3, 4, 5, 6}; (A ∪ B)′ = {7, 8, 9, 10} (1) A′ = {1, 3, 5, 7, 8, 9, 10}; B′ = {4, 6, 7, 8, 9, 10} A′ ∩ B′ = {7, 8, 9, 10} (2) From (1) and (2) it is verified that (A ∪ B)′ = A′ ∩ B′ (i) A ∩ B = {2 } (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9, 10} (3) A′ = {1, 3, 5, 7, 8, 9, 10}; B′ = {4, 6, 7, 8, 9, 10} A′ ∪ B′ = {1, 3, 4, 5, 6, 7, 8, 9, 10} (4) From (3) and (4) it is verified that (A ∩ B)′ = A′ ∪ B′ Example 3: If A = {a, b, c, d, e, f, g, h}; B = {a, b, e, f} and C = {a, c, e, g, h, k} verify the DeMorgan’s laws (i) A – (B ∪ C) = (A – B) ∩ (A – C) and (ii) A – (B ∩ C) = (A – B) ∪ (A – C) Solution: (i) B ∪ C = {a, b, c, e, f, g, h, k} A – (B ∪ C) = {d} (1) A–B = {c, d, g, h}; A – C = {b, d, f} (A – B) ∩ (A – C) = {d} (2) From (1) and (2) we get A – (B ∩ C) = (A – B) ∪ (A – C) Example 4: Verify DeMorgan’s law (A ∪ B)′ = A′ ∩ B′ using Venn diagrams: Solution: (A B)’ A B
B
A
A’
A
B
A
Fig.3.27
Fig.3.28 B’
B
B
A
Fig.3.29 Fig.3.30 From Fig.3.28 and Fig.3.31 we find that (A ∪ B)′ = A′ ∩ B′ 39
A’
B’
B
A
Fig.3.31
Example 5: Verify the DeMorgan’s law (A ∩ B)′ = A′ ∪ B′, using Venn diagrams: Solution: (A B)’
A B
B
A
B
A
Fig.3.32
Fig.3.33
A’
A’
B’
B
A
B
A
Fig.3.34
B’
B
A
Fig.3.35
Fig.3.36
From Fig.3.33 and Fig.3.36 we find that (A ∩ B)′ = A′ ∪ B′ Example 6: Verify the Demorgan’s law A – (B ∪ C) = (A – B) ∩ (A – C) using Venn diagrams: A - (B C) B C Solution:
A
B
A
C
C
Fig.3.37
Fig.3.38
A-B
C
(A - B)
A-C B A
A
B
(A - C)
B A
C
B
C
Fig.3.39 Fig.3.40 Fig.3.41 From Fig.3.38 and Fig.3.41 we find that A – (B ∪ C) = (A – B) ∩ (A – C) 40
Example 7: Verify the DeMorgan’s law A – (B ∩ C) = (A – B) ∪ (A – C) using Venn diagrams: Solution:
A - (B
B C
A
B
A
A
B
C Fig.3.43
C Fig.3.42
(A B) (A - C)
A-C
A -B B
C)
B A
A
B
C
C C Fig.3.44 Fig.3.45 Fig.3.46 From Fig.3.43 and Fig.3.46 we find that A – (B ∩ C) = (A – B) ∪ (A – C) In class IX we have learnt to solve problems involving two sets using formula and by using Venn diagrams. Here we are going to solve problems involving 3 sets using Venn diagrams and formula. n(A ∪ B ∪ C) = n(a) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C) Verify this formula if A = {2, 3, 4} B = {2, 3, 5, 6} C = {3, 4, 5, 7, 8} Example 8: In a cultural programme, 24 students, took part in dance 11 in drama, 25 in group songs, 7 in dance and drama, 4 in drama and group songs, 12 in dance and group songs and 3 participated in all the three. If total of 50 students were there in the class, find how many did not participate in the programme. Solution: Let the sets A, B and C denote those who participate in dance, drama and group songs respectively. Total number of students = n(ξ) = 50 The number of students who participate in the dance = n(A) = 24 The number of students who participate in drama = n(B) = 11 The number of students who participate in group songs = n(C) = 25 The number of students who participate in dance and drama = n(A ∩ B) = 7 The number of students who participate in drama and group songs = n(B ∩ C) = 4 The number of students who participate in dance and group songs = n(A ∩ C) = 12 Number of students who participated in all the three = n(A ∩ B ∩ C) = 3 ∴ Number of students who took part in any one of the three programme is n(A∪B∪C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) = 24 + 11 + 25 – 7 – 4 – 12 + 3 = 63 – 23 = 40 41
Those who did not participate in any one of the three programmes = n(ζ) – n(A ∪ B ∪ C) = 50 – 40 = 10. Using Venn diagrams we solve this problem: No.of students who participate in any one of the three programmes = 8 + 4 + 3 + 9 + 3 + 1 + 12 = 40 No.of students who did not participate in any one of the three programmes = 50 – 40 = 10.
B
A
11247-3 (4+3+9) = 4 (4+3+1) =3 =8 3 4-3 12-3 =9 =1 25(9+3+1) = 12 C
Fig.3.47 Example 9: In a class of 35 students, 18 speak Tamil, 12 speak Hindi and 15 speak English. 2 students speak Tamil and Hindi, 4 Hindi and English and 5 speak English and Tamil. Calculate the number of students who speak all the three languages. Also find the number of students who speak Hindi and English but not Tamil. Solution: Let A, B, C be the sets those who speak Tamil, Hindi and English respectively. Let the number of students who speak all the three languages = x. No.of students who speak Tamil = 18 B No.of students who speak Hindi = 12 A 11 + x 2-x 6+x No.of students who speak English = 15 No.of students who speak Tamil and Hindi = n(A∩B) =2 x No.of students who speak Hindi and English = n(B∩C) =4 4-x 5-x No.of students who speak English and Tamil = n(C∩A) =5 Let the no.of students who speak all three = n(A∩B∩C) = x 6+x No.of students who speak only Tamil 18 – [2 – x + x + 5 – x] C = 18 – (7 – x) = 11 + x No.of students who speak only Hindi = 12 – (2 – x + x + 4 – x) = 6 + x Fig.3.48 No.of students who speak only English 15 – [5 – x + x + 4 – x] = 15 – (9 – x) = 6 + x Strength of the class = 35. No.of students who speak any one of the languages = 11 + x + 2 – x + 6 + x + 5 – x + x + 4 – x + 6 + x = 34 + 4x – 3x = 34 + x 34 + x = 35, x = 35 – 34 = 1 No.of students who speak all three languages = x = 1 No.of students who speak Hindi and English but not Tamil = 4 – x = 4 – 1 = 3. Example 10: In a higher secondary class, 66 play football, 56 play hockey, 63 play cricket, 27 play both football and hockey, 25 p lay hockey and cricket, 23 play cricket and football and 5 do not play any game. If the strength of the class is 130. Calculate (i) the number who play only two games, (ii) the number who play only football and (iii) number of students who play all the three games. Solution: Let F, H, C be the sets those who play football, hockey and cricket respectively. Strength of the class n(ξ) = 130 No.of students who play football = n(F) = 66 No.of students who play hockey = n(H) = 56 No.of students who play cricket = n(C) = 63 No.of students who play football and hockey = n(F∩H) = 27 42
No.of students who play hockey and cricket = n(H ∩ C) = 25 No.of students who play cricket and football = n(C ∩ F) = 23 Let the no.of students who play all the three games = n(F ∩ H ∩ C) = x No.of students who do not play any game = 5
F
16 + x
27-x
4+x
H
x 23 - x
25 - x
15 + x
No.of students who play only football = 66 – (27 – x + x + 23 + x) = 66 – (50 – x) = 66 – 50 + x = 16 + x No.of students who play only hockey = 56 – (27 – x + x + 25 – x) = 56 – (52 – x) = 4 + x No.of students who play only cricket = 63 – (23 – x + x + 25 – x) = 63 – (48 – x) = 15 + x
C Fig.3.49
No.of students who play atleast one game is n(F∪H∪C) = n(ξ) – No.of students who do not play = 130 – 5 = 125 From the diagram, n (F ∪ H ∪ C) = 16 + x + 27 – x + 4 + x + 23 – x + x + 25 – x + 15 + x = 110 + x ∴ 110 + x = 125, x = 125 – 110 = 15. i)
No.of students who play only two games
= 27 – x + 23 – x + 25 – x = 75 – 3x = 75 – 3(15) = 75 – 45 = 30 ii) No.of students who play only football = 16 + x = 16 + 15 = 31 iii) No.of students who play all the three games = x = 15
Exercise 3.1 1.
2.
3.
Verify the commutative laws of union and intersection of the following sets (i) A = {1,2,3,4,5,6}, B = {5,6,7,8} (ii) A = {a, e, i, o, u}, B = {a, u} (iii) A = {3, 7, 9, 11}, B = {4, 5, 6, 8} (iv A = {1, 3, 5, 7, 9}, B = {2,4,6,8} (v) A = {0, 1, 3, 4}, B = {1, 2, 3, 5} Verify the associative laws for the following sets (i) A = {a, b, c, d, e}, B = {b, d, f, g}, C = {b, e, f, h} (ii) A = {1, 2, 3, 4}, B = {5, 6, 7, 8}, C = {8, 9, 10} (iii) A = {3, 4, 5,6}, B = {2, 5, 6,7}, C = {1, 3, 6, 7} (iv) A = {p, q, r, s}, B = {2, 3, 4}, c = {2, 4, p, r} (iv) A = {5, 6, 7, 8}, B = {4, 5, 6}, C = {6, 7, 8, 9} If A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9} 43
A ∩ B = {1, 2} and A = {1, 2, 3, 4, 5} find the set B. 4.
If A ∪ B = {2, 3, 4, 5, 6, 8, 9, 11}, A ∩ B = {5, 8} and B = {2,5, 8, 9} find the set A − B.
5. 6.
If A = {p, q, r, s} find A ∩ A and A ∪ A. Verify the distributive in the following problems: (i)
A = {a, b, c, d}, B = {c, d, e, f}, C = {b, d, g, f}
(ii)
A = {3, 4, 5, 6}, B = {2, 3, 5, 7, 9}, C = {3, 4, 7, 8, 10}
(iii) A = {3, 5, 6, 7, 9}, B = {1, 2, 3, 4, 7}, C = {3, 6, 7, 8, 9} (iv) U = {x/0 < x < 10, x is an integer},
A = {x/0 < x < 9, x is an even integer}
B = {x/3 < x < 8, x is an odd integer} C = {x/1 < x < 3, x is an integer} 7.
8.
Verify DeMorgan’s laws (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′, using the following sets: (i)
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 7, 8};
B = {3, 4, 6, 7, 8}
(ii)
U ={4, 5, 6, 7, 8, 9, 10}
A = {4, 7, 9};
B = {6, 7, 1}
(iii) ξ = {5, 6, 9, 11, 13, 17, 18}
A = {6, 9, 13, 17};
B = {5, 9, 17, 18}
(iv) ξ = {x/5 < x < 18, x ∈ N}
A = {8, 12, 16};
B = {6, 12, 14}
Verify DeMorgan’s laws A – (B ∪ C) = (A – B) ∩ (A – C) and A – (B ∩ C)) = (A – B) ∪ (A – C): (i)
A = {2, 4, 8},
B = {1, 2, 6,8}
C = {1, 5, 6, 8}
(ii)
A = {4, 7, 9}
B = {8, 10}
C = {6, 7, 10}
B = {2, 5, 6, 7}
C = {1, 3, 6, 7}
(iii) A = {3, 4, 5, 6}
(iv) A = {x/-4 < x < 6, x∈ z), B = {x/0 < x < 4, x∈ z), C = {x/-2 < x < 3, x ∈ z} 9.
In a class of 50 students the number of students who passed in the various subjects is as follows: English 25, Mathematics 18, Science 14, English and Mathematics 8, Maths and Science 5, English and Science 7, all the three subjects 3. Find the no.of students who failed in all the examination.
10. 200 persons lived in a street, of them 120 read English newspapers, 80 read Tamil and 30 read Telugu newspapers. If 60 read and both English and Tamil, 20 read both Tamil and Telugu, 15 read Telugu and English newspapers and 9 read all the papers. Find how many do not read any of the three newspapers. 11. 65% of the people in a city speak Tamil, 52% speak Hindi and 40% speak Malayalam. If 30% speak both Tamil and Hindi, 32% speak both Tamil and Malayalam 25 speak both Hindi and Malayalam and 10% of the people in the city speak other different languages find the percentage of the people who can speak all the three. 12. In a birthday party attended by 100 children, ice cream was served in three different flavours of chocolate, pista and vanilla. 54 children had chocolate 32 pista and 36 vanilla flavour. 16 had both chocolate and pista, 16 had both pista and vanilla, 14 both vanilla and chocolate. 18 children did not take ice cream at all (i) How many had all the three? (ii) How many had chocolate only? 44
13. When a sweet company interviewed 150 students in a school, it was found out that 10 students like product A only, 16 like product B only, 20 like product C only, 40 like both A and B, 20 like both B and C and 50 like both A and C. 3 like all the three products. Find (i) How many like A and B but not C. (ii) How many like B and C but not A (iii) the most popular product and (iv) how many like none. 14. In a street, there are 150 persons are living. They are using 3 brands of soaps 30 persons use only the 1st and the 2nd brand, 31 use only the 2nd and the third brand, 20 use only the 1st and the third brand and 9 use all the three. If same number of people use only one brand find the number of persons who use only one brand. Answers Exercise 3.1 (3) B = {1, 2, 6, 7,8,9} (4) A − B = {3, 4, 6, 11} (5) A ∩ A = {p, q, r, s} A ∪ A = {p, q, r, s} (9) 10 (10) 56 (11) 20% (12) 630 (13) 37, 17, A none (14) 87
45
4. ALGEBRA 4.0
INTRODUCTION
Algebra has been studied for many centuries. Babylonian, ancient Chinese and Egyptian mathematicians proposed and solved problems in words. However it was not until 3rd century that algebraic problems began to be considered in a form similar to those studied today. In the 3rd century, the Greek Mathematician Diophantus of Alexandria wrote his book Arithmetica which provides earliest record of an attempt to use symbols to represent unknown quantities. Several Indian Mathematicians carried out important work in the field of algebra in the 6th and 7th centuries. These include Aryabhatta whose book entitled Aryabhatiya included work on linear and quadratic equations. Sridhara wrote Patiganita Sara, a book on Algebra in 750 A.D. A book on Algebra written in the 9th century by Arabic mathematician AlKhwarizmi developed methods for solving 6 different types of quadratic equations and contained the first systematic consideration of the subject separately from number theory. Brahmagupta gave solutions of quadratic equation and indeterminate equation Nx2 + 1 = My2 called Varaga-Prakriti equation. He is also the founder of the branch of higher mathematics knwon as ‘Numerical Mathematics’. In about 1100, the Persian Mathematician Omar Khayam wrote a treatise on Algebra based on Euclid’s methods. He identified 25 types of equations and made the first formal distinction between arithmetic and Algebra. Bhaskaracharya (1114 AD - 1185 AD) was the first to declare that any number divided by zero is infinity and the sum of any number and infinity is also infinity. His famous book Siddhanta Sironmani is divided into 4 sections, one of which is Bijaganita (algebra) which means ‘the other mathematics’. He introduced Chakrawal or the cyclic method to solve algebraic equations. Six centuries later European mathematicians like Galois, Euler and Lagrange rediscovered this method and called it ‘inverse cyclic’. Bhaskara’s work was mainly in Algebra. Indian Algebra and Trignometry reached Europe through a cycle of translations traveling from the Arab world to Spain and Sicily and eventually all of Europe. In 13th century Leonardo Fibonacci wrote some books on Algebra. In the 14th century Narayana made important contribution to algebra and magic squares. Other good works were from Italian Luca Pacioli (1445-1517) and of English mathematician Robert Recorde (1510-1558). Rules for solving the cubic equations were discovered in 1515 by Scipione del Ferro (1465-1526) and for quartic equation by Ludovico Ferrai (1522-1565) in 1545. Kerala’s Chitrabanu of 16th century gave integral solutions to 21 types of systems of two algebraic equations. In 1799 Karl Friedrich Gauss proved the fundamental theorem of Algebra which has been proposed as early as 1629. In 1824 Niels Henric Abel (1802-1829) proved that in general, it is not possible to give general rules for solving equations of the 5th degree or higher.
4.1
SIMULTANEOUS EQUATIONS
A set of linear equations having a common solution set is called a system of simultaneous linear equations. In earlier classes, you have learnt different methods of solving simultaneous linear equations in two variables. A linear equation in three unknowns say x, y, 46
z is a statement of equality of the form ax + by + cz + d = 0 where a, b, c, d are real numbers with a ≠ 0, b ≠ 0 and c ≠ 0. For example 2x – 3y + 6z = 5 is a linear equation in 3 variables. In this section we are going to learn how to solve linear equations in three variables. To find the values of three unknowns, we need to be given three linear equations in the three unknown variables. 4.1.1
Procedure of solving three given linear equations in x, y, z Three equations are given. Take any two say the first two equations. Eliminate one variable say z. Similarly eliminate z from the second and the third (or first and the third equation). We get two linear equations in x, y. Solve them in the usual way learnt in early cases. Substitute the values of x and y in any one of the three equations to get the value of z. Thus the values of x, y and z are obtained.
Example 1: Solve the equations : x + 2y + 3z = 14, 3x + y + 2z = 11, 2x + 3y + z = 11. Solution: Let the given equations be identified as follows : x + 2y + 3z = 14 (1) 3x + y + 2z = 11 (2) 2x + 3y + z = 11 (3) Consider the equations (1) and (3) (1) ⇒ x + 2y + 3z = 14 (3) × 3 ⇒ 6x + 9y + 3z = 33 –5x – 7y 5x + 7y
= –19 = 19
Consider the equations (2) and (3) (2) ⇒ 3x + y + 2z = 11 (3) × 2 ⇒ 4x + 6y + 2z = 22 –x – 5y x + 5y
(4)
subtracting
= –11 = 11
Consider the equations (4) and (5) (4) ⇒ 5x + 7y = 19 (5) × 5 ⇒ 5x + 25y = 55 –18y
subtracting
= –36 ;
(5)
subtracting ∴y = 2
Substitute y = 2 in (5) we get x + 5(2) = 11; x + 10 = 11; ∴x = 1 Substitute x = 1, y = 2 in (3) we get 2(1) + 3(2) + z = 11; 2 + 6 + z = 11 ⇒ z = 3 The solution is x = 1, y = 2, z = 3. 47
Example 2: Solve : 3x – 3y + 4z = 14; –9x – 6y + 2z = 1; 6x + 3y + z = 5 Solution: Let the given equations be identified as follows: 3x – 3y + 4z = 14 –9x – 6y + 2z = 1 6x + 3y + z = 5
(1) (2) (3)
Consider the equations (1) and (2) (1) × 3 ⇒ 9x – 9y + 12z (2) ⇒ –9x – 6y + 2z –15y + 14z
= 42 = 1 adding = 43
Consider the equations (1) and (3) (1) × 2 ⇒ 6x – 6y + 8z = 28 6x + 3y + z = 5 – 9y + 7z =
subtracting
23
Considering the equations (4) and (5) (4) ⇒ –15y + 14z = 43 (5) × 2 ⇒ –18y + 14z = 46 3y
(4)
= –3
(5)
subtracting ⇒
y = –1
Substituting y = –1 in (4) –15(–1) + 14z = 43 or 14z = 43 – 15 = 28 ⇒ z = 2 Substituting y = –1 and z = 2 in equation (1) we get 3x – 3(–1) + 4(2) = 14 or 3x = 14 – 11 = 3 or x = 1 ∴ The solution is x = 1, y = –1, z = 2 Example 3: Solve : x+ y = 3, y + z = –5, z + x = 2. Solution: Let the equations be identified as x+y = 3 y + z = –5 z+x = 2
(1) (2) (3)
Adding all the three equations, we get 2x + 2y + 2z = 3 + (–5) + 2 or 2(x + y + z) = 0 or x + y + z = 0 Substituting y + z = –5 in equation (4) we get x + (–5) = 0 ∴x = 5. Substituting z + x = 2 in equation (4) we get y + 2 = 0 ∴ y = –2. Substituting x = 5 in equation (3) we get z + 5 = 2 or z = 2–5 = –3 The solution is x = 5, y = –2, z = –3. 48
(4)
Example 4: Solve : 2/x + 3/y – 4/z = – 20; 2/y – 4/x + 3/z = 45; 3/x – 4/y + 2/z = 5 Solution: Let 1/x = a, 1/y = b, 1/z = c Then 2a + 3b – 4c = –20 –4a + 2b + 3c = 45 3a – 4b + 2c = 5
(1) (2) (3)
Consider equations (1) and (2) (1) (2)
× 3 ⇒ 6a + 9b – 12c = –60 × 4 ⇒ –16a + 8b + 12c = 180 –10a + 17b
adding
= 120
(4)
Consider equations (1) and (3) (1) (3)
⇒ 2a + 3b – 4c = –20 × 2 ⇒ 6a – 8b + 4c = 10 8a – 5b
(4) (5)
× 8 ⇒ –80a + 136b × 10 ⇒ 80a – 50b 86b
adding
= –10
(5)
= 960 = –100
adding
=
∴ b = 10
860
Substituting b = 10 in equation (5) we get 8a – 5(10) = –10; 8a = –10 + 50 = 40 ∴ a = 5. Substituting a = 5, b = 10 in equation (3) we get 3(5) – 4(10) + 2c = 5 or –25 + 2c = 5 or 2c = 5+25 = 30 ∴ c = 15. Since a = 5, b = 10, c = 15, we get x = 1/5, y = 1/10, z = 1/15. Exercise 4.1.1 Solve the following simultaneous equations. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.
3x – 2y + z = 0; 2x – 2y + 4z = –12; 2x + 3y – z = 5; x + 2y + 3z = 10; x + y = 7; x + y = –3; x – y = 2; x + 2z = 7; 3/x – 4/y – 6/z = –3; 2/x + 3/y + 1/z = 4; 1/x + 1/y = 1; 4/x + 3/y = 1;
4x + 6y – 3z = 13; 3x + 2y + 2z = 19; 4x + y + 3z = 5; x – 2y + 4z = 3; y + z = 4; y + z = 1; 3x + 2y – 3z = 13; 2x – y + 3z = 9; 2/y – 6/x + 3/z = 1; 4/x – 6/y + 3/z = –5; 1/y + 1/z = 2; 9/y – 8/z = 5;
49
x – 2y + 2z = –4 –x + y – z = 3 3x + 2y + 2z = 5 x + y – 3z = 2 z+x=1 z + x = –8 x – 3y + 5z = 3 y–z=1 9/z – 9/x + 1/y = 7/2 3/x – 5/y + 2/z = –4 1/z + 1/x = 4 1/z + 6/x = 2
4.1.2 Problems leading to simultaneous equations In this section we shall discuss applications of simultaneous linear equations in solving problems related to our day-to-day life. When the equations are not directly given, we have to form the equations from the given data and solve the equations formed. Example 5: In Δ ABC, m∠C is 20° greater than m∠A. The sum of m∠A and m∠C is twice m∠B. Find three angles. Solution: Let x, y, z be the angles m∠A, m∠B, m∠C respectively in Δ ABC. According to the given data. z = x + 20 ; x + z = 2y Since the sum of three angles in triangle is 180°, x + y + z = 180° The above equations can be written as –x + z = 20° x – 2y + z = 0° x+y+z = 180° Considering (2) and (3) (2) ⇒ x – 2y + z = 0 (3) × 2 ⇒ 2x + 2y + 2z = 360° 3x + 3z x+z
(1) (2) (3)
adding
= 360° = 120°
Considering (1) and (4) (1) ⇒ –x + z (4) ⇒ x+z 2z
(4)
= 20° = 120°
adding
= 140°
(or)
∴ z = 70°
Substitute z = 70° in (4) we get x + 70° = 120° or x = 50° Substitute x = 50° and z = 70° in (3) we get 50° + y + 70° = 180° or y = 60° Hence the angles of a triangle are m∠A = 50°; m∠B = 60°; m∠C = 70° Example 6: In a shop three persons A, B and C purchased the following quantites of rice, wheat and sugar. A B C
Rice (Kg) 3 2 4
What (Kg) 2 3 5
50
Sugar (Kg) 4 2 4
If they have paid respectively Rs.140, Rs. 104 and Rs.196 for the purchases made, find the cost of 1 kg of rice, wheat and sugar. Solution: Let tyhe sale price of 1 kg of rice be Rs. x, wheat by Rs.y and sugar be Rs.z, then we get 3x + 2y + 4z = 140 (1) 2x + 3y + 2z = 104 (2) 4x + 5y + 4z = 196 (3) Considering the equations (1) and (2) (1) (2)
⇒ 3x + 2y + 4z = 140 × 2 ⇒ 4x + 6y + 4z = 208 – x – 4y x + 4y
subtracting
= –68 = 68
(4)
Considering equations (2) and (3) (2) (4)
×2⇒ ⇒
4x + 6y + 4z = 208 4x + 5y + 4z = 196 y
Substituting
=
subtracting
12
y = 12 in (4) we get x + 4(12) = 68 or x = 68 – 48 = 20
Substituting
x = 20, y = 12 in (1) we get 3(20) + 2(12) + 4z = 140 or 84 + 4z = 140 z = 56/4 = 14 ∴ x = 20, y = 12, z = 14
∴ Rice = = Rs. 20 / Kg ; Wheat = Rs. 12 / Kg ; Sugar = Rs. 14/Kg. Example 7: A bag contains ten, five and two rupees currencies. The total number of currencies is 20 and the total value of money is Rs.125. If the second and third sorts of currencies are interchanged the value will be decreased by Rs. 6. Find the number of currency in each sort. Solution: Let x, y, z be the number of Rs.10, Rs.5 and Rs.2 currencies respectively. The total number of currencies is 20 ⇒ x + y + z = 20 (1) If the total value of money is Rs. 125 ⇒ 10x + 5y + 2z = 125
(2)
If the II and III sorts of currencies are interchanged the value will be decreased by Rs.6 ⇒ 10x + 5z + 2y = 125 – 6 = 119 10x + 2y + 5z = 119
(3)
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10 × (1) – (2) ⇒ (2) – (3) ⇒
5y + 8z 3y – 3z y–z
Let us solve (4) and (5) (4) ⇒ 5y + 8z 8 × (5) ⇒ 8y – 8z 13y
= = =
= =
75 16
=
91
75 6 2
(4) (5)
y=7
Substituting y = 7 in (5) we get z=5 Substituting z = 5,
y = 7 in (1) we get
x + 7 + 5 = 20 ⇒ 20 – 7 – 5 or x=8 Number of Rs. 10, Rs. 5 and Rs. 2 currencies are respectively 8, 7 and 5 Example 8: The sum of three numbers is 24. Among them one number is equal to half of the sum of other two numbers but four times the difference of them. Find the numbers. Solution: Let the numbers be x, y, z. Sum of numbers is 24 ⇒ x + y + z = 24 (1) One number is equal to half of the sum of the other two ⇒ x = ½ (y + z) (2) The same number is four times the difference of them ⇒ x = 4(y–z) or x–4y+4z=0 (3) (1) + (2) ⇒ 3x = 24 or x = 8 (1) ⇒ y + z = 24 – x = 24 – 8 = 16 or y + z = 16 (4) (3) ⇒ 4y – 4z = x = 8 or y – z = 2 (5) (4) + (5) ⇒ 2y = 18 or y = 9. (4) ⇒ 9 + z = 16 or z = 7 ∴ The numbers are 8, 9 and 7. Exercise 4.1.2 1.
2. 3.
4.
In a workshop the pay bill of workers in 3 successive weeks were Rs. 1200, Rs. 1130 and Rs. 1160. If 5 men, 5 women and 6 children worked in the first week, 4 men, 6 women and 5 children worked in the second week, 4 men, 7 women and 4 children worked in the third week find the wage paid for each worker. 4 pens, 12 note books, 6 ball pens cost Rs. 160, 3 pens, 4 note books and 1 ball pen cost Rs. 66, 3 pens, 6 note books and 4 balls pens cost Rs. 94. Find the cost of each. The sum of the digits in a three digit number is 24. Twice the tenth digit is equal to sum of the digits in the other two places. If 198 is added with the number, then the digits will be in the reversed order. Find the number. In an auditorium, there are 500 seats. The cost of 1st, 2nd, 3rd class tickets were respectively Rs.100, Rs.50 and Rs.30. On a houseful day, the gate collection was Rs. 25000. On a charitable show day, which was again a houseful day, the tickets were sold at Rs.200, Rs. 100 and Rs.50 and the total collection was Rs.47,500. Find the number of seats in each class.
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5. 6.
7.
4.2
In ΔABC, the sum of the first two angles is twice the third angle m∠B is 5° greater than m∠C. Find the angles. 100 pencils are to be kept inside three types of boxes A, B and C. If 5 boxes of type A, 3 boxes of type B, 2 boxes of type C are used 6 pencils are left out. If 3 boxes of type A, 5 boxes of type B. 2 boxes of type C are used, 2 pencils are left out. If 2 boxes of type A, 4 boxes of type B and 4 boxes of type C are used, there is space for 4 more pencils. Find the number of pencils that each box can hold. The sum of the digits in a three digit number is 15. If 99 is subtracted the digits will be in the reverse order. The digit in the ten’s place is equal to 2/3 times the sum of the digits in the hundred’s place and unit’s place. Find the number.
POLYNOMIAL
In class IX we have learnt about polynomials and operations on polynomials in detail. Recall all you have learnt by answering the following. •
Write a polynomial of degree n
•
Give examples for monomial, binomial and trinomial.
•
What is the degree of 7x4 – 2x3 + 5x + 6?
•
What is a zero polynomial?
•
What is a constant term?
•
Find the quotient and remainder on dividing x3 – 3x2 + 6x + 5 by x – 2.
What is your answer for the last problem? The quotient is x2 – x + 4 and the remainder is 13. You got the answer by long division method. The procedure of long division is shortened by a method known as synthetic division. 4.2.1 Synthetic Division In this method, the dividend is first arranged in descending powers of x. If any power of x in between missing the term is written as 0. The calculations are carried out with the successive coefficients of the dividend as shown. 1 Dividend : x3 – 3x2 + 6x – 5 Divisor : x – 2 2 0 Quotient is = x2 – x + 4; remainder is = 13 1
–3 2 –1
6 –2 4
5 8 13 = Remainder
Coefficients of the quotient Example 9: Find the quotient and the remainder when x3 + x2 – 2x + 7 is divided by x + 4 1 1 −2 7 Solution:
−4
− 4 + 12 − 40 1 − 3 + 10 − 33
Quotient
= x2 – 3x + 10
remainder
= –33
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Example 10: Find the quotient and the remainder when 6x4 – 11x3 + 5x2 – 7x + 9 is divided by 2x – 3. 3⎞ ⎛ Solution: Consider the divisor 2x – 3 = 2 ⎜ x − ⎟ 2⎠ ⎝ 3 6 −11 5 −7 9 2 +9 −3 3 −6 6
−2
+2 −4
3
1 (6x3 – 2x2 + 2x – 4) = 3x3 – x2 + x – 2 2
Quotient
=
Remainder
=3
Exercise 4.2.1 Find the quotient and the remainder using synthetic division 1. 3. 5. 7.
x3 + x2 – 3x + 5 ÷ x – 1 5x3 + 4x2 – 6x – 8 ÷ x + 2 3x3 + 4x2 – 10x + 6 ÷ 3x – 2 8x4 – 2x2 + 6x – 5 ÷ x + 1
4.2.2
2. 4. 6. 8.
4x3 – 3x2 + 2x – 4 ÷ x – 3 x3 + x2 – 10x + 8 ÷ x + 4 4x3 – 6x2 – 8x + 7 ÷ 2x + 5 3x3 – 4x2 – 5 ÷ x – 1
Remainder Theorem
We see that when (x–2) divides the polynomial P(x) = 2x4 + x2 – 7x + 3, the remainder is 25. Now calculating P(2), we find that P(2) = 2(2)4 + 22 – 7(2) + 3 = 25. We observe that when the polynomial P(x) is divided by (x–2), the remainder is 25 = P(2). Taking a few polynomials of our own and dividing them by the binomial (x − a) of our choice, we will find the remainder P(a). This is generalized by the remainder theorem. Remainder Theorem : Let P(x) be any polynomial of degree greater than or equal to one and let a be any real number. When P(x) is divided by the binomial (x–a) the remainder is P(a). Proof : P(x) is divided by (x – a), let the quotient be q(x) and the remainder be r(x). Then we have p(x) = (x–a) q(x) + r(x) where degree of r(x) < degree of divisor or r(x) = 0. Since degree of x – a is 1, r(x) is a constant say r, as discussed earlier.
Hence for all values of x, P(x) = (x – a) q(x) + r. In particular for x = a, we have from above P(a) = (a – a) q(a) + r = 0 × q(a) + r ⇒ P(a) = r.
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Note : If (x + a) divides P(x), the remainder is P(–a). If (ax + b) divides P(x), the remainder is P (–b/a). Example 11: Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1). Solution: P(x) = x3 – 5x2 + 7x – 4. By the remainder theorem when P(x) is divided by (x – 1), the remainder is P(1).
∴ The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1. Example 12: Find the remainder when 3x3 + 4x2 – 5x + 8 is divided by x + 2. Solution: When P(x) is divided by (x + 2), the remainder is P(–2).
∴ The remainder P(–2) = 3(–2)3 + 4(–2)2 – 5(–2) + 8 = 3(–8) + 4(4) + 10 + 8 = 10 Example 13: Find m if 5x5 – 9x3 + 3x + m leaves a remainder 7 when divided by (x+1). Solution: Remainder is P(–1) = 5(–1)5 – 9(–1)3 + 3(–1) + m = –5 + 9 – 3 + m = 1 + m But remainder is 7 ∴ 1 + m = 7 ⇒ m = 7 – 1 = 6. Example 14: The polynomial x2 + ax + b gives remainder 18, when divided by x – 2 and leaves a remainder –2 when divided by (x + 3). Find the values of a and b. Solution: P(x) = x2 + ax + b. When x – 2 divides P(x), the remainder is P(2). ∴P(2) = 4 + 2a + b.
But remainder = 18 ⇒ 4 + 2a + b = 18 ; 2a + b = 14
(1)
When (x + 3) divides P(x), the remainder is P(–3). ∴ P(–3) = (–3)2 + a(–3) + b = 9 – 3a + b. But remainder = –2 ; ∴ 9 – 3a + b = –2 ; ⇒ –3a + b = –11 (1) ⇒ 2a + b = 14 (2) ⇒ –3a + b = –11 subtracting 5a = 25 (or) a = 5 Substituting
(2)
a = 5 in equation (1) we get 10 + b = 14; b = 4, ∴ a = 5, b = 4
Example 15: If a quadratic polynomial is divided by (x–1), (x+1) and (x–2) leaves the remainders 2, 4, 4 respectively, find the quadratic polynomial. Solution: Let the quadratic polynomial be p(x) = ax2 + bx + c. By the given data, we have p(1) = 2, p(–1) = 4 and p(2) = 4.
p(1) = 2 ⇒ p(–1) = 4 ⇒ p(2) = 4 ⇒ Considering (1) and (2)
a+b+c a–b+c 4a + 2b + c
= 2 = 4 = 4
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(1) (2) (3)
(1) ⇒ (2) ⇒
a+b+c a–b+c 2b
=2 =4
subtracting
= –2 ; b = –1
Substitute b = –1 in (1) and (3) we get (1) ⇒ a – 1 + c = 2; ⇒∴a+c=3 (3) ⇒ 4a – 2 + c = 4; ⇒ ∴ 4a + c = 6
(4) (5)
Considering (4) and (5) we have (5) ⇒ 4a + c = 6 (4) ⇒ a + c = 3 subtracting 3a
= 3
;
a = 1
put a = 1 in (4) (or) 1 + c = 3 ⇒ c = 2 ∴ The required quadratic polynomial is q(x) = x2 – x + 2. Example 16: Given that px2 + qx + 6 leaves a remainder 1 on division by 2x + 1 and 2qx2 + 6x + p leaves a remainder 2 on division by 3x – 1 find p and q. Solution: f(x) = px2 + qx + 6
When 2x + 1 divides f(x), the remainder is f (–1/2) p q p − 2q + 24 ⇒ f( –1/2) = − +6 = 4 2 4 p − 2q + 24 But remainder = 1 (given) ∴ = 1 4 p – 2q + 24 = 4 (or) p – 2q = – 20
(1)
When (3x – 1) divides g(x) = 2qx2 + 6x + p, the remainder is g(1/3) g (1/3) = 2q (1/3)2 + 6(1/3) + p 2q + 18 + 9p 2q 6 + + p = 9 3 9 2q + 18 + 9p But remainder is 2 (given) ∴ = 2 9 2q + 18 + 9p = 18 (or) 9p + 2q = 0 =
(1) + (2) ⇒
p – 2q
⇒
9p + 2q 10p Substituting
= –20 = 0
∴ p = –2 = –20 p = –2 in (2) we get
– 18 + 2q = 0 ⇒ ∴ q = 9, p = –2 ; q=9
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(2)
Example 17: Find the value of a and b if ax3 + bx2 + 7x + 9 and x3 + ax2 – 2x + b – 4 when divided by x + 2 leave remainders –13 and –16 respectively. Solution: f(x) = ax3 + bx2 + 7x + 9 When (x+2) divides f(x), the remainder is –13
⇒ f(–2) = –13 or –8a + 4b – 14 + 9 = –13 –8a + 4b = –8 ⇒ 8a –4b = 8 ⇒ 2a – b = 2 g(x) = x3 + ax2 – 2x + b – 4.
(1)
When x + 2 divides g(x), the remainder is –16.
⇒ g(–2) = –16 or –8 + 4a + 4 + b – 4 = –16 (or) 4a + b = –8
(2)
Let us solve (1) and (2) 2a – b = 2 4a + b = –8 adding 6a
= –6 (or)
Substituting
a = –1
a = –1 in (1) we get –2 – b = 2 (or) ⇒ b = –4
The solution is a = –1, b = –4 Exercise 4.2.2 1. Find the remainder using remainder theorem a) 4x3 – 5x2 + 2x – 6 ÷ x – 2 b) 5x3 – 6x2 + 3x – 4 ÷ x + 2 d) x3 – 17x – 21 ÷x + 4 c) x4 + 2x3 – 5x2 – 6x – 4 ÷x – 3 4 3 2. When a polynomial 3x + mx – 2x – 8 is divided by x + 2, the remainder is 20. Find the value of m. 3. When x + 2 divides 4x3 + 5x2 + px – 2 without remainder find p. 4. If (x–2) divides 3x3 – 2x2 + mx – 20 without remainder find m. 5. Find the value of a if 10x2 + ax – 10 leaves a remainder 2 when divided by 2x – 3. 6. Find the value of m if 2x3 + 3x2 + mx + 5 leaves a remainder – 15 when divided by 2x + 5. 7. Find a and b if x3 + 7x2 + ax + b leaves a remainder 40 when divided by x–2 and a remainder 25, when divided by x + 3. 8. When ax2 + bx + c is divided by (x+1), (x–2), (x+3), the remainders are –8, 13, 8 respectively. Find a, b and c. 9. Find the values of a and b if x3 + ax2 + bx + 8 leaves a remainder 2 when divided by (x–1) and (x–2). 10. Given that px3 + 9x2 + qx + 1 leaves remainder 4 on division by 2x + 1 and 9x3 + qx2 + px + 1 leaves the remainder 3 on division by 3x – 1. Find p and q.
57
4.2.2
Factor Theorem If p(x) is a polynomial of degree n > 1 and a is any real number then (i) (x–a) is a factor of p(x) if p(a) = 0 and (ii) p(a) = 0 if (x–a) is a factor of p(x). Example 18: Determine whether (x–3) is a factor of the polynomial p(x) = x3 – 3x2 + 4x – 12 Solution: For (x–3) to be a factor of p(x), p(3) should be zero by the factor theorem. Now p(3) = 33 – 3(3)2 + 4(3) – 12 = 27 – 27 + 12 – 12 = 0 Hence (x–3) is a factor of the given polynomial. Example 19: Determine the value of m if x + 1 is a factor of x3 + mx2 + 19x + 12 Solution: Let P(x) = x3 + mx2 + 19x + 12 P(–1) = (–1)3 + m(–1)2 + 19(–1) + 12 = –8 + m By factor theorem since (x + 1) is a factor P(–1) = 0 or –8 + m = 0 or m = 8 Example 20: Find the values of a and b if 3x4 + x3 + ax2 + 5x + b is exactly divisible by x + 2 and x – 1. Solution: Let P(x) = 3x4 + x3 + ax2 + 5x + b
(x+2) and (x–1) are factors of P(x) and hence both P(–2) and P(1) = 0 P(–2) = 3(–2)4 + (–2)3 + a(–2)2 + 5(–2) + b = 30 + 4a + b Since P(–2) = 0 we get 4a + b = –30 4
3
(1) 2
P(1) = 3(1) + (1) + a(1) + 5(1) + b = 9 + a + b Since P(1) = 0 we get 9 + a + b = 0 or a + b = –9
(2)
(1) – (2) gives 4a + b = –30 a + b = –9 3a = –21 (or) Substituting
a = –7
a = –7 in equation (2) we get –7 + b = –9 (or) ∴ b = –2
∴ a = –7, b = –2. Example 21: If (x – 1), (x + 2), (x – 3) exactly divide a cubic polynomial with leading coefficient unity, then find the polynomial. Solution: Since the leading coefficient of the cubic polynomial is 1, let the cubic polynomial be f(x) = x3 + ax2 + bx + c. Given that x – 1, x + 2 and x – 3 exactly divide f(x) ⇒ they are factors of f(x). By using factor theorem, we have f(1) = f(–2) = f(3) = 0.
f(1) = 0 ⇒ 1 + a + b + c = 0 (or) a + b + c = –1 f(–2) = 0 ⇒ –8 + 4a – 2b + c = 0 (or) 4a – 2b + c = 8 f(3) = 0 ⇒ 27 + 9a + 3b + c = 0 (or) 9a + 3b + c = –27 58
(1) (2) (3)
Considering the equations (1) and (2) (2) ⇒ 4a – 2b + c (1) ⇒ a + b + c 3a – 3b (3) ⇒ 9a + 3b + c 4a – 2b + c 5a + 5b
= = = = =
8 –1 9 (or) –27 8
= –35 (or)
subtracting a–b=3
(4)
subtracting a + b = –7
(5)
Considering the equations (4) and (5) we have (4) ⇒ a – b = 3 a+b = –7 2a
= –4 (or)
a = –2
Substituting
a = –2 in (5) we get –2 + b = –7 (or) b = –5
Substituting
a = –2, b = –5 in (1) we get –2 –5 + c = –1 (or) c = 6
Hence the required cubic polynomial is f(x) = x3 – 2x2 – 5x + 6 Exercise 4.2.3 1.
2.
3. 4. 5. 6.
Determine whether (x – 1) is a factor of b. x3 – 27x2 + 8x + 18 a. x3 + 8x2 – 7x – 2 4 3 c. 8x – 12x + 18x + 14 d. 8x4 + 12x3 – 16x – 4 Find a in the following functions: a. x3 – 3x + 3a is exactly divisible by x + 3 b. (x + 1) is a factor of 8x4 – ax3 – x2 – 3x + 4 c. 3x4 + ax2 + 58x + 40 is divisible by x +5 d. x3 + 8x2 + ax – 2 is divisible by (x – 1) If x2 – 5x + 6 is a factor of 3x3 + ax2 + bx + 12 find a and b. If (x – 2) is a common factor of x3 – 4x2 + ax + b and x3 – ax2 + bx + 8, find the values of a and b. If (x – 7) and (x – 4) are the factors of px3 + qx2 – 5x + 84, find the values of p and q. If x – 1, x + 2 and x – 2 are the factors of x3 + ax2 + bx + c, find a, b and c.
4.2.4 Factorisation
We have already learnt in Class IX to factorise the quadratic polynomial expressions. Let us learn in this class, to factorise polynomial expressions of degree three or more using factor theorem and synthetic division. Note : 1. If the sum of all coefficients in a polynomial including the constant term is zero, then x – 1 is a factor. 2. If the sum of the coefficients of even powers together with the constant term is the same as the sum of the coefficients of odd powers, then x + 1 is a factor.
59
Example 22: Factorise 2x3 + x2 – 5x + 2 Solution: Since the sum of the coefficients of all the terms: 2 + 1 – 5 + 2 = 5 – 5 = 0 we guess that (x – 1) is a factor.
By synthetic division 1 2 +1 − 5 + 2 2 +3 −2 2
3
−2
0 = Remainder
Remainder is 0. Quotient is 2x2 + 3x – 2 To find other factors, factorise the quotient, 2x2 + 3x – 2 = 2x2 + 4x – x – 2 = 2x (x + 2) – 1 (x + 2) = (x + 2) (2x – 1) 3 2 ∴ 2x + x – 5x + 2 = (x – 1) (x + 2) (2x – 1) Example 23: Factorise 3x3 – 4x2 – 13x – 6. Solution: Sum of the coefficients of all terms: 3 – 4 – 13 – 6 = 20 ≠ 0. ∴ (x – 1) is not a factor. Sum of the coefficients of even degree terms = –4 – 6 = –10. Sum of the coefficients of odd degree terms = 3 – 13 = –10. Since they are equal (x + 1) is a factor. By synthetic division −1 3 −4 −13 −6 −3
+7
+6
3 −7
−6
0
Remainder is 0. ∴(x + 1) is a factor. Factorising the quotient 3x2 – 7x – 6
= 3x2 – 9x + 2x – 6 = 3x (x–3) + 2 (x–3) = (x–3) (3x+2) 3 2 ∴ 3x – 4x – 13x – 6 = (x+1) (x–3) (3x+2) Example 24: Factorise x3 – 3x2 – 10x + 24 Solution: Sum of the coefficients of terms: 1–3–10 + 24 = 12 ≠ 0. ∴ (x–1) is not a factor. Sum of the coefficients of even degree terms = –3 + 24 = 21 Sum of the coefficients of odd degree terms = 1 – 10 = –9 Since they are not equal we guess that (x + 1) is also not a factor. Let us check whether x – 2 is a factor. By synthetic division.
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2 1 − 3 − 10 + 24 +2 1
−2
− 1 − 12
− 24 0 Remainder
Since the remainder is 0, (x – 2) is a factor. To find other factors x2 – x – 12 = x2 – 4x + 3x – 12 = x (x–4) + 3 (x–4) = (x + 3) (x – 4) ∴ x3 – 3x2 – 10x + 24 = (x–2) (x–4) (x+3)
Exercise 4.2.4 Factorise the following 1. 4. 7. 10. 13.
x3 – 23x2 + 142x – 120 x3 – 3x2 + 4 3x3 – 10x2 + 11x – 4 x3 + 6x2 + 11x + 6 2x3 + 3x2 – 2x – 3
2. 5. 8. 11. 14.
x3 + 13x2 + 32x + 20 x3 + 2x – 3 x3 + 2x2 + 2x + 1 x3 – 6x2 + 11 x –6 x3 – 3x2 – x + 3
4.3
G.C.D AND L.C.M
4.3.1
Greatest Common Divisor (G.C.D.)
3. 6. 9. 12. 15.
x3 – 7x + 6 x3 + 4x2 + 5x + 2 x3 – x2 + x – 6 2x3 – x2 – 8x + 4 x3 – 5x + 4
The greatest common divisor G.C.D. or Highest Common Factor (HCF) of two or more polynomials is that common divisor which has highest degree among all common divisors and in which the coefficient of highest degree term is positive. For example consider x3 y2 z p, x3 y7 z3 p2, x2 y z2. We observe that x, x2, y, z, xy, x2y, xz, yz, x2z, x2yz are common divisors of the given terms. Among all the common divisors x2yz is the common divisor of highest degree. Hence the G.C.D. of x3y2 zp, x3y7z3p2, x2yz2 is x2 yz. Example 25: Find the G.C.D. of (1) 25, 35, 45 (2) 36, 48, 144. Solution: 1) 25 = 52 ; 35 = 5 × 7 ; 45 = 32 × 5 ⇒ G.C.D. is 5 2) 36 = 22 × 32 48 = 24 × 3 144 = 24 × 32 ⇒ G.C.D = 22 × 3 = 12 Example 26: Find the G.C.D. of x5, x7 and x10 Solution: G.C.D. is x5 Example 27: Find the G.C.D. of 3x2 y2 z2, 6x2 yz2, 9xyz2 Solution: Given 3x2 y2 z2, 2 × 3x2 yz2, 32 xyz2 ⇒ G.C.D = 3xyz2 Example 28: Find the G.C.D. of a3 – 1 and a2 – 1 Solution: a3 – 1 = a3 – 13 = (a – 1) (a2 + a + 1) and a2 – 1 = a2 – 12 = (a+1) (a–1) The common factor is a – 1 ∴ G.C.D. = a – 1
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Example 29: Find the G.C.D. of x2 + 2xy + y2, (x + y)3, 25 (x2 – y2) Solution: x2 + 2xy + y2 = (x + y)2 (x + y)3 = (x + y)3 25 (x2 – y2) = 25 (x + y) (x – y) x + y is the only common factor ∴ G.C.D. = (x + y) G.C.D. by long division method
When the polynomials are not easily factorisable the G.C.D. can be found by long division method as follows: Let f(x) and g(x) be two polynomials with deg g(x) < deg f(x). Then leave out the common numerical factor of f(x) and g(x) and divide f(x) by g(x) to get the remainder r(x). If r(x) is zero, then g(x) is the G.C.D. If not leave out the common numerical factor of r(x) and then divide g(x) by r(x) to get the remainder s(x). If s(x) is zero then r(x) is the G.C.D. If not repeat the above process again and again till we get the remainder as zero. The divisor of the last polynomial is called the G.C.D. of f(x) and g(x). Example 30: Find the G.C.D. of the following polynomials x3 – 9x2 + 23x – 15 and 4x2 – 16x + 12 Solution: Let f(x) = x3 – 9x2 + 23x – 15 and g(x) = 4x2 – 16x + 12 = 4 (x2 – 4x + 3)
x–5 x2 – 4x + 3
x3 – 9x2 + 23x – 15 x3 – 4x2 + 3x – 5x2 + 20x – 15 – 5x2 + 20x – 15 0
G.C.D. =
x2 – 4x + 3
Example 31: Find the H.C.F. of the polynomials 2x3 + 2x2 + 2x + 2 and 6x3 + 12x2 + 6x + 12 Solution: Let f(x) = 2x3 + 2x2 + 2x + 2 = 2 (x3 + x2 + x + 1) and
g(x) = 6x3 + 12x2 + 6x + 12 = 6(x3 + 2x2 + x + 2) 1 x3 + x2 + x + 1
x3 + 2x2 + x + 2 x3 + x2 + x + 1
≠0
x2 + 1
62
Dividing x3 + x2 + x + 1 by x2 + 1, we have x+1 x2 + 1
x3 + x2 + x + 1 +x x3 x2 + 1 x2 + 1 0
∴ H.C.F. of the polynomials = 2 (x2 + 1) (since the G.C.D. of 2 and 6 is 2). Example 32: Find the G.C.D. of the polynomials x3–3x2+4x–12 and x4 + x3 + 4x2 + 4x Solution: Let f(x) : x3–3x2 + 4x – 12 and g(x) = x4 + x3 + 4x2 + 4x = x (x3 + x2 + 4x +4)
1 x3 + x2 + 4x + 4
x3 – 3x2 + 4x – 12 x3 + x2 + 4x + 4 – 4x2
– 16
= – 4 (x2+4) ≠ 0
x+1 x2 + 4
x3 + x2 + 4x + 4 + 4x x3 x2 + 4 x2 + 4 0
∴ G.C.D. = x2 + 4 Exercise 4.3.1 1.
2.
Find the G.C.D. of the following terms a) 48x9, 64x5 b) 78a3, 52a10 c) 24m6, 36m d) 12y4 20y5 e) p8, p9, p11 f) m6, m12, m18 n n+1 n+2 4 6 7 g) x , x , x h) 3a , 9a , 12a i) 3a2bc, 6ab2c, 9abc2 j) 4x3y3z2, 6xy2 z3, 8xyz2 k) 4p2q3r, 8p3q2r2, 16p2q4r3 l) 14m2n, 28mn2, 21m3n3 Find the G.C.D. of the following a) 3x – 6, 5x – 10 b) 4x + 10, 6x – 15 c) 4x2 – 3x, 3x2 + 2x d) (a – b)2, a2 – b2 e) x2 – 4, x + 2 f) x3 + 1, x2 – 1 g) x2 – 4, x3 – 8 h) 2x2 – 11x – 40, 2x2 – 9x – 35
63
3.
4.3.2
i) 6a2 – 7a – 3, 10a2 – 11a – 6 j) a2 – 8a+16, (a+3) (a–4) (a2–a–12) k) 6x2 + 7x – 3, 10x2 + 11x – 6, 2x2 – 11x – 21 l) 64(x2–9), 36(x2 + 10x + 21), 24 (x2 + 2x – 3) Find the G.C.D. of the following pair of polynomials by successive division method. a) x4 + 2x3 + x2 – 1; x4 + x2 + 1 b) x3 + 6x2 + 11x + 6 ; x3 + 9x2 + 27x + 27 c) 3x2 + 13x + 10 ; 3x3 + 18x2 + 33x + 18 d) x3 + 4x2 – 5 ; x3 – 3x + 2 e) 24x4 – 2x3 – 60x2 – 32x ; 18x4 – 6x3 – 39x2 – 18x Least Common Multiple (L.C.M.)
The least common multiple of two or more polynomials is the polynomial of the lowest degree which is exactly divisible by the given polynomials and whose coefficient of the highest degree term has the same sign as the sign of the coefficient of the highest degree term in their product. Example 33: Find the L.C.M of 12(x–1)3 and 15(x–1) (x+2)2 Solution: 12(x–1)3 = 22 × 3 (x –1)3 15 (x–1) (x + 2)2 = 5 × 3 (x – 1) (x + 2)2 2 L.C.M. = 2 × 3 × 5 (x – 1)3 (x + 2)2 = 60 (x – 1)3 (x + 2)2 Example 34: Find the L.C.M. of 28 p and 98 q. Solution: 28p = 4 × 7p = 22 × 7p ; 98q = 2 × 49q = 2 × 72 q ∴ L.C.M. = 22 × 72 pq = 4 × 49 pq = 196 pq Example 35: Find the L.C.M. of 6x2y, 9x2yz, 12x2y2z Solution: 6x2y = 2 × 3 × x2 y 9xy2z = 32 x2 yz 12x2y2z= 22 × 3x2 y2z L.C.M. = 22 × 32 × x2y2z = 4 × 9 x2y2z L.C.M. = 36x2y2 z Example 36: Find the L.C.M. of (x–1) (x+2) and (x+2) (x+3) Solution: L.C.M. = (x–1) (x+2) (x+3) Example 37: Find the L.C.M. of x3 + 1, x2 – 1, (x + 1)2 Solution:
x3 + 1 = (x + 1) (x2 – x+ 1) x2 – 1 = (x + 1) (x – 1) ; (x + 1)2 = (x + 1)2 L.C.M. = (x + 1)2 (x – 1) (x2 – x + 1)
Example 38: Find the L.C.M. of x3 + y3, x3 – y3 and x4 + x2y2 + y4 Solution:
x3 + y3 = (x + y) (x2 – xy + y2) 64
x3 – y3 = (x – y) (x2 + xy + y2) x4 + x2y2 + y4 = (x2 + xy + y2) (x2 – xy + y2) L.C.M = (x + y) (x2 – xy + y2) (x – y) (x2 + xy + y2) = (x3+y3) (x3–y3). Relation between L.C.M. and G.C.D. Consider two polynomials f(x) and g(x). By inspection we find that L.C.M. and G.C.D. are connected by the following relations:
f(x) × g(x) = L.C.M. of f(x) and g(x) × G.C.D. of f(x) and g(x). If we know any three of the four functions in the above product, we can find the fourth one. For example consider, f(x)
=
x2 – 2x + 1 = (x – 1)2
g(x)
=
x2 + x – 2 = (x – 1) (x + 2)
G.C.D.
=
(x – 1)
L.C.M.
=
(x – 1)2 (x + 2)
Now, f(x) × g(x)
=
(x – 1)2 × (x – 1) (x + 2) = (x – 1)3 (x + 2) and
L.C.M. × G.C.D.
=
(x – 1)2 (x + 2) × (x – 1) = (x – 1)3 (x + 2)
∴ f(x) × g(x)
=
L.C.M. × G.C.D.
Example 39: The L.C.M. of two polynomials x3 – 12x2 + 44x – 48 and x2 – 10x + 24 is x3 – 12x2 + 44x – 48 find the G.C.D. Solution: Let f(x) = x3 – 12x2 + 44x – 48 ; g(x) = x2 – 10x + 24
L.C.M. = x3 – 12x2 + 44x – 48
∴
G.C.D. = =
f (x) × g(x) L.C.M. ( x 3 − 12x 2 + 44 x − 48) ( x 2 − 10x + 24) ( x 3 − 12 x 2 + 44 x − 48)
= x 2 −10x + 24
Example 40: The G.C.D. and L.C.M. of two polynomials are x + 1 and x6 – 1 respectively. If one of the polynomials is x3 + 1, find the other. Solution: Given G.C.D. = x + 1; L.C.M. = x6 + 1 ; f(x) ; x3 + 1
∴ g(x) = =
L.C.M.× G.C.D. f (x)
=
( x 3 + 1) ( x 3 − 1) ( x + 1) ( x 3 + 1)
(x 6 − 1) × (x + 1) x3 + 1
=
( x 3 − 1) ( x + 1)
Example 41: Find the L.C.M. of the following two polynomials. x3 – x2 – 25 x – 30 and x3 + 4x2 – 5. Solution: Let f(x) = x3 – x2 – 25 x – 30 g(x) = x3 + 4x2 – 5. We first find the G.C.D. of f(x) and g(x).
65
1 x3 + 4x2 – 5
x–1
x3 – x2 – 25x – 30 – 5 x3 + 4x2
x2 + 5x + 5
– 5x2 – 25x – 25 – 5(x2+5x+5) ≠ 0
x3 + 4x2 – 0 – 5 x3 + 5x2 + 5x – x2 – 5x – 5 – x2 – 5x – 5 0
∴ G.C.D. = x2 + 5x + 5. Also g(x) = x3 + 4x2 – 5 = (x – 1) (x2 + 5x + 5) ∴ L.C.M. =
f (x) × g(x) G.C.D.
=
(x 3 − x 2 − 25x − 30) (x 3 + 4x 2 − 5) (x 2 + 5x + 5)
(x 3 − x 2 − 25x − 30) (x − 1) (x 2 + 5x + 5) (x 2 + 5x + 5) ∴ L.C.M. = (x – 1) (x3 – x2 – 25x – 30) =
Exercise 4.3.2 1.
Find the L.C.M. of the following terms a) a8, a10 b) x7, x4 d) 15a6 75a5 e) 20x3,36x6 2 2 g) x y, y z h) ab, bc k) 12xy, 18yz, 24 zx j) am+2,am+3, am+4+
c) f) i) l)
m5, m9 45p9, 100p8 14x3y, 21xyz2, 28x2y2z 8p2qr, 12p2r2, 24pqr
2.
Find the L.C.M. of the following a) (x+2)3, (x+2)4 b) x2–16, 4x2 + 16x c) a3–b3, a – b d) 4x–12, 3x–9 e) a2b+ab2, a2+ab f) x2–x–12, x2 – 9x+20 2 2 2 2 g) 8x –10x–3, 16x –1 h) x +x–6, x –5x+6 i) 4(x3+1), 6(x4–1), 12(x2–1) j) x2 + 5x + 6, x2 + 8x + 15, x2 + 7x + 10 k) 6x2 – x – 2, 2x2 – 7x – 4, 3x2 – 14x + 8 1) 9x2 + 12x – 5, 3x2 + 14x – 5, 4x2 + 21x + 5
3.
The G.C.D. and L.C.M. of two polynomials are 5x2 + x and (x3 – 4x) (5x + 1) respectively. One of the polynomials is 5x3 – 9x2 – 2x. Find the other.
4.
The L.C.M. and H.C.F. of two polynomials are (x–1) (x–2) (x2–3x+3) and (x–1) respectively. If one of the polynomials is x3 – 4x2 + 6x – 3, find the other.
5.
The H.C.F. and L.C.M. of two polynomials are (x+1) and 2(x+1) (x2–4) and one of the polynomials is (x+1) (x–2). Find the other.
6.
Find the L.C.M. of x3 – x2 – 4x – 6 and x2 – 2x + 3 with the help of their H.C.F.
4.4
RATIONAL EXPRESSIONS
An expression of the form P(x) / Q(x) where P(x) and Q(x) are two polynomials over the set of real numbers and Q(x) ≠ 0 is called a rational expression. For example 2/x2 , (x4+x3+x+1)/(x+5), 5/(x+7), (x–2)/(x+2) are rational expressions. 66
4.4.1
Simplification of Rational expressions
A rational expression P(x) / Q(x) can be reduced to its lowest term by dividing both numerator P(x) and denominator Q(x) by the G.C.D. of P(x) and Q(x). Example 42: Simplify Solution:
5x + 20 7 x + 28 5x + 20 5 ( x + 4) 5 = = 7 x + 28 7 ( x + 4) 7
Example 43: Simplify
3x + 9 3x + 15
Solution:
3x + 9 3(x + 3) x+3 = = 3x + 15 3(x + 5) x +5
Example 44: Simplify Solution:
Example 45: Simplify Solution:
x2 − x − 6 x 2 + 5x + 6 ( x − 3) ( x + 2) x2 − x − 6 x −3 = = 2 ( x + 2) ( x + 3) x+3 x + 5x + 6 x−y x 3 − y3 x−y x−y 1 = = 2 3 3 2 2 x −y ( x − y) ( x + xy + y ) x + xy + y 2
Exercise 4.4.1 Simplify the following
1)
2x+10 2x -6
2)
3a 4 b a3b2
3)
5x+20 6x+24
4)
9x 2 - 25y 2 3x 2 - 5xy
5)
x4 - x2 y 2 y4 - x2 y 2
6)
xy 3 - x 3 y y 3 - x3
7)
x 2 + 7x + 10 x2 - 4
8)
2x 2 + x - 3 2x 2 +5x + 3
9)
6x 2 - 54 x 2 +7x+12
8x 3 - 27y 3 16x 4 - 81y 4
11)
64a 3 - 125b3 4a 2 b + 5ab 2
12)
4x 2 +17x+5 8x 2 + 6x - 5
4.4.2
10)
Multiplication and division of rational expressions
If p(x) / q(x) and g(x) / h(x) are the two rational expressions, then their product is p( x ) g ( x ) p ( x ) . g ( x ) × = q( x ) h ( x ) q( x ) . h ( x ) The resulting expression is then reduced to its lowest form.
67
If p(x) / q(x) and g(x) / h(x) are the two rational expressions, then their quotient is p ( x ) g ( x ) p( x ) h ( x ) p ( x ) . h ( x ) ÷ = × = q( x ) h ( x ) q( x ) g( x ) q( x ) . g( x ) The resulting expression is then reduced to its lowest form. 5ab 4cb 16ac × × 15cd 32ad 2bc ab 5 × a × b × 4 × c × b × 16 × a × c 5ab 4cb 16ac Solution: × × = = 2 15cd 32ad 2bc 5 × 3 × c × d × 2 × 16 × a × d × 2 × b × c 3d Example 46: Simplify :
Example 47: Multiply : Solution:
a 3 + b3 a 2 − b2 × a−b a 2 + 2ab + b 2
a 3 + b3 a 2 − b 2 (a + b) (a 2 − ab + b 2 ) × (a + b) (a − b) 2 = × = a – ab + b2 a −b (a + b ) ( a + b ) ( a − b ) a 2 + 2ab + b 2
x 2 − 2x + 1 3x − 6 × Example 48: Multiply 2 6x − 6 x − 3x + 2 2 2 Solution: x – 2x + 1 = (x–1) x2 – 3x + 2 = (x – 2) (x – 1); 6x – 6 = 6(x –1); 3x – 6 = 3( x – 2) 2 x − 2x + 1 3x − 6 (x − 1) 2 × 3(x − 2) 1 ∴ 2 × = = x − 3x + 2 6x − 6 (x − 2)(x − 1) × 6 (x − 1) 2 Example 49: Divide
x 2 − 25 (x + 5) 2 by x +3 x2 − 9
(x + 5) (x − 5) (x + 5) 2 ÷ x +3 (x + 3) (x − 3) ( x + 5) ( x − 5) ( x + 3) ( x − 3) ( x − 5) ( x − 3) = × = x +3 x +5 ( x + 5) 2
Solution: ⇒
( x − 1) 2 x2 − 1 by 2x 2 + x − 3 2 x 2 + 5x + 3 Solution: 2x2 + x – 3 = 2x2 + 3x – 2x – 3 = x(2x + 3) –1 (2x + 3) = (2x + 3)(x–1) 2 2x + 5x + 3 = 2x2 + 2x + 3x + 3 = 2x (x + 1) + 3 (x + 1) = (2x + 3) (x+1) 2 (x − 1) x 2 −1 (x −1) 2 (x + 1) (x − 1) Hence ÷ = ÷ 2 2 2x + x − 3 2x + 5x + 3 (2x + 3) (x − 1) (2x + 3) (x + 1) (x − 1) (x −1) (x − 1) (2x + 3) ÷ = × =1 = 2x + 3 (2x + 3) (2x + 3) (x − 1) Example 50: Divide
68
Example 51: Divide Solution:
x 2 + 3x + 2 x 2 − 4x −12 by x 2 − 3x −18 x 2 − 2x − 3
x 2 − 4x −12 x 2 + 3x + 2 ÷ 2 x 2 − 3x −18 x − 2x − 3 (x − 6) (x + 2) (x + 2) (x + 1) x + 2 x − 3 x −3 = = ÷ = × (x − 6) (x + 3) (x − 3) (x + 1) x + 3 x + 2 x +3
Exercise 4.4.2 1.
Multiply
a)
a+b a 3 - b 3 × a - b a 3 +b3
(b)
x 2 - 9y 2 x 2 -y 2 × 2 3x - 3y x +4xy+3y 2
(c)
x 2 - 4x -12 x 2 - 2x - 3 × x 2 - 3x -18 x 2 +3x+2
(d)
x 2 - 3x - 10 x 2 - 4x+16 × x 2 - x - 20 x 3 +64
(e)
x 2 -16 x2 - 4 × 3 x-2 x + 64
(f)
x+7 x 2 +8x+7 × x+1 x 2 +14x+49
(g)
x 2 - 5x + 6 4x - 8 × 2 6x + 6 x - 4x+3
(h)
p 2 -1 p3 1 × × p p -1 p+1
2.
Divide
(a)
px - 2p ax - 2a ÷ qx - 3q bx - 3b
(b)
a2 4a 4 ÷ 2 2a+3 6a +9a
(c)
x 2 - 4 (x + 2)2 ÷ 2 x +3 x -9
(d)
x 2 - 16 2x + 8 ÷ 2 3x - 9 x - 8x + 16
(e)
3x 2 - 7x + 2 9x 2 - 6 +1 ÷ 2x 2 - 5x - 3 x2 - 9
(f)
⎛ 1 1 ⎞ ⎛ 1 1⎞ ⎜ 2 - 2 ⎟÷⎜ - ⎟ b ⎠ ⎝a b⎠ ⎝a
(g)
x 2 - 2x - 8 4x - 8 x 2 - 7x + 12 × 2 ÷ x-2 x - 4x - 12 x 2 - 9x + 18
(h)
2x 2 + 13x + 15 2x 2 - x -6 ÷ x 2 + 3x - 10 x 2 - 4x + 4
4.4.3
Addition and subtraction of Rational Expression
While adding or subtracting two rational expressions p(x) / q(x) and g(x) / h(x) we follow the same rules as in the case of rational numbers. If p(x) / q(x) and g(x) / h(x) are two rational p(x) × h(x) ± q(x) × g(x) p(x) g(x) expression we define their sum or difference as ± = . q(x) h(x) q(x).h(x) 69
p( x ) g( x ) are two rational expressions having the same denominator then and q( x ) q( x ) p(x) g(x) p(x) ± g(x) ± = q(x) q(x) q(x)
If
Example 52: Add
4x + 3 x+2 and 2 x + 3x + 2 x + 3x + 2 2
4x + 3 x+2 4x + 3 + x + 2 5x + 5 5 + 2 = = = 2 (x + 1) (x + 2) x+2 x + 3x + 2 x + 3x + 2 x + 3x + 2 a b and 3 Example 53: Add 3 3 a +b a + b3 a b a+b a+b 1 + 3 = 3 Solution: 3 = = 2 3 3 3 2 2 a +b a +b a +b (a + b) (a − ab + b ) a − ab + b 2 Solution:
2
x 2 + 3x − 4 x 2 + 5x + 6 + x 2 + 6x + 8 x 2 − x − 12 (x + 4) (x − 1) (x + 2) (x + 3) x 2 + 3x − 4 x 2 + 5x + 6 + 2 = + Solution: 2 (x + 4) (x + 2) (x − 4) (x + 3) x + 6x + 8 x − x − 12 Example 54: Simplify
=
2x 2 − x + 8 (x − 1) (x + 2) (x − 1) (x − 4) + (x + 2) 2 + = = (x + 2) (x − 4) (x + 2) (x − 4) (x + 2) (x − 4)
Example 55: Simplify
Solution:
a3 b3 a3 b3 a3 b3 + = + = − a − b b − a a − b − (−b + a ) a − b a − b =
a 3 − b3 (a − b) (a 2 + ab + b 2 ) = = a 2 + ab + b 2 a−b (a − b)
Example 56: Simplify
Solution:
a3 b3 + a−b b−a
x2 y2 − x 4 − y4 x 4 − y4
x 2 − y2 (x 2 − y 2 ) = x 4 − y 4 (x 2 ) 2 − ( y 2 ) 2
=
(x 2 − y 2 ) 1 = 2 2 2 2 2 (x + y ) (x − y ) x + y 2
Example 57: What rational expression should be added to Solution: Given
x3 −1 2x 3 − x 2 + 3 ? to get x2 + 2 x2 + 2
x3 −1 2x 3 − x 2 + 3 + Rational expression = . x2 + 2 x2 + 2 70
Required rational expression =
2x 3 − x 2 + 3 x 3 − 1 2x 3 − x 2 + 3 − x 3 + 1 x 3 − x 2 + 4 = − 2 = x2 + 2 x +2 x2 + 2 x2 + 2
Example 58: What rational expression should be added to Solution: Required rational expression =
x2 +1 x 4 − 3x + 1 to get ? x−2 x+3
x 2 + 1 x 4 − 3x + 1 − x−2 x+3
=
(x 2 + 1) (x + 3) − (x 4 − 3x + 1) (x − 2) x 3 + 3x 2 + x + 3 − ( x 5 − 2x 4 − 3x 2 + 6x + x − 2) = ( x − 2) ( x + 3) (x − 2) (x + 3)
=
x 3 + 3x 2 + x + 3 − x 5 + 2x 4 + 3x 2 − 7 x + 2 − x 5 + 2x 4 + x 3 + 6x 2 − 6x + 5 = ( x − 2) ( x + 3) ( x − 2) ( x + 3)
Example 59: Which rational expression should be subtracted from
4x 3 − 7 x 2 + 5 to get 2x − 1
2x2 – 5x + 1? 4x 3 − 7 x 2 + 5 − (2 x 2 − 5x + 1) 2x − 1
Solution: Required expression =
=
(4x 3 − 7x 2 + 5) − (4x 3 − 10x 2 + 2x − 2x 2 + 5x − 1) 5x 2 − 7x + 6 = 2x − 1 2x − 1
Example 60: If P =
x −1 x +1 express (P+Q)2, (P–Q)2 as rational expressions. ,Q= x +1 x −1
⎛ x −1 x +1⎞ Solution: (P+Q) = ⎜ + ⎟ ⎝ x +1 x −1⎠ 2
⎡ x 2 − 2x + 1 + x 2 + 2x + 1 ⎤ =⎢ ⎥ (x + 1) (x − 1) ⎣ ⎦
2
2
⎡ ( x − 1) 2 + ( x + 1) 2 ⎤ = ⎢ ⎥ ⎣ ( x + 1) ( x − 1) ⎦
⎡ 2x 2 + 2 ⎤ ⎢ 2 ⎥ ⎣ x −1 ⎦
2
⎡ 2(x 2 + 1) ⎤ = ⎢ ⎥ 2 ⎣ x −1 ⎦
⎡ (x − 1) 2 − (x + 1) 2 ⎤ ⎛ x −1 x +1⎞ and (P–Q) = ⎜ − = ⎟ ⎢ ⎥ ⎝ x +1 x −1⎠ ⎣ (x + 1) (x − 1) ⎦ 2
2
⎛ − 4x ⎞ = ⎜ 2 ⎟ ⎝ x −1⎠
2
2
2
2
=
4(x 2 + 1) 2 (x 2 − 1) 2
⎡ x 2 − 2x + 1 − (x 2 + 2x + 1) ⎤ ⎢ ⎥ (x 2 − 1) ⎣ ⎦
16 x 2 = ( x 2 − 1) 2
Exercise 4.4.3 I.
Simplify x 5x a) + 16 16
b)
x y + x+y x+y 71
c)
8 3 + 2 2 x y xy
2
1 2b + 2 2 a+b a -b y x f) 2 2 x-y x -y d)
x-2 x+3 + 2 x -7x+10 x -2x-15 5 4 k) − 2 2 4a − 9 2a − a − 3
i)
m)
2
x+2 x-3 + 2 x +3x+2 x -2x-3 3 a 64 g) a-4 a-4
e)
2
x4 81 h) x-3 x-3
x x - 2 x +5x+6 x +7x+12 x+1 (x − 1) − l) (x − 2)(x − 1) (x+1)(x − 2) j)
m 1 1 + + 2 m+1 m+1 m − 1
n)
x+a x-a 4ax − + 2 x-a x+a a − x 2
p)
2
1 1 4x − 2 + 2x+3y 2x − 3y 4x − 9y 2
x x x − 2 + 2 x − 9x+20 x − 8x+15 x − 7x+12 2 x +x-3 3x 2 +x+4 to get II. 1. Which rational expression should be added to ? 2x-1 x+2 o)
2
x 4 − 7x 3 +4 1 − 5x − x 2 to get ? x+1 x2 − 2 5x 2 − x+2 3x 2 +1 3. Which rational expression should be subtracted from ? to get 3x+4 2x-3 x+1 x −1 III.1. If P = and Q = express (P+Q)2 and (P–Q)2 as a rational expression. x −1 x+1 x y 1 2Q find − 2 ,Q= 2. If P = x+y x+y P − Q P − Q2 2. Which rational expression should be added to
4.5
SQUARE ROOT
We recall that by a square root of a positive number x, we mean a number which when multiplied by itself gives x. The square root of a polynomial is defined similarly, as that polynomial when multiplied with itself gives the original polynomial. For definiteness, we consider positive square roots, wherever the case may be. For example let us find the square root of 784. 784 = 7 × 7 × 4 × 4 = 72 × 42 ∴
784 = 7 2 × 42 = 7 × 4 = 28
As in the case of numbers we denote P( x ) , for the square root of the polynomial P(x). In finding square roots, we use the following methods namely (i) factorisation method, (ii) division method. 4.5.1 Square root by factorisation method Example 61: Find the square root of (i) x2 y4 z8 (ii) 256 z2 y6 (iii) 36a2 (b–c)4 (c+a)8 (iv) (p+q)2 – 4pq
72
Solution: i)
x2 y4 z8 =
x2(y2)2 (z4)2 = (xy2z4)2
x 2 y 4 z8 = (xy 2 z 4 ) 2 = (xy 2 z 4 )
Hence ii)
256z 2 y 6 = (16zy3 ) 2 = 16zy3
iii)
36a 2 (b − c) 4 (c − a ) 8 = [6a (b − c) 2 (c − a ) 4 ] 2 = 6a (b − c) 2 (c + a ) 4
iv)
(p + q) 2 − 4pq =
(p − q) 2 = p − q (i) x2 + 10x + 25 (ii) 4a2 + 20ab + 25b2
Example 62: Find the square root of :
(iii) (x2 – 4) (x2 + x – 6) (x2 + 5x + 6) Solution: i)
x2 + 10x + 25 = x2 + 5x + 5x + 25 = x (x+5) + 5 (x+5) = (x+5) (x+5) = (x + 5)2 ∴
ii)
x 2 + 10x + 25 = ( x + 5) 2 = x + 5
4a 2 + 20ab + 25b 2 =
(2a + 5b) 2 = (2a + 5b)
iii) x2 – 4 = x2 – 22 = (x–2) (x+2) x2 + x – 6 = x2 + 3x – 2x – 6 = x (x+3) – 2 (x+3) = (x+3) (x–2) x2 + 5x + 6 = x2 + 3x + 2x – 6 = x (x+3) + 2 (x+3) = (x+3) (x+2) Hence (x2 – 4) (x2 + x – 6) (x2 + 5x + 6) = (x + 2) (x + 3) (x – 2) (x + 3) (x + 2) (x – 2) = (x–2)2 (x + 2)2 (x + 3)2 = [(x – 2) (x + 2) (x + 3)]2 ∴ The required square root is (x – 2) (x + 2) (x + 3) Exercise 4.5.1 I.
II.
Find the square root by factor method 1. 576 2. 676 3. 2025 0.0144 1 8. 7 6. 0.0169 7. 0.0625 9
4. 9801 21 9. 4 25
5. 11025 14 10. 11 25
Find the square root of the following
1. 169 a8b6c4
2. 9x2y4z8
3. 49 (2a–4b)2c2
4. 36 (2–x)4 (3–x2)6
5. 81 (a–b)2 (x+y)4
6. (x–y)2 + 4xy
7. 121z2y4 ÷ 49x6
8. 50(x+y)2 ÷128(x–y)4
9.
25(a+b)8 (x+y)4 9(x-y)6 (a+b)16
10.
(x 2 -y 2 )2 36(x-y)2 (x+2y)6
73
III.
Find the square root of the following 1) 9x2 + 30x + 25 2) 16x2 – 24x + 9 3) (x2 – 4) (x2 – 3x – 10) (x2 – 7x + 10) 4) (x2 – 1) (x2 – 4x + 3) (x2 – 2x – 3) 2 2 2 2 2 2 5) (a – b ) (a – 4ab + 3b ) (a – 2ab – 3b ) 6) a2 + b2 + c2 – 2ab + 2bc – 2ca x2 y2 8) 2 + 2 + 2 7) x2 + y2 + z2 + 2xy – 2yz – 2zx y x 1 1 9) x 2 + 2 -2 10) x 4 +2+ 4 x x 2 2 2 11) (6x + 7x – 20) (3x + 23x – 36) (2x + 23x + 45) 12) (2x2 – x – 1) (3x2 – 2x – 1) (6x2 + 5x + 1) 4.5.2 Finding square root by division method
The division method can be used to find the square root of polynomials which cannot be easily reduced to factors. Recall the method of finding the square root of a number by division method. 125 1
1, 56, 25 1
22
56 44
245
12 25 12 25
∴ Square root of 15625 is 125.
0 We can use the same procedure to find the square root of a polynomial Example 63: Find the square root of 4x4 – 4x3 + 5x2 – 2x + 1 Solution:
2x2 – x + 1 2x2
Procedure Square root of the first term
4x4 – 4x3 + 5x2 – 2x + 1 4x4
4x2 – x
4x 4 = 2x 2
2 × 2x2 = 4x2, –
– 4x3 + 5x2
4x 3 =− x 4x 2
– 4x3 + x2 4x2 – 2x + 1
2 × (2x2–x) = 4x2 – 2x 4x 2 =1 4x 2
4x2 – 2x + 1 4x2 – 2x + 1 0
∴ Square root = ± (2x – x + 1) 2
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Example 64: Find the square root of a4 – 2a3 +
3 2 1 1 a − a+ 2 2 16
Solution:
a2 – a + a2
1 4
a4 – 2a3 +
3 2 1 1 a − a+ 2 2 16
a4 3 2 a 2 – 2a3 + a2
2a2 – a
2a2 – 2a +
– 2a3 +
1 4
1 2 1 1 a − a+ 2 2 16 1 2 1 1 a − a+ 2 2 16 0 2
∴
3 1 1 a − 2a + a 2 − a + = ± 2 16 2 4
3
1⎞ 1⎞ ⎛ 2 ⎛ 2 ⎜a − a + ⎟ = ± ⎜a − a + ⎟ 4⎠ 4⎠ ⎝ ⎝
Example 65: If p + qx + 10x2 + 12x3 + 9x4 is a perfect square, find the value of p and q. Solution:
3x2 + 2x + 1 3x2 6x2 + 2x 6x2 + 4x + 1
9x4 + 12x3 + 10x2 + qx + p 9x4 12x3 + 10x2 12x3 + 4x2 6x2 + qx + p 6x2 + 4x + 1 0
because given expression is a perfect square ∴ p = 1 q = 4 Exercise 4.5.2
1.
Find the square root by division method (a) 7225 (b) 8649 (c) 18225 (d) 524176 (e) 287296 (f) 186624 (g) 2819041 (h) 1708249.
2.
Find the square root by division method (b) 4x4 + 8x3 + 8x2 + 4x + 1 (a) x4 – 4x3 + 10x2 – 12x + 9
75
3 2 1 1 x + x+ 2 2 16 4 3 2 4 3 2 (f) 16x – 24x – 31x + 30x + 25 (e) x – 6x + 11x – 6x + 1 1 4 13 1 1 4 (g) x − 3x 3 + 13x 2 − 24 x + 16 (h) x 4 − x 3 + x 2 − 2x + 9 3 3 4 4 Find the value of a in the following given that the polynomials are perfect squares (b) 4x4 + 12x3 + 13x2 + ax + 1 (a) 9x4 – 6x3 + 7x2 – 2x + a 4 3 2 (c) x – 6x + 11x + ax + 1 Find the values of a and b in the following given that the polynomials are perfect squares (b) 9x4 + 12x3 + 40x2 + ax + b (a) 25x4 – 40x3 – 34x2 + ax + b 4 3 2 (c) 4x + 12x + x + ax + b (c) 9x4 – 18x3 + 33x2 – 24x + 16
3.
4.
4.6
(d) x4 – 2x3 –
QUADRATIC EQUATION
An equation of the form ax2 + bx + c = 0, where a, b, c ∈ R and a ≠ 0 is called a quadratic equation. It is an equation with degree 2. For example, 5x2 + 6x + 7 = 0, 9x2 – 4 = 0, 2x2 – 3x = 0 are quadratic equations. If p(x) = 0 is a quadratic equation, then the zeros of the polynomial p(x) are called the roots of equation p(x) = 0. Finding the roots of a quadratic equation is known as solving the quadratic equations. 4.6.1
Solution of quadratic equations by factorisation
In this section, we apply the method of factorization to solve quadratic equation. This method is used when the quadratic equation is expressible as the product of two linear equations. Example 66: Solve the quadratic equation 2x2 + 3x – 5 = 0 Solution: 2x2 + 3x – 5 = 2x2 + 5x – 2x – 5 = x (2x + 5) – 1(2x + 5) = (x – 1) (2x + 5) Since 2x2 + 3x – 5 = 0 we get (x – 1) (2x + 5) = 0 ⇒ x – 1 = 0 or 2x + 5 = 0 ⇒ x = 1, –5/2 ⎧ − 5⎫ The solution set = ⎨1, ⎬ ⎩ 2 ⎭ Example 67: Solve 64x2 – 36 = 0 Solution: 64x2 – 36 = 0 or (8x + 6) (8x – 6) = 0 or x = –6/8 = –3/4 or x = 6/8 = 3/4 ⎧− 3 3 ⎫ , ⎬ The solution set = ⎨ 4⎭ ⎩ 4 Example 68: Solve 3x2 – 4x = 0 Solution: 3x2 – 4x = x (3x – 4) Since 3x2 – 4x = 0, we get x (3x – 4) = 0 or x = 0 or 3 x –4 = 0 ⇒ 3x = 4 ⇒ x = 4/3 4⎫ ⎧ The solution set = ⎨0 , ⎬ 3⎭ ⎩
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Example 69: Solve (x + 3) (x + 5) = 1 – x Solution: (x + 3) (x + 5) = 1 – x x2 + 8x + 15 – 1 + x = 0 or x2 + 9x + 14 = 0 or (x + 2) (x + 7) = 0 or x = –2 x = –7 The solution set = {–2, –7}. Example 70: Solve (2x + 1) (x – 2) = 0 Solution: (2x + 1) (x – 2) = 0 or 2x + 1 = 0 or x – 2 = 0 or x = –1/2, x = 2 The solution set = {-1/2, 2}. Example 71: Solve x2 – x – 12 = 0 Solution: Factorising the given quadratic equation we have x2 – x – 12 = 0 (x + 3) (x – 4) = 0 or x + 3 = 0, x – 4 = 0 or x = –3 x = 4 The solution set = {− 3, 4} 1 15 = Example 72: Solve x − x 4 2 x −1 1 15 15 Solution: x − = or = x 4 x 4 or 4(x2 – 1) = 15x or 4x2 – 4 – 15x = 0 or 4x2 – 15x – 4 = 0 or 4x2 – 16x + x – 4 = 0 or −1 4x (x – 4) + 1 (x – 4) = 0 or x – 4 = 0, 4x + 1 = 0 or x = 4 , x = 4 ⎧ −1 ⎫ Solution set = ⎨4 , ⎬ 4⎭ ⎩ Example 73: Solve 3 x 2 + 11x + 6 3 = 0 Solution: Factorising the equation we have 3 x 2 + 11x + 6 3 = 0 or
x
(
3x+2
) +3 3(
)
3 x + 2 = 0 or
3 x + 2 = 0 , x + 3 3 = 0 or x =
(
3 x 2 + 2x + 9x + 6 3 = 0 3 x+ 2
) (x + 3 3) = 0
−2 or x = − 3 3 3
⎧ −2 ⎫ The solution set = ⎨ , − 3, 3 ⎬ ⎩3 ⎭ Example 74: Solve a2b2x2 – (a2 + b2) x + 1 = 0 Solution: a2b2x2 – (a2 + b2) x + 1 = 0 or a2b2x2 – a2x – b2x + 1 = 0 or 1 1 (a2x – 1) (b2x – 1) = 0 or x = 2 , x = 2 a b ⎧1 1⎫ The solution set = ⎨ 2 , 2 ⎬ ⎩a b ⎭ Example 75: Solve 2 (x + 1)2 – 5(x + 1) = 12 Solution: Put x + 1 = t then we get 2t2 – 5t – 12 = 0 or 2t (t – 4) + 3t – 12 = 0
2t (t – 4) + 3 (t – 4) = 0
or
(2t + 3) (t – 4) = 0
77
or
t=−
3 ,t= 4 2
3 3 5 ⇒ x + 1 = − ⇒ x = − and t = 4 ⇒ x + 1 = 4 ⇒ x = 3 2 2 2 ⎧ −5 ⎫ The solution set = ⎨ , 3⎬ ⎩2 ⎭ t=−
Example 76: Solve x + 2 + x + 3 = 3x + 7 Solution: Squaring both sides we get x + 2 + x + 3 + 2 ( x + 2 ( x + 3) = 3x + 7 2 (x + 2) (x + 3) = 3x + 7 − 2x − 5 = x + 2
Squaring both sides we get 4 (x + 2) (x + 3) = (x + 2)2 ⇒ (x + 2) [(x + 2) – 4(x + 3)] = 0 ⇒ (x + 2) (–3x –10) = 0 − 10 ⇒ (3x + 10) (x + 2) = 0 ⇒ x = , x = −2. 3 10 When x = − , x + 2 is not real 3 10 (Hence x = − is inadmissible) ⇒ the solution of the given equation is x = –2. 3 Solution by the method of completion of squares
Consider the quadratic equation ax2 + bx + c = 0. b c We write this equation as x 2 + x + = 0 a a b c ⎛ ⎞ or x2 + 2⎜ ⎟ x + = 0 or a ⎝ 2a ⎠
c ⎛ b ⎞ x2 + 2 ⎜ ⎟ x = − a ⎝ 2a ⎠
or
b2 b2 c ⎛ b ⎞ x +2⎜ ⎟x+ 2 = 2 − a 4a 4a ⎝ 2a ⎠
b 2 − 4ac b ⎞ ⎛ ⎟ = ⎜x + 2a ⎠ 4a 2 ⎝
or
x+
2
b =± 2a
b 2 − 4ac 2a
or
2
or x=
−b ±
b 2 − 4ac 2a
Thus we get the roots of the quadratic equation using the method of completion of squares. This method was given by Indian mathematician Sridharacharya (A.D 1025). Example 77: What should be added with x2 + 12x to get a perfect square? What is that square? Solution: Comparing with the expression a2 + 2ab + b2, here 2ab = 12x where a = x. 12x ∴ b= = 6 and b 2 = 62. So the term to be added is 36. 2x The required square = x2 + 12x + 62 = (x + 6)2.
78
Example 78: Solve the equation by completing the square x2 + 6x – 7 = 0. 2
⎛6⎞ Solution: x2 + 6x = 7. The term to be added = ⎜ ⎟ = 9. ⎝2⎠ 2 Adding 9 on both sides we get x + 6x + 9 = 7 + 9 or (x + 3)2 = 16 Taking square root on both sides x + 3 = ± 16 = ± 4 x + 3 = 4 ⇒ x = 1, x + 3 = –4 ⇒ = –7 The solution set = {1, –7} Example 79: Solve x2 + 2x – 1 = 0 Solution: x2 + 2x = 1 or x2 + 2x + 1 = 1 + 1 or (x + 1)2 = 2 x + 1 = + 2 ⇒ x = –1 + 2 ∴ x = –1 + 2 , –1 – 2 The solution set = {1 + 2, − 1 − 2} Example 80: Solve 6x2 + x –1 = 0 Solution: 6x2 + x = 1. Multiply by 6 on both sides we get 1 1 1 (6x)2 + (6x) = 6 ; (6x)2 + 2 × 6x × + = 6 + 2 4 4 2
1⎞ 25 1 5 ⎛ ⇒ 6x + = ± ⎜ 6x + ⎟ = 2⎠ 4 2 2 ⎝ 1 5 1 5 ⇒ 6x = + or 6x = − − 2 2 2 2 ⇒ 6x = 2 or 6x = –3 2 3 1 1 ⇒ x= or x = − ⇒ x = or x = − 6 6 3 2 The solution set = {− 1/2, 1/3} Solution by formula method
Consider the quadratic equation ax2 + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. Its solution is given by the formula : x=
−b ±
b 2 − 4ac
2a Note : We derived this formula in the method of completion of squares. Example 81: Solve the equation using formula x2 – 7x + 12 = 0. Solution: Comparing with ax2 + bx + c = 0, we get a = 1, b = –7, c = 12.We know,
x=
−b ±
b 2 − 4ac − (−7) ± = 2a
79
(−7) 2 − 4(1)(12) 2 ×1
=
+7 ± 49 − 48
2
=
7 ±1 = 4,3 2
The solution set = {4, 3} Example 82: Solve x2 + 2x – 1 = 0. Solution: Comparing the given equation x2 + 2x – 1 = 0 with ax2 + bx + c = 0, we get a = 1.
b = 2, c = –1. By formula x =
−b ±
b 2 − 4ac −2 ± = 2a =
{
}
22 − 4 × 1 × (− 1) 2 ×1
−2 ± 8 = −1 ± 2 2
∴The solution set − 1 + 2, − 1 − 2 . Exercise 4.6.1 I.
Solve by factorisation 1) (x – 5) (x – 2) = 0 4) (x + 7)2 = 0 7) x2 – 64 = 0 10) 9x2 – 64 = 0
2) 5) 8) 11)
13) 3x2 – 5x + 2 = 0
14) 9x2 – 15x – 14 = 0
(x + 3) (x + 7) = 0 x (x – 8) = 0 4x2 – 25 = 0 x2 + 10x + 21 = 0
3) (x – 9)2 = 0 6) x (3x – 4) = 0 9) 3x2 – 75 = 0 12) x2 – 10x – 24 = 0 15) x −
3 =2 x
1 26 = 17) (x – 2) (2x + 3) = 3 (x – 4) (x + 8) x 5 18) (5x – 2) (x + 1) = 3x (3x – 1) 19) 4a2x2 – 5abx + b2 = 0 16) x +
20) II.
III.
5 x 2 + 8x + 3 5 = 0
1. What should be added to make the following a perfect square. b) x2 + 5x c) x2 – 10x d) x2 – 7x a) x2 + 16x f) 2x2 + 9x g) ax2 + bx h) 5x2 – 4x e) 3x2 – 8x 2. Solve the equations using completion of square method b) 2x2 – x – 10 = 0 c) 3x2 – 4x + 1 = 0 a) x2 + 2x – 3 = 0 e) x2 + x = 72 d) x2 + 5 = –6x d) 5x2 – 7x – 6 = 0 g) x2 – 4x – 7 = 0 h) x2 + 14x + 40 = 0 Solve the equations using formula 1) 6x2 + x – 1 = 0 2) 4x2 + 17x + 4 = 0 3) x2 – 28x + 160 = 0 4) 15x2 – 11x + 2 = 0 5) 3x2 – 6x + 2 = 0 7) x2 – 10x + 9 = 0
8) x2 – 5x + 5 = 0
10) x2 + 7x – 6 = 0
11)
x + 5 = 2x + 3
80
1 1 =2 x 2 2 9) 2 (x + 2) = x + 4
6) x +
12) 3a2x2 – abx – 2b2 = 0
4.6.2
Problems leading to quadratic equation
In some problems the quadratic equation will not be given directly. In such cases we form the quadratic equation first from the given data and then solve it. In this section we are going to form the equation translating the given statement and then solve the equation either by factorization method or by formula method. Example 83: The sum of a number and its square is 90. Find the number. Solution: Let the required number be x. Its square is x2. By the given data, we know that x + x2 = 90 or x2 + x – 90 = 0 or (x + 10) (x – 9) = 0 x + 10 = 0 , x – 9 = 0 ⇒ x = –10, x = 9. So the required number is –10 or 9. Example 84: The height of right circular cone is 7 cm greater than its radius. The slant height is 8 cm greater than its radius. Find the curved surface area of the cone. Solution: Let the radius be x cm. Then the height h = x + 7 cm ; the slant height l = x + 8 cm. We know that in a cone, ⇒ x2 + (x + 7)2 = (x + 8)2 r2 + h2 = l2 2x2 + 14x + 49 – x2 – 16x – 64 = 0 x2 + x2 + 14x + 49 = x2 + 16x + 64 ⇒ x2 – 2x – 15 = 0 ⇒ (x – 5) (x + 3) = 0 ⇒ x = 5 and x = –3 The radius cannot be negative ∴ x = 5 cm ; h = x + 7 = 12 cm ; l = x + 8 = 13 cm Curved surface area of the cone = πrl = π x 5 x 13 = 65πcm2 Example 85: A two digit number is such that the product of the digits is 8. When 18 is added to the number, they interchange their places. Determine the number. Solution: Let the number in the tens place be x and the number in the unit place be y. Hence the value of the number is 10x + y. Since the product of the digits is 8 we get. xy = 8 (1) When 18 is added to the number, the digits get interchanged ∴ 10x + y + 18 = 10y + x ⇒ 9x – 9y = –18 ⇒ x – y = –2 ⇒ y = x + 2 (2) Substitute y = x + 2 in (1) we get (x + 4) (x – 2) = 0 x (x + 2) = 8 or x2 + 2x – 8 = 0 or x + 4 = 0, x – 2 = 0 or x = –4, x = 2. 8 8 Since the digits are +ve integers x = 2 ⇒ y = = = 4 ∴ The two digit number is 24. x 2 Example 86: A farmer wishes to start a 100 sq.m. rectangular vegetable garden. Since he has only 30 m. barbed wire, he fences the sides of the rectangular garden letting his house compound wall act as the fourth side fence. Find the dimensions of the garden. Solution: Let the breadth and length of the garden be x meters and y meters. Area = 100m2 ∴ xy = 100 (1) Given length of wire = 30 m
81
∴ 2x + y = 30 y = 30 – 2x (2) Substituting (2) in (1) we get x (30 – 2x) = 100 30x – 2x2 – 100 = 0 ⇒ – 2x2 + 30x – 100 = 0 x2 – 15x + 50 = 0 ⇒ (x – 5) (x – 10) = 0 x = 5, x = 10 when x = 5, (2) ⇒ y = 30 – 2 x 5 = 30 – 10 = 20, when x = 10, (2) ⇒ y = 30 – 2 x 10 = 30 – 20 = 10. The dimensions of the garden is 5m × 20m. Example 87: The age of the father is square of the age of his daughter Ramya. Five years hence, the father is three times as old as Ramya. Find their present ages. Solution: Let the present age of Ramya be x years and that of her father's age be x2 years. Five years hence the age of Ramya will be (x + 5) years and that of her father will be (x2 + 5) years. By the given data x2 + 5 = 3 (x + 5) ⇒ x2 + 5 = 3x + 15 ⇒ x2 – 3x – 10 = 0 (x + 2) (x – 5) = 0 ⇒ x = – 2 or x = 5 Since the age cannot be negative, we get x = 5. ∴ The present age of Ramya x = 5 years and the present age of father x2 = 25 years. N
Example 88: Two trains leave Chennai Central Railway Station. The first train travels due west and the second train due North. The first train travels 5 km per hour faster than the second train. If after 2 hours they are 50 km apart, find B the average speed of each train. Solution: Let the speed of the second train be x km/hour. 50 2x Then the speed of the first train will be (x + 5) km/hour. Let C denote the Chennai Railway Station. After two hours, let W E C 2(x+5) A the first train be at A and the second train be at B. Then CA = 2 (x + 5) km and CB = 2x km. Given AB = 50 km. S Fig.4.1 In right angled triangle ABC, CA2 + CB2 = AB2 ⇒ 4 (x + 5)2 + 4x2 = 2500 ⇒ 4(x2 + 10x + 25) + 4x2 – 2500 = 0 ⇒ 4x2 + 40x + 100 + 4x2 – 2500 = 0 ⇒ 8x2 + 40x – 2400 = 0 ⇒ x2 + 5x – 300 = 0 (x + 20) (x – 15) = 0 ⇒ x = – 20 or x = 15 Since the speed of the train is positive, x = 15. Hence the speed of the first train = x + 5 = 15 + 5 = 20 km/hour and the speed of the second train x = 15 km/hour. Example 89: Two pipes can together fill a tank in 11 1/9 minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would fill the tank. Solution: Pipe A fills the tank in x minutes. Pipe B fills the tank in x + 5 minutes. In 1 minute 2x + 5 1 1 1 both pipes fill = part of the tank. The full tank is filled in 11 minutes + x (x + 5) x x +5 9 x ( x + 5) x ( x + 5) 100 = minutes or = 2x + 5 9 2x + 5 9x2 + 45x = 200x + 500 or 9x2 – 155x – 500 = 0 or 9x2 – 180x + 25x – 500 = 0 − 25 or x = 20 ⇒ Pipe A fills 9x (x – 20) + 25 (x – 20) = 0 or (9x + 25) (x – 20) = 0 or x = 9 the tank in 20 minutes, pipe B fills the tank in 25 minutes.
82
x Example 90: The outer dimensions of a bordered table are 72 cm x and 108 cm. If the area of the table, excluding the border is 6400 cm2, how wide is the border? Solution : Length of the rectangle excluding the border = 108–2x x Breadth of the rectangle excluding the border = 72 – 2 x x Area of the table excluding the border = (108 – 2x) (72–2x) = 6400 or 4x2 – 360 x + 1376 = 0 or x2 – 90x + 344 = 0 or x2 – 86x – 4x + 344 = 0 or x (x – 86) – 4 (x – 86) = 0 or (x – 86) (x – 4) = 0 ⇒ x = 86 or x = 4. x = 86 is not admissible. Hence x = 4. The width of the border = 4 cm.
x x 72 x x 108
Fig.4.2
Example 91: A train travels a distance of 300 km at a constant speed. If the speed of the train is increased by 5 km per hour, the journey would have taken 2 hours less. Find the speed of the train. Solution: speed = s km/hour and time = t hour speed x time = distance or s × t = 300 ; Again (s + 5) × (t – 2) = 300 st + 5t – 2s – 10 = 300 or 300 + 5t – 2s – 10 = 300 or 5t – 2s – 10 = 0 300 5× − 2s − 10 = 0 or 2s2 + 10s – 1500 = 0 or s2 + 5s – 750 = 0 s (s + 30) (s – 25) = 0 ⇒ s = – 30 or s = 25. Speed of the train = 25 km/hr. Example 92: A motor boat whose speed is 5 km/hr in still water goes 30 km downstream and comes back in 4 hours 30 minutes. Determine the speed of water. Solution: Let the speed of water be x km/hr. Since the speed of motor boat in still water is 15 km/hr, its speed downstream is (15 + x) km/hr and the speed upstream is (15 – x) km/hr. 30 Time taken for 30 km downstream = hours 15 + x 30 Time taken for 30 km upstream = hours 15 − x 1 9 The total time is given to be 4 hours 30 minutes = 4 hours = hours 2 2 ⎛ 30 ⎞ ⎡ 15 − x + 15 + x ⎤ 30 30 9 9 9 ∴ + = or 30 ⎢ = = or 30 ⎜ 2 ⎥ 2 ⎟ 15 + x 15 − x 2 2 2 ⎣ (15 + x) (15 − x) ⎦ ⎝ 15 − x ⎠
1800 = 9 (225 – x2) or 200 = 225 – x2 or x2 = 25 Since x is positive, the speed of the water = 5 km/hour.
⇒ x=5
Exercise 4.6.2
1. 2. 3. 4.
65 . Find the number. 8 Sum of the squares of 3 consecutive number is 194. Find them. The product of two consecutive odd number is 323. Find them. The perimeter of a rectangle is 36 cm and its area is 80 sq.cm. Find its dimensions. The sum of a number and its reciprocal is
83
5. 6. 7. 8. 9. 10. 11.
12.
13.
The area of a rectangular land is 240 m2. If 8 m is decreased from its length it will become a square. Determine the length and breadth of the land. The sides of a right angle triangle are (x – 1.8) cm, (x + 1.8) cm and (x + 1) cm. Find the area of the triangle. The sum of ages of a father and the son is 45 years. Five years hence the product of their ages will be 600. Find their present ages. The product of age of a man, 6 years before and 10 years later is 960. Find his present age. Five times a certain number is equal to three less than twice the square of the number. Find the number. Find two consecutive positive even integers whose product is 224. The length of a verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter. Find the length and breadth of the verandah. A trader bought a number of articles for Rs.1200. Ten were damaged and he sold each of the rest at Rs.2 more than what he paid for it thus getting a profit of Rs.60 on the whole transaction. Find the number of articles bought. In ca class of 60 students each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total contribution then collected is Rs.1600. How many boys are there in the class?
4.6.3 Nature of roots
The roots of the quadratic equation ax2 + bx + c = 0 are
−b ±
b 2 − 4ac
. The nature 2a of the roots depends on the value b2 – 4ac. The value of the expression b2 – 4ac discriminates the nature of the roots and so its is called the discriminant of the quadratic equation. It is denoted by the symbol Δ. Discriminant Δ = b2 – 4ac
Nature of roots
1.
Δ > 0, but not a perfect square
Real, unequal and irrational
2.
Δ > 0 and a perfect square
Real, unequal and rational
3.
Δ=0
Real, equal and rational
Δ<0
Unreal (imaginary)
Sl.No.
Note: Consider x2 + 1 = 0. Its roots are given by x2 = –1 or x = + −1 which are not real since the square root of negative quantity is not real. Such roots are called imaginary roots. Example 93: Determine the nature of the roots of the equation x2 – 11x – 30 = 0. Solution: Herea = 1 b = –11 c = –30
84
Discriminant Δ = b2 – 4ac = (–11)2 – 4 (1) (–30) = 121 + 120 = 241. Δ > 0, but not a perfect square. Hence the roots are real, unequal and irrational. Example 94: Determine the nature of the roots of the equation 5x2 – 2x – 7 = 0. Solution: Herea = 5 b = –2 c = –7 Discriminant Δ = b2 – 4ac = (–2)2 – 4(5) (–7) = 4 + 140 = 144 = 122. Δ > 0, but not a perfect square. Hence the roots are real, unequal and rational. Example 95: Determine the nature of the roots of the equation 4x2 – 28x + 49 = 0. Solution: Herea = 4 b = 28 c = 49 2 2 Discriminant Δ = b – 4ac = (28) – 4 (4 x 49) = 784 – 784 = 0. Δ = 0 Hence the roots are real, equal and rational. Example 96: Determine the nature of the roots of the equation x2 – 2x + 5 = 0. Solution: Herea = 1 b=2 c=5 Discriminant Δ = b2 – 4ac = (2)2 – 4 (1) (5) = 4 – 20 = –16 < 0. Δ < 0. Hence the roots are not real. Example 97: Prove that the roots of the equation (a – b + c) x2 + 2 (a–b) x + (a–b–c) = 0 are real. Solution: HereA = a – b + c , B = 2 (a – b), C = a – b – c. Δ = B2 – 4AC = 4 (a – b)2 – 4 (a – b + c) (a – b – c) = 4 (a – b)2 – 4 [(a – b) + c] [(a–b) – c] = 4 (a – b)2 – 4 [(a – b)2 – c2] = 4 [(a – b)2 – (a – b)2 + c2] = 4c2 > 0.
∴ The roots are real. Example 98: Find the value of K such that the equation (2K + 3) x2 + 2 (K + 3) x + (K + 5) = 0 has equal roots. Solution: Here A = 2K + 3, B = 2 (K + 3), C = K + 5. Since the roots are equal, Δ= 0. B2 – 4AC = 0 ⇒ 4 (K + 3)2 – 4 (2K + 3) (K + 5) = 0 ⇒ 4[K2 + 6K + 9 – (2K2 + 10K + 3K + 15) = 0 ⇒ K2 + 6K – 9 – 2K2 – 13K – 15 = 0 ⇒ K2 + 7K + 6 = 0 ⇒ (K + 1) (K + 6) = 0 ⇒ K = –1, K = –6. Example 99: If b = a + c, show that the equation ax2 + bx + c = 0 has rational roots. Solution: Consider Δ = b2 – 4ac = (a+c)2 −4ac = a2 + 2ac + c2 − 4ac = a2 − 2ac + c2 = (a − c)2 = a perfect square ∴ The roots are rational. Example 100: Show that the roots of the equation x2 + 2(a + b) x + 2(a2 + b2) = 0 are not real. Solution: Comparing with Ax2 + Bx + c = 0, we get A = 1, B = 2 (a + b), C = 2(a2 + b2) Now Δ
85
= B2 − 4AC = 4 (a + b)2 − 4(1) 2 (a2 + b2) = 4 (a2 + b2 + 2ab) − 8a2 − 8b2 = 4a2 + 4 b2 + 8ab − 8a2 − 8b2 = − 4a2 − 4b2 + 8ab = − 4(a2 + b2 − 2ab) = − [2(a−b)]2 ∴Δ < 0. The roots are not real. Example 101: If the equation (1 + m2) x2 + 2mcx + c2 − a2 = 0 has equal roots, prove that c2 + a2 = (1 + m2). Solution: Comparing with Ax2 + Bx + C = 0, we get A = 1 + m2, B = 2mc, C = c2 − a2. Since the roots are equal, Δ = 0. B2 − 4AC = 0. (2mc)2 − 4(1 + m2) (c2 − a2) = 0 or 4m2 c2 − 4(c2 − a2 + m2c2 − m2a2) = 0 or 4 [m2c2 − c2 + a2 − m2c2 + m2a2) = 0. −c2 + a2 + m2a2 = 0 or a2 + m2 a2 = c2 or a2 (1+ m2) = c2 or ∴c2 = a2 (1+m2). Example 102: If a, b, c, x are all real numbers and (a2 + b2) x2 − 2b (a + c)x + (b2 + c2) = 0, then prove that a, b, c are in G.P. and x is the common ratio. Solution: We have (a2 + b2) x2 − 2b (a + c) x + (b2 + c2) = 0 or (a2x2 − 2abx + b2) + (b2 x2 − 2bc x + c2) = 0 or (ax − b)2 + (bx − c)2 = 0 Since a, b, c, x are all real ⇒ (ax − b)2 ≥ 0, (bx − c)2 ≥ 0
⇒ ax − b = 0 and bx − c = 0 or x =
b c = ⇒ a, b, c are in G.P. with common ratio x. a b
Example 103: If the roots of the equation (b − c) x2 + (c −a)x + (a − b) = 0 be equal, prove that a, b, c are in A.P. Solution: If the roots are equal Δ = 0 or B2 − 4AC = 0 or (c − a)2 − 4(b−c) (a − b)= 0 or (c2 + a2 − 2ca − 4ab + 4ac + 4b2 − 4bc) = 0 or (c2 + a2 + 2ca) + 4b2 − 4b (c + a ) = 0 or (c + a)2 + (2b)2 − 2 x 2b (c + a) = 0 or [(c + a) − 2b]2 = 0 or c + a = 2b or c − b = b − a ⇒ a, b, c are in A.P. Exercise 4.6.3
1.
Determine the nature of the roots of the equation. (a) 6x2 − 2x − 1 = 0 (b) 9x2 + 12x + 4 = 0 (d) x2 − 8x + 12 = 0
2.
3. 4. 5.
x + 5 = 2x + 3 (h) 9x2 − 16x + 25 = 0
(e)
(c) 2x2 − 3x + 4 = 0 (f) x2 + 9 = 0
(g) x2 + 4x + 7 = 0 (i) x2 − 10x + 25 = 0 (j) 7x2 − 8x + 1 = 0 Find the value of K such that the following equations may have equal roots (a) 3x2 − 4x + K = 0 (b) x2 − K(2x−17) = 12 (c) 9x2 − Kx + 4 = 0 (d) x2 + Kx + 1 = 0 (e) x2 + K(7x − 8) + 65 = 0 (f) Kx2 − 24x + 9 = 0 If 2b = (a + c), show that the equations (b − c)x2 + (c − a)x + (a − b) = 0 has equal roots. If a, b, c are real, show that the equation a (b − c)x2 + b(c − a) x + a (c − b) = 0 has real roots. Prove that the roots of x2 − 2ax + a2 − b2 − c2 = 0 are always real. 86
6. 7. 8.
Show that the roots of equation x2 + 2(a + b) x + 2(a2 + b2) = 0 are unreal. Show that the roots of the equation 3 p2x2 − 2pq x + q2 = 0 are not real. Show that the roots of x2 − 2px + p2 + q2 + r2 = 0 are not real.
ANSWERS Exercise 4.1.1
1 2
(1) 1, , −2
(2) 5, 5, −3 (3) 1, 1, 0 (4) 3, 2, 1
(7) 4,2,1
(8) 3, 3, 2
(9) 3, 2,
(5) 2, 5, −1
(6) −6, 3, −2
3 2 2 (10) 1, 1, −1 (11) , − 2, (12) 2,−3, −1 2 3 5
Exercise 4.1.2
(1) Rs. 100, Rs.80, Rs.50
(2) Rs.10, Rs.8, Rs.4 (3) 789
(5) 55°, 65°, 60° Exercise 4.2.1
(6) 8, 10, 12
1) 2) 3) 4) 5) 6) 7) 8)
Quotient x2 + 2x − 1 4x2 + 9x + 29 5x2 − 6x + 6 x2 − 3x + 2 x2 + 2x − 2 2x2 − 8x + 16 8x3 − 8x2 + 6x 3x2 − x − 1
Remainder 4 83 −20 0 2 −73 −5 −6
(4) 100, 150, 250
(7) 564 Exercise 4.2.2
(1) (a) 10 (b) −74 (c) −4 (d) −17 (2) m = 3 (3) p = −7 (4) m = 2 (5) a = −7 (6) m = 3 (7) a = 3, b = −2 (8) a = −3, b =10, c =5 (9) a = 0, b = −7 (10) p = 6, q = −3 Exercise 4.2.3 (1) (a) Yes (b) Yes (c) No (d) Yes (2) (a) 6 (b) −16 (c) −65 (d) −7 (3) a = −13 (b) = 8 (4) a = 4, b = 0 (5) p = 1, q = −8 (6) a = −1, b = −4, c = 4
Exercise 4.2.4
(1) (4) (7) (10) (13)
(x − 1) (x − 10) (x − 12) (x + 1) (x − 2)2 (x − 1) (x − 1) (3x − 4) (x + 1) (x + 2) (x + 3) (x − 1) (x + 1) (2x + 3)
(2) (5) (8) (11) (14)
(x + 1) (x + 2 ) (x + 10) (x − 1) (x2 + x + 3) (x + 1) (x2 + x + 1) (x − 2) (x − 3) (x − 1) (x − 1) (x − 3) (x + 1)
(3) (6) (9) (12) (15)
(x − 1) (x + 3) (x − 2) (x + 1) (x + 1) (x + 2) (x − 2) (x2 + x + 3) (x − 2) (2x − 1) (x+2) (x − 1) (x2 + x − 4)
Exercise 4.3.1 1) (a) 16x5 (b) 26 a3 (c) 12m (d) 4y4 (e) p8 (f) m6 n 4 2 2 2 (g) x (h) 3a (i) 3abc (j) 2xyz (k) 4p q r (l) 7 mn (c) x (d) a − b (e) x + 2 (f) x + 1 2) (a) x − 2 (b) 2x + 5 (i) 2a − 3 (j) a − 4 (k) 2x + 3 (l) 4(x+3) (g) x − 2 (h) 2x + 5 2 (b) x + 3 (c) x + 1 (d) x − 1 (e) x (3x + 2) 3) (a) x + x + 1
87
Exercise 4.3.2
1) (a) a10 (b) x7 (g) x2y2z (h) abc 2) (a) (x+2)4 (b) 4x (x+4) (x−4) (f) (x + 3) (x − 4) (x − 5) (i) 12 (x4 − 1) (x2 − x + 1) (k) (x − 4) (2x +1) (3x − 2) 3) x(x + 2) (5x + 1) 5) 2(x + 1) (x + 2)
(c) m9 (d) 75a6 (e) 180x6 (f) 900 p9 (i) 84x3y2z2 (j) a m+4 (k) 72 xyz (l) 24p2 qr2 (c) a3 − b3 (d) 12 (x−3) (e) ab (a+b) (g) (4x + 1) (4x − 1) (2x − 3) (h) (x − 2) (x − 3) (x + 3) (j) (x + 2) (x + 3) (x + 5) (l) (x + 5) (3x − 1) (3x + 5) (4x + 1) 4) (x − 1) (x − 2) 6) (x3 − x2 − 4x − 6) (x + 1)
Exercise 4.4.1
x+ 5 1) x−3 6)
10)
xy ( x + y)
7)
y 2 + yx + x 2
x+5 x−2
4 x 2 + 6 xy + 9 y 2
8)
x −1 x+ 1
2
5)
− x2 y2
6( x − 3) x+4
9)
16a 2 − 20ab + 25b 2 11) ab
( 4 x + 9 y ) ( 2 x + 3y ) 2
3x + 5y 4) x
5 3) 6
3a 2) b
12)
x+ 3 2x − 1
Exercise 4.4.2
1) (a)
(e)
2. (a) (e)
a 2 + ab + b 2
(b)
a 2 − ab + b 2 ( x − 4) ( x + 2) x − 4 x + 16 (b)
(c)
(f) 1
2
bp aq
x − 3y 3
3 4a
( x − 2) ( x + 3) (2x + 1) (3x − 1)
(g)
x−3 x+3
(d)
2( x − 2) 2 3( x − 1) 2
(c)
( x − 2) ( x − 3) x+ 2
(d)
(f)
b+a ab
(g) 4
x+2
(x + 4)2
(h) p2
3( x − 3) 2( x − 4) (h) 1
Exercise 4.4.3
3x 1 2 8 y + 3x y − x 2 − xy I. (a) (b) 1 (c) (d) (e) (f) (g) a2 + 4a + 16 2 2 2 2 8 a−b x +1 x y x −y (h) (x + 3)(x2 + 9)
(l)
(i)
4x ( x − 1) ( x + 1) ( x − 2)
2 x −5 (m)
(j)
2x ( x + 2) ( x + 3) ( x + 4)
m2 − 2 m2 −1
(n) 0
88
(o) 0
(k)
− 3a − 7 (2a − 3) (2a + 3) (a + 1)
(p)
x ( x − 2) ( x − 3) ( x − 4) ( x − 5)
II. (1)
5x 3 − 4 x 2 + 8x + 2 ( x + 2) (2 x − 1)
(2)
− x 5 + 5x 4 + 2x 3 + 3x 2 + 6x − 6 (x + 1) (x 2 − 2)
x 3 − 29x 2 + 4x − 10 (3) (3x + 4) (2x − 3)
III. (1)
4( x 2 + 1) 2 ( x 2 − 1) 2
16 x 2
,
(2) 1
( x 2 − 1) 2
Exercise 4.5.1
I. (1) 24 (8)
(2) 26
8 3
(9)
II. (1) 13a4b3c2 (6) x+ y
(3) 45
11 5
(10)
(7)
7x III. (1) (3x + 5)
3
(9) x −
1 x
17 5
(8)
5( x + y) 8( x − y)
(2) (4x − 3)
(5) (a−b) (a+b) (a−3b) (10) x2 +
(5) 105
(3) 7c (2a − 4b)
(2) 3xy2 z4
11zy 2
(4) 99
3( x − y) (a + b) 3
(7) x + y − z
(7)
12 25
(5) 9 (a−b) (x+y)2
5(a + b) 4 ( x + y) 2
(3) (x−2) (x+2) (x−5)
(6) a − b − c
1
(4) 6(2−x)2 (3−x2)3 (9)
2
(6) 0.13
4
(10)
x+y 6(x + 2y)3
(4) (x−1) (x+1) (x−3) (8)
x y + y x
(11) (2x + 5) (3x − 4) (x + 9) (12) (x −1) (2x+1) (3x +1)
x2
Exercise 4.5.2
1) (a) 85
(b) 93 (c) 135
(d) 724
(e) 536
(f) 432
2) (a) x2 − 2x + 3
(b) 2x2 + 2x + 1
(c) 3x2 − 3x + 4
(e) x2 − 3x + 1
(f) 4x2 − 3x − 5
(g)
3) (a) a = 1 (b) a = 6 4) (a) a = 40, b = 25
(c) a = −6 (b) a = 24, b = 36
1 2 x −3x +4 2
(g) 1679 (d) x2 − x − (h)
(c) a = −12, b = 4
(2) −3, −7
(3) 9
(4) −7
(5) 0,8
4 −5 5 (6) 0, (7) −8, 8 (8) (9) −5, 5 , 3 2 2 2 −8 8 (11) – 3, –7 (12) –2, 12 (13) , 1 , (10) 3 3 3
89
14)
1 4
1 2 1 x − 2x + 3 2
Exercise 4.6.1
I. (1) 2,5
(h) 1307
−2 7 , 3 3
(15) –1, 3 (16)
−3 5
(20) − 5 , II. 1. (a) 64 (h)
4 5
(b)
1 ,5 5
25 4
(c) 25
5 , −2 2 (g) 2 + 11, 2 − 11
2. (a) –3, 1
(f) –5, –1
(17) – 18, 5
(b)
(18) 1,
1 2
49 4 1 (c) , 1 3 (d)
(19)
b b , 4a a
b2 16 81 (f) (g) 3 8 4a 2 3 (d) , − 2 (e) –9, 8 5 (e)
(h) –10, –4
1 −1 2 1 3+ 3 3− 3 1 , (2) − , − 4 (3) 20, 8 (4) , (5) , 3 2 5 3 3 3 4 1 5+ 5 5− 5 − 7 + 17 − 7 − 17 (6) 2, (7) 1, 9 (8) (9) , , 2 2 2 4 4 − 7 + 73 − 7 − 73 − 11 + 57 − 11 − 57 b − 2b (10) (11) (12) , , , 2 2 8 8 a 3a
III. (1)
Exercise 4.6.2
(1) 8 (2) 7,8,9 (3) 17, 19 (4) 10 cm, 8 cm (5) 20m, 12m (6) 3.2 cm, 6 cm, 6.8 cm 9.6 sq.cm (7) 35 years, 10 years (8) 30 years −1 (10) 14, 16 (11) 6m, 3m (12) 100 (13) 40 or 20 (9) 3, 2 Exercise 4.6.3
1. (a) Real, unequal and irrational (c) Unreal or imaginary (e) Real, unequal and irrational (g) Real, unequal and irrational (i) Real, equal and rational 2. (a)
4 3
(b) 8 or 9
(c) ± 12
(b) Real, equal and rational (d) Real, unequal and irrational (f) Unreal or imaginary (h) Unreal or imaginary (j) Real, unequal and rational (d) ±2
90
(e) 2
(f) 16
5. APPLIED MATHEMATICS 5.0 INTRODUCTION The needs of Science and Technology, Social Sciences, Humanities and Computers posed new and challenging problems. Such problems could be effectively studied for analytical and exact solution only with the help of Mathematics. Mathematics interacted well with all other branches of Science and Social Sciences and new fields such as Operations Research, Industrial Mathematics, Mathematics of Computation, Mathematical Statistics, Mathematical Physics, Mathematical Biology, Mathematical Modelling, Cryptology, Mathematical Economics and so on and so forth. Today all these fields of sciences have been growing exponentially thanks to mathematical treatment. More than 60% of the Nobel prizes in Economics have gone to those economists who have done their research work purely based on Mathematics. We should observe that Applied Mathematics represents a great stream gushing out irresistibly in many directions overcoming all resistances and all along retaining its originality.
5.1 LINEAR PROGRAMMING The Russian Mathematician L.V.Kantorovich first of all applied mathematical models to solve linear programming problems. He pointed out in 1939 that many classes of problems which arise in production can be defined mathematically and therefore can be solved numerically. This decision making technique was later on developed by George B. Dantziz and he formulated the general linear programming problem and developed the simplex method (1947) for solving such problems. Linear programming today is one of the best developed optimization techniques from theory, application and computation point of view. Linear Inequations We have already learnt to draw the graph of a linear equation of the form ax + by = c. The resulting line divides the plane into three sets of points. They are i) ii) iii)
the set of points on the line the set of points on one side of the line the set of points on the other side of the line.
Each of these three sets of points are generated by the relations ax + by = c, ax + by > c and ax + by < c which are linear in x and y. Here ax + by = c is called a linear equation and ax + by > c and ax + by < c are called linear inequations or linear inequalities. When we combine the equation ax + by = c with ax + by > c we get ax + by > c. Similarly by combining ax + by = c and ax + by < c we get ax + by < c. Here also ax + by > c and ax + by < c are called linear inequations or linear inequalities. 91
We can understand the above discussions by an example. x + y = 2. Its graph is drawn as follows. x y
x+ y=2 ⇒ y =2 − x
0 2
Consider the equation
2 0
3 –1
y 5 4 3 2 1 x’
-7 -6 -5 -4 -3 -2 -1
o 1 2 3 4 5 6 7 -1 -2 -3
x
x+
x+y<2
x+y>2
y= 2
y’
Fig.5.1
In the above graph the straight line represents the equation x + y = 2. Choose a point on one side of the line. Let us take (0, 0). The inequation x + y < 2 is satisfied by (0,0). In Fig.5.1 the shaded portion containing (0,0) represents the relation x + y < 2. The unshaded portion on the other side of the line represents x + y > 2. The relations x + y < 2 and x + y > 2 are called the inequations or inequalities. Note: 1. x + y < 2 represents the points on the line and the points to the left of the line. 2. x + y > 2 represents the points on the line and to the right of the line. Method to draw the graph of an inequation
The following steps are used to draw the graph of inequations ax + by > c or ax + by < c. Step 1: Treat the inequation as if it were an equation. That is ax + by = c. Step 2: Draw the graph of ax + by = c. Step 3: The line ax + by = c divides the plane into two regions. Take any point in one of the regions (usually the origin (0, 0) is taken). Substitute this point in the given inequation. If the inequation is satisfied then the region containing the point is the desired region and it is shaded. If the point does not satisfy the given inequation then the region not containing the point is the desired region and it is shaded. The shaded region represents the given inequation. All the points in this region satisfy the inequation and they are the solutions of the inequation. 92
Example 1: Draw the graph of the inequation 2x + y > 10. Solution: y 2x + y > 10 (-1,12)
x’
-7 -6 -5 -4 -3 -2 -1
12 10 8 6 4 2
(3,10)
(6,2) x
o 1 2 3 4 5 6 7 -2 -4 -6
y’
Fig.5.2
Draw the graph of 2x + y = 10.
2x + y =10 ⇒ y =10 − 2x
x y
1 8
5 0
0 10
Plot the points (1,8), (5,0) and (0,10) in the graph and join them to get the straight line 2x + y = 10. Take the point (0,0) and substitute in the inequation 2x + y > 10. 0 + 0 > 10 which is false. So the region not containing the point (0,0) is the desired region and it is shaded. The shaded region in the graph represents the inequation 2x + y > 10. All the points in the shaded region satisfy 2x + y > 10 and they are the solutions of 2x + y > 10. Note: We can choose points arbitrarily from the shaded region and show the points satisfy 2x + y > 10.
Let us choose the points (6, 2), (3,10) and (–1, 12) and substitute in 2x + y > 10. 2(6) + 2 = 14 > 10 2(3) + 10 = 16 > 10 2(–1) + 12 = 10 Example 2: Sketch the graph 3x + 4y < 12 Solution: Draw the graph of 3x + 4y = 12 3x + 4y = 12 ⇒ y =
12 − 3x 4
93
x y
0 3
4 0
6 –1.5
y 5 4 3 3x + 4y 2 = 12 1
3x + 4y < 12
x’
x
o 1 2 3 4 5 6 7 -1 -2 -3
-7 -6 -5 -4 -3 -2 -1
y’
Fig.5.3 Plot the points (0, 3), (4,0) and (6, –1.5) in the graph and join them to get the dotted line 3x + 4y = 12. The given inequation 3x + 4y < 12 does not contain equality sign. So the points on the line will not satisfy 3x + 4y < 12 and they should be excluded. In order to show this the dotted line is drawn. Take the point (0,0) and substitute it in the inequation 3x + 4y < 12, 0 + 0 < 12 which is true. Therefore the region containing the point (0,0) is the desired region and it is shaded. The shaded region represents the inequation 3x + 4y < 12. All the points in the shaded region are the solutions of 3x + 4y < 12.
Example 3: Draw the graph of x – 4y > 0. Solution: 5 4 3 2 1 x’
x - 4y
=0
x
o 1 2 3 4 5 6 7 -1 -2 -3
-7 -6 -5 -4 -3 -2 -1
x - 4y > 0 y’ Fig.5.4
Draw the graph of x – 4y = 0
x − 4y = 0 ⇒ y =
x y
x 4
0 0
4 1
6 1.5
The straight line passes through (0,0). So take any point which does not lie on the line. The point (5, –1) satisfies the inequation x – 4y > 0. 94
Thus the region containing (5, –1) is the desired region and it is shaded. The shaded region represents the inequation x – 4y > 0. All points in this region are the solutions of x – 4y > 0. Solving the system of inequations by graph
Two or more inequations considered simultaneously are called a system of inequations. To solve them, shade the regions represented by the inequations. The intersection of the shaded regions is the solution set of the inequations. Example 4: Draw the graph of the system of linear inequations: x – 2y > 3 and 2x + 3y < 6 and find the solution set. Solution:
y
5 4 3 2 1 x’
x
o 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 y’
-7 -6 -5 -4 -3 -2 -1
Fig.5.5
x 3 0 1 i) x – 2y > 3: Draw the graph of x – 2y = 3 y 0 – 1.5 – 1 x −3 x − 2y = 3 ⇒ y = 2 The point (0,0) does not satisfy the inequation x – 2y > 3. So, shade the region not containing the point (0,0). ii) 2x + 3y < 6: Draw the graph 2x + 3y = 6
2x + 3y = 6 ⇒ y =
6 − 2x 3
x y
3 0
0 2
1.5 1
The point (0,0) satisfies the inequation 2x + 3y < 6. So shade the region containing the point (0,0). The intersection of the shaded regions (darkly shaded) is the solution set of the inequations. This region extends infinitely.
95
Example 5: Draw the graph of the following system of linear inequations and find the solution set. x – 2y > – 8, 3x + y < 18, x > 0 y > 0. Solution: y 18 16 14 12 10 8 6 4 2 x’
-10 -8 -6 -4 -2
= 2y
x-
-8
x
3x +
o 2 4 6 8 10 -2 -4 -6 y=
y’
18
Fig.5.6
Since x > 0 and y > 0, the solution set is restricted to the first quadrant. i) x – 2y > –8 Draw the graph of x – 2y = –8
x − 2y = − 8 ⇒ y =
x +8 2
x y
0 4
–8
0
4 6
The point (0,0) satisfies the inequation x – 2y > –8. The region containing (0,0) represents x – 2y > –8 ii) 3x + y < 18 Draw the graph of 3x + y = 18
3x + y =18 ⇒ y =18 − 3x
x y
6 0
0 18
2 12
The point (0,0) satisfies the inequation 3x + y < 18. The region containing (0,0) represents 3x + y < 18. The intersection of the two regions is shaded and it is the solution set of the inequations. Example 6: Find the solution set of the system of inequations x – 2y > 0, 2x – y < –2 x > 0, y > 0.
96
y
x’
-y 2x
6 5 4 3 2 1
=-
2
Solution:
x
-2
y=
0
x
o 1 2 3 4 5 6 7 -1 -2 -3
-7 -6 -5 -4 -3 -2 -1
y’
Fig.5.7 Since x > 0 and y > 0, the solution set is restricted to the first quadrant. i) x – 2y > 0 Draw the graph of x – 2y = 0
x − 2y = 0 ⇒ y = x / 2
x y
0 0
2 1
1 0.5
The line passes through (0,0). So we choose a point (4,1). The point (4,1) satisfies the inequation x – 2y > 0. The region containing (4,1) represents x – 2y > 0. ii) 2x – y < –2 Draw the graph 2x – y = –2
x 0 –1 2 y 2 0 6 The point (0,0) does not satisfy the inequation 2x – y < –2. The region not containing (0,0) represents 2x – y < –2. The two regions are not intersecting. Therefore there is no solution for the system of inequations. 2x − y = − 2 ⇒ y = 2x + 2
x+
Example 7: Find the solution set of the system of linear inequations x + y < 8, 2x – 3y < 1, x > 2, x > 0 and y > 0 y Solution: 8 7 C 6 1 y= 5 3 2x 4 3 B 2 1 A x’ x -3 -2 -1 o 1 2 3 4 5 6 7 8 -1 -2 -3 y
=
x=2
8
y’
Fig.5.8
97
i) x + y < 8 Draw the graph of x + y = 8
x + y = 8 ⇒ y = 8− x
x y
0 8
8 0
4 4
The point (0,0) satisfies the inequation x + y < 8. The region containing (0,0) represents x + y < 8. ii) 2x – 3y < 1 Draw the graph of 2x – 3y = 1
2x − 3y =1 ⇒ y =
2x − 1 3
x y
5 3
2 1
0.5 0
The point (0,0) satisfies the inequation 2x – 3y < 1. So, the region containing (0,0) represents 2x – 3y < 1. iii) x > 2 Draw the graph of x = 2
This line passes through the point (2,0) and is parallel to the y axis. The point (0,0) does not satisfy x > 2. So the region not containing (0,0) represents x > 2. The intersection of the three regions is shaded and it is the solution set of inequations. Linear Programming Problem
Linear programming is a mathematical technique of resource planning. It is a method used in making decisions in the areas of agriculture, industry, military, transportation, economics, health system etc. In these fields the available resources are limited. These resources can be of any type such as manpower, materials, money, storage space, machine time etc. The problem is to make the best use of these resources so as to yield the maximum production or to minimise the cost of production or to give the maximum profit. Linear programming refers to a plan that allocate the limited available resources for various uses. The plan helps to maximise the profit or minimise the cost subject to the limitations. The profit or the cost is made up of unknown quantities. These unknowns are variables of degree one and they are called as decision variables. The profit or the cost is a linear function of these variables. This linear function is aimed at maximising the profit or minimising the cost. This function is called the objective function. The limitations are generally expressed in the form of linear inequations. These inequations are called as constraints. The linear programming problem (LPP) is to maximise or minimise the objective function subject to the constraints. Graphical method of solving a LPP • •
Draw the graph of the constraints. Determine the region which satisfies all the constraints and non-negative constraints (x > 0, y > 0). This region is called the feasible region. 98
• • •
Determine the co-ordinates of the corners of the feasible region. Calculate the values of the objective function at each corner. Select the corner point which gives the optimum (maximum or minimum) value of the objective function. The co-ordinates of that point determine the optimal solution.
Example 8: Use graphical method to solve the following linear programming problem.
x=5
Maximise Z = 2x + 10 y Subject to the constraints 2 x + 5y < 16, x < 5, x > 0, y > 0. Solution: y 5 4 2x+ C(0,3.2) 3 5y= 16 2 B(5,1.2) 1 A(5,0) x’ o 1 2 3 4 5 6 7 8 9 -1 -2 -3
x
y’
Fig.5.9
Since x > 0 and y > 0 the solution set is restricted to the first quadrant. i) 2x + 5y < 16 Draw the graph of 2x + 5y = 16
16 − 2x 5 Determine the region represented by 2x + 5y < 16 2x + 5y = 16 ⇒ y =
x y
8 0
0 3.2
3 2
ii) x < 5 Draw the graph of x = 5
Determine the region represented by x < 5. Shade the intersection of the two regions. The shaded region OABC is the feasible region B(5, 1.2) is the point of intersection of 2x + 5y = 16 and x = 5. The corner points of OABC are O(0,0), A(5,0), B(5,1.2) and C(0,3.2). Corners Z = 2x + 10y
O(0,0) 0
A(5,0) 10
Z is maximum at x = 0, y = 3.2 Maximum value of Z = 32. 99
B(5,1.2) C(0,3.2) 22 32
Example 9: Use graphical method to solve the following linear programming problem. Maximise Z = 20 x + 15y Subject to 180x + 120y < 1500, x + y < 10, x > 0, y > 0 Solution: y 12 C(0,10) 10 8 6 B(5,5) 4 2 x’ o 2 4 6 8 10 -2 -4 -6
x
/3, 25 A(
x
+
+2 3x 0)
y
=
10
y= 25
y’
Fig.5.10
Since x > 0 and y > 0, the solution set is restricted to the first quadrant. i) 180x + 120 y < 1500
180x + 120y < 1500 => 3x + 2y < 25. Draw the graph of 3x + 2y = 25 3x + 2y = 25 ⇒ y =
25 − 3x 2
x y
0 25/3 25/2 0
5 5
Determine the region represented by 3x + 2y < 25. ii) x + y < 10 Draw the graph of x + y = 10
x + y = 10 ⇒ y =10 − x
x y
0 10
10 0
5 5
Determine the region represented by x + y < 10. Shade the intersection of the two regions. The shaded region OABC is the feasible region. B(5,5) is the point of intersection of 3x + 2y = 25 and x + y = 10. The corner points of OABC are O(0,0), A(25/3, 0), B (5,5) and C(0,10). Corners Z = 20x + 15y
O(0,0) 0
A(25/3,0) 166.67
B(5,5) 175
Z is maximum at x = 5 and y = 5. Maximum value of Z = 175. 100
C(0,10) 150
Exercise 5.1 1. Solve the following inequations graphically: i) x – y > 2, 3x + 2y < 21 ii) 5x + 2y < 25, y < 5, iii) 2x + y > 4, 3x + 5y > 15 2x + y < 4 iv) x + 2y > 0,
x > 0, y > 0 x > 0, y > 0 x > 0, y > 0
2. Use graphical method to solve the following linear programming problems. i) Maximise Z = 6x + 10y ii) Maximise Z = 30x + 20 y Subject to 2x + y < 800, Subject to 2x + y > 1, 5x + 10y > 4, x + 2y < 1000, x > 0, y > 0 x > 0, y > 0
5.2 NETWORK PROGRAMMING The problems in set theory which cannot be easily solved using algebraic equations can be easily solved using Venn diagrams. Similarly network diagram is used to determine the project completion time. A project will consist of a number of jobs and particular jobs can be started only after finishing some other jobs. There may be jobs which may not depend on some other jobs. Network scheduling is a technique which helps to determine the various sequences of jobs concerning a project and the project completion time. There are two basic planning and control techniques that use a network to complete a predetermined schedule. They are Program Evaluation and Review Technique (PERT) and the Critical Path Method (CPM). The critical path method (CPM) was developed in 1957 by J.E. Kelly of Ramington Rand and M.R. Walker of Dupont to help schedule maintenance of chemical plants. CPM technique is generally applied to well known projects where the time schedule to perform the activities can exactly be determined. Activity: An activity is a task or item of work to be done, that consumes time, effort, money or other resources. It lies between two events, called the starting event and ending event. An activity is represented by an arrow indicating the direction in which the events are to occur.
Activity 1
2
Starting event
Ending event Fig.5.16
Event: An event represents the start or completion of some activity. It has no time duration and does not consume any resource or time. An event is represented on the network by a circle. Activities are identified by the numbers of their starting event and ending event. In the Fig.5.11, 1 is the starting event and 2 is the ending event. The activity is denoted by 1-2. Network: A network is a diagrammatic representation of various activities concerning a project arranged in a logical manner. A path of a network is the sequence of activities starting from the initial event to the final event proceeding in the direction of arrows. The duration of a path is the sum of the durations of activities coming along the path.
101
7
2
4
5
8
1
6
6
8
5
Starting event
4
3
Ending event
5
Fig.5.12
In the Figure 5.12, the initial event of the project is 1 and the final event is 6. The duration of activity 1-2 is 5, duration of activity 1-3 is 8, duration of activity 2-4 is 7 etc. The various paths and their durations in the above network diagram are given below. Paths Duration 1-2-4-6 5 + 7 + 8 = 20 1-3-5-6 8 + 4 + 5 = 17 1-3-4-6 8 + 6 + 8 = 22 The path that takes the longest duration is called the critical path. The duration of this path is called the project duration. The critical path method is used to calculate the total project duration and to check the actual progress of the project against the scheduled duration of the project. Illustration Following are the activities which are to be performed to construct a building, the duration for each activity is given. Activity Duration (days) Activity Duration (days) Foundation 15 Wood work 7 Brick work 10 Grill work 5 Electrical work 5 Cementing 8 Plumbing 4 Flooring 5
Foundation 1
Brick work 2
15
e le
c tri
3 10
p lu
cal
4
7
5
dw o rk
6
4 mb
woo
5
in g
5
g r il
r l wo
Flooring 8
7
Painting 5
8
k
Fig.5.13
Paths 1. 1-2-3-4-6-7-8 2. 1-2-3-5-6-7-8
Duration 15+10+5+7+8+5 = 50 15+10+4+5+8+5 = 47
The first path takes the longest duration. ∴ The critical path is 1-2-3-4-6-7-8. The duration is 50 days. Note: The activities plumbing and grillwork will go on simultaneously along with electrical and wood work. That is why the activities 3 – 5 – 6 do not appear in the critical path.
102
Example 10: The following table gives the activities of a project and their duration in days.
Activity Duration
1-2 5
1-3 8
2-4 7
2-3 6
3-4 5
3-5 4
4-5 8
Determine the critical path and duration. Solution : 7 2
4 8
5 6
5
1
3
Starting event
5
4
8
Ending event
Fig.5.19
The critical path is 1-2-3-4-5. The project duration is 24 days. Example 11: A project has the following time schedule Activity
1-2
1-6
2-3
2-4
3-5
4-5
6-7
5-8
7-8
Duration in days
7
6
14
5
11
7
11
4
18
Draw the network diagram and find the critical path. Solution 7
14
1
11
2
3 5
5
7
4
4
6
11
6
7
Fig.5.15
Paths 1-2-3-5-8 1-2-4-5-8 1-6-7-8
Duration 7+14+11+4 = 36 7+5+7+4 = 23 6+11+18 = 35
The critical path is 1-2-3-5-8 The project duration is 36 days. 103
18
8
Example 12: A projects consists of 12 jobs. Draw the project network and determine the critical path.
Job
Activity
Duration (weeks)
a b c d e f g h i j k l
1-2 2-3 2-4 3-4 3-5 4-6 5-8 6-7 6-10 7-9 8-9 9-10
2 7 3 3 5 3 5 8 4 4 1 7
Solution: 7 1
2 a
b
2
5 e
3
8
3 d
c 3
5 g
5
k 1 f
4
h 6
j 7
3
8
9 4 l
i
7
4
10
Fig.5.16
Path Duration 1-2-3-5-8-9-10 2+7+5+5+1+7 = 27 1-2-3-4-6-7-9-10 2+7+3+3+8+4+7 = 34 1-2-3-4-6-10 2+7+3+3+4 = 19 1-2-4-6-7-9-10 2+3+3+8+4+7 = 27 1-2-4-6-10 2+3+3+4 = 12 The critical path is 1-2-3-4-6-7-9-10. The duration is 34 weeks. Exercise 5.2 1. The following table gives the activities in a construction project and relevant information Activity Duration in days
1-2 22
1-3 27
2-3 12
2-4 14
3-4 6
4-5 12
Draw the network for the project and find the critical path. Compute the project duration.
104
2. A project has the following schedule Activity
1-2
2-3
2-4
3-5
4-6
5-6
Duration in weeks
6
8
4
9
2
7
i) Construct the network ii) Find the critical path and project duration. 3. A small maintenance project consists of the following jobs whose activities and durations are given below. Activity
1-2
1-3
2-3
2-4
3-4
3-5
4-5
Duration in days
20
25
10
12
5
8
10
i) Draw the network diagram ii) Find the critical path and project duration. 4. The following table gives the characteristics of a project Activity
1-2
1-3
2-3
3-4
3-5
4-6
5-6
6-7
Duration in days
5
10
3
4
6
6
5
5
i) Draw the network diagram ii) Find the critical path and project duration 5. Find the critical path and project duration of the following 3
3
6
4 2 1
5 9
8 5 2 7 4
Fig.5.17
105
ANSWERS Exercise 5.1 2. i) x = 0 y = 1 Z = 10
(ii) x = 200 y = 400 Z 14000
Exercise 5.2 22
1
1.
14
2
12
4
5
12 27
6
3
Critical path is 1-2-3-4-5; Duration 52 days 9 3
8
5
2. 7
6
1
2
4
2
4
6
Critical path is 1-2-3-5-6; Duration 30 weeks 3.
2
20
12 4
10
1
10
5 25
5
3 8
The critical path is 1-2-3-4-5; Duration 45 days 2
5
4. 4
3
1
4 10
6
3
6
7 5
6
5 5
The critical path is 1-3-5-6-7; Duration 26 days 5. Critical path is 1-2-4-6; Duration 21 days.
106
6. GEOMETRY 6.0
INTRODUCTION
Third century saw the culmination of Greek Geometry. Euclid belongs to this period. Later Appolonius extended Euclid's work on geometry to conics. Greeks are known for their acumen in Geometry and a significant contribution was made to geometry by Greeks. Vedic period findings in Geometry (Rekha - Ganit) include construction of a square, construction of a square equal in area to a given rectangle, construction of a square equal in perimeter to a circle. Mathematicians of vedic period knew sutras related to construction of mounds for yogas and altars for keeping fire. These sutras indicated knowledge of the form a2 + b2 = c2 which we call Pythagoras theorem. Thus we find that the Greek mathematician Pythagoras (6 BC) was not the first to discover this theorem. Pythagoras was familiar with Upanishads and learnt his basic geometry from the Sulvasutras. History of Indian Mathematics used to begin by describing the geometry contained in sulvasutras. An early statement of what is commonly known as Pythagoras theorem is to be found in Baudhayana's sutra (800 BC): "The chord which is stretched across the diagonal of a square produces an area of double the size!" Ramanujan showed interest in construction of a square very nearly equal to a given circle in area. Squaring a circle is a famous problem in Mathematics having its origin in antiquity. Even today many mathematicians are crazy about squaring the circle. Ramanujan's intuitive flash gave a rider to the theorem of Pythagoras which is a very interesting property of the right angled triangle.
6.1
LOCUS
Let us observe the motion of the seconds hand of a watch. The tip of the seconds hand occupies several positions in the plane of the dial. We observe that the tip of the seconds hand traces a circle, namely a closed curve each of whose points is equidistant from the point about which the second hand is rotating. The curve so traced by the tip of the seconds hand is called its locus. Locus is a path pursued by a moving point which satisfies certain geometrical conditions. From these we learn that i) points satisfying the conditions lie on the locus ii) all points on the locus satisfy the conditions. P Q We can find the locus by plotting the points. Mark two points P,Q on a sheet of paper. Fold the paper such that P and Q fall on each other. Every point on Fig. 6.1 the fold is at the same distance from both points P and Q. The mid point of the segment PQ is also on the fold. So, we find the fold as the perpendicular bisector of the segment PQ (Fig.6.1). We state this in the following theorem:
107
Theorem 1. The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points. l1 úUÛm l2 Gu\ C n a sheet of paper. Fold the paper such that the lines fall on each other. Every point on the fold is equidistant from the two lines. So we conclude that the locus of a point equidistant from two given parallel lines is the parallel line situated midway between the lines (Fig. 6.2). Fig.6.2
1
d d 2
Let us now take an angle to find the locus of a point equidistant from the arms of an angle. Draw an ∠AOB on a sheet of paper fold the paper such that A OA falls on OB. Every point on the fold is at the same distance from M OA and OB. So, we conclude that the locus of a point equidistant from the arms of an angle is the angular bisector (Fig.6.3). P O
Let us now take two intersecting lines l1 and l2 and find the N locus of a point equidistant from both the lines. Let the lines l1 and l2 intersect at O. Four angles ∠AOB, ∠BOC, ∠COD, ∠DOA are B Fig. 6.3 formed. The locus of a point equidistant from OA and OB is OP, the angular bisector of ∠AOB. Similarly the angular bisector of ∠BOC, ∠COD, ∠DOA are OQ, OR and OS respectively. (Fig.6.4). ll P
B From the figure O o we have m∠AOB + m∠BOC = 180 (Adjacent angles) m∠POB + m ∠BOQ = ½ {m∠AOB + m∠BOC} Q m∠POQ = ½ × 180o = 90o Similarly m∠QOR = m ∠ROS = m∠SOP = 90o C Hence m∠POQ + ∠QOR = 180o R Hence PR is a line Fig. 6.4 Likewise QS is also a line. PR is the angular bisector of ∠AOB and ∠COD. QS is the angular bisector of ∠BOC and ∠DOA So, we conclude that all points on PR and QS are equidistant from the lines l1 and l2 We state this in the following theorem.
11
A
l2
S
D l2
Theorem 2 : The locus of a point equidistant from two intersecting lines is the pair of bisectors of the angles formed by the given lines. Exercise 6.1 1. 2. 3. 4. 5.
Find the locus of the point equidistant from three non collinear points A,B and C. PQ is a line, X and Y are two given points. If PQ is not parallel to XY find the point equidistant from X and Y which lies on PQ. Show that the points on the diagonals of a square are equidistant from its sides. Are the points on the diagonals of rectangle are equidistant from its sides? Show that the points on the diagonals of a rhombus are equidistant from its sides. 108
6.2 CIRCLES The locus of a point, which moves such that its distance from a fixed point always a constant is called a Circle. The fixed point is called its centre and the constant distance is called its radius. The perimeter or boundary of a circle is called its circumference. A line segment whose end points lie on the circumference of a circle is called a chord. A chord passing through the centre of a circle is called its diameter. A diameter is the longest chord of a circle. Recall: We will now learn the relationship between the centre and chords in a circle. The congruency of two triangle can be ensured by establishing any one of the following criterion. i) Side - Angle - Side (SAS) ii) Angle - Side - Angle (ASA) iii) Side - Side - Side (SSS) In a right angled triangle congruency is established by iv) Right angle - Hypotenuse - Side (RHS) Theorem 3: Perpendicular from the centre of a circle to a chord bisects the chord. Given : AB is a chord in a circle with centre O. OC ⊥ AB. To prove: The point C bisects the chord AB. Construction: Join OA and OB (Fig.6.5). Proof: In triangles OAC and OBC m∠OCA = m∠OCB = 90o (Given); OA = OB (Radii) OC = OC (common side) ΔOAC ≡ ΔOBC (RHS) CA ≡ CB (corresponding sides) ∴The point C bisects the chord AB. Hence the theorem is proved. Theorem 3A: (converse of Theorem 3) The line joining the centre and the mid point of a chord is perpendicular to the chord. Given : AB is a chord in a circle with centre O. C is the mid point of AB. To prove : OC ⊥ AB Construction : Join OA and OB (Fig.6.6) Proof: In triangles OAC and OBC OA = OB (Radii); AC = BC (Given); OC = OC (Common side) ΔOAC ≡ ΔOBC (SSS) m∠OCA ≡ m∠OCB (corresponding angles) But m∠OCA + m∠OCB = 180o m∠OCA + m∠OCA = 180
o
(ACB is a line) (m∠OCB = m∠OCA) 109
O
A
C
B
Fig. 6.5
O
A
C Fig. 6.6
B
= 180o; ∴OC ⊥ AB Hence the theorem is proved. 2m∠OCA
m∠OCA = 90o
Let us find how many circles can be drawn through three given non-collinear points. Let A,B and C be three given non collinear points. Draw line segments AB and BC. Draw perpendicular bisectors R PQ and RS of AB and BC respectively. AB is not parallel to BC. PQ will not be parallel to RS. Hence they will intersect at some point say O. Join OA, OB and OC (Fig.6.7). O lies on PQ (perpendicular bisector of AB). OA = OB (by theorem 1) (1) similarly OB = OC (2) From (1) and (2) OA = OB = OC = r, (say)
P C
O
S Q
A
B
Fig. 6.7
Taking r as radius O as centre draw a circle. It will pass through the points A, B and C. We shall now prove that this is the only circle passing through A,B and C. If possible there will be another circle with centre O′ and radius r′ passing through A, B and C. Then O′ must lie on perpendicular bisectors PQ and RS. Since two lines can not intersect at more than one point, O′ must coincide with O. Therefore OA = O′A = r (= r′). So we state this in the following theorem. Theorem 4 : There is one and only one circle passing through three given non-collinear points.
C
Now let us examine another result concerning two equal chords of a circle. Let us draw a circle with centre O on a tracing paper. Draw two equal chords AB and CD. We now draw OP ⊥ AB and OQ ⊥ CD (Fig.6.8). Let us now fold the paper so that A falls on C and B falls on A D. Because of the equal chords AB and CD this is possible. Then we observe that P falls on Q and the line of fold passes through O. Thus OP = OQ that is AB and CD are equidistant from centre O. We may repeat this activity by drawing several circles and equal chords. Every time we shall obtain the same result. So, we state this in the following C theorem. Theorem 5: Equal chords of a circle are equidistant from the centre.
D
Q O P
B
Fig. 6.8 D
Q O
The converse of theorem 5 can be seen in the following manner. A P B Let us draw a circle with centre O on a tracing paper. Draw a chord AB. From O draw OP ⊥ AB. Now let us mark a point Q in the circle Fig. 6.9 such that OP = OQ. Now through Q draw a chord CD such that OQ ⊥ CD (Fig.6.9). By folding the paper, so that OP falls on OQ. we can easily see that AB falls completely on CD. That is AB = CD. We may repeat this activity by drawing several circles and chords with equal distances. Every time we shall obtain the same result. So we get this in the following theorem. 110
Theorem 5A: (Converse of theorem 5). In a circle, chords equidistant from the centre are equal. Example 1: How far away is a chord of length 10 cm from the centre of a circle of radius 13 cm. Solution: AB is a chord of length 10 cm C in the mid point of AB (Fig.6.10) AC = ½ AB = ½ × 10 = 5 cm ; radius OA = 13 cm A In a right angled triangle OAC (Fig.6.10) OC2 = OA2 – AC2 = 132 – 52 = 169 – 25 = 144 OC = 144 = 12 cm The chord is 12 cm away from the centre.
O
C
B
Fig. 6.10
Example 2: AB and CD are equal chords of a circle whose centre is O. OM ⊥ AB and ON ⊥ CD. Prove that m∠OMN = m∠ONM. Solution: C A Given : In a circle with centre O chords AB and CD are equal OM ⊥ AB, ON ⊥ CD (Fig.6.11). To prove : ∠OMN = ∠ONM M N Proof : AB = CD (given) OM ⊥ AB (given); ON ⊥ CD (given) O OM = ON (equal chords equidistant from the centre) D B In triangle OMN m∠OMN = m∠ONM (Δ OMN is isosceles) Fig. 6.11
Example 3 : In two concentric circles, chord AB of the outer circle cuts the inner circle at C and D. Prove AC = BD. Solution : O Given : Chord AB of the outer circle cuts the inner circle at C and D (Fig.6.12). To prove : AC = BD A C D B M Construction: Draw OM ⊥ AB Proof : Since OM ⊥ AB (by construction) OM is also ⊥ CD (ACDB is a line) Fig. 6.12 In the outer circle, AM = BM (OM ⊥ AB, perpendicular bisects the chord) (1) In the inner circle, CM = DM (OM ⊥ CD, perpendicular bisects the chord) (2) From (1) and (2) D AM – CM = BM – DM or AC = BD B Q Example 4 : AB and CD are two chords of a circle with centre O. The M O two chords intersect each other at the point M. If OM bisects ∠AMD P C prove that AB = CD. Solution : A Given : AB and CD are any two chords of circle with centre O. They Fig. 6.13 intersect at M. OM bisects ∠AMD (Fig.6.13) To prove : AB = CD 111
Construction : Draw OP ⊥ AB; OQ ⊥ CD Proof : In triangles OMP, OMQ ∠OPM = ∠OQM = 90o (construction) ∠OMP = ∠OMQ (given) OM = OM (common side) ΔOMP ≡ ΔOMQ (AAS) ∴OP = OQ (corresponding sides) Hence AB = CD (chords equidistant from the centre are equal) A Example 5 : If two circles intersect in two points prove that the line through the centres is the perpendicular bisector of the common chord. Solution: O O' M Given : Two circles C (O, r) and C′ (O′, r′) intersect at A and B so that AB is the common chord of the two circles (Fig.6.14). To prove : OO′ is the perpendicular bisector of the chord AB. B Construction : Draw the line segments OA, OB, O′A and O′B. Let M be the point of intersection of AB and OO′. Fig. 6.14 Proof : In triangles OAO′ and OBO′ OA = OB (radius of the circle with centre O) O′A = O′B (radius of the circle with centre O′) OO′ = OO′ (common side) ∴ By SSS criterion Δ OAO′ ≡ ΔOBO′ ⇒ m∠AOO′ = m∠BOO′ Now in triangle AOM and BOM. OA = OB m∠AOM = m∠BOM (Q m∠AOM = m∠AOO′ and m∠BOM = m∠BOO′) OM = OM (common side) By SAS criterion ΔAOM ≡ ΔBOM ⇒ AM = MB and m∠AMO = m∠BMO
But m∠AMO + m∠BMO = 180o ∴m∠AMO = m∠BMO = 90o ∴OM ⊥ AB Since M is the mid point of AB, and OO′ ⊥ AB ⇒ OO′ is the perpendicular bisector of the chord AB. Exercise 6.2 1. How far is a chord of length 12 cm away from the centre of a circle of radius 10 cm. 2. A chord is 15 cm away from the centre of a circle of radius 17 cm. Find the length of the chord. 3. AB and CD are equal chords of a circle whose centre is O. When produced the chords meet at P. Prove that PB = PD. 4. AB and CD are two equal chords of a circle with centre O. If P is the mid point of AB, Q is the mid point of CD, prove that ∠APQ = ∠CQP. 5.
In ΔABC, AB = AC , prove that the circumcentre of Δ ABC lies on the angular bisector of ∠BAC.
112
6.3
ANGLES IN A CIRCLE
Any two points say A and B of a circle divide the circumference of the circle into two parts called the arcs of the circle. If the two parts are unequal, the smaller part is called the minor arc and the larger one is called the major arc. Each of these may be named as arc AB or AB. Sometimes a third point is taken on the arc to distinguish it from the other arc. C
C
O
O
B
A
B
A
D
D
Fig. 6.16
Fig. 6.15
In the Figs.6.15 and 6.16 ACB is the major arc and ADB is the minor arc. The minor arc ADB subtends ∠AOB at the centre, ∠ACB at the circumference. Likewise the major arc ACB subtends reflex m∠AOB at the centre and m∠ADB at the circumference. We know
m∠AOB + m reflex ∠AOB = 360o.
We will establish a relationship between the angles subtended by an arc at the centre and at the circumference. Recall, the angles opposite to equal sides of a triangle are equal. If a side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles. Theorem 6 : Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. C Q O
O P A
B
A
B
D Fig. 6.18
Fig. 6.17
Given : In a circle with centre O, AB is a minor arc, C is a point on the major arc (Fig.6.17). AB is a major arc, D is a point on the minor arc (Fig.6.18). To prove : m∠AOB = 2m∠ACB, m reflex ∠AOB = 2m∠ADB Construction : In Fig. 6.17 join CO and extend it to P. In Fig.6.18 join DO and extend it to Q. Proof : In Δ AOC OA = OC (Radii of a circle) m∠OCA = m∠OAC (Angles opposite to the equal sides) In Δ AOC
113
side CO, extended it to P (construction) ∠AOP is the exterior angle m∠AOP = m∠OAC + m ∠OCA (exterior angle = sum of interior opposite angles) m∠AOP = m∠OCA + m ∠OCA as ∠OAC = ∠OCA = 2m ∠OCA = 2m ∠OCB similarly m∠BOP = m∠AOP + m∠BOP (parts of angle) m∠AOB = 2m∠OCA + 2m∠OCB = 2{m∠OCA + m ∠OCB} = 2m∠ACB Likewise we can show that m reflex ∠AOB = 2m∠ADB Hence the theorem is proved. Let us now study the angles subtended by an arc at more than one point on the circumference. C
D
D
C A
O
O
B
B
A Fig. 6.19
Fig. 6.20
In a circle with centre O and ∠ACB, ∠ADB are two angles formed in the same segment of the circle (Fig.6.19 and Fig.6.20). = ½ m∠AOB (by theorem 6) m∠ACB = ½ m∠AOB (by theorem 6) m∠ADB ∴m∠ACB = m∠ADB We state this in the following theorem. Major Segment
Theorem 7: Angles in the same segment of a circle are equal.
O Segments of circle: The region in the interior of a circle is divided into two parts by the chord AB (see the Fig.6.22). These two regions are Minor called segments. The segment which contains the centre is the major A Segment segment and the smaller region is the minor segment. A diameter divides the circumference into two equal parts. Each of these is called a Fig. 6.21 semicircle. C Let us now examine the case when an arc AB is a semicircle (see the Fig.6.22). As done in theorem 6, we join CO and produce it to P. We note that.
m∠AOP + m∠BOP m∠AOB But m∠AOB m∠ACB
= = = =
180o 180 o 2m∠ACB (by theorem 6) ½ m∠AOB
114
A
O P Fig. 6.22
B
B
= ½ × 180 o =
180 o = 90o = a right angle. 2
This fact is expressed as
Theorem 8: The angle in the semicircle is a right angle.
P
To know the converse of the theorem 8 let us take a right angled triangle PQR right angled at P. Now let us draw a circle with hypotenuse QR as diameter (see the Fig.6.23). Let us recall, if we join the mid point of the hypotenuse to the vertex of the right angle then Q the line segment so formed is half of the hypotenuse. Thus in this case OP = OQ = OR. Hence the circle passes through P. We state this in the following theorem:
Theorem 8A: (converse of theorem 8). The circle drawn with hypotenuse of a right angled triangle as diameter passes through its opposite vertex at any point of the semicircle.
R
Fig. 6.23
A
In a circle with centre O, AB and CD are equal chords (See the Fig.6.24). In triangle AOB and COD. OA = OC (radii); OB = OD (radii) ; AB = CD (equal chords) ΔAOB ≡ ΔCOD . (SSS) m∠AOB = m∠COD (corresponding angles) We state this in the following theorem.
Theorem 9 : Equal chords subtend equal angle at the centre. In a circle with centre O, m∠AOB = m∠COD (see the Fig.6.25) In triangles AOB and COD. OA = OC (radii) m∠AOB = m∠COD (equal angles) OB = OD (radii) ΔAOB ≡ Δ COD (SAS) AB = CD (corresponding side) We state this in the following theorem.
O
O
D
B
C
Fig. 6.24
A
O
B
D
C
Fig. 6.25
Theorem 9A: (converse of theorem 9). If the angles subtended by the chords at the centre are equal then the chords are equal. Cyclic quadrilateral If all the vertices of a quadrilateral lie on a circle then it is called a cyclic quadrilateral. We now get an important relation regarding the angles of a cyclic quadrilateral.
Theorem 10 : The sum of the opposite angles of a cyclic quadrilateral is 180o. Given : ABCD is a cyclic quadrilateral in a circle with centre O. To prove : m∠A + m∠C = 180o, m∠B + m∠D = 180o
115
Construction : Join B and D to the centre O (Fig.6.26). C Proof : m∠BCD = ½ m∠BOD (Angle at the centre = Twice angle at any point on the circumference) O = ½ reflex m∠BOD(same theorem) m∠BAD m∠BCD + m ∠BAD = ½ m∠BOD + ½ reflex m∠BOD = ½ (m∠BOD + reflex m∠BOD) D = ½ × 360o = 180o A m∠A + m∠C = 180o m∠A + m∠B + m∠C + m∠D = 360o Fig. 6.26 (The sum of angles of a quadrilateral is 360) But m∠A + m∠C = 180o (proved) ⇒ m∠B + m∠D = 180o Hence the theorem is proved. Therefore the opposite angles in a cyclic quadrilateral are supplementary.
B
Theorem 10A (Converse of theorem 10). If the sum of opposite angles of a quadrilateral is 180o, then it is cyclic. C
E
D
D
E
C
B
A
B
A Fig. 6.27
Fig. 6.28
Given : ABCD is a quadrilateral in which m∠A + m∠C =180o, m∠B + m∠D = 180o To prove : ABCD is cyclic Construction : Draw the circumcircle of ΔABD if C does not lie on it. Let BC or its extension meet the circle at E. Proof: ABED is cyclic (construction) m∠A + m∠E = 180o (opposite angles) But m∠A + m∠C = 180o (given) m∠A + m∠E = m∠A + m∠C = m∠C m∠E m∠BED = m∠BCD In ΔDEC (Fig.6.27) m∠BCD is the exterior angle m∠BCD > m∠BED (exterior angle = sum of interior opposite angles) In Δ DEC (Fig.6.28) ∠BED is the exterior angle m∠BED > m∠BCD (exterior angle = sum of interior opposite angles) Hence we arrive at a contradiction ⇒ C and E will coincide That is the circle through A, B and D will also pass through C That is ABCD is cyclic. Hence the theorem is proved.
116
Corollary : The exterior angle formed by extending a side of a D cyclic quadrilateral is equal to the internally opposite angle. C In a cyclic quadrilateral ABCD, the side AB is extended to E O (Fig.6.29). By theorem 10, m∠ABC + m∠ADC = 180o E A B m∠ABC + m∠CBE = 180 o (ABE is a line) Fig.6.29 m∠ABC + m∠ADC = m∠ABC + m∠CBE m∠ADC = m∠CBE ∠CBE is the exterior angle and ∠ADC is the internally opposite angle. C Example 6: In the figure O is the centre. m∠OAC = 35o and m∠OBC = 45o find m∠AOB O 45 Solution : Join OC (Fig.6.30) 35 OA = OB = OC; m∠OCA = m∠OAC = 35o B A m∠OCB = m∠OBC = 45o; m∠ACB = m∠OCA + m∠OCB = 35o + 45o = 80o; m∠AOB = 2m ∠ACB = 2 × 80o = 160o Fig.6.30 o
o
Example 7: Prove that angle in a minor segment is obtuse. Solution: Let ∠ACB be an angle formed in a minor segment of a circle with centre O. We have to prove that ∠ACB is obtuse. Now m∠ACB = ½ m reflex ∠AOB but m reflex ∠AOB > 180o ½ m reflex ∠AOB > ½ × 180o or ½ m reflex ∠AOB > 90o Hence m∠ACB > 90o. That is m ∠ACB is an obtuse angle in a minor segment.
O B
A
C
Fig.6.31
Example 8: In the figure PQ is a diameter. If m∠OPS = 50o find m∠QOS and m∠QRS. Solution: OP = OS (radii); m∠OSP = m∠OPS = 50o (Angles opposite to equal sides) m∠QOS = m∠OSP + m∠OPS (Exterior angle = sum of interior R S opposite angles) = 50o + 50o = 100o; PQRS is cyclic ⇒ m∠QRS+ m∠QPS = 180o P Q O0 (opposite angles of a cyclic quadrilateral supplementary) m∠QRS + 50o = 180o, m∠QRS = 180o – 50o = 130o Fig.6.32 Example 9: ABCD is a cyclic trapezium with AD⎥⎥ BC. If ∠B = 65o, determine other three angles of the trapezium. Solution: ABCD is a cyclic trapezium, AD⎥⎥ BC, AB transversal. m∠BAD+ m∠ABC =180o [sum of interior angles) m∠BAD+65o = 180o; m∠BAD = 180o – 65o = 115o; m∠ADC+ m∠ABC = 180o (ABCD is cyclic) m∠ADC+65o = 180o; m∠ADC = 180o – 65o = 115o m∠BCD+ m∠BAD =180o (ABCD cyclic) ∠BCD+ 115o =180o ; m∠BCD =180o –115o = 65o 117
C
D
A
B
Fig.6.33
Exercise 6.3 1. Find the angle marked as x in the following figures: (a) (b) (c)
(d)
x
(e)
x x O
O
o
33
o
O
240 O
x
o
130
O
o
52
o
110
Fig.6.34
x
Fig.6.35
Fig.6.36
Fig.6.37
Fig.6.38
C
2. In the figure m∠OAC =30o m ∠OBC = 25o find m ∠AOB
O
A
B
3. 4. 5. 6.
6.4
Fig.6.39 Prove that angle in a major segment is acute. Prove that parallelogram inscribed in a circle is a rectangle. Two circles cut each other at A and B. Lines CAD, and EBF are drawn through A and B to cut the circles at C,D,E and F. Prove that CE || DF. Two diameters of a circle intersect each other at right angles. Prove that the quadrilateral formed by joining their end points is a square.
CIRCLES AND TANGENTS
Given a circle and a line in a plane there exists three possibilities. In Fig.6.40 line AB does not intersect the circle. In Fig.6.41 line AB intersects the circle in two distinct points and is called a secant. In Fig.6.42 line AB touches the circle in exactly one point that is P. In this case line AB is a tangent to the circle at that point P. The point P is called the point of contact of AB with the circle. A
A
B
P
O
O
O
A
B
B
Fig.6.40 Fig.6.41 Fig.6.42 In the Fig.6.43 given a circle with centre O. TAT′ is the tangent and A is the point of contact. You find that all other points on the tangent lie outside the circle. Let B be any point on TAT′, OB > OA . You know that the perpendicular from a point to a line is the shortest among the lines joining the point and the point on the lines. ⎯→
ThereforeOA is perpendicular to TAT′ and OA is the radius at A. Hence we get the following theorem: 118
T
O
A
P
B
Fig.6.43
T’
Theorem 11: A tangent at any point on a circle is perpendicular to the radius through the point of contact. You know that one and only one perpendicular can be drawn from a point to a line. Since the tangent is perpendicular to the radius at the point of contact, there can be only one tangent at a point on the circle. This is stated as the following theorem:
Theorem 12: There is one and only one tangent at any point on the circle. If a point lies on the circle only one tangent can be drawn through that point to the circle. If a point lies inside the circle no tangents can be drawn to the circle. If a point lies outside the circle two tangents can be drawn through that point to the T circle. A Theorem 13: The lengths of two tangents drawn from an external point to a circle are equal. Given: PT and PT′ are the tangents from a point P outside O P the circle with centre O. A and B are the points of contact. PA and PB are tangent segments (Fig.6.44). To prove : PA = PB B T’ Construction: Join OP. Fig.6.44 o Proof : In triangles OAP, OBP m ∠OAP = m∠OBP = 90 (radius is perpendicular to the tangent at the point of contact). OP = OP (Common side); OA = OB (radii)
Δ OAP ≡ ΔOBP (RHS); PA = PB (Corresponding sides). Hence the theorem is proved. Angles in the alternate segment Let TAT′ be a tangent and AB be a chord to the circle at A. (Fig.6.45). Chord AB makes two angles with the tangent TAT′ namely∠TAB and ∠T′AB. Let C and D be any two points on either side of AB. Then ∠ADB is said to be the angle in the alternate segment of ∠TAB. Similarly ∠ACB is said to be the angle in the alternate segment of ∠T′AB.
B C
O D A
T
T’
Fig. 6.45 Theorem 14: The angle between the tangent and a chord passing through the point of contact is equal to the angle in the alternate segment. Given: TAT′ is a tangent and AB is a chord through the point of contact A. To prove: m ∠TAB = m∠ADB, m∠T′AB = m∠ACB (Fig.6.48). Construction: Draw diameter AE and join BE Proof : m∠T′AE = 90o ( OA ⊥AT′); m∠T′AE = m ∠T′AB + m∠BAE (Parts of angle) 119
∴ m∠T′AB +m∠BAE = 90o (1) m∠EBA = 90o (Angle in the semicircle) m∠AEB+m∠BAE = 90o (ABE right angled Triangle) (2) From (1) and (2) m∠T′AB+m∠BAE = m∠AEB+m∠BAE m∠T′AB = m∠AEB (Removing common∠BAE) (3) m∠AEB =m∠ACB (Angles in the same segment) (4) From (3) and (4) m∠T′AB = m∠ACB m∠TAB+m∠T′AB = 180o (TAT′ is a line) (5) o m∠ACB +m∠ADB = 180 (ABCD is cyclic) (6) From (5) an d(6) m∠TAB + m∠T′AB = m∠ACB +m∠ADB But m∠T′AB =m∠ACB ∴m∠TAB =m∠ADB Hence the theorem is proved. Touching Circles: Let us see how circles can locate themselves.
E B C
O D
A
T
T’
Fig.6.46
A P B
Fig.6.47
Fig.6.48
Fig.6.49
O
Q
Fig.6.50
O’
Fig.6.51
O
Fig.6.52
In Fig.6.47 each one of the circle is outside the other. There are no common points between them. In Fig.6.48 the circles touch each other externally. There is one common point (say P). In Fig.6.49 the circles cut each other and they have two common points (say A and B). In Fig. 6.50 the circles tough each other internally. There is one common point (say Q). In Fig. 6.51 one circle is inside the other circle and they have no common point. But they have common centre (say O ). They are called concentric circles. In Fig.6.52 one circle is inside the other and they have no common point. They have different centres. T
T
P
P
B
A
T’
AB
T’
Fig.6.53
Fig.6.54 120
The above circles are called touching circles. The common point in called the point of contact. A and B are the centres of touching circles and P is the point of contact. Let TPT′ be the common tangent to the circles. In Fig. 6.53 m∠APT = 90o (AP⊥PT); m∠BPT = 90o (BP ⊥ PT); m∠APT + m∠BPT = 180o ∴ APB is a line. In Fig. 6.54 m∠APT = m∠BPT Since only one perpendicular can be drawn at P to PT, PA and PB lie on the same line. ∴PAB is a straight line. So, we state this in the following theorem : Theorem 15 : When two circles touch each other, their centres and the point of contact lie on a line. A Example 10: Determine the length of tangent to a circle of radius 3 3cm cm from a point at a distance of 7.8 cm from the centre of the circle. P O 7.8 cm Solution: O is the centre of the circle. P is a point outside the circle such that OP = 7.8 cm. PA is a tangent, OA is radius = 3 cm. OA ⊥ PA (radius is perpendicular to the tangent at the point of contact). ∴ In a right angled triangle OAP. Fig.6.55 PA2 = OP2 – 32 = (7.8)2 – 32 = (7.8+3) (7.8–3) = (10.8) (4.8) = 51.84 or PA = 51.84 = 7.2 cm Example 11: In the Fig.6.56 AB is the diameter of the circle, D ∠BAC = 42o. Find ∠ACD o Solution: In the given figure m∠ACB = 90 (Angle in the semicircle) C m∠ACB+ m∠BAC +m∠ABC = 180o (Sum of three angles of a triangle) A 42o o m∠ABC = 180 – (m∠ACB + m∠BAC) O = 180o – (90o+42o) = 48o But m∠ACD = m∠ABC (Angle between the chord and tangent = Angle B in the alternate segment).
Hence m∠ACD = 48o Fig.6.56 Example 12: If all the sides of a quadrilateral touch a circle prove P that the sum of a pair of opposite sides is equal to the sum of the other D pair of sides. S Solution: PQRS is a quadrilateral touching the circle with centre O at A A,B,C and D. O C To prove: PQ+ RS = PS +QR (Fig.6.59) Proof: Length of tangents from an external point are equal Q B ⇒. PA = PD; QA = QB; RC = RB ; SC = SD R Adding we get PA+ QA+ RC +SC = PD +SD +QB+RB ⇒ PQ + RS = PS + QR Fig.6.57 Example 13 : Three circles with centres at A, B and C touch each other externally. C AB = 4 cm BC = 6 cm CA = 8 cm. Find their radii. r Solution: There circles with centres A,B,C touch one another. Let r1, r2, r r r3 be the radii of the circles. A r Given: AB = r1+ r2 = 4; BC = r2 + r3 = 6; CA = r3 + r1 = 8 (Fig.6.58). r r Adding we get 2r1+ 2r2 + 2r3 = 18; 2 ( r1 + r2 + r3) = 18; r1 + r2 + r3 = 9 B r1 = (r1+r2+r3)– (r2+r3) = 9–6 = 3 r2 = (r1+r2+r3)– (r1+r3) = 9–8 = 1; Fig.6.58 r3 = (r1+r2+r3)– (r1+r2) = 9–4 = 5 ⇒ radii are 3 cm, 1 cm and 5 cm respectively. 3
3
1
1
2
2
121
Example 14: In the Fig.6.59 PA, PB, CD are tangents to a circle. Prove PC + PD + CD = PA + PB. A C Solution: In the given figure let Q be the point of contact of the tangent CD to the circle. Q CA = CQ (Tangent from an external P point C are equal) (1) DB = DQ (Tangent from an external point D D B are equal) (2) Fig.6.59 PA+PB = PC+CA+ PD+DB = PC+CQ+ PD+DQ (from (1) and (2)) = PC +PD+CQ +DQ
= PC+ PD+CD, ( CQ+DQ=CD) Example 15: AB is a tangent at P to the circumcircle of ΔPQR. If AB⎥⎥ QR, prove that P ΔPQR is isosceles. B A Solution: Given: AB is a tangent at P to the circumcircle of ΔPQR and AB⎥⎥ QR Qa R To prove: Δ PQR is isosceles (Fig.6.60) Proof : AB ⎥⎥ QR (Given). PQ transversal (1) Fig.6.60 m∠APQ = ∠PQR (alternative angle) AB is a tangent and PQ is a chord. ∠APQ = ∠PRQ (Angle between a tangent and a chord = Angle in the alternate segment) (2) From (1) and (2) we get ∠PQR = ∠PQR. Base angles of triangle PQR are equal. ∴PQ = PR ∴ Triangle PQR is isosceles. Example 16: If all the sides of a parallelogram touch a circle show that the parallelogram is a rhombus. D Solution: R Given: All the sides of a parallelogram ABCD touch a circle with S C centre O. To prove ABCD is a rhombus O a Q Proof: Since the lengths of the tangents from an external point to a A given circle are equal (Fig.6.61) P AP = AS; BP = BQ; CR = CQ; DR = DS B Adding (AP+BP)+ (CR+DR) = (AS+DS) + (BQ+CQ) ⇒ AB + CD = AD + BC Fig.6.61 ⇒ AB + AB = AD+AD ( ABCD is a parallelogram ⇒ CD= AB, BC=AD) ⇒ 2AB = 2AD ⇒ AB= AD
But AB= CD and AD = BC ( opposite sides of a parallelogram are equal) ∴ AB = BC = CD = AD. Hence ABCD is a rhombus.
122
Exercise 6.4 1. 2.
3. 4. 5. 6. 7. 8.
9.
A
R
Determine the length of the tangent to a circle of radius 6 cm from a point at a distance of 10cm from the centre of the circle. O In the Fig.6.62 APB is a tangent O is the centre P aQ (a) If m∠POQ = 70o find m∠QPB, m∠QPA b) If m∠QPB = 80o find m∠POQ. In the Fig.6.63 TPT′ is a tangent m∠PBA = 38o find m∠APT, Fig.6.62 m∠ACP B Prove that the tangent at the ends of a diameter are parallel. In triangle ABC, D,E,F are points of contact of in-circle. If BD =4 cm, CE = 7 cm AF = 2cm, find the perimeter of the triangle. Three circles with centres at A,B,C, touch one another. AB = 8 cm, BC= 5cm, and AC= 7 cm. Find the radii of the circles. Two circles touch each other internally at A. Two lines ABC and ADE meet the circle in B,D and C,E respectively. Prove that CE || BD B A ABCD is a cyclic quadrilateral, PQ is a tangent at D. If AC is a diameter m∠CAD= 27o and m∠ACB = 63o find m∠BAC, m∠PDA. Three circles with equal radii touch one another externally. C Prove that the triangle formed by joining their centres is an P T T’ equilateral Triangle. Fig.6.63
6.5 SIMILAR TRIANGLE In standard IX we have stated that the geometric figures are congruent if they have the same shape and the same size. In this chapter we shall learn some properties of geometric figures that are of the same shape but not necessarily of the same size. Such figures are said to be similar. It is obvious that the congruent figures are similar but the converse is not necessarily true. Let us consider the following figures
(i)
Fig.6.64. Equilateral triangles
(ii)
Fig.6.65. Circles
123
(iii)
Fig.6.66 Squares
(iv) Fig.6.67 Segments
We note that any two equilateral triangles are similar, any two circles are similar, any two squares are similar, and any two segments are similar. Photographs of different sizes printed from the same negative , different print of movies for projection on different screens etc are some examples for similar figures that we come across in our day-to-day life. We define the similarity of two triangles as follows. Two triangles are similar if the angles of one triangle are equal to the corresponding angles of the other and their corresponding sides are in the same ratio (sides are proportional). Thus if in two triangles ABC and DEF (See Figs. 6.68 and 6.69)
D
A
B
C
E
F
Fig.6.68
Fig.6.69
AB BC CA , then the triangles = = DE EF FD ABC and DEF are similar. We write ΔABC⎥⎥⎥ DEF. (The students are advised to observe carefully which pair of angles are equal and which pairs of sides are in the same ratio in the similarity of triangles ABC and DEF. It is for this reason that in the above case we would not write Δ ABC⎟⎟⎟ ΔEDF or ΔABC⎟⎟⎟ ΔFDE). D E F
m∠A = m∠D, m ∠B = m ∠E, m ∠C = m∠F and
Some basic results on proportionately
Let us draw a line AP. From the point A let us step off five equal segments on the line AP as shown in the Fig.6.70. Let us denote 2nd and 5th points by B and C respectively. Through A let us draw a line AD. Through B and C let as draw lines BE, CF both parallel to the line AD.
Q H
G
A
124
1
2 B
3
Fig.6.70
4
5 C
P
Let us draw a transversal AQ to intersect the lines BE and CF at the points G and H respectively. AB 2 AG 2 AB AG (By Construction) and (By equal intercepts) ⇒ . We = = = BC 3 GH 3 BC GH AB AG observe in the figure that ACH is a triangle BG is parallel to the line CH and = BC GH we state that in the following theorem. Then
Theorem 16 (Basic proportionality theorem): If a line is drawn parallel to one side of a triangle the other two sides are divided in the same ratio. We also note that (see the Fig. 6.70) AB AG In ΔACH BG⎥⎥ CH . Then = . Adding 1 on both sides BC GH AB AB + BC AG + GH AG AC AH or +1 = + 1 or = = BC GH BC GH BC GH
Similarly we can get the result
AB AG = AC AH
Theorem 16A : (Converse of Theorem 16)
If a line divides any two sides of a triangle in the same ratio the line is parallel to the third side. Given: ABC is a triangle DE is a line which meets AB at D and AC at E such that AD AE A = DB EC To prove: DE⎥⎥ BC E Construction: If DE is not parallel to BC, D F draw DF⎥⎥ BC which meets AC at F. AD AE = (Given) (1) B Proof: C DB EC DF⎥⎥ BC (By construction) Fig.6.71 AD AF (By basic proportionality theorem) (2) ∴ = DB FC AE AF = From (1) and (2) we get EC FC
This means both the points E and F divide AC in the same ratio. There can be only one point dividing a segment in a given ratio. ∴ E and F will coincide. But DF ⎢⎢ BC That is DE ⎢⎢BC Hence the theorem is proved
125
A-A-A-Similarity
A
D P
B
Q
Fig.6.72
C
E
F
Fig.6.73
Consider two triangles ABC and DEF such that m∠A = m∠D, m∠B =m∠E, m∠C = m∠F. Take points P and Q respectively on the sides AB and AC such that AP = DE, AQ = DF. Join P and Q (see the Fig.6.72). ΔAPQ≡ Δ DEF (SAS) m∠APQ= m∠DEF (corresponding angles)
(1)
But
(2)
m∠DEF = m∠ABC.
By (1) and (2) we get m∠APQ = m ∠ABC Then PQ ⎪⎪BC (corresponding angles are equal) ∴
PB QC = (By basic proportionality Theorem) AP AQ
Adding 1 on both sides
PB QC +1 = +1 AP AQ
PB + AP QC + AQ = AP AQ AB AC = AP AQ AB AC (AP = DE, AQ= DF) = DE DF AB BC = Similarly we can show that DE EF AB BC AC = = From (3) and (4) If follows that DE EF DF
(3) (4)
We conclude this by the following theorem. Theorem 17: If in two triangles the corresponding angles are equal then their corresponding sides are proportional.
126
S-S-S Similarity
A
D P
Q
C
B
E
F
Fig.6.74
Fig.6.75 AB BC AC . Take points P and Q Consider two triangles ABC and DEF such that = = DE EF DF respectively on the sides AB and AC such that AP = DE and AQ = DF Join P and Q (see Fig.6.74, Fig.6.75).
AB AC AB AC = (given) ; = (AP = DE, AQ=DF by construction) DE DF AP AQ AB − AP AC − AQ PB QC = Subtract 1 from both sides and obtain = ; AP AQ AP AQ Thus PQ ⎟⎟ BC (By Basic proportionality theorem) ∠APQ =∠ABC (corresponding angles) ∠AQP=∠ACB (same)
(1)
From (1) we note that all the angles of Δ APQ are equal to the corresponding angles of ΔABC. AB BC = (2) ∴ AP PQ AB BC (given) = DE EF AB BC = (3) or AP EF From (2) and (3) we get
BC BC = . PQ EF
That is PQ = EF ΔAPQ ≡ ΔDEF (SSS) so that ∠PAQ = ∠D, ∠APQ = ∠E, ∠AQF = ∠F. Hence ∠A = ∠D, ∠B = ∠E, ∠C = ∠F (from (1)) We conclude this in the following theorem. Theorem 17A: (Converse of theorem 17). If the sides of two triangles are proportional, the triangles are equiangular.
127
S-A-S Similarity
A
D P
B
Q
C
E
Fig.6.76
F
Fig.6.77
AB AC = and ∠A = ∠D. Let us take points P and Q on DE DF sides AB and AC respectively such that AP = DE and AQ = DF. Join P and Q (see the Fig.6.76). Consider triangles in which
AP = DE and AQ = DF (by construction). AB AC = (by basic proportionality). AP AQ Consequently PQ || BC. Thus ∠ABC = ∠APQ ∠ACB = ∠AQP (corresponding angles are equal). ΔABC ||| ΔAPQ (AAA). Also ΔAPQ ≡ ΔDEF (SAS). Hence ΔABC ||| ΔDEF. From this we get the following theorem. Theorem 18: If an angle of one triangle is equal to one angle of the other and the sides including the angles are proportional then the triangles are similar. A
D
B
C E
M
Fig.6.78
N
F
Fig.6.79
Let ABC, DEF be two similar triangles. From the points A and D let us draw AM and DN perpendicular to the sides BC and EF respectively (Fig.6.78 and Fig6.79). ΔABC ||| ΔDEF (given)
128
∴
AB BC AC = = DE EF DF
(1)
In triangles ABM and DEN m∠ABM = m∠DEN (Q ABC ||| ΔDEF) By construction m∠AMB = m∠DNE = 90o Remaining m∠BAM = m∠EDN. ∴ ΔABM ||| ΔDEN. Thus
AB AM AM BC = = . Thus (from (1)) DE DN DN EF
(2)
1 × BC × AM Area of Δ ABC 2 BC BC BC2 Now (from (2)) = = × = 1 Area of Δ DEF EF EF EF2 × EF × DN 2 Area of ΔABC AC2 AB2 = = Similarly we can show that Area of ΔDEF DF2 DE 2 From thus we get the following theorem. Theorem 19: The ratio of the areas of similar triangles is equal to the ratio of the squares of corresponding sides.
E
Theorem 20 (Angle bisector theorem): The bisector of any angle of a triangle divides the opposite side in the ratio of the corresponding adjacent sides. Given: In Triangle ABC, AD bisects ∠BAC BD AB = To Prove: DC AC Construction: Through C, draw CE || DA, meeting BA produced at E. Proof: DA || CE BE transversal m∠BAD = m∠AEC (corresponding angles) DA || CE AC transverse m∠CAD = m∠ACE (alternate angles) But m∠BAD = m∠CAD (given) m∠AEC = m∠ACE Base angles of ΔACE are equal. ∴ AC = AE In ΔBCE, DA || CE
129
A
B
C
D
Fig.6.80
BD AB (by basic proportionality). But AE = AC (just proved). = DC AE BD AB ∴ = DC AC Hence the theorem is proved. ∴
A
Theorem 21: If a perpendicular is drawn from the vertex of a right angle to a hypotenuse, the triangle on each side of the perpendicular are similar to the whole triangle and to each other. Given: A triangle ABC right angled at A and AD is perpendicular to hypotenuse BC. To prove: (i) ΔDBA ||| Δ ABC B (ii) ΔDAC ||| Δ ABC (iii) ΔDBA ||| Δ DAC Proof: In triangles DBA and ABC m∠ADB = m∠BAC = 90o m∠ABD = m∠ABC (common angle). Remaining m∠BAD = m∠ACB. Thus triangles DBA and ABC are equiangular. Hence ΔDBA ||| ΔABC (1) In triangles DAC and ABC m∠ADC = m∠BAC = 90 m∠ACD = m∠ACB (common angle) Remaining m∠DAC = m∠ABC. Thus triangles DAC and ABC are equiangular. Hence ΔDAC ||| ΔABC (2) From the results (1) and (2) we obtain ΔDBA ||| ΔDAC Hence the theorem is proved. Theorem 22 (Pythagoras theorem): In a right angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides. Given: A triangle ABC right angled at A. To Prove: BC2 = AB2 + AC2 Construction: Draw AD perpendicular to BC B Proof: In triangles ABC and DBA o m∠BAC = m∠ADB = 90 m∠ABC = m∠ABD (common angle) Remaining m∠ACB = m∠BAD ΔABC ||| ΔDBA (by AAA similarity). BC AB ∴ Corresponding sides are proportional = AB BD AB2 = BC. BD In triangles ABC and DCA m∠BAC = m∠ADC = 90o
130
D
C
Fig.6.81
A
D
Fig.6.82
(1)
C
m∠ACB = m∠ACD (common angle) Remaining m∠ABC = m∠DAC ΔABC ||| ΔDCA (by AAA similarity). Corresponding sides are proportional
BC AC = AC CD
(2) AC2 = BC. CD 2 2 Adding (1) and (2), we get AB + AC = BC.BD + BC. CD = BC (BD + CD) = BC.BC = BC2 Hence the theorem is proved. R
C b
A
b
a
c
a
Q
B P
Fig.6.83
Fig.6.84
Consider a triangle ABC in which c2 = a2 + b2 where BC = a, AC = b and AB = c (Figs.6.83). Let us construct a right triangle PQR such that QR = a, PR = b and ∠PRQ = 90o Let PQ = r In triangle PQR (Fig.6.84) m∠PRQ = 90o (by construction). Thus r2 = a2 + b2 (by Pythagoras theorem) (1) But c2 = a2 + b2 (given) (2) From (1) and (2) we get r2 = c2 That is r = c. (3) In triangles ABC and PQR BC = a = QR (by construction). AC = b = PR (by construction). AB = c = PQ (by 3). ∴ ΔABC ||| ΔPQR (SSS similarity). So that m∠ACB = m∠PRQ = 90o. From this we get the theorem. Theorem 23: In a triangle the square on one side is equal to the sum of the squares on the remaining two, the angle opposite to the first side is the right angle (converse of Pythagoras theorem). Intersection of two chords of a circle Consider a line segment AB and a point P on it. Then the product of PA. PB represents the area of the rectangle whose sides are PA and PB. This product is referred to as the area A of the rectangle contained by two parts PA and PB of the line segment AB.
131
P Fig.6.85
B
Consider two chords AB and CD of a circle with centre O. Let these chords intersect at a point P inside or outside the circle. D B O In triangles DAP, BCP (see the Fig.6.86). P m∠PAD = m∠PCB (angles in the same segment) A C m∠ADP = m∠CBP (same). Fig.6.86 m∠APD = m∠CPB (vertically opposite angles). ΔDAP ||| Δ BCP. PA PD = ; That is PA. PB = PC. PD. PC PB A B In triangles ACP and BDP (see the Fig.6.87). P m∠P = m∠P (common); m∠PAC = m∠PDB (in a cyclic D C quadrilateral exterior angle = the interior opposite angle). m∠PCA = m∠PBD (same reason) ⇒ ΔACP ||| ΔBDP Fig.6.87 PA PC ⇒ = PD PB That is PA. PB = PC.PD. From this we get the following theorem. Theorem 23: If two chords of a circle intersect either inside or outside the circle the area of the rectangle obtained by the parts of a chord is equal in area to the rectangle by the parts of the other. Example 17: In triangle ABC, DE || BC A AD = 6, DB = 10, AE = 3. Find AC. 6 3 AD AE = (DE || BC) Solution: D DB EC E 6 3 10 x = ; 6x = 30 10 x C B 30 Fig.6.88 x= = 5 ; AC = x + 3 = 5 + 3 = 8 6 P Example 18: In triangle PQR; PA = 5, AQ = 10, PB = 4, BR = 6, verify AB || QR. PA 5 1 = = Solution: A AQ 10 2 B PB 4 2 1 2 PA PB = = . Since ≠ , ∴ ≠ BR 6 3 2 3 AQ BR Q R AB is not parallel to QR ( by converse of BPT) Fig.6.89 Example 19: Chord AB and CD cut at P inside the circle. AB = 11, AP = 3 B C CP = 6. Find CD. Solution: Given AB = 11, AP = 3, CP = 6, Then PB = 11 – 3 = 8. We know that AP × PB = CP × PD, 3 × 8 = 6 × PD P A D 3×8 PD = = 4 ; CD = PC + PD = 6 + 4 = 10 Fig.6.90 6
132
Example 20: Chords AB and CD cut at P outside the circle such that PC = 15 CD = 7 PA = 12 find AB. A Solution: Given PC = 15, CD = 7, PA = 12, PD = PC - CD B = 15 – 7 = 8. We know that PA. PB = PC. PD or 12 × PB = 15 × 8 D 15 × 8 PB = = 10 ; AB = PA – PB = 12 – 10 = 2. C 12 Fig.6.91 A Example 21: A ladder 17 m long touches a window of a house 15 m above the ground. Determine the distance of the foot of the 15m ladder from the house. Solution: AC = ladder = 17 m. AB = height the ladder touches = 15 m B In a right angled triangle ABC AC2 = AB2 + BC2; BC2 = AC2 – AB2 = 172 - 152 = (17 + 15) (17 – 15)
P
17m
C
Fig.6.92
= 32 × 2 = 64 ; BC = 64 = 8 m
Distance of the foot of the ladder from the house = 8 m. Example 22: The length of the diagonal of a square is 4 2 m. Find the length of the side.
Let a be the side of the square given. AC = 4 2 . In a right angled triangle ABC
A
D 4
a
B
2
a
C
Fig.6.93
AB + BC = AC ; a + a = ( 4 2 ) ; 2a = 16 × 2 2
a2 =
2
2
2
16 × 2 = 16; a = 2
2
2
2
16 = 4. Side of square is 4 m.
Example 23: ABC is a triangle in which AB = AC and D is a point on the side AC such that BC2 = AC. CD. Prove that BD = BC. Given: ABC is a triangle in which AB = AC and D is a point on the side AC such that BC2 = AC . CD. To prove: BD = BC. A Proof: BC2 = AC. CD (given) AC BC = (1) BC CD D In triangles ABC and DBC ∠C is common and the sides containing ∠C are proportional B C from (1) ∴ ΔABC ||| ΔDBC (by S-A-S- similarity). Fig.6.94 AC AB AB AB Thus = ; = (since AB = AC given). BC BD BC BD 1 1 That is = ; That is BC = BD. BC BD
133
A
Example 24: Prove that the medians of two D similar triangles are proportional to the corresponding sides. Solution: ΔABC ||| Δ DEF In ΔABC, AP is the median through A. Q C E F B P In ΔDEF, DQ is the median through D. AP AB To prove = Fig.6.95 Fig.6.96 DQ DE AB BC AC Proof: ΔABC ||| ΔDEF; = = (corresponding sides are proportional). DE EF DF AB 2BP ⎡ P is the midpo int of BC ⎤ AB BP = ; = DE 2EQ ⎢⎣ Q is the mid po int of EF⎥⎦ DE EQ In triangles ABP and DEQ, ∠B = ∠E (corresponding angles of similar triangles ABC and DEF). AB BP = (Proved); ∴ ΔABP ||| ΔDEQ (SAS similarity). DE EQ AB BP AP AB AP = = . Therefore = . P DE EQ DQ DE DQ Example 25: PQR is an equilateral triangle and PA is perpendicular from P to QR. Prove that PA2 = 3QA2 Solution: Given PQR is an equilateral triangle in which PA ⊥ QR. R Q A To Prove: PA2 = 3QA2 Proof: Since PA ⊥ QR in an equilateral triangle PQR Fig.6.97 PA is the median. ∴A is the mid point of QR. That is QA = RA = QR/2 that is QR = 2QA (1) In a right angled triangle PQA PQ2 = PA2 + QA2; PA2 = PQ2 – QA2 = QR2 – QA2 (since PQR is equilateral PQ = QR) = (2QA)2 – QA2 (From (1)) = 4QA2 – QA2 = 3QA2 Example 26: ABCD is a quadrilateral with ∠B = 90o. If AD2 = AB2 + BC2 + CD2, prove that ∠ACD = 90o. Solution: D Given: ABCD is a quadrilateral in which m∠B = 90o. 2 2 2 2 AD = AB + BC + CD . A To prove: m∠ACD = 90o Construction: Join AC Proof: In triangle ABC (by construction). (1) AC2 = AB2 + BC2 (By Pythagoras theorem) 2 2 2 2 But AD = AB + BC + CD (given) C B AD2 = AC2 + CD2 (by (1)) Fig.6.98 By the converse of Pythagoras theorem. angle opposite to AD in ΔADC is 90o That is m∠ACD = 90o
134
Example 27: P and Q are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AQ2 + BP2 = AB2 + PQ2 A Solution: Given: ABC is a triangle right angled at C. P and Q are any two points on the sides AC and BC respectively. P To prove: AQ2 + BP2 = AB2 + PQ2 Proof: In right angled ΔAQC B C Q (1) AQ2 = AC2 + QC2 (by Pythagoras theorem) In right angled ΔBPC Fig.6.99 2 2 2 BP = BC + PC (by Pythagoras theorem) (2) Adding (1) + (2) we get (3) AQ2 + BP2 = AC2 + QC2 + BC2 + PC2 But in right angled ΔPQC PQ2 = PC2 + QC2 (By Pythagoras theorem) (4) 2 2 2 In right angled ΔABC; AB = AC + BC (By Pythagoras theorem) (5) Substituting (4) and (5) in (3) we get AQ2 + BP2 = AC2 + BC2 + QC2 + PC2 = AB2 + PQ2 Example 28: In the figure, DEFG is a square and m∠C = 90o, AD FE Prove that (i) ΔADG ||| ΔGCF (ii) ΔADG ||| ΔFEB (iii) = (iv) DE2 = AD × EB. DG EB Solution: Given: DEFG is a square inscribed in ΔACB which is right angled at C. C Proof: (i) DE || GF (Q opposite side of square). ⇒ AB || GF and AC is a transversal. G F ∴ m∠DAG = m∠FGC (Q corresponding angles are equal).
Also m∠GDE + m∠GDA = 180o (straight angle). But m∠GDE = 90o (Q angle of a square). ∴ 90o + m∠GDA = 180o; m∠GDA = 90o Now in triangles ADG and GCF
A
D
Fig.6.100 o
m∠DAG = m∠FGC (Proved); m∠GDA = m∠GCB (each is 90 ). ∴ By AAA criterion, we have ΔADG ||| GCF. (ii) Similarly ΔFEB ||| ΔGCF. Since triangles ADG and FEB are both similar to ΔGCF. ∴ ΔADG ||| ΔFEB. (iii) Since ΔADG ||| ΔFEB ⇒
AD DG AD FE = ⇒ = FE EB DG EB
(iv) Since ΔADG ||| ΔFEB ⇒
AD DG AD DE = ⇒ = (Q FE = DG = DE sides of a square) FE EB DE EB
∴ DE2 = AD × EB 135
E
B
Example 29: ABC is a right angled triangle right angled at C. If p is the length of the perpendicular from C to AB and AB = c, BC = a, CA = b, then prove that (i) pc = ab 1 1 1 (ii) 2 = 2 + 2 p a b Solution: (i) In right angled ΔABC, AB2 = AC2 + CB2 c2 = b2 + a2 (1) In right angled triangles ADC and ACB m∠A = m∠A m∠ADC = m∠ACB = 90o ∴ By AAA criterion DC AC p b ΔADC ||| ΔACB ⇒ = ⇒ = ⇒ pc = ab CB AB a c ab 1 c 1 c2 (ii) Now from (i) we have p = ⇒ = ⇒ 2 = 2 2 c p ab p a b
=
A
c b p
C
D
a
B
b2 + a 2 (Q By (1)) a 2 b2
b2 a2 + a 2 b2 a 2 b2 1 1 1 = 2 + 2 2 p a b
=
Example 30: Prove that four times the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Solution: Given: ABCD is a rhombus with diagonals AC and BD. D To Prove: 4AB2 = AC2 + BD2 C Proof: Since the diagonals of a rhombus bisect each other at right angles. 0 ∴ OA = OC = ½ AC; OB = OD = ½ BD o m∠AOB = m∠BOC = m∠COD = m∠DOA = 90 (1) In right angled ΔAOB AB2 = OA2 + OB2 (2) In right angled ΔBOC BC2 = OB2 + OC2 2 2 2 B A In right angled Δ COD CD = OC + OD (3) 2 2 2 In right angled Δ DOA DA = OD + OA (4) Fig.6.102 Adding (1), (2), (3) and (4) we have AB2 + BC2 + CD2 + DA2 = 2OA2 + 2OB2 + 2OC2 + 2OD2 = 2OA2 + 2OB2 + 2OA2 + 2OB2 (Q OC = OA; OD = OB) = 4OA2 + 4OB2 = (2OA)2 + (2OB)2 AB2 + BC2 + CD2 + DA2 = AC2 + BD2 (Q OA = ½ AC ; OB = ½ BD) Since in a rhombus ABCD, all the sides are equal, that is AB = BC = CD = DA. 4AB2 = AC2 + BD2.
136
R
Example 31: AB is a line segment and M is r its mid point. Semicircles are drawn with AM, BM and AB as diameters on the same side of O the line AB. A circle with centre O and r r radius r is drawn so that it touches all the 1 Q P semicircles. Prove that r = AB. 6 Solution: Given: AM, MB and AB are diameters of the L M N B semicircles and r is the radius of the circle as A Fig.6.103 shown in the figure. To prove: r = AB/6. Proof: Let L, N be the mid points of AM, MB respectively. Let the circle (O, r) touch the semicircle with centres L,M,N at P,R,Q respectively. Then the points 0, P, L are collinear, the points O, Q, N are collinear, the points R, O, M are collinear. Let AB = x, then OL = OP + PL = r + LM = r + x/4 and ON = r + x/4. ∴ Triangle OLN is isosceles; M is the midpoint of base LN. ∴ OM ⊥ LN. Now in right angled triangle OML 2 2 ⎛ x⎞ ⎛x⎞ OL2 = OM2 + LM2 ⇒ ⎜ r + ⎟ = (RM − OR) 2 + ⎜ ⎟ ⎝ 4⎠ ⎝4⎠ 2
rx x 2 ⎛ x x2 ⎞ ⇒r + + = ⎜ − r⎟ + 2 16 ⎝ 2 16 ⎠ 2 2 rx x x x2 ⇒ r2 + + = − rx + r 2 + 2 16 4 16 rx x 2 3 x2 ⇒ = − rx ⇒ rx = 2 4 2 4 3 x ⇒ 2 r = 4 ( x ≠ 0) x 2 1 1 ∴ r = × = x ∴ r = AB 4 3 6 6 Example 32: From a point O in the interior of a ΔABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that (i) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2 (ii) AF2 + BD2 + CE2 = AE2 + BF2 + CD2 A Solution: E Given: ABC is a triangle in which OD, OE and F OF are drawn perpendicular to the sides BC, CA and AB respectively. O To prove: (i) AF2 + BD2 + CE2 = OA2 + OB2 2 2 2 + OC – OD – OE – OF 2 2 2 C D B (ii) AF + BD + CE = AE2 + BF2 + CD2 Construction: Join OA, OB and OC. Fig.6.104 Proof: i) In right angled ΔOAF, AF2 + OF2 = OA2 ; In right angled ΔOBD, BD2 + OD2 = OB2 In right angled ΔOCE, CE2 + OE2 = OC2 Adding we get AF2 + BD2 + CE2 + OF2 + OD2 + OE2 = OA2 + OB2 + OC2 ∴ AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OF2 – OD2 – OE2 2
137
ΔOCD, OD2 + CD2 = OC2
ii) In right angled ΔOBD, OD2 + BD2 = OB2, in right angled Subtracting we get BD2 – CD2 = OB2 – OC2
(1)
Similarly by considering right angled triangles OCE and OAE CE2 – AE2 = OC2 – OA2
(2)
By considering right triangles OAF and OBF AF2 – BF2 = OA2 – OB2
(3)
Adding (1), (2) and (3) we get BD2 + CE2 + AF2 – CD2 – AE2 – BF2 = 0 ∴ AF2 + BD2 + CE2 = AE2 + BF2 + CD2. Example 33: A vertical stick 15 cm long casts its shadow 10 cm long on the ground. A flag pole casts a shadow 60 cm long at the same place. What is the height of the flag pole. Solution: Let AB = 15 cm be the stick and BC = 10 cm be its shadow. Let PQ be the flag pole and RQ = 60 cm be its shadow. In ΔABC and ΔPQR m∠ABC = m∠PQR = 90o m∠BCA = m∠QRP (angle of elevation same) ∴ ΔABC ||| ΔPQR (by AAA criterion). AB PQ 15 PQ ⇒ = ⇒ = BC QR 10 60
SUN
P
A 15 cm C 10 cm
R
B
60 cm
Q
Fig.6.105
∴ PQ =
15 × 60 = 90 cm 10
∴ The height of the flag pole = 90 cm. Example 34: Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Solution: Let AB = 11m and CD = 6 m be the two poles such that BD = the distance between their feet = 12 m. We have to find AC. Draw CE ⊥ AB. ∴CE = DB = 12 m AE = AB – BE = AB – CD = 11 – 6 = 5 m In right angled triangle AEC, by Pythagoras theorem, AC2 = CE2 + AE2 = 122 + 52 = 144 + 25 = 169 ∴ AC = 169 = 13 cm
138
A 5m C
E 12m
6m
6m B
D 12m
Fig.6.106
11m
C
D
Example 35: A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that E OB2 + OD2 = OC2 + OA2. Solution: A point O is an interior point of a rectangle ABCD. Through O draw a line parallel to AB meeting AD and BC in A
O
F B
Fig.6.107
E and F respectively. Then in rectangle ABFE opposite sides are equal. ∴ AE = BF. Similarly in rectangle EFCD ED = FC. In right angled triangle OFB OB2 = OF2 + BF2 In right angled triangle OED OD2 = OE2 + ED2 Adding (1) and (2) we get OB2 + OD2 = OF2 + BF2 + OE2 + ED2 In right angled triangle OFC OC2 = OF2 + FC2 or OC2 = OF2 + ED2 (Q FC = ED) In right angled triangle OEA OA2 = OE2 + AE2 or OA2 = OE2 + BF2 (Q AE = BF) Adding (4) and (5) we have OC2 + OA2 = OF2 + ED2 + OE2 + BF2 From (3) and (6) we get OB2 + OD2 = OC2 + OA2.
(1) (2) (3) (4) (5) (6) A
Example 36: D, E, F are the mid points of the sides BC, CA and AB respectively of a triangle ABC. Prove that ΔABC = 4ΔDEF (areas). F E Solution: Given: ABC is a triangle in which D, E, F are the mid points of the sides BC, CA and AB respectively. DE, EF, FD are joined forming ΔDEF. B C D To prove: ΔABC = 4ΔDEF (areas). Proof: D and E are mid points of BC and CA respectively. Fig.6.108 1 ⇒ DE = AB (since the line joining the mid points of two sides of a triangle is parallel to 2 the third side and half its length). DE EF FD 1 Similarly EF = ½ BC, and FD = ½ CA. ∴ = = = ⇒ Δ DEF ||| ΔABC AB BC CA 2 Area of ΔABC AB2 Area of ΔABC ⎛ AB ⎞ ⎛2⎞ = =⎜ = 4. ⎟ =⎜ ⎟ ; ∴ 2 Area of ΔDEF Area of ΔDEF DE ⎝ DE ⎠ ⎝1⎠ Area of ΔABC = 4 Area of ΔDEF. Example 37: Prove that the diagonals of a trapezium divide each D other proportionally. Solution: Given: ABCD is a trapezium in which AB || DC and whose diagonals AC and BD intersect at 0. AO BO To Prove = A OC OD 2
2
∴
C
O
B
Fig.6.109
139
E
Construction: Through 0, draw a line parallel to AB and CD meeting BC at E. Proof: In triangle CAB, OE || AB AO BE ∴ By basic proportionality theorem, we have (1) = OC EC In triangle BCD; OE || DC BO BE ∴ By basic proportionality theorem, we have (2) = OD EC AO BO From (1) and (2), we get = OC OD Example 38: AD is a median of a ΔABC. The bisectors of m∠ADB and m∠ADC meet AB and AC in E and F respectively. Prove that EF || BC. Solution: Given: AD is a median of ΔABC. The bisectors of angles of m∠ADB and m∠ADC meet AB and AC in E and F respectively. A To prove: EF || BC Proof: In ΔADB, DE is the bisector of ∠ADB. AE AD ∴ By angle bisector theorem, we have (1) = EB DB E F In ΔADC, DF is the bisector of m∠ADC AF AD ∴ By angle bisector theorem = FC DC B D C But DC = BD since AD is a median. AF AD Fig.6.110 ∴ = (2) FC DB AE AF From (1) and (2) we get = EB FC ∴By converse of basic proportionality theorem. EF || BC. Exercise 6.5 1.
In ΔABC; DE || BC. Find what is required (a) AD = 3 DB = 5 AE = 6 EC = ? (b) DB = 6 EC = 8 AE = 5 AD = ? (c) AE = 3 EC = 7 AD = 6 AB = ? (d) AB = 12 AD = 5 AE = 6 AC = ? (e) AC = 15 AE = 3 DB = 9 AB = ?
2.
In ΔABC; PQ || BC; find for what sets of measurements will PQ parallel to BC (a) AB = 18, AP = 8, AQ = 12, QC = 15 (b) AP = 5, BP = 6, AQ = 6, CQ = 5 (c) AP = 4, BP = 4.5, AQ = 4, QC = 4.5 (d) AB = 1.28, AC = 2.56, AP = 0.16 AQ = 0.32 Chords AB and CD cut at P inside the circle. Given the following measurements find the length of the segment asked. (a) AP = 8, AB = 17, CP = 12 CD = ? (b) AP = 2x, PB = x, CP = 10 PD = 5, AP = ?
3.
140
4.
Chords AB and CD cut at P outside the circle. Given the following measurements, find what is asked. (a) AB = 8, BP = 4 CD = 8 Find DP. (b) AP = 9x, BP = 4x, CP = 16, DP = 9 find BP.
5.
Which are the sides of a right triangle (a) 7,24, 25
(b) 6, 9, 12
(c) 50, 80, 100
(d) 5, 5, 5 2
6.
A ladder reaches a window which is 12 metres above the ground on one side of the street. Keeping its foot on the same point the ladder is turned to the other side of the street to reach a window 9 metres high. Find the width of the street if the length of the ladder is 15 metres.
7.
P and Q are the mid points of the sides CA and CB respectively of ΔABC right angled at C. Prove that (i) 4AQ2 = 4AC2 + BC2 (ii) 4BP2 = 4BC2 + AC2
ANSWERS Exercise 6.2
(1) 8 cm (2) 16 cm Exercise 6.3
(1) (a) 65o (b) 220o (c) 26o (d) 60o (e) 57o (2) 110o Exercise 6.4
(1) 8 cm (2) (a) 35o, 145o (8) 27o, 63o
(b) 160o (3) 38o, 142o (5) 26 cm (6) 2 cm, 3 cm, 5 cm
Exercise 6.5
(1) (a) 10 (2) (a) PQ || BC (3) (a) 18 cm (5) (a) right triangle (6) 21 metres
(b) 6.25 (c) 20 (d) 14.4 (e) 11.25 (c) PQ || BC (d) PQ || BC (b) PQ || BC (b) 10 cm (4) (a) 4 cm (b) 8 cm (b) not a right triangle (c) not a right triangle (d) right triangle
141
7. ANALYTICAL GEOMETRY 7.0 INTRODUCTION Rene Descartes (1596-1650) whose “La Geometric” was published in 1637 as an appendix to his “Discours de la methods” is regarded as the inventor of Analytical Geometry and Modern Analytical Geometry is called “Cartesian Geometry” after him. However the fundamental principles and methods of analytical geometry were already discovered by Pierre de Fermat (1601-1665) earlier. The work of Fermat was mentioned by him in correspondence but his treatese on the subject Ad locus Planos et Solidos Isagoge was published in 1679 only after his demise. So Descartes came to be regarded as the inventor of the subject. Descartes′ procedure in geometry was to begin with a geometrical problem to convert it into an algebraic equation simplify it and then solve the equation geometrically. However, even the usual formulae for distance between two points, slope and angle between two straight lines are not to be found in the work of Decartes. These are due to later mathematicians like Clairaut (1729) Monge (1781) and Lacroin (1765-1843). Newton used several types of co-ordinates including polar and bipolar. An important text book on analytical geometry was written by L’Hospital around 1700.
7.1
RATIO FORMULA
Let A and B be two given points. Let P be a point on the line segment AB or on AB produced. Then P divides AB into two segments AP and PB . The lengths of AP and PB AP m = . are AP and PB. These lengths are in some ratio m : n; that is AP : PB = m : n or PB n If P lies inside AB we say that P divides AB internally in the ratio m : n. If P lies outside AB , that is, P lies on AB produced, then we say that P divides AB externally in the ratio m : n. With a given ratio m : n, AB can be divided either internally or externally. Example 1: Divide the line segment AB of length 16 units in the ratio 3:5 Solution: D
A
C
B
Fig.7.1
AC 3 = . Since the numerator is smaller than the CB 5 denominator, C is closer to A than to B (See Fig.7.1). Then 5 AC = 3 BC or 5 AC = 3 (AB – AC) or 8 AC = 3 (16) = 48 AC = 6 units and so CB = AB – AC = 16 – 6 = 10 units.
i) Let C be the point inside AB such that
142
Hence C lies inside AB 6 units distance from A and 10 units distance from B. The point C is unique and it divides AB internally in the given ratio 3:5. AD 3 = . Since the numerator is smaller than DB 5 the denominator, D is closer to A than to B (See Fig.7.1). Now we have 5 AD = 3DB or 5 AD = 3(AD + AB) or 5 AD = 3AD + 3AB 2 AD = 3AB 3(16) = 48 or AD = 24 Then DB = DA + AB = 24 + 16 = 40 ∴ D lies outside AB 24 units distance from A and 40 units distance from B. The point D is unique and it divides AB externally in the given ratio 3:5.
ii) Let D be the point outside AB such that
Example 2: Divide the line segments AB of length 16 units in the ratio 3:1. Solution: E A
F B
Fig.7.2
AE 3 = . Since the numerator is greater than the EB 1 denominator, the point E is closer to B than to A (See Fig.7.2). Then AE = 3 EB or AE = 3(AB – AE) or AE = 3AB – 3AE or 4AE = 3AB = 3(16) = 48
i) Let E be the point inside AB such that
or AE = 12. Then EB = AB – AE = 16 – 12 = 4 ∴ E lies inside AB , 12 units distance from A and 4 units distance from B. The point E is unique and it divides AB internally in the ratio 3:1. AF 3 ii) Let F be the point outside AB such that = . Since the numerator is greater than FB 1 the denominator, F is closer to B than to A (See Fig.7.2). Then AF = 3FB or AF = 3(AF – AB) or AF = 3AF – 3AB 2AF = 3AB = 3(16) = 48 or AF = 24 ∴ FB = AF – AB = 24 – 16 = 8 ∴ F lies outside AB 24 units distance from A and 8 units distance from B. The point F is unique and it divides AB externally in the ratio 3:1. From the above two examples, we understand that a given line segments can be divided either internally or externally by a point in a given ratio. Section Formula
To find the coordinates of the point which divides internally the line segment joining two given points (x1, y1) and (x2, y2) in the given ratio m : n. 143
Let A and B be the given points (x1, y1) and (x2, y2) respectively. Let C (x3, y3) be the point which divides the line segment AB internally in the ratio m : n. Then we have AC m = CB n
y B C T A
Draw AP, BQ and CR perpendicular to the
S
x-axis. Draw AS perpendicular to CR and CT perpendicular to BQ (See Fig.7.3). We observe that the triangles ΔASC and ΔCTB are AS CS AC similar. So we have = = CT BT CB But AS = x3 – x1; CT = x2 – x3; x 3 − x1 y −y m ∴ = 3 1 = x 2 − x 3 y 2 − y3 n n(x3 – x1) (or) nx3 – nx1 (or) (m+n) x3 mx 2 + nx1 = m+n
Cross multiplying
∴
x3
0
P
R
Q
x
Fig.7.3 CS = y3 – y1; BT = y2 – y3 y3 − y 1 x −x m m ∴ 3 1 = = and x 2 − x3 n y 2 − y3 n = m(x2 – x3) = mx2 – mx3 = mx2 + nx1
and
and n(y3 – y1) = m(y2 – y3) and ny3 – ny1 = my2 – my3 and (m+n) y3 = my2 + ny1 my 2 + ny1 y3 = m+n
That is the point which divides AB internally in the ratio m : n is given by ⎡ mx 2 + nx1 my 2 + ny 1 ⎤ (1) ⎢ m+ n , m+ n ⎥ ⎣ ⎦ Note: We have obtained the above derivation by taking A and B in the 1st quadrant as in the (Fig.7.3). However, the same derivation holds as well for other positions of A and B. The students are advised to do the derivation in other cases. Next we proceed to obtain the coordinates of the point D which divides AB externally in the ratio m : n (m > n). Then D lies AD m m -n:n outside but closer to B and = DB n A(x 1 ,y 1 ) B(x 2 , y 2 ) D(x 4 , y 4 ) (See Fig.7.4). We observe that B AB Fig.7.4 divides AD internally in the ratio . BD AD m Since = , we get nAD = mBD or n(AB + BD) = mBD or nAB = (m–n) BD or BD n AB m − n = . That is B divides AD internally in the ratio (m – n): n. If D is (x4, y4) then BD n (m − n) x4 + nx1 (m − n) y4 + ny1 x2 = , y2 = ( by using (1)) (m − n) + n (m − n) + n
144
and my2 = (m – n) y4 + ny1 mx2 = (m – n) x4 + nx1 mx 2 − nx1 my 2 − ny1 (or) x4 = , y4 = m−n m−n Hence the point which divides AB externally in the ratio m : n (m > n) is given by ⎡ mx 2 - nx1 my 2 - ny 1 ⎤ ⎢ m- n , m- n ⎥ ⎣ ⎦ (or)
(2)
Let us now consider the situation where a point E divides AB externally in the ratio m : n AE m (m < n) then = < 1 or AE < EB and so the point E is closer to A (See Fig.7.5). We EB n EA . observe that A divides EB internally in the ratio AB AE m Since = , we get nAE = mEB or nEA = m(EA + AB) EB n (or) (n – m) EA = mAB m:n-m EA m (or) = E(x5, y5) A(x1, y1) B(x2, y2) AB n − m Fig.7.5 That is A divides EB internally in the ratio m : n – m mx 2 + (n − m) x 5 mx 2 + (n − m) x 5 ⇒ x1 = ∴ x1 = m+n−m n and y1 =
my 2 + (n − m) y5 m+n−m
⇒ y1 =
my 2 + (n − m) y5 n
nx1 - mx 2 ny − my 2 , y5 = 1 n-m n−m Thus we get the point which divides AB externally in the ratio m : n (m < n) as ∴ x5 =
⎡ nx1 - mx 2 ny1 - my 2 ⎤ ⎢ n-m , n-m ⎥ ⎣ ⎦
(3)
The formulae (1), (2) and (3) are called Ratio formulae. Let F be the middle point of the line segment AB . Then we observe that F divides AB internally in the ratio 1:1 since AF = FB (see Fig.7.6).
A
F
Fig.7.6
∴ The point F, the middle point of AB is given by ⎡1 × x 1 + 1 × x 2 1 × y1 + 1 × y 2 ⎤ ⎡ x 1 + x 2 y1 + y 2 ⎤ =⎢ , ⎥=⎢ 2 , 2 ⎥ 1+1 1+1 ⎣ ⎦ ⎣ ⎦
145
B
This result is known as the middle point formula or simply the mid-point formula. We x1 + x 2 is the average of the observe that 2 A(x , y ) x-co-ordinates of the end points of the segments. y +y Similarly, 1 2 is the average of the y-co-ordinates 2 of the end points of the line segment. 1
1
G
We are now able to find the coordinates of the centroid of the triangle whose vertices are the given points (x1, y1), (x2, y2) and (x3, y3). First we recall that a median of a triangle is a line segment joining a vertex to the mid point of the opposite side. So there are three medians of a triangle and they are concurrent at a point G, called the centroid of the triangle.
B(x 2 , y 2 )
D
C(x 3 , y 3 )
2:1 A(x 1 , y 2 )
G x2 + x3 2
D y2 + y3 , 2
Fig.7.7
The centroid of the triangle divides each median internally in the ratio 2:1. Consider the median AD, G divides AD in the ratio 2:1 (See Fig.7.7). ⎡ ⎤ ⎛ x 2 + x3 ⎞ ⎛ y 2 + y3 ⎞ ⎢ 2 × ⎜ 2 ⎟ + 1× x1 2 × ⎜ 2 ⎟ + 1 × y1 ⎥ ⎛ x1 + x 2 + x 3 y1 + y 2 + y 3 ⎞ ⎝ ⎠ ⎝ ⎠ The point G is ⎢ , ⎥= ⎜ ⎟ , 3 3 ⎠ ⎢ ⎥ ⎝ 2 +1 2 +1 ⎢ ⎥ ⎣ ⎦
Example 3: Find the point which divides the line segment joining the points (–1, 2) and (4,–5) internally in the ratio 2:3. Solution: The required point is given by ⎛ mx 2 + nx1 my 2 + ny1 ⎞ ⎛ 2 × 4 + 3 × (−1) 2 × (−5) + 3 × 2 ⎞ , = ⎜ , ⎜ ⎟ ⎟ m+n ⎠ 2+3 2+3 ⎝ m+n ⎝ ⎠ ⎛ 8 − 3 −10 + 6 ⎞ ⎛ 5 −4 ⎞ = ⎜ =⎜ , (x1, y1) Æ (–1, 2) (x2, y2) Æ (4,–5) , ⎟ 5 ⎟⎠ ⎝ 5 5 ⎠ ⎝ 5 m:n=2:3 − 4 ⎛ ⎞ = ⎜1, Formula (1) is applied. ⎟ 5 ⎝ ⎠ Example 4: Find the point which divides the line segment joining the points (2,1) and (3,5) externally in the ratio 2:3. Solution: The required point is given by ⎛ 3 × 2 − 2 × 3 3 ×1 − 2 × 5 ⎞ ⎛ nx1 − mx 2 ny1 − my 2 ⎞ , = ⎜ , ⎟ ⎜ ⎟ n−m ⎠ 3− 2 ⎠ ⎝ 3− 2 ⎝ n−m ⎛ 6 − 6 3 − 10 ⎞ = ⎜ (x1, y1) Æ (2,1), (x2 , y2) Æ (3,5), , ⎟ 1 ⎠ ⎝ 1 m:n=2:3 = (0,–7) Here m < n formulae (3) is applied.
146
Example 5: Find the points which divides the line segment joining (–3, –4) and (–8, 7) internally and externally in the ratio 7:5. Solution: i) Internal ratio 7:5. The required point is ⎛ mx 2 + nx1 my 2 + ny1 ⎞ , ⎜ ⎟ m+n ⎠ ⎝ m+n
⎛ 7 × (−8) + 5 × (−3) 7 × 7 + 5 × (−4) ⎞ =⎜ , ⎟ 7+5 7+5 ⎝ ⎠ ⎛ −56 − 15 49 − 20 ⎞ =⎜ , 12 ⎟⎠ ⎝ 12 ⎛ −71 29 ⎞ = ⎜ , ⎟ ⎝ 12 12 ⎠
(x1, y1) Æ (–3, –4) (x2, y2) Æ (–8, 7) m : n = 7.5 Formula (1) is applied.
ii) External ratio 7:5 The required point is ⎛ 7 × (−8) − 5 × (−3) 7 × 7 − 5 × ( −4) ⎞ =⎜ , ⎟ 7−5 7 −5 ⎝ ⎠
⎛ mx 2 − nx1 my 2 − ny1 ⎞ , ⎜ ⎟ m−n ⎠ ⎝ m−n
⎛ −56 + 15 49 + 20 ⎞ =⎜ , ⎟ 2 2 ⎠ ⎝ ⎛ −41 69 ⎞ = ⎜ , ⎟ ⎝ 2 2 ⎠
(x1, y1) Æ (–3, –4) (x2 y2) Æ (–8, 7) m:n = 7:5 Here m > n formula (2) is applied.
Example 6: In what ratio is the line segment joining the points A(4,4) and B(7,7) divided by the point C(–1, –1)? Solution: First, we have to know whether the points A, B, C are collinear and if so whether the point C lies inside or outside AB . For this, we find the lengths AC = (−1 − 4) 2 + ( −1 − 4) 2 =
25 + 25 C
= CB = = AB = =
50 = 5 2
A
Fig.7.8
(−1 − 7) 2 + (−1 − 7) 2 = 64 + 64
128 = 8 2 (7 − 4) 2 + (7 − 4)2 =
3 2
5 2
9+9
18 = 3 2
147
(x1, y1) Æ A(4,4) (x2, y2) Æ B(7,7) (x3, y3) Æ C(–1, –1)
B
We observe that AC + AB = CB. So the points A, B and C are collinear and C lies outside AB but closer to A (See Fig.7.8). AC 5 2 5 Now the required ratio is = = = 5:8 CB 8 2 8
⇒ The point C(–1,–1) divides AB externally in the ratio 5:8 We verify this result with the ratio formula (3), since 5 < 8 the point which divides AB externally in the ratio 5 : 8 is given by ⎛ nx1 − mx 2 ny1 − my 2 ⎞ ⎛ 8× 4 − 5× 7 8× 4 − 5× 7 ⎞ , , ⎜ ⎟ ⎜ ⎟ n−m ⎠ 8−5 ⎠ ⎝ n−m ⎝ 8−5 ⎛ 32 − 35 32 − 35 ⎞ = ⎜ , 3 ⎟⎠ ⎝ 3 ⎛ −3 −3 ⎞ = ⎜ , ⎟ 3 ⎠ ⎝ 3 = (−1, − 1)
(x1, y1) Æ (4, 4) (x2, y2) Æ (7, 7) m:n=5:8 m
Formula (3) is applied
Alternately, let C lie on AB and divide it in the ratio m : n. Then C is given by ⎛ m(7) + n(4) m(7) + n(4) ⎞ , ⎜ ⎟ m+n ⎠ ⎝ m+n
⎛ mx 2 + nx1 my 2 + ny1 ⎞ , ⎜ ⎟ m+n ⎠ ⎝ m+n
⎛ 7m + 4n 7m + 4n ⎞ , = ⎜ ⎟ m+n ⎠ ⎝ m+n
(x1, y1) Æ (4, 4) (x2, y2) Æ (7, 7)
But C is given to be (–1, –1). (x3, y3) Æ (–1, –1)
∴ Equating the co-ordinates we get
7m + 4n = − 1 or m+n
m : n = 5 : 8, m < n Formula (1) is applied 7m + 4n = –m – n or
8m = –5n or m/n = –5/8
⇒ C divides AB in the ratio (–5) : 8 (or) C divides AB externally in the ratio 5 : 8 and C is closer to A than to B. Note: If C divides AB externally in the ratio m : n where m and n are positive numbers with m < n, then we can consider C as the point which divides AB internally in the ratio (–m): n. Since C is ⎛ nx1 − mx 2 ny1 − my 2 ⎞ ⎛ (− m)x 2 + nx1 (− m)y 2 + ny1 ⎞ , , ⎜ ⎟=⎜ ⎟ n − m ⎠ ⎝ (−m) + n (−m) + n ⎠ ⎝ n−m
148
Similarly, if C divides AB externally in the ratio m : n with m > n, then C can be considered as the point which divides AB internally in the ratio m : (–n), since C is ⎛ mx 2 − nx1 my 2 − ny1 ⎞ ⎛ mx 2 + (−n)x1 my 2 + (−n)y1 ⎞ , , ⎜ ⎟=⎜ ⎟ m − n ⎠ ⎝ m + (−n) m + (−n) ⎠ ⎝ m−n Thus, a ratio of the form m : (–n) or (–m) : n where m and n are positive numbers, corresponds to external division. For example, if we say that the point C divides AB “internally” in the ratio (–5): 9 then we mean that C divides AB externally in the ratio 5:9 and that C is closer to A than to B. Example 7: Find the co-ordinates of the mid point of the line segment joining the points A(–3,2) and B(7,8) Solution: The required mid point is ⎛ (−3) + 7 2 + 8 ⎞ =⎜ , (x1, y1) Æ (–3,2) 2 2 ⎟⎠ ⎝ ⎛ 4 10 ⎞ = ⎜ , ⎟ = (2,5) (x2, y2) Æ (7,8) ⎝2 2 ⎠ ⎛ x +x y +y ⎞ formula ⎜ 1 2 , 1 2 ⎟ is applied. 2 ⎠ ⎝ 2 Example 8: The centre of a circle is (–6,4). A diameter of the circle has its one end at the origin. Find its other end. Solution: Let the diameter be OA where O is the origin (0, 0). Let the other end A be (x1, y1). The mid point of a diameter is the centre of the circle. So the mid point of OA is the centre (–6,4). But by the mid point formula, the mid point of OA is ⎛ 0 + x1 0 + y1 ⎞ ⎛ x1 y1 ⎞ ⎜ 2 , 2 ⎟=⎜ 2 , 2 ⎟ ⎝ ⎠ ⎝ ⎠ x y But this point is the centre (–6,4). So 1 = − 6 and 1 = 4 (or) x1 = –12, y1 = 8 2 2 ⇒ the other end A of the diameter is (–12, 8). Example 9: If the points A(2, –2), B (8,4), C(5,7) are the three vertices of a parallelogram ABCD taken in order, find the fourth vertex D. Solution: Let D(a,b) be the fourth vertex (See Fig.7.9) since ABCD is a parallelogram, the diagonals AC and BD bisect each other. That is the mid point of AC is the same as the D C mid point of BD . 2 + 5 ( − 2) + 7 ⎛ ⎞ ⎛7 5⎞ , But the mid point of AC is ⎜ ⎟ = ⎜2, 2⎟ 2 2 ⎠ ⎝ ⎠ ⎝ 8 + a 4 + b ⎛ ⎞ and the mid point of BD is ⎜ , ⎟ 2 ⎠ ⎝ 2 A B 8+a 7 4+b 5 Equating the co-ordinates, we get = , = Fig.7.9 2 2 2 2
=> 8 + a = 7, 4 + b = 5 => a = –1, b = 1
∴ The fourth vertex is D(–1, 1) 149
Example 10: Find the centroid of the triangle whose vertices are the points (8, 4), (1,3) and (3,–1). Solution: The centroid of the triangle is
⎛ 8 + 1 + 3 4 + 3 + (−1) ⎞ ⎜ 3 , ⎟ 3 ⎝ ⎠ ⎛ 12 6 ⎞ =⎜ , ⎟ ⎝ 3 3⎠ = (4,2)
(x1 y1) Æ (8,4) (x2 y2) Æ (1,3) (x3 y3) Æ (3, –1) ⎛ x + x + x 3 y1 + y 2 + y3 ⎞ , formula ⎜ 1 2 ⎟ is applied. 3 3 ⎝ ⎠
Example 11: If a triangle has its centroid at (4,3) and two of its vertices are (2,–1) and (7,8) find the third vertex. Solution: Let the third vertex be (a,b) then the centroid of the triangle is ⎛ 2 + 7 + a (−1) + 8 + b ⎞ , (x1, y1) Æ (2, –1) ⎜ ⎟ 3 3 ⎝ ⎠ ⎛ 9+a 7 + b ⎞ (x2, y2) Æ (7,8) =⎜ , ⎟ 3 ⎠ ⎝ 3 (x3, y3) Æ (a, b ⎛ x + x + x 3 y1 + y 2 + y3 ⎞ formula ⎜ 1 2 , ⎟ is applied. 3 3 ⎝ ⎠ But the centroid is given to be (4,3) so equating the coordinates, we get
9+a 7+b = 4 and = 3. 3 3
⇒ 9 + a = 12 and 7 + b = 9
⇒ a = 3 and b = 2
⇒ the third vertex is (3,2) m . If we denote this real number by Note: The ratio m : n corresponds to the real number n m λ λ (Greek letter called lambda) then = λ = . So m : n = λ : 1 n 1 5 5 For example, the ratio 5 : 3 is same as the ratio : 1 and conversely the ratio : 1 is same as 3 3 the ratio 5 : 3. Exercise 7.1
1. Find the co-ordinates of the point which divides the line segment joining the points i) A(–2,6) and B(3,6) in the ratio 3 : 2 internally. ii) P(3,4) and Q(–6,2) in the ratio 1 : 3 externally. iii) A(4, –7) and B(–1,5) in the ratio 5:2 externally iv) (–1, –2) and (5,3) in the ratio 2 : –1 internally v) (3, –4) and (–2,6) in the ratio –2:3 internally vi) (a+b, a–b) and (a–b, a+b) in the ratio 3 : 2 internally. 150
2. Find in what ratio the point P divides the line segment joining the two points A and B where P, A and B are respectively given by i) (11,7) (13,4) and (7,13) ii) (–5,3, (–3, –1) and (–8, 9) iii) (1,12) (5,6) and (7,3) 3. Find the ratio in which the x-axis divides the line segment joining the points. i) (1,2) and (–2,5) ii) (–3, –2) and (–1, 4) iii) (3, –2) and (–7, –1) iv (5, –4) and (9, 1) 4. Find the ratio in which the y-axis divides the line segment joining the points iii) (3, –4) and (–6,2) iv) (–1, 2) and (5, –2) i) (3,0) and (–3,5) ii) (–2,6) and (3,4) 5. Find the mid point of the line segment joining the points. ii) (8, –2) and (3, –4) i) (–1, –3) and (–5, –7) 6. ABC is a triangle whose vertices are A(2,–1), B(–4, 2) and C(2,5). Find the length of the median AD 7. The mid point of the line segment PQ is (5,1). If P is (8,4), find the point Q. 8. The centre of a circle is (4,–1). If one end of a diameter of the circle is (9,7), find the other end. 9. Prove that the following points form a parallelogram. i) (1,2), (–2, 2), (–4, –3) and (–1, –3) (ii) (–2, –1), (1, 0), (4,3) and (1,2) iii) (2, –2), (8,4), (5,7) and (–1, 1) (iv) (0, 3), (4,4), (6,2) and (2,1) 10. Find the fourth vertex of the parallelogram three of whose vertices are given by i) (1, 1), (2,3) and (–2,2) taken in order ii) (–1,0), (5,2) and (7,4) taken in order iii) (2,3), (3,8) and (10,–1) taken in order iv) (–2,–5), (4, –5) and (4,7) taken in order 11. Find the centroid of the triangle whose vertices are given by i) (1,10), (–7, 2) and (–3, 7) ii) (–1, –3), (2, 1) and (2, –4) iii) (1, 1), (2,3) and (–2, 2) iv) (1, 3), (2,7) and (12, –16) 12. The centroid of a triangle ABC is (3, –2). Find the vertex C if A and B are given by i) (1, –2) and (7, 4) ii) (3, 4) and (–1, 9) iii) (5, –1) and (–2, –7) iv) (–11, 1) and (2, –5)
7.2 AREA OF A TRIANGLE We have already learnt in practical geometry that the area of a trapezium is one half the sum of the lengths of two parallel sides multiplied by the perpendicular distance between them. Using this formula, we shall now derive an algebraic formula for the area of a triangle when the coordinates of its vertices are given. Let ABC be the triangle. The vertices A, B and C be respectively (x1,y1), (x2, y2) and (x3, y3). Draw AL , BM and CN perpendicular to the x-axis (See Fig.7.10).
A
y
C B
O
M
N
L
Fig.7.10
Area of the ΔABC = Area of the trapezium ABML + Area of the trapezium ALNC – Area of the trapezium BMNC 151
X
1 1 1 ML (AL + BM) + LN (AL + CN) − MN (BM + CN) 2 2 2 1 1 1 = (x1 − x 2 ) (y1 + y 2 ) + (x 3 − x1 ) (y3 + y1 ) − (x 3 − x 2 ) (y 2 + y3 ) 2 2 2 1 = ⎡⎣ ( x1 y1 + x1 y 2 − x 2 y1 − x 2 y 2 ) + ( x 3 y3 + x 3 y1 − x1 y3 − x1 y1 ) 2 − ( x 3 y 2 + x 3 y3 − x 2 y 2 − x 2 y3 ) ⎤⎦ =
=
1 ⎡ x1 y1 + x1 y 2 − x 2 y1 − x 2 y 2 + x 3 y3 + x 3 y1 − x1 y3 − x1 y1 2 ⎣ − x 3 y 2 − x 3 y3 + x 2 y 2 + x 2 y3 ⎤⎦
We denote the area of the triangle by Δ. 1 Thus we have Δ = [ x1 (y 2 − y 3 ) + x 2 (y 3 − y1 ) + x 3 (y1 − y 2 )] square units. This is called 2 the area formula for a triangle. Note: i) All areas are positive. So Δ (area) must be positive. In order that Δ should be positive, the points (x1, y1), (x2, y2) and (x3, y3) taken in order, should be in the counter clockwise direction (See Fig.7.11). If the points are taken in order are in the clockwise direction then the formula for the area yields a negative number (See Fig.7.12). C
A
B
A
B
C
Counter Clockwise Direction Fig.7.11
Clockwise Direction Fig.7.12
ii) If the three points (x1, y1) (x2, y2) and (x3, y3) are collinear, then the area of the triangle formed by them is 0, that is D C 1 [ x1 (y2 − y3 ) + x 2 (y3 − y1 ) + x 3 (y1 − y2 )] = 0 2 (or) x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0 ∴ The condition for the collinearity of three given points (x1, y1), (x2, y2) and (x3, y3) is x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0
A
B
Fig.7.13 iii) Using the area formula for a triangle, we can calculate the area of four sided geometrical figures such as quadrilateral by dividing the quadrilateral ABCD into two triangles ΔABD and ΔBCD having one diagonal BD as the common side (See Fig.7.13).
152
Example 12: Find the area of the triangle whose vertices are (5, 2), (–9, –3) and (–3, –5). Solution: Let the vertices be A, B and C respectively. ∴Area of the Δ ABC 1 1 = [5(–3+5) + (–9) (–5 –2) + (–3) (2+3)] Δ = [x1 (y2–y3) + x2 (y3–y1) + x3 (y1–y2)] 2 2 58 1 = = 29 square units (x1,y1) → (5,2) [10 + 63 – 15] = 2 2 (x2,y2) → (–9, –3)
(x3,y3) → (–3, –5) Note: Since the area is positive, the given points taken in order are in counter clockwise direction. If we consider (–9, –3) as A, (5,2) as B and (–3, –5) as C, then, 1 Area of Δ ABC = [(–9) (2+5) + 5 (–5+3) + (–3) (–3–2)] 2 1 1 1 [–63–10+15] = (–63+5) = (–58) = –29 = 2 2 2 We get a negative number for the area since the points (–9, –3), (5,2) and (–3,–5) taken in order, are not in counter clockwise direction (i.e. the points are in clockwise direction). We can plot the points in the Cartesian plane and confirm ourself the above observation. Example 13 : Find the area of the triangle whose vertices are (4,5), (4,2) and (–2,2). Solution : From the rough sketch (See Fig. 7.14) we take y C (4,5) (–2,2) as A, (4,2) as B and (4,5) as C. So that A,B and C taken in order are in counter clockwise direction. Then by the area formula Area of 1 Δ = [x1(y2–y3) + x2 (y3–y1) + x3 (y1–y2)] 2 1 Δ ABC = [(–2) (2–5) +4(5–2) +4 (2–2)] A B 2 (-2,2) (4,2) 1 1 = [6+12+0] = (18) = 9 sq.units. x 2 2 0 Fig.7.14 Note : In case, if we take (4,5) as A, (4,2) as B and (–2,2) as C then we get the area of the ΔABC as –9 square units. Omit the minus sign and say that the area of ΔABC is 9 units. Example 14: Prove that the points (a,b+c), (b, c+a) and (c,a+b) are in a straight line. Solution: Let the given points be P,Q and R respectively. 1 Area of Δ PQR = [a{(c+a) – (a+b)} + b{(a+b) – (b+c)}+c {(b+c) – (c+a)}] 2 1 1 1 = [a(c–b) + b(a–c) +c(b–a)] = [ac–ab + ba – bc + cb – ca] = (0) = 0 2 2 2
∴ P,Q,R are collinear,
that is they lie on a straight line.
153
Example 15: If the points (1,4), (r, –2) and (–3, 16) are collinear find r. Solution: Let the points be A,B and C respectively. Then the area of ΔABC is 1 1 1 = [1(–2–16) + r(16 – 4) + (–3) (4+2)] = [(–18+12r–18)] = [–36 +12r] = –18 + 6r. 2 2 2 If A,B and C are collinear, then the area of ΔABC is 0 and so. 18 –18 + 6r = 0 or 6r = 18 or r= or r=3 y 6 D(3,9)
Example 16 : Find the area of the quadrilateral formed by the points A(-1,6) (–3,–9), (–1,6), (3,9) and (5,–8). Solution: Plotting the points roughly in the Cartesian plane (See Fig.7.15) we find that (–1,6) (–3, –9), (5,–8) and (3,9), taken in order, x’ x O are in counter clockwise direction. Let these points be A, B, C and D respectively. Then the area of ΔABC 1 = [(–1) (–9 +8) + (–3) (–8 –6) + 5(6+9)] C(5,-8) 2 B(-3,-9) 1 y’ [(–1) (–1) + (–3) (–14) + 5(15)] = 2 Fig.7.15 1 = [118] = 59 square units 2 1 Area of ΔACD = [(–1) (–8–9) +5(9–6)+3(6+8)] 2 1 = [(–1) (–17) + 5(3) + 3(14)] 2 1 1 = [17 + 15 + 42] = [74] = 37 square units. 2 2 ∴Area of the quadrilateral ABCD = Area of ΔABC + Area of ΔACD = 59 + 37 = 96 square units. Exercise 7.2
1. Find the area of the triangle whose vertices are i) (0,4), (3,6) and (–8,–2) ii) (3,4), (2,–1) and (4,–6), iii) (5,6), (2,4) and (1,–3) iv) (1,3), (–7,6) and (5,–1) v) (1,1), (3,4) and (5,–2) 2. Show that the following points are collinear 1 i) (3,1), (–1,–7), (5,5) ii) (– ,3), (–5,6), (–8,8) iii) (9,0), (1,4), (11,–1) 2 3. If the points (2,5), (4,6) and (a,b) are collinear, find the relation between a and b. 4. If (x,–11), (2,3) and (4,–1) lie on a straight line find x. x y 5. If the point (x,y) is collinear with the points (a,0) and (0,b), then prove that + = 1 . a b 6. Find the area of the quadrilateral whose vertices are i) (–1,6), (–3,–9), (5,–8) and (3,9) ii) (1,2), (–3,4), (–5,–6) and (4,–1) iii) (5,8), (6,3), (3,1) and (2,6) iv) (3,4), (5,–2), (4,–7) and (1,1).
154
7.3
STRAIGHT LINE y
A linear equation or an equation of the first degree in x and y represents a straight line. The equation of a straight line is satisfied by the co-ordinates of every point lying on the straight line and not by any other point outside the straight line. If a straight line l cuts the x-axis at A and y-axis at B then the x coordinate of A that is OA is called the x-intercept and the y coordinate of B that is OB is called the y-intercept of the straight line (Fig.7.16).
B
x
x’ A
0 y’
Fig.7.16
Equation of a straight line in various forms. Equation of x-axis is y = 0 and Equation of y-axis is x = 0
We know that on the x-axis, the y co-ordinate of any point is zero, that is y = 0, ⇒ the equation of the x-axis is y = 0. Similarly on the y-axis, the x-co-ordinate of any point is zero. y Hence the equation of y-axis is x = 0.
Let l be a straight line parallel to the y-axis at a distance `a’ from it. Then the x-coordinate of any point on l is clearly `a’. Hence the equation of the straight line l is x = a (Fig.7.17).
a
x’
x=a
Equation of a straight line parallel to the y-axis and at a distance `a’ from it is x = a
x
O
Fig.7.17 Equation of a straight line parallel to the x-axis at a distance `b’ from x-axis is y = b
y
Let l be a straight line parallel to the x-axis at a distance ‘b’ from it. Then the y-coordinate of every point on l is clearly b. Therefore the equation of the straight line l is y = b (Fig.7.18). The angle made by a straight line l with the positive direction of x-axis is called the inclination of a line.
y=b b x’
x O
y’
Results:
o
i) The inclination of x-axis is 0 . ii) The inclination of every line parallel to x-axis is 0o. iii) The inclination of y-axis is 90o. iv) The inclination of every line parallel to y-axis is 90o.
155
Fig.7.18
y
y
θ
θ x’
x
x’
x
0 y’ y’ Fig.7.19
Fig.7.20 y
Slope of a straight line
The tangent of the angle made by the straight line l with positive direction of the x-axis in the counter clockwise sense) is called the slope or gradient of the straight line. From the figure 7.21, slope of the straight line = m = tan θ.
θ x
x’ 0
Slope of the straight line joining the two points A(x1,y1) and B (x2,y2) y Let BN be drawn perpendicular to the x-axis and AM perpendicular to BN meeting it at M. The slope m of the straight line joining the points A (x1,y1) and B(x2,y2) is given by m = tan θ BM BN − MN BN − AC y 2 − y1 = = = = AM CN ON − OC x 2 − x 1 y − y2 y − y1 x’ or m = 1 ∴ m= 2 x1 − x 2 x 2 − x1
y’
) y2 x 2, ( B ) ,y 1 x1 ( A
θ M
θ C
x
N
Fig.7.22
y’
Condition for parallelism and perpendicularity
Fig.7.21
y 2
Two lines are parallel if and only if their slopes are equal. Let l1 and l2 be two parallel lines with slopes m1 and m2 and inclinations θ1 and θ2 respectively. Then m1 = tan θ1 and m2 = tan θ2 (Fig.7.23). Given l1 || l2 l1 || l2 ⇒ θ1 = θ2 [corresponding angles] ⇒ tan θ1 = tan θ2 ⇒ m1 = m2 Conversely, let m1 = m2 ⇒ tan θ1 = tan θ2 ⇒ θ1 = θ2 ⇒ l1 || l2
156
1
θ2
x’
θ1
0
Fig.7.23 y’
x
[Q θ1 and θ2 are corresponding angles]. Q l1 || l2 ⇔ m1 = m2 If the two lines l1 and l2 are perpendicular with slopes m1 and m2 then m1m2 = –1 Let the lines l1 and l2 intersect at B (x2, y2). A(x1, y1) and C(x3, y3) be any two points on the lines l1 and l2 respectively. y Now ∠ABC = 90o since l1 and l2 are perpendicular (Fig.7.24). y − y3 Slope of l1 = m1 = 2 (1) A (x ,y ) x2 − x3 y − y2 (2) Slope of l2 = m2 = 1 C (x ,y ) x1 − x 2 x’ x 0 Since ABC is a right angled triangle. ∴AB2 + BC2 = AC2 y’ (x1 – x2)2 + (y1– y2)2 + (x2 – x3)3 + (y2 – y3)2 = (x1 – x3)2 + (y1 – y3)2 Fig.7.24 2
1
1
3
3
B
(x 2 ,y 2)
1
x12 + x 22 − 2x1 x 2 + y12 + y 22 − 2y1 y 2 + x 22 + x 32 − 2x 2 x 3 + y 22 + y32 − 2y 2 y3 = x12 + x 32 − 2x1 x 3 + y12 + y32 − 2y1 y3 2y22 – 2y1y2 – 2y2y3 + 2y1y3 = –2x22 + 2x1 x2 + 2x2 x3 – 2x1 x3 ( ÷ by 2)
y22 – y2y3 – y1y2 + y1y3 = –x22 + x2 x3 – x1x3 + x1x2 y2 (y2–y3) – y1 (y2 – y3) = x2 (x3 – x2) – x1 (x3 – x2) (y2 – y3) (y2 – y1) = (x3 – x2) (x2 –x1) –(y2 – y3) (y1 – y2) = (x2 – x3) (x1 – x2) ⎡ y1 − y 2 ⎤ ⎡ y 2 − y 3 ⎤ ⎥ = –1 ⎢ ⎥ ⎢ ⎣ x1 − x 2 ⎦ ⎣ x 2 − x 3 ⎦
∴ m1 . m2 = –1
by (1) & (2)
∴ If the two lines are perpendicular with the slopes m1 and m2 then m1 . m2 = –1. Slope – intercept form y
The equation of straight-line with slope m and having a y-intercept `c’ is y = mx + c. Let a line l have slope m and y-intercept c. Let θ be the angle made by the line with the positive direction of x-axis. Given that the straight line cuts off an intercept c on the y-axis implies that the line passes through A(0, c). Let P(x,y) be any point on the line. Draw PN perpendicular to x-axis and AM perpendicular to PN meeting it at M. Since the slope of the line is m = tan θ, from ΔAPM we get
P(
) 0,C A( θ
θ x’
M
C N
0 y’
Fig.7.25
157
) x,y
x
m = tan θ = =
PM PN − NM = AM ON
PN − OA y − c = ON x
⇒ y – c = mx
(or) y = mx + c
Hence the equation of a straight line with slope m and having y-intercept c is y = mx+ c. Note : If the straight line passes through the origin, its equation is y = mx.
For a straight line passing through the origin, y intercept is zero ⇒ c = 0. Therefore its equation is y = mx. y
Slope – Point form P(x,y)
The equation of a straight line with slope m and passing through a point (x1, y1) is y – y1 = m (x – x1). Let the line l make an angle θ with the x-axis as shown in the Fig.7.26 and pass through the given point Q (x1, y1). Let P (x,y) be any point on the given straight y − y1 line, then the slope of the straight line is . But the x − x1 y − y1 slope is given to be m ⇒ m = x − x1 (or) y – y1 = m (x – x1).
Q(x1,y1)
θ
x’
x
0
Fig.7.26
y’
Hence the equation of a straight line with slope m and passing through the point Q (x1, y1) is y – y1 = m (x – x1). Note: If the straight lines passes through the origin, that is x1 = 0, y1 = 0, then the equation is y – 0 = m(x – 0). ∴ y = mx. y
Two-points form
The equation of a straight line passing through two y − y1 x − x1 points (x1,y1) and (x2,y2) is given by . = y 2 − y1 x 2 − x 1 Let l be a straight line passing through A(x1,y1) and B(x2,y2) Fig. 7.27. Then the slope of the straight line is y − y1 m= 2 . Since the straight line passes through x 2 − x1 158
x 2, B(
) y2
x 1, A(
x’
) y1
x 0
y’
Fig.7.27
A(x1, y1) its equation is y-y1 = m (x–x1). ⎡ y − y1 ⎤ y − y1 x − x1 y–y1 = ⎢ 2 = ⎥ (x − x 1 ) (or) y 2 − y1 x 2 − x 1 ⎣ x 2 − x1 ⎦ Hence the equation of the straight line l passing through two points (x1,y1) and (x2,y2) is y − y1 x − x1 = y 2 − y1 x 2 − x 1 Aliter: The equation of a line l passing through two points (x1, y1) and (x2, y2) is given by y − y1 x − x1 = y y 2 − y1 x 2 − x 1
Let l be the line passing through Q(x1,y1) and R(x2,y2). Let P(x,y) be any point on QR. The points Q,P,R are collinear (Fig. 7.28). y − y1 y 2 − y1 ∴ Slope PQ = Slope of QR ⇒ = x − x1 x 2 − x1 y − y1 x − x1 . = y 2 − y1 x 2 − x 1 Aliter : From the figure 7.29. QS = x – x1 PS = y – y1 QT = x2 – x1 RT = y2 – y1 ΔQSP ||| ΔQTR PS QS PS RT ∴ RT QT QS QT y − y1 y 2 − y1 (or) = x − x1 x 2 − x1 y − y1 x − x1 = y 2 − y1 x 2 − x 1 x’ Hence the equation of the straight line passing through two given points is y − y1 x − x1 = . y 2 − y1 x 2 − x 1 Intercept form
R(x2,y2) P(x,y)
Q(x1,y1)
x
x’
Equation of a straight line which makes intercepts a and x y b on the co-ordinate axes is + = 1. a b Let l represents the given straight line which make intercepts a and b with x and y axes respectively. If it intersects x-axis at A and y-axis at B then we have OA = a and OB = b, so that A is (a,0) and B is (0,b). The two points form of the equation is given by
0
Fig.7.28 y’
y x 2, R( P(x Q (x
,y)
,y 1)
1
S
T
x
O
Fig.7.29
y’ y
B(0,b)
b
x’
a 0
y’
159
y 2)
Fig.7.30
) a,0 A(
x
y − y1 x − x1 = y 2 − y1 x 2 − x 1
Substituting (a,0) for (x1,y1) and (0,b) for (x2,y2) we get y−0 x−a y x−a = = b−0 0−a b −a y x a y x = − = − +1 b −a −a b a x y ∴ + =1 a b x y + = 1. a b Note : A general first degree equation in two variables always represents a straight line. Hence we can take general equation of a straight line as ax + by + c = 0 with atleast one of a or b different from 0. a c Further, this gives by = –ax – c or y = − x − b b Now comparing this with the equation y = mx + c we get slope = m = –a/b coefficient of x ∴m=– coefficient of y
Hence the equation of a line having x-intercept `a’ and y-intercept `b’ is given by
Example 17 : Find the equation of a straight line parallel to x-axis and passing through the point (–3,2). Solution : Equation of a line parallel to x-axis is given by y=b (1) It is given that the straight line (1) passes through (–3,2). ∴(–3,2) should satisfy the equation (1). Substituting y = 2 in (1) we get b = 2 ∴y = 2 Hence the required equation of the line is y – 2 = 0. Example 18 : Find the equation of a straight line parallel to y-axis and passing through the point (–7,5). Solution : Equation of a straight line parallel to y-axis is given by x = a (1) It is given that the straight line (1) passes through (–7,5). ∴(–7,5) should satisfy the equation (1) substituting x = –7 in (1), we get a = –7. Hence, x + 7 = 0 is the required straight line. Example 19 : Find the equation of a straight line whose inclination with x-axis 60o and passes through the origin. Solution : Slope = m = tan θ = tan 60o =
3.
Since the straight line passes through the origin its y-intercept is zero. ∴c = 0. Equation of the straight line in slope intercept form is y = mx + c
(or)
y=
3 x+0
(or)
160
y=
3 x.
Example 20 : Find the equation of a straight line whose inclination is 30o with the x-axis and whose y-intercept is –3. 1 . Solution : Slope of the given straight line is m = tan 30o = 3 The y-intercept is c = –3. The equation of the straight line in slope intercept form is 1 y = mx + c. ∴ The required equation is y = x – 3 (or) 3 y = x – 3 3 (or) 3 x – 3 y–3 3 = 0. Example 21 : Find the equation of a straight line whose slope is Solution : Slope m =
3 and y-intercept is – 4. 4
3 , c = –4 4
Equation of the straight line is y = mx + c, y =
∴The required equation is 3x – 4y – 16 = 0.
3 x–4 (or) 4y = 3x–16 (or) 3x – 4y – 16 = 0. 4
Example 22 : The equation of a straight line is 3x + 2y + 1 = 0 find its slope and y-intercept. 1 3 Solution : The equation is 3x + 2y + 1 = 0 (or) 2y = –3x – 1 (or) y = – x − 2 2 3 1 Comparing this with the equation y = mx + c, m= − , c=− . 2 2 3 1 ∴ Slope = − , y-intercept = − . 2 2 Example 23 : Find the equation of a straight line whose slope is 4 and which passes through the point (5,–7). Solution : Equation of a straight line with the slope 4 and passing through the point (5,–7) is y – (–7) = 4 (x–5) (x1,y1) = (5,–7) (or) y + 7 = 4x–20 m = 4 (or) 4x – y – 27 = 0 y – y1 = m (x–x1) in the slope – point form Example 24: Find the equation of a straight line parallel to the line joining the points (7,5) and (1,3) and passing through the point (–3,4). Solution : Slope of the straight line joining the points (7,5) and (1,3) is y − y1 3−5 −2 1 = = m= 2 m= x 2 − x1 1− 7 − 6 3 1 m= 3 1 If two straight lines are parallel then their slopes are equal. ∴Slope of the required line is 3 1 (x1,y1) = (–3,4) y–4 = (x + 3) 3 3y – 12 = x+3 y – y1 = m(x – x1) x – 3y + 15 = 0 Hence the required equation is x – 3y + 15 = 0.
161
Example 25 : Find the equation of the straight line passing through the point (3,2) and perpendicular to the straight line joining the points (4,5) and (1,2). Solution : Slope of the straight line joining the points (4,5) and (1,2) is ⎛ y 2 − y1 ⎞ 2−5 −3 = m= =1 ⎜Q m = ⎟ x 2 − x1 ⎠ 1− 4 − 3 ⎝ m=1 If the two straight lines are perpendicular m1 × m2 = –1
Slope of the perpendicular straight line is = –1/m = –1/1 = –1. ∴Slope of the required straight line is –1. ∴ The required equation is y – 2 = –1 (x – 3) y – y1 = m(x – x1) (or) y – 2 = –x + 3 (or) x + y – 5 = 0 The required equation of the line is x + y – 5 = 0. Example 26 : Find the equation of the straight line through (1,3) and parallel to a straight line 5x – 3y + 1 = 0. ⎛ −coefficient of x ⎞ Solution : Equation of the given straight line is 5x – 3y + 1 = 0 ⎜Q m = ⎟ coefficient of y ⎠ ⎝ ⇒ Slope of the line = –5/–3 = 5/3 The straight line passes through (1,3). ⇒ The equation of the straight line with slope 5/3 and passing through the point (1,3) is y – 3 = 5/3 (x – 1) (x1,y1) = (1,3) 3y – 9 = 5x – 5 y – y1 = m(x – x1) 5x – 3y + 4 = 0 Example 27 : Find the equation of a straight line through (0,7) and perpendicular to a straight line 5x + 4y + 11 = 0. Solution : Equation of the given straight line is 5x + 4y + 11 = 0. Slope of the given straight line = –5/4; −coefficient of x m= ∴ Slope of a perpendicular line = 4/5 coefficient of y The straight line passes through (0,7) If two lines are perpendicular ∴Equation of the line is y – 7 = 4/5 (x – 0) m1m2 = –1 5y – 35 = 4x (x1,y1) = (0,7) 4x – 5y + 35 = 0 y–y1 = m(x–x1) Example 28 : The vertices of a triangle are (1,2), (–3,4) and (5,–3). Find the equation of the altitude from (1,2) to the opposite side. Solution : Let the vertices be A(1,2), B(–3,4) and C(5,–3). ⎛ y 2 − y1 ⎞ −3−4 7 Slope of BC = =− ⎜Q m = ⎟ x 2 − x1 ⎠ 5+3 8 ⎝ If two lines are perpendicular m1m2 = –1.
Since the altitude through A is perpendicular to BC, Slope of the altitude through A = 8/7. 162
Altitude passes through the point A(1,2). ⇒ The equation of the straight line is y – 2 = 8/7 (x–1) 7y – 14 = 8x – 8 (or) 8x – 7y + 6 = 0 ( y – y1 = m(x – x1)) Example 29 : Find the equation of the perpendicular bisector of the line joining the points A(1,7) and B(–3,3). ⎡ x + x 2 y1 + y 2 ⎤ ⎡− 3 +1 3 + 7⎤ Solution : Mid point of AB = ⎢ Mid point = ⎢ 1 , = (−1,5) , ⎥ 2 ⎦ 2 ⎥⎦ ⎣ 2 ⎣ 2 y − y1 m= 2 x 2 − x1 3−7 −4 Slope of AB = = =1 − 3 −1 − 4 ∴ Slope of a perpendicular line = –1
Now equation of the perpendicular bisector passing through the mid point (–1,5) with slope –1 is y – 5 = –1 (x+1) m = –1 y – 5 = –x – 1 (x1,y1) = (–1,5) x+y–4=0 y – y1 = m(x – x1) Example 30 : Find the equation of the straight line which joins the points A(5,1) and B(–2,2). Solution : The given points are A(5,1) and B(–2,2). ∴The equation of the straight line joining A and B is y − y1 x − x1 y −1 x −5 = = 2 −1 − 2 − 5 y 2 − y1 x 2 − x 1 y −1 x − 5 (or) –7y + 7 = x – 5 = +1 −7 A(-2,8) ∴The required equation is x + 7y – 12 = 0.
Example 31 : The vertices of a triangle ABC are A(–2,8), B(1,2) and C(7,–8). Find the equation of the median through A. Solution : Given that the vertices of the triangle ABC are A(–2,8), B(1,2) and C(7, –8). Let D be the mid point of BC.
B(1,2)
D
Fig.7.31
⎡ x + x 2 y1 + y 2 ⎤ Mid point = ⎢ 1 , 2 ⎥⎦ ⎣ 2
⎡1 + 7 2 − 8 ⎤ , = (4, –3) ∴ Mid point of BC = ⎢ 2 ⎥⎦ ⎣ 2
The median AD passes through A (–2,8) and D(4,–3). ∴ The equation of AD is 163
C(7,-8)
y − y1 x − x1 = y 2 − y1 x 2 − x 1
y−8 x+2 = −3−8 4+ 2 y −8 x + 2 = 6 − 11 6y – 48 = –11x – 22 11x + 6y – 26 = 0
(x1, y1) = (− 2, 8) (x2, y2) = (4, –3)
Example 32 : Show that the points (4,2), (7,5) and (9,7) are collinear. Solution : The given points are A(4,2), B(7,5) and C(9,7). y − y1 x − x1 y−2 x−4 Equation of AB is = = 5−2 7−4 y 2 − y1 x 2 − x 1 y−2 x-4 = 3 3 ∴x–y–2=0
Substituting (9,7) in the equation (1) x – y – 2 = 0 (or) 9–7–2=0
(1)
(or) 0 = 0
∴ (9,7) satisfies the equation of AB. Hence C(9,7) lies on the straight line AB ⇒ The points are collinear. Example 33 : If the straight line 7x – 5y = k passes through the point (1,1) what is k? Solution : Equation of the given straight line is 7x – 5y = k (1) This straight line passes through (1,1). Substituting A(1,7) (1,1) in the equation (1) 7(1) – 5(1) = k 2 = k ∴k=2 Example 34 : Find the equation of the sides of the triangle ABC whose vertices are A(1,7), B(0,–2) and C(3,3). y−7 x −1 Solution : Equation of AB is = − 2 − 7 0 −1
y − 7 x −1 = −9 −1 or 9x – y – 2 = 0 or
B(0,-2)
or –y + 7 = –9x + 9
Equation of BC is
y+2 x−0 = 3+ 2 3−0
Equation of AC is
y − 7 x −1 (or) = 3 − 7 3 −1
(or)
y+2 x = 5 3
(or) 3y + 6 = 5x
D
Fig.7.32
(or) 5x – 3y – 6 = 0
y − 7 x −1 = −4 2
2y –14 = –4x + 4 (or) 4x + 2y – 18 = 0 (or) 2x + y – 9 = 0.
164
C(3,3)
Example 35 : The mid points of three sides of a triangle are (5,–3), (–5,3) and (6,6). Find the equation of the sides of the triangle. Solution : Let the triangle be ABC. Let D,E and F be the mid points of BC, CA and AB respectively. The co-ordinates of D, E and F are (5,–3), (–5,3) and (6,6) respectively. The line joining the mid points of two sides of a triangle is parallel to the third side ∴ BC || EF 6−3 3 ∴ Slope of BC = Slope of EF = = . 6 + 5 11
A
F(6,6)
B
E(-5,3)
D(5,-3)
C
Fig.7.33 D (5,–3) is a point on the line BC. ∴ Equation of BC with slope 3/11 and passing through the point D(5,–3) is 3 y+3= (x – 5) or 11y + 33 = 3x – 15 or 3x – 11y – 48 = 0 11 ∴Equation of BC is 3x – 11y – 48 = 0
Again, DF || CA ∴Slope of AC = slope of DF =
E(–5,3) is a point on AC. ∴Equation of AC is y – 3 = 9 (x + 5) (or) Lastly, DE || AB
6+3 9 = =9 6−5 1
y – 3 = 9x + 45.
∴9x –y + 48 = 0
3+3 6 −3 = = − 5 − 5 − 10 5 3 ∴Equation of AB is y – 6 = − (x – 6) (or) 5y – 30 = –3x + 18, ∴ 3x + 5y – 48 = 0 5 Example 36 : Show that the straight lines x – 2y = 0 and 2x + y + 1 = 0 are perpendicular to each other. −1 1 − coefficient of x m= Solution : Slope of the straight line x – 2y = 0, is m1 = = −2 2 coefficient of y −2 Slope of the line 2x + y + 1 = 0 is m2 = = –2. 1 1 Product of the slopes = m1 m2 = × (–2) = –1. 2 Hence the two straight lines are perpendicular to each other. Slope of AB = slope of DE
=
Example 37 : Is the straight line x = 2y parallel to 2x – 4y + 7 = 0? Solution : Slope of the straight line x – 2y = 0 is −1 1 m1 = = −2 2 Slope of the straight line 2x – 4y + 7 = 0 is
165
−2 1 1 = ⇒ m1 = m2 = 2 −4 2 Since the slopes are equal the two straight lines are parallel.
m2 =
Example 38 : Find the equation of the straight line whose intercepts on the axes are given by 1/3 and 2/5 respectively. Solution : a = x – intercept = 1/3, b = y – intercept = 2/5 x y x y ∴Equation of the straight line is given by + =1 Q + =1 1/3 2/5 a b 5y or 3x + = 1 ∴6x + 5y – 2 = 0 2 Example 39 : Find the intercepts made by the straight line 3x –2y – 6 = 0 on the axes of co–ordinates. Solution : Equation of the straight line is 3x – 2y = 6 3x 2y Dividing the equation by 6 (to get 1 in the R.H.S.), − =1 6 6 x y x y The equation can be rewritten as + = 1 and comparing with the equation + = 1 2 −3 a b ∴ we get x–intercept = 2, y – intercept = –3 Example 40 : Find the equation of the straight line passing through (6,5) with intercepts on the axes are equal in magnitude but opposite in sign. Solution : Let the intercepts on the axes be a and –a. ∴The required equation of the straight line in the intercept form is x y x – y = a ...... (1) + =1 a −a
This straight line passes through the point (6,5) ⇒ (6,5) should satisfy (1) ∴ 6–5 = a, a = 1 ∴From (1) the equation of the required straight line is x – y = 1 or x – y – 1 = 0 Example 41: Find the equation of the straight line passing through (–3,10) and whose sum of the intercepts is 8. x y Solution: Equation of the straight line in the intercept form is + = 1 (1) a b Given, sum of the intercepts = 8 ⇒ a + b = 8, b = 8 – a y x Substituting in (1) we get + =1 (2) a 8−a
This straight line passes through (–3,10) ⇒ (–3,10) should satisfy (2) −3 10 −3(8 − a) + 10a + = 1, =1 a 8−a a(8 − a) 166
–24 + 3a + 10a = 8a – a2 (a – 3) (a + 8) = 0
or or
a2 + 5a – 24 = 0 a = 3, a = – 8
when a = 3, b = 8 – a = 8 – 3 = 5, when a = –8, b = 8 – (–8) = 8 + 8 = 16 x y when a = 3 and b = 5 the equation of the straight line is + = 1 ⇒ 3 5 5x + 3y = 15 or 5x + 3y – 15 = 0 x y when a = –8 and b = +16, the equation of the straight line is + =1 − 8 16 − 2x + y = 1 , –2x + y = 16, 2x – y + 16 = 0 16 Example 42 : Find the equation of the straight line through the point (4,–5) and having x and y intercepts in the ratio 3:5. Solution : Let the intercepts on the axes be a and b It is given a : b = 3:5 ⇒ a = 3k and b = 5 k x y Now the equation becomes + = 1, 5x + 3y = 15k 3k 5k Substituting (4,–5) in the equation 5x + 3y = 15k we get 5(4) + 3(–5) = 15k or 20 – 15 = 15k or 5 = 15k ∴ k = 1/3 Hence the equation of the line is 5x + 3y = 15 (1/3); 5x + 3y – 5 = 0 y
Example 43: Find the equation of the straight line the portion of which between the axes is divided by the point (4,3) in the ratio 2:3. Solution : Let the equation of the straight line be x y + =1 a b (1) This straight line meets the x and y axes at A(a,0) and B(0,b) respectively. The straight line BA is divided by the point (4,3) internally in the ratio 3:2. 3× a + 2 × 0 20 ⇒4= ⇒ 20 = 3a ∴a = 3+ 2 3 3 ×0 + 2× b 15 ⇒3= ⇒ 15 = 2b ∴ b = 3+ 2 2 B
(0,b)B 3 (4,3) 2 x’
y’
Fig.7.34
2
3 (4.3)
(0,b)
Hence by (1) the required straight line is given by x y + = 1, (20/3) (15/2) 3x 2y (or) + = 1 or 9x + 8y = 60. 20 15 167
x
(a,0)A
0
Fig.7.35
A (a, 0)
Exercise 7.3 1. (i) Find the equation of a straight line parallel to x-axis and which passes through the point (2,–3). (ii) Find the equation of a straight line parallel to y-axis and which passes through the point (–3,5). 2. Find the equation of a straight line whose inclination is 60o and y-intercept – 3. 3. Find the equation of a straight line whose slope is 3 and y-intercept – 2/3. 4. The equation of a straight line is 2x–2 3 y – 3 = 0. Find (i) the slope of the straight line (ii) the inclination of the straight line. 5. Find the slope and y-intercept of the straight line. (i) 3x + 2y = 4 (ii) 2x = y 6. Find the equation of a straight line (i) Whose slope is –3 and which passes through the point (–2,3) (ii) Whose slope is –5/3 and which passes through the point (–3,5) (iii) Whose slope is 2/3 and which passes through the point (3,1) 7. Find the equation of the straight line parallel to the straight line joining the point. (i) (0,7) and (2,1) and which passes through (–9,5) (ii) (2,3) and (4,5) and which passes through (2,–4). 8. Find the equation of the straight line perpendicular to the straight line joining (i) (3,2) and (6,–4) and which passes through (–3,4) (ii) (0,7) and (2,1) and which passes through (–9,5). 9. Find the equation of the straight line parallel to the straight line given by the equation (i) 5x – 3y +1 = 0 and which passes through (2,1) (ii) 3x + 2y – 1 = 0 and which passes through (1,3). 10. Find the equation of the straight line perpendicular to the straight line given by the equation (i) x – 2y + 1 = 0 and which passes through (2,4) (ii) 4x – 3y + 2 = 0 and which passes through (–2,3). 11. (i) Find the equation of the altitude AD of triangle ABC where A,B and C are points (1,–3) (–2,5) and (–3,4) respectively. (ii) The vertices of a triangle are (1,3), (–2,4) and (3,–5). Find the equation of the altitude from (1,3) to the opposite side. 12. (i) Find the equation of the perpendicular bisector of PQ where Pis (5,–6) and Q is (5,–4). (ii) Find the equation of the perpendicular bisector of the straight line joining (5,6) and (2,–2). 13. Find the equation of the straight line which joins the points. (i) A(–1,3) and B(4,–2) (ii) (3,6) and (–2,1) (iii) (0,0) and (3,2) (iv) (0,8) and (3,1) 14. (i) ABC is a triangle with vertices A(3,2), B(0,4) and C(6,2). Find the equation of the median through A. (ii) Find the equation of the median through R if P(0,–6), Q(–8,2) and R(8,5) are the vertices of the triangle PQR. (iii)Find the equation of the median from A in triangle ABC with vertices A(5,6), B(–11,2) and C(6,–11). 15. Show that the following points are collinear. (i) (3,–2), (1,4) and (–3,16) (ii) (2,–3), (4,1) and (1,–5).
168
16. Check whether (i) (ii) (iii) (iv)
(2,0) lies on the straight line 4x – 3y – 8 = 0 (1,0) lies on the straight line x + 3y – 1 = 0 (1,2) lies on the straight line 2x – y + 3 = 0 (b,a) lies on the straight line ax + by – 2ab = 0
17. Find the value of k if the following point (i) (1,1) lies on the straight line 2x + ky + 1 = 0 (ii) (1,–2) lies on the straight line 3x + 12y + k = 0 (iii) (1,2) lies on the straight line x – ky = 5 18. If the vertices of a triangle ABC are A(1,2), B(3,5) and C(2,–5). Then find the equation of the sides. 19. (i) Find the equation of the sides of triangle ABC whose three sides BC, CA and AB have mid points at (2,1), (5,3) and (3,–4) respectively. (ii) Find the equation of the side BC of triangle ABC if the mid points of the sides BC, CA and AB are (–5,7), (–5,–5) and (2,1) respectively. 20. Find the equation of the straight line whose intercepts on the axes of co–ordinates are given below (i) 3,2 (ii) 4,–3
(iii) –2,3/4
21. Find the intercepts made by the following straight lines on the axes of co-ordinates (i) 4x + 3y + 12 = 0 (ii) 5x + y + 3 = 0 22. Find the equation of the straight line passing through (4,5) and making equal intercepts on the axes of coordinates. 23. Find the equation of the straight line passing through (3,–4) and making equal intercepts on the axes of coordinates. 24. Find the equation of the straight line passing through (5,2) and having its y-intercept twice as its x-intercept. 25. Find the equation of the straight line passing through (2,–1) and whose intercepts on the axes of coordinates are equal in magnitude but opposite in sign. 26. Find the equation of the straight line which makes intercepts of 2a on the x-axis and 3a on the y-axis, given that the straight line passes through the point (14,–9). 27. Find the equation of the straight line which passes through the point (3,4) and makes intercepts on the axes of coordinates such that their sum is 14. 28. Find the equation of the straight line passing through (1,4) and has intercepts which are in the ratio 3:5. 29. Show that the straight line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other. 30. Prove that the straight lines x + 2y + 1 = 0 and 2x – y + k = 0 are perpendicular for all values of k. 31. Show that the straight lines 3x + 4y+ 7 = 0 and 21x + 28y + 50 = 0 are parallel. 32. Find the value of k if the straight lines x + 2y +1 = 0 and 3x + ky + 5 = 0 are parallel.
169
7.4
SOME PROPERTIES OF STRAIGHT LINES
7.4.1
Intersection of two straight lines
If two straight lines are not parallel then they will meet at a point. This common point for both straight lines is called the point of intersection. If the equations of two intersecting straight lines are given, then their point of intersection is obtained by solving the equations simultaneously. Example 44 : Find the point of intersection of the straight lines 2x – 2y = 6 and x + y = 3 Solution: Let us solve the equation 2x – 3y = 6 (1) x+ y=3 (2) We have, 2x – 3y = 6 (1) (2) × 3 ⇒ 3x + 3y = 9 (3) (1) + (3) ⇒ 5x = 15 ∴ x = 3 Substituting x = 3 in equation (2) we get 3 + y = 3 or y = 0 Hence the point of intersection of the straight lines is (3,0). Example 45: Where does the straight line 6x – 3y – 30 = 0 meet (i) x – axis (ii) y – axis. Solution : Equation of the given straight line is 6x – 3y – 30 = 0 (1) i) Let the straight line meet the x-axis at A. Then the y-coordinate of A is y = 0 ∴ Substituting y = 0 in the equation (1) we get, 6x – 3(0) – 30 = 0 or 6x = 30 or x = 30/6 = 5 ⇒ The straight line meets the x-axis at A (5,0).
y
B
6x - 3y - 30 = 0
x
x’ A
O y’
ii) Let the straight line meets the y-axis at B. Fig.7.36 Then the x-coordinate of B is 0 Substituting x = 0 in the equation (1) we get, 6(0) –3y – 30 = 0, –3y = 30 or y = 30/–3 = –10 ⇒ The straight line meets the y-axis at B(0,–10) Example 46 : Find the equation of the straight line passing through the point of intersection of the straight lines. 2x + y = 3 and 3x – y = 7 and is parallel to 2x – y + 5 = 0. Solution : Solving the two equations of the intersecting straight lines, 2x + y = 3 (1) 3x – y = 7 (2) (1) + (2) 5x =10 x =2
Substituting the value of x = 2 in (1) we get 2(2) + y = 3, y = 3 – 4 or y = –1 The point of intersection of two straight lines is (2,–1). The required straight line passes through (2,–1) and parallel to the straight line 2x – y + 5 = 0. Slope of the line 2x – y + 5 = 0 is = –2/–1 = 2 170
⇒ Slope of the straight line parallel to 2x – y + 5 = 0 is 2.
Hence the equation of the straight line passing through (2,–1) with slope m = 2 is y – (–1) = 2 (x – 2) or y + 1 = 2x – 4 or 2x – y – 5 = 0 Hence the required equation of the straight line is 2x – y – 5 = 0. Example 47 : Find the equation of the straight line passing through the point of intersection of the straight lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and perpendicular to the straight line 4x – 2y = 3. Solution : To find the point of intersection of the straight lines, let us solve 5x – 8y + 23 = 0 (1) 7x + 6y – 71 = 0 (2) (1) x 6 ⇒ 30x – 48y + 138 = 0 (3) (2) x 8 ⇒ 56x + 48y – 568 = 0 (4) 430 (or) x = (3) + (4) ⇒ 86x – 430 = 0 (or) 86x = 430 =5 86 Substituting x = 5 in (1) 5(5) – 8y + 23 = 0, –8y = –48, y = –48/–8 = 6 ⇒ Point of intersection of the straight lines is (5,6). Slope of straight line 4x –2y = 3 is –4/–2 = 2 ⇒ Slope of the straight line perpendicular to the straight line 4x – 2y = 3 is –1/2 Hence the equation of the straight line passing through the point (5,6) with slope –1/2 is y – 6 = –1/2 (x – 5) or 2y – 12 = –x + 5 or x + 2y = 17 Hence the required equation of the straight line is x + 2y – 17 = 0 Example 48 : The straight lines x + y – 5 = 0 and 2x + 3y = 13 are diameters of a circle. Find the radius of the circle if the circle passes through the point (2,1). Solution : The centre of the circle is the point of intersection of the diameters. Hence solving x+y–5 =0 (1) 2x + 3y = 13 (2) (1) x 2 ⇒ 2x + 2y = 10 (3) (2) – (3) ⇒ y =3
Substituting y = 3 in (1) x + y – 5 = 0, we get x + 3 – 5 = 0 or x = 2. ⇒ The point of intersection is (2,3) (or) the centre of the circle is (2,3). Given that the point (2,1) lies on the circle. ⇒ Radius = Distance between the centre (2,3) and the point (2,1). Radius =
(2 − 2) 2 + (1 − 3) 2 =
02 + (−2) 2 =
4 units
Q d = (x 2 − x1 ) 2 + (y 2 − y1 ) 2
Radius = 2 units Example 49 : A point is collinear with the points (3,4) and (8,5). It also lies on the straight line 2x + y + 1 = 0. Find the coordinates of the point.
171
Solution : The required point is collinear with (3,4) and (8,5) and lies on the straight line 2x + y + 1 = 0. The equation of the straight line joining the points (3,4) and (8,5) is
y − y1 x − x1 y−4 x −3 y−4 x −3 or = = = 5−4 8−3 1 5 y 2 − y1 x 2 − x 1 5y – 20 = x – 3, x – 5y + 17 = 0 B(8,5) The required point is the point of intersection of x – 5y + 17 = 0 and 2x + y + 1 = 0. Hence solving x – 5y + 17 = 0 (1) 2x + y + 1 = 0 (2) (2) × 5 ⇒ 10x + 5y + 5 = 0 (3) (1) ⇒ x – 5y + 17 = 0 (1) + (3) ⇒ 11x + 22 = 0 11x = –22 or x = –22/11 = –2 Substituting the value of x in (1) x = –5y + 17 = 0 or –2 –5y + 17 = 0 or –5y + 15 = 0 A(3,4) or –5y = –15, y = 3. The required point is (–2,3). Fig.7.37 Example 50: Find the length of the straight line segment joining the point (3,1) and the point of intersection of the straight lines 2x – y + 5 = 0 and x + y + 1 = 0. Solution : Let the straight lines 2x – y + 5 = 0 and x + y + 1 = 0 P intersect at P. Let us solve the equations 2x – y + 5 = 0 (1) x+y+1=0 (2) (1) + (2) ⇒ 3x + 6=0 3x = – 6 (or) x = –6/3 or x = –2 substituting the value of x in (1) 2x – y + 5 = 0 A(3,1) ⇒ 2(–2) – y + 5 = 0, –4 – y + 5 = 0, –y + 1 = 0, –y = –1, y = 1 ⇒ The straight lines intersect at (–2,1). Fig.7.38 Let A(3,1) be the given point. The distance between the points A(3,1) and P(–2,1). ⇒
2x
+y
+
1=
5
+
-y
1=
0
2x
y+
=
0
0
x+
= (−2 − 3) 2 + (1 − 1) 2 = 25 = 5 units Hence the length of the required straight line segment is 5 units. Example 51 : Find the equation of the straight line joining the point (4,5) and the point of intersection of the straight lines 5x – 3y = 8 and 2x – 3y = 5. Solution : In order to get the point of intersection of the straight lines, let us solve : 5x – 3y = 8 (1) 2x – 3y = 5 (2) (1) – (2) ⇒ 3x = 3 (or) x = 1 Substituting the value of x in (1) 5x – 3y = 8, 5(1) – 3y = 8, –3y = 8 – 5 =3, y = 3/–3 = –1 ∴ y = –1 Hence the point of intersection of the straight lines (1) and (2) is (1, –1). Now we can find the equation of the straight line joining the points (1,–1) and (4,5) as follows.
172
y +1 x −1 or = 5 +1 4 −1 6x – 3y – 9 = 0 2x – y – 3 = 0
y +1 x −1 = 6 3
or
3y + 3 = 6x – 6 ( Dividing by 3)
Hence the required equation of the straight line is 2x – y – 3 = 0. Example 52: Find the equation of the straight line through the point of intersection of the straight lines x + y = 3 and 2x + y = 5 and bisecting the line segment joining the points (1,5) and (–5,1). Solution : In order to get the point of intersection of the straight lines, let us solve x+y =3 (1) 2x + y = 5 (2) Subtracting –x = –2 (or) x = 2 Substituting x = 2 in (1), x + y = 3 or 2 + y = 3 or y = 3 – 2 = 1. ⇒ The point of intersection of (1) and (2) is (2,1). The mid point of the straight line joining the points (1,5) and (–5,1) ⎡1 − 5 5 + 1⎤ = (–2,3) , =⎢ 2 ⎥⎦ ⎣ 2 The required equation of the straight line through the points (2,1) and (–2,3) y −1 x − 2 y −1 x − 2 or = = 3 − 1 −2 − 2 2 −4 –4y + 4 = 2x – 4 (or) 2x + 4y – 8 = 0 (or) x + 2y – 4 = 0 Hence the required equation is x + 2y – 4 = 0. Exercise 7.4.1 1. Find the point of intersection of the straight lines (i) 2x + 3y = 8, 2x – 3y = 4, (ii) 3x + 5y = 6, 5x – y = 10. 2. Where does the straight line 4x + 3y – 12 =0 meet the axis of coordinates? 3. Find the equation of the straight line passing through the point of intersection of the straight lines 2x –y + 5 = 0 and x + y + 1 = 0 and parallel to the straight line 3x – y + 1 = 0. 4. Find the coordinates of the vertices of a triangle the equations of whose sides are x + 4y = 9, 9x + 10y + 23 = 0 and 7x + 2y = 11. 5. Find the point of intersection of the straight lines 2x + y – 1 = 0 and x – 3y + 3 = 0 and also find the equation of the straight line parallel to x-axis which passes through the point of intersection of these lines. 6. Find the equation of the straight line parallel to y-axis and passing through the point of intersection of the straight lines x + y – 5 = 0 and 2x + 3y = 13. 7. Find the coordinates of the point of intersection of the following pairs of straight lines (i) x= –4, y = 0 (ii) y = 0, x = 3. 8. The straight lines x + y – 5 = 0 and 3x – y + 1 = 0 are the diameters of a circle. Find the radius of the circle if the point (0,1) lies on the circle.
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9. Find the equation of the straight line passing through the point of intersection of 4x–y–3=0 and x + y – 2 = 0 and perpendicular to 2x – 5y + 3 = 0. 10.A point is collinear with the points (7,5) and (1,1). It also lies on the straight line x – 3y + 2 = 0. Find the coordinates of the point. 11. Find the length of the straight line segment joining the point (3,4) and the point of intersection of the straight lines 2x + 5y – 25 = 0 and 5x + 4y – 20 = 0. 12. Find the equation of the straight line joining the point (2,3) and the point of intersection of the straight lines x + y – 5 = 0 and 3x – y + 1 = 0. 13. Find the equation of the straight line through the point of intersection of the straight lines 2x + y – 5 = 0 and x + y – 3 = 0 and bisecting the straight line segment joining the points (2,3) and (–4,1). 14.Find the length of the straight line segment which joins the point of intersection of the straight lines 2x + y – 3 = 0 and 5x +y – 6 = 0 and the mid point of the straight line joining the points (7,2) and (3,2). 15.Find the equation of the straight line passing through the point of intersection of the straight lines 2x + y – 3 = 0 and 5x + y – 6 = 0 and perpendicular to the line joining the points (1,2) and (2,1). 7.4.2
Concurrency of straight lines
If three or more straight lines passes through the same point then that common point is called the point of concurrency. Steps to find out whether the three given straight lines are concurrent i. Solve any two equations of the straight lines and obtain their point of intersection. ii. Substitute the co-ordinates of the point of intersection in the third equation. iii. Check whether the third equation is satisfied. iv. If it is satisfied, the point lies on the third line and so the three straight lines are concurrent. Example 53: Show that the straight lines 2x – 3y + 4 = 0, 9x + 5y = 19 and 2x – 7y + 12 = 0 are concurrent. Find the point of concurrency. Solution : The given equations are 2x – 3y + 4 = 0 (1) 9x + 5y = 19 (2) 2x – 7y + 12 = 0 (3) Let us solve equations (1) and (2) (1) x 5 ⇒ 10x – 15y + 20 = 0 (4) (2) x 3 ⇒ 27x + 15y – 57 = 0 (5) (4) + (5) ⇒ 37x – 37 = 0 37 x= =1 37 Substituting x = 1 in (1) 2x – 3y + 4 = 0 we get 2(1) – 3y + 4 = 0, or 3y = 6 6 y= =2 3 ∴The point of intersection of (1) and (2) is (1,2).
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Now substituting x = 1 and y = 2 in equation (3) 2x –7y + 12 = 0, we get 2(1) – 7(2) + 12 = 0 or 2 – 14 + 12 = 0 or –12 + 12 = 0, 0 = 0. ⇒ The third equation is satisfied. So the point (1,2) lies on the third straight line. Hence the three straight lines are concurrent. The point of concurrency is (1,2). Example 54 : Find the value of `a’ for which the straight lines 2x + y – 1 = 0, 2x + ay – 3 = 0 and 3x + 2y – 2 = 0 are concurrent. Solution: Let us solve the equations 2x + y – 1 = 0 (1) 3x + 2y – 2 = 0 (2) 2x + ay – 3 = 0 (3) (1) × 2 ⇒ 4x + 2y – 2 = 0 (4) 3x + 2y – 2 = 0 (2) (4) – (2) ⇒ x = 0
Substituting x = 0 in (1) 2x + y – 1 = 0, we get 2(0) + y – 1 = 0 or y = 1. ⇒ The point of intersection of (1) and (2) is (0,1). Since the three lines are concurrent, (0,1) lies on the straight line 2x + ay – 3 = 0 ⇒ 2(0) + a(1) – 3 = 0 or a – 3 = 0 or a = 3. Example 55 : A straight line is concurrent with the straight lines x + y – 5 = 0 and 3x –y + 1 = 0 and is parallel to 5x – y + 2 = 0. Find its equation. Solution : Let us find the point of intersection of the straight lines x+y–5 = 0 (1) 3x – y + 1 = 0 (2) (1) + (2) ⇒ 4x – 4 = 0 x= 1 Substituting x = 1 in equation (1) x + y – 5 = 0 we get 1 + y – 5 = 0 or y = 4. ⇒ The required point of intersection is (1,4). The required straight line is parallel to the straight line 5x – y + 2 = 0. Slope of 5x – y + 2 = 0 is –5/–1 = 5. ⇒ Slope of the required line = 5. Hence the equation of the line with slope 5 and passing through (1,4) is y – 4 = 5(x – 1) or y – 4 = 5x – 5 or 5x – y – 1 = 0 Example 56 : Find the equation of the straight line which is concurrent with the straight lines x – y – 2 = 0 and 3x + 4y + 15 = 0 and also concurrent with the straight lines x – 3y + 3 = 0 and 2x + y = 8. Solution : We can get the point of intersection by solving the equations. x–y–2 = 0 (1) 3x + 4y + 15 = 0 (2) (1) × 4 ⇒ 4x – 4y – 8 = 0 (3) 3x + 4y + 15 = 0 (2) (3) + (2) ⇒ 7x + 7 = 0 x = –1 Substituting x = –1 in x – y – 2 = 0, we get –1 –y – 2 = 0 or –y = 3 or y = –3. ⇒ The point of intersection is (–1, –3). Again we can get the point of intersection by solving the equations x – 3y + 3 = 0 (4)
175
2x + y – 8 = 0 x – 3y + 3 = 0 (5) × 3 ⇒ 6x + 3y – 24 = 0 (4) + (6) ⇒ 7x – 21 = 0 x = 3 Substituting the value of x in x – 3y + 3 = 0, we get 3 – 3y + 3 = 0 or –3y + 6 = 0 or y = 2. ⇒ The point of intersection is (3,2).
(5) (4) (6)
Now to find the equation of a straight line passing through the points (–1, –3) and (3,2). y + 3 x +1 y + 3 x +1 , = = 2 + 3 3 +1 5 4 4y + 12 = 5x + 5 or 5x – 4y – 12 + 5 = 0 5x – 4y – 7 = 0 ∴ The required straight line is 5x – 4y – 7 = 0 Exercise 7.4.2
1. Show that the following set of lines are concurrent. Find their point of concurrency (i) x + y = 7; 2x + y = 16; 3x + 8y = 11 (ii) x + y – 3 = 0; x + 2y – 5 = 0 and x + 3y – 7 = 0 2. Find the value of m for which the lines are concurrent (i) 3x + y + 2 = 0; 2x – y + 3 = 0; x + my – 3 = 0 (ii) 3x – 4y + 5 = 0; 7x – 8y + 5 = 0 and 4x + my – 45 = 0 3. Obtain the equation of the line which passes through the origin and is concurrent with the lines x – y – 4 = 0 and 7x + y + 20 = 0. 4. Find the equation of the line which is concurrent with the lines x + y – 3 = 0 and 3x + 2y + 1 = 0 and also concurrent with the lines y – x = 1 and 2x + y + 2 = 0. 5. Find the equation of the line which passes through the intersection of the lines x+ y – 2 = 0, 2x + y – 3 = 0 and bisects the line joining the points (4,2) and (–6,4). 6. Find the equation of the line which is concurrent with the lines 9x + 4y = 1 and 2x – y = 4 and perpendicular to 3x – y + 7 = 0 7. Obtain the equation of the line which is concurrent with the lines x – y – 2 = 0 and 3x + 4y + 15 = 0 and is perpendicular to the line joining the points (2,3) and (1,1). 7.4.3 Incentre, circumcentre and orthocentre of a triangle Incentre : The internal bisectors of the three vertical angle of a triangle are concurrent. This point of concurrency is called the incentre of the triangle. The incentre is deonoted by I. To find the coordinates of the incentre of the triangle formed by the points (x1,y1), (x2,y2) and (x3, y3) : Let ABC be a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3). Let AD, BE and CF be the internal bisectors of the angles of the triangle ΔABC. The incentre I of ΔABC is the point of intersection of AD, BE and CF.
176
A(x1,y1) x x
F
E
y y B(x2,y2)
z z D
Fig.7.39
C(x1,y1)
If AB = c, BC = a and CA = b, by angle bisector theorem,
BD AB c = = DC AC b
(1)
⇒ D divides BC internally in the ratio c : b. DC b ⎛ cx + bx 2 cy3 + by 2 ⎞ Hence D is ⎜ 3 from (1) , = ⎟ c+b ⎠ BD c ⎝ c+b BC BD + DC c + b = = BD BD c cBC ca ⇒ BD = = b+c b+c b+c AI AB c Again, from triangle ABD, = = = ca a ID BD b+c ⇒ I divides AD in the ratio b + c : a ⎡ cx + bx 2 ⎤ + ax 1 (b + c) ⎢ 3 ax + bx 2 + cx 3 b + c ⎥⎦ ⎣ = 1 ⇒ x co-ordinates of I is b+c+a a+b+c ay + by 2 + cy 3 Similarly the y co-ordinates of I is 1 a+b+c ⎡ ax + bx 2 + cx 3 ay1 + by 2 + cy 3 ⎤ , Hence the incentre I of a triangle is given by ⎢ 1 ⎥ a+b+c a+b+c ⎣ ⎦ Example 57 : Find the coordinates of the incentre of the triangle whose vertices are A(1,1), B(2,1) and C(2,2). Solution : The vertices of the triangle are A(1,1), B(2,1) and C(2,2). a = BC =
(2 − 2) 2 + (2 − 1) 2 =
b = CA =
(1 − 2) 2 + (1 − 2) 2 =
c = AB =
(2 − 1) + (1 − 1) 2
2
0 + 12 = 1
(−1) 2 + (−1) 2 = 1 + 1 =
2
= 1 +0 = 1=1 2
⎡ ax + bx 2 + cx 3 ay1 + by 2 + cy 3 ⎤ Incentre I, of the triangle is given by ⎢ 1 , ⎥ a+b+c a+b+c ⎣ ⎦ ⎡1(1) + 2 (2) + 1(2) 1(1) + 2 (1) + 1(2) ⎤ , I is ⎢ ⎥ 1+ 2 +1 1+ 2 +1 ⎣ ⎦ ⎡3 + 2 2 3 + 2 ⎤ ∴ Incentre I is ⎢ , ⎥ 2 2 + 2+ 2⎦ ⎣ Circumcentre: The point of concurrency of the perpendicular bisectors of the sides of a triangle is called the circumcentre of the triangle. The circumcentre is denoted by S. Let ABC be the given triangle and D, E and F are the mid points of BC, CA and AB respectively. Find the slopes of the perpendicular bisectors of BC, CA and AB. Then find the equation of the perpendicular bisectors. By solving any two equations of the perpendicular bisectors we can get the circumcentre. Example 58 : Find the co-ordinates of the circumcentre of a triangle whose vertices are (2,–3), (8,–2) and (8,6).
177
Solution : Let A(2,–3), B(8,–2) and C(8,6) be the vertices of the triangle ABC. Let S be the circumcentre which is the point of intersection of the perpendicular bisectors of the sides AB, BC. Let D and E are the mid points of AB and BC respectively. ⎡ 2 + 8 − 3 − 2⎤ ⎡ − 5⎤ The co-ordinates of D = ⎢ = 5, , 2 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎣ 2 −2+3 1 = Slope of AB = A(2,-3) 8−2 6 Slope of the perpendicular bisector of AB = Slope of DS = –6 Equation of the perpendicular bisector through D D is y + 5/2 = –6 (x–5) or y + 5/2 = –6x + 30 S or 6x + y – 55/2 = 0 or 12x + 2y – 55 = 0 (1) The co-ordinates of E = ⎡ 8 + 8 − 2 + 6 ⎤ ⎡16 4 ⎤ ⎢⎣ 2 , 2 ⎥⎦ = ⎢⎣ 2 , 2 ⎥⎦ = (8,2) E C(8,6) B(8,-2) 6+2 8 Slope of BC = = =∞ Fig.7.40 8−8 0 −1 =0 Slope of the perpendicular bisector through E = ∞ Equation of the perpendicular bisector SE is y – 2 = 0 (x – 8), y – 2 = 0, y = 2 (2) By solving the equations (1) and (2) 12x + 2y – 55 = 0, y = 2 we get 12x + 4 – 55 = 0, or 12x – 51 = 0 or 12x = 51 x = 51/12 = 17/4 ⎡17 ⎤ Hence the circumcentre S is ⎢ ,2⎥ ⎣4 ⎦ Orthocentre: It can be shown that the altitudes of a triangle are concurrent and the point of concurrence is called the orthocentre of the triangle. The orthocentre is denoted by O. Let ABC be the given triangle. AD, BE and CF are three altitudes from A,B and C to BC, CA and AB respectively. Find the slopes of altitudes of AD, BE and CF. Now find the equation of AD, BE and CF by using slope point form. By solving any two altitudes we can get the orthocentre. Example 59 : Find the coordinates of the orthocentre of the triangle whose vertices are (3,1), (0,4) and (–3,1). A(3,1) Solution : Let the vertices of the triangle be A(3,1), B(0,4) and C(–3,1). Let AD and BE be the altitudes. 1− 4 −3 Slope of BC = = =1 −3−0 −3 E ⇒ Slope of the altitude AD = –1 Equation of altitude AD is y – 1 = –1 (x – 3) O or y – 1 = –x + 3 x+y–4=0 (1) D 1−1 0 C(-3,1) B(0,4) = =0 Slope of AC = Fig.7.41 −3−3 −6
178
⇒ Slope of the altitude BE = –1/0 = ∞ ⇒ Equation of BE is y – 4 = –1/0 (x – 0) or 0 = –x + 0 or x = 0 (2) Solving the equations (1) and (2) x + y – 4 = 0 and x = 0, we get y – 4 = 0 or y = 4 Hence the orthocentre is (0,4) Example 60: Obtain the coordinates of the orthocentre of the triangle whose vertices are the points (1,–2), (3,1) and (–2,3). 3 −1 +2 2 Solution : BC is the straight line joining (3,1) and (–2,3). Its slope = = =− −2−3 −5 5 If AD is drawn perpendicular to BC then its slope is 5/2. ⇒ Equation of AD is y + 2 = 5/2 (x – 1) A(1,-2) or 5x – 2y = 9 (1) AC is the line joining the points (–2,3) and (1,–2). −2−3 5 =− Its slope = 1+ 2 3 E If BE is drawn perpendicular to AC then its slope is 3/5 ⇒ Equation of BE is y – 1 = 3/5 (x – 3) i.e. 3x – 5y = 4 (2) The orthocentre is the point of intersection of D C(-2,3) B(3,1) altitudes AD and BE. (1) x 3 ⇒ 15x – 6y = 27 (3) Fig.7.42 (2) x 5 ⇒ 15x –25y = 20 (4) 19y = 7 y = 7/19 Substituting y = 7/19 in (1) wet get 5x – 2 (7/19) = 9, x = 37/19. ⎡ 37 7 ⎤ The orthocentre is ⎢ , ⎥ . ⎣ 19 19 ⎦ Exercise 7.4.3 1. Find the coordinates of the incentre of the triangle whose vertices are (i) (3,1), (0,4) and (–3,1) (ii) (–36,7), (20,7) and (0,–8) 2. Find the coordinates of the circumcentre of the triangle whose vertices are (i) (3,1), (2,2) and (2,0) (ii) (0,0), (–4,0) and (0,4) 3. Find the coordinates of the orthocentre of the triangle whose vertices are (i) (1,2), (2,3) and (4,3) (ii) (0,1), (1,–2) and (2,–3)
ANSWERS Exercise 7.1
⎡ 5a − b 5a + b ⎤ , (iii) (–13/3, 13) (iv) (11,8) (v) (13,–24) (vi) ⎢ 5 ⎥⎦ ⎣ 5 (2) (i) 1:2 internally (ii) 2:3 internally (iii) 2:3 externally (3) (i) 2:5 externally (ii) 1:2 internally (iii) 2:1 externally (iv) 4:1 internally (4) (i) 1:1 internally (ii) 2:3 internally (iii) 1:2 internally (iv) 1:5 internally (1) (i) (1,6)
(ii) (15/2, 5)
179
3 13 (7) (2,–2) (8) (–1,–9) 2 (9) (i) parallelogram (ii) parallelogram (iii) parallelogram (iv) parallelogram (10) (i) (–3,0) (ii) (1,2) (iii) (9,–6) (iv) (–2,7) (11) (i) (–3,19/3) (ii) (1,–2) (iii) (1/3,2) (iv) (5,–2) (12) (i) (1,–8) (ii) (7,–19) (iii) (6,2) (iv) (18,–2) (5) (i) (–3,–5)
(ii) (11/2, –3)
(6)
Exercise 7.2
(1) (i) 1 square units (iv) 10 square units (6) (i) 96 square units Exercise 7.3
(ii) 7.5 square units (v) 9 square units (ii) 43 square units
(iii) 9.5 square units (3) a – 2b + 8 = 0 (4) 9 (iii) 17 square units (iv) 41/2 square units.
(1) (i) y + 3 = 0 (ii) x + 3 = 0 (2) y = 3 x – 3 (3) 3 3 x – 3y – 2 = 0 o (4) m = 1/ 3 , θ = 30 (5) (i) m = –3/2, c = 2 (ii) m = 2, c = 0 (6) (i) y + 3x + 3 = 0 (ii) 3y + 5x = 0 (iii) 2x – 3y – 3 = 0 (7) (i) 3x + y + 22 = 0 (ii) x + 3y + 10 = 0 (8) (i) x – 2y + 11 = 0 (ii) x – 3y + 24 = 0 (9) (i) 5x – 3y – 7 = 0 (ii) 3x + 2y – 9 = 0 (10) (i) 2x + y – 8 = 0 (ii) 3x + 4y – 6 = 0 (11) (i) x + y + 2 = 0 (ii) 5x – 9y + 22 = 0 (12) (i) y = –5 (ii) 6x + 16y – 53 = 0 (13) (i) x + y – 2 = 0 (ii) x – y + 3 = 0 (iii) 2x – 3y = 0 (iv) 7x + 3y – 24 = 0 (14) (i) x = 3 (ii) 7x – 12y + 4 = 0 (iii) 7x – 5y – 5 = 0 (16) (i) Yes (ii) Yes (iii) No (iv) Yes (17) (i) k = –3 (ii) k = 21 (iii) k = –2 (18) 7x + y – 9 = 0; 10x –y – 25 = 0; 3x –2y + 1 = 0 (19) (i) 2x – 3y – 18 = 0, 7x – 2y – 12 = 0, 5x + y – 28 = 0 (ii) 6x – 7y + 79 = 0 (20) (i) 2x + 3y – 6 = 0 (ii) 3x – 4y – 12 = 0 (iii) 3x – 8y + 6 = 0 (21) (i) (–3,–4) (ii) (–3/5, –3) (22) x + y – 9 = 0 (23) x + y + 1 = 0 (24) 2x + y – 12 = 0 (25) x – y – 3 = 0 (26) 3x + 2y – 24 = 0 (27) 4x + 3y = 24 and x + y – 7 = 0 (28) 5x + 3y – 17 = 0 (32) k = 6 Exercise 7.4.1 (1) (i) (3,2/3) (ii) (2,0) (4) (–7,4), (3,–5), (1,2) (9) 5x + 2y – 7 = 0 (13) x + 3y – 5 = 0 Exercise 7.4.2 (1) (i) (9,–2) (ii) (1,2) (5) x + y – 2 = 0 Exercise 7.4.3
(2) x-axis at (3,0) and y-axis at (0,4) (3) 3x – y + 7 = 0 (5) (0,1), y = 1 (6) x = 2 (7) (i) (–4,0) (ii) (3,0), (8) 10 (10) (1,1) (11) 10 (12) x + y – 5 = 0 (14) 17 (15) x – y = 0 (2) (i) 4(ii) 5 (3) 3x – y = 0 (4) 5x + 3y + 5 = 0 (6) x + 3y + 5 = 0 (7) x + 2y + 7 = 0
⎡ 2 + 4⎤ (1) (i) ⎢0, ⎥ (ii) (–1,0) 2 +1⎦ ⎣
(2) (i) (2,1) (ii) (–2,2) (3) (i) (1,6) (ii) (–7,–6)
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8. TRIGONOMETRY 8.0
INTRODUCTION
Indian Mathematicians have shown keen interest in Trigonometry. The famous Indian Mathematician astronomer Aryabhata's work entitled Aryabhatiya (499 A.D) contains a part named Ganitabhaga devoted to Mathematics. This part contains a knowledge of the trigonometric ratio of sine (jya). Greek Mathematician Ptolmey, Father of trigonometry proved the equation sin2A+cos2A=1 using geometry involving a relationship between the chords of a circle. But ancient Indians used simple algebra to calculate sin A and cos A and proved this relation. Brahmagupta was the first to use algebra in trigonometry. Bhaskaracharya II (114 A.D) was very brilliant and most popular Mathematician. His work known as Siddhantasironmani is divided into four parts, one of which is Goladhyaya (spherical trigonometry). The genius of the 20th century Srinivasa Ramanujan at his age 12, borrowed the book Trigonometry (Part II) by Loney from his friend studying at Government College, Kumbakonam. Ramanujan had not only finished mastering this book at one reading but he had taught himself how to do every problem in it as mental sums. It is mention worthy that this book on trigonometry has some of the advanced topics of mathematics in it. This book Ramanujan's first contact led him towards learning easily advanced topics of mathematics with his intuitive, superactive and creative brain! But the first trigger - book for Ramanujan was Carr's synopsis of elementary results in pure and applied mathematics. In IX standard we have studied basic trigonometry such as angles, trigonometrical ratios and trigonometrical identities. Further we found the values of trigonometric ratios of angles 0°, 30°, 45°, 60°, 90°. We recall the tabulation of those values. θ
0°
30°
45°
sin θ
0
1
1
cos θ
1
tan θ
0
cot θ
∞
3
sec θ
1
2 3
2
cosec θ
∞
2
2
2
2
1
3
2
2 1
1
3
1
60° 3 2 1
2
3 1
3
90° 1 0 ∞ 0
2
∞
2
1
3
From the above table we observe that when values of θ are increasing their corresponding trigonometric values sin θ, tan θ, sec θ are also increasing whereas those of cos θ, cot θ, cosec θ are decreasing. 181
In this chapter we are going to learn in the section 8.1 how to solve problems by using the trigonometric tables and in section 8.2. We are going to solve a few problems on heights and distances.
8.1
USE OF TRIGONOMETRIC TABLES
We have seen with the help of geometry how to calculate the trigonometric ratios of certain angles. We computed trigonometric ratios for angular measures 30°, 45°, 60° and 90°. Mathematicians have found these values for all angular measures ranging from 0° to 90° and formed tables (see Appendix). Before we see how to use them let us define a few terms. A degree is subdivided into 60 equal parts. Each part is called a minute and it is denoted by 1′. Again a minute is subdivided into 60 equal parts. Each part is called a second and it is denoted by 1″. Then we have 60 seconds = 1 minute; 60 minutes = 1 degree using symbols for seconds, minutes and degrees we write 60'' = 1' ; 60' = 1° Following is the method given in detail to find the values of trigonometric ratios for any values of θ from 0° to 90°. The values of sin θ, cos θ, tan θ can be easily taken from the table. Since 0.1°=6' the table increases only by 6'. If the given angle is not given as a multiple of 6', then we have to split the given angle the following way. For example 38° 15' = 38°+12'+3' Example 1: Find from the table sin 40° 38′ Solution: Since sine value increases as the degree measure increases from 0° to 90°. 40° 38′ = 40° 36′ + 2′ sin 40° 36′ = 0.6508 Difference of 2′ = 0.0004 sin 40° 38′ = 0.6508 + 0.0004 = 0.6512 Example 2: Find from the table cos 63° 29′ Solution: Since cosine value decreases as the degree measure increases from 0° to 90° 63° 29′ = 63° 24′ + 5′ cos 63° 24′ = 0.4478 Difference of 5′ = 0.0013 cos 63° 29′ = 0.4478 – 0.0013 = 0.4465 Example 3: Find from the table tan 25° 15′. Solution: 25° 15′ = 25° 12′ + 3′ tan 25° 12′ = 0.4706 Difference of 3′ = 0.0011 tan 25° 15′ = 0.4706 + 0.0011 = 0.4717 182
Example 4: Find the value of θ i) sin θ = 0.9409 ii) cos θ = 0.8131 iii) tan θ = 2.9714 iv) sin θ = 0.0987 v) tan θ = 1.091 Solution: i) From the table of sines we find 70°12′ against 0.9409 ∴ sin 70° 12′ = 0.9409, θ = 70° 12′ ii) From the table of cosines we find 35°36′ against 0.8131 ∴ cos 35° 36′ = 0.8131, θ = 35° 36′ iii) From the table of tangents we find 71°24′ against 2.9714 ∴ tan 71° 24′ = 2.9714, θ = 71° 24′ iv) From the table of sines we find 5° 42′ against 0.0993 and 2′ against 0.0006 0.0987 = 0.0993 – 0.0006 = 0.0987 That is 5° 40′ against 0.0987 ∴ sin 5° 40′ = 0.0987, θ = 5° 40′ v) From the table of tangent we find 1.0913 is nearer to 1.091 that is we find 47° 30′ against 1.091 ∴ tan 47° 30′ = 1.091, θ = 47° 30′ Example 5: Use trigonometric tables to find i) sin 64° 42′ + cos 42° 20′ ii) tan 36° 40' + cot 63° 20' Solution: From the table we find the values of the following i) sin 64° 42′ + cos 42° 20′ = 0.9041 + 0.7392 = 1.6433 ii) tan 36° 40′ + cot 63° 20′ = tan 36° 40′ + cot (90–26° 40′) = tan 36° 40′ + tan 26° 40′ = 0.7445 + 0.5022 = 1.2467
5
Example 6: Find the length of the chord of a circle of radius 5 cm subtending at the centre the angle of 144°. Solution: Let AB be a chord of a circle with centre at 0 of radius 5 cm. Draw OC⊥AB. Then C is the mid point of AB. and ∠AOB = 144° ⇒ ∠COB = 72° O In right angled triangle OCB cm 72 5 BC = sin 72° A B C OB BC = 5 sin 72° cm Fig.8.1 = 5 × 0.9511 = 4.7555 cm ∴ Length of chord AB = 2 × BC = 2 × 4.7555 cm = 9.5110 cm
8
A
cm
cm
183
O 8
Example 7: Find the length of a side of a regular polygon of 25 sides inscribed in a circle of radius 8 cm. Solution: Let AB be a side of a regular polygon with 25 sides. From the centre O of the circle draw OC ⊥AB, then C is the mid point of AB and 360° ∠AOB = 25
cm
o
C
Fig.8.2
B
360° 36° 1 ∠AOB = = = 7° 12′ 2 50 5 In right angled triangle OCB BC sin 7° 12′ = OB BC = 8 × sin 7° 12′ cm = 8 × 0.1253 cm = 1.0024 cm Length of side AB = 2 × BC = 2 ×1.0024 cm = 2.0048 cm
∴∠COB =
Example 8: Find the radius of the incircle of a regular hexagon each side of length 6 cm. Solution: Let AB be the side of the regular hexagon. From the centre O of the inscribed circle, 1 1 draw OM⊥AB. Then M is the mid point of AB and ∠MOB = ∠AOB = × 60° = 30°. 2 2 Let r be the radius of the circle. Then OM = r. 1 1 Also MB = AB = × 6 cm = 3 cm 2 2 In right angled triangle OMB. O MB 3 = tan 30° = 30 OM r n r 3 1 = r B 3 A M o
r
= 3 3 cm = 3 × 1.732 cm = 5.196 cm
Fig.8.3
Example 9: Find the area of an isosceles triangle with base 16 cm and vertical angle 57°. Solution: Let ABC be an isosceles triangle with AB = AC. BC = 16 cm, and ∠A = 57o Draw AD ⊥ BC. Then AD bisects BC so that BD = DC = 8 cm. Also AD bisects the vertical angle ∠A. A 1 o o ∴∠BAD = ∠DAC = × 57 = 28 30' 28 30' 2 In right angled triangle ADC h AD o cot 28 30' = 8 cm DC 8 cm o o B C D AD = 8 × cot 28 30' cm = 8 tan 61 30′ b = 16 cm h = 8 × 1.842 cm = 14.736 cm 1 bh Area of triangle ABC = 2 1 = × 16 × 14.736 cm2 Fig.8.4 2 = 8 × 14.736 cm2 = 117.888 cm2 o
Example 10 : Find the area of a right angled triangle with hypotenuse 10 cm and one of the acute angle is 66o48'
184
Solution: Let ABC be the right angled triangle with ∠B = 90o, ∠C = 66o48' and AC = 10 cm BC = cos66 o 48' AC BC = 10 × 0.3939 cm = 3.939 cm AB = sin66 o 48' AC AB = 10 × 0.9191 cm = 9.191 cm 1 Area of the right angled triangle = × BC × AB 2 1 × 3.939 × 9.191 cm2 = 1.9695 × 9.191 cm2 = 2 = 18.1016745 cm2 Area of the triangle = 18.10 cm2 (approximately)
A
10 cm
o
66 48' B
Fig.8.5
C
Exercise 8.1 1.
9.
Find the values from the tables (a) sin 22o (b) cos 35o (c) tan 56o (d) sin 36o24' (e) cos 24o48' (f) tan 62o12' (g) sin 48o56' (h) sin 52o17' (i) cos 46o34' (j) tan 37o45' Find the value of θ (a) sin θ = 0.1736 (b) cos θ = 0.9063 (c) tan θ = 6.3188 (d) sin θ = 0.8221 (e) cos θ = 0.4115 (f) tan θ = 1.5697 Use trigonometric tables to find (a) sin 29o20' + cos 57o40' (b) sin 44°36' + tan 49o40' (c) cot 48o42' + tan 70o20' Find the length of a side of a regular polygon inscribed in a circle of radius 6 cm if it has 24 sides. Find the area of an isosceles triangle with base 20 cm and vertical angle 48o40'. Find the radius of the in circle of a regular polygon of 36 sides, each of length 10 cm. Find the area of a right angled triangle with hypotenuse 10 cm and one of the acute angle is 66o33'. Find the length of each side of a regular polygon of 12 sides is 20 cm. Find the radius of its circumscribing circle. If sin θ = cos θ where θ is an acute angle. Find the value of 2tan2θ – sin2θ – 1.
8.2
HEIGHTS AND DISTANCES
2.
3. 4. 5. 6. 7. 8.
Trigonometric ratios are very useful in solving problems on heights and distances. Many times we are required to find the distance between two objects and to find the height of a building, tree, tower, distance of a ship from a light house, width of a river etc. Though we can not measure them easily we can determine those by using trigonometric ratios. The problem of finding heights and distances is solved by observations of angles subtended by those objects at the eye of the observer. 185
Suppose we wish to determine the height of a tower without actually measuring it. We could stand on the ground at a point P at some distance say 8 m from the foot Q of the tower. Our height is negligible compared to the height of the tower (see Fig.8.6) suppose the measure of ∠QPR = 60o. Then we can find the height of QR of the tower by using trigonometric ratios QR QR tan 60o = ⇒ 3= PQ PR QR = 3PQ
R
60o P
Q
8m
= 8 3m
Fig.8.6
Thus we have been able to find the height of the tower using trigonometric ratios. So, in a right angled triangle if one side and one angle are known we can find the remaining sides of the triangle. Before we proceed to solve problems of the above type let us first define a few terms. Suppose we are viewing an object. The line of sight or line of vision is a straight line from our eye to the object we are viewing. A
If the object is above the horizontal level from the eye we have to lift up our head to view the object. In the process, our eyes move through an angle. This angle is called the angle of elevation. If the object is below the horizontal level from the eye, we have to move downwards our head to view the object. In the process our eyes move through an angle. This angle is called the angle of depression (see Fig.8.8). B
θ2
D
θ1
Fig.8.7
C
It should also be noted that the angle of elevation of one position as seen from the other is equal to the angle of depression of the latter as seen from the former. θ2 = angle of depression; ∴ θ1 = θ2 θ1 = angle of elevation;
Fig. 8.8
186
Example 11 : The angle of elevation of the top of a tower at a distance of 60 m from its foot on a horizontal plane is found to be 30o. Find the height of the tower. Solution: Let CA be the tower equal to h metres in height and B a point at a distance of 60 metres from its foot C. It is given that A ∠ABC = 30o From the right angled triangle ABC AC h tan30° = = BC 60m 1 h o 30 or 3 h = 60 m = 3 60m B C 60 m
h=
60m
=
60 3m
=
60 3m
= 20 3m 3 3 3. 3 = 20 × 1.732 = 34.64 m The height of the tower = 34.64 m (approximately).
Fig.8.9
Example 12 : The angle of depression of a stone on the ground from the top of a building is 60o. If the stone is at a distance of 50 metres away from the building, find the height of the building. P Solution: PQ = height of building equal to h metres 60 S RQ = Distance of the stone from the building = 50m PS is the line of vision (horizontal direction) h Angle of depression ∠SPR = 60o Angle of elevation ∠QRP = Angle of depression ∠SPR 60 = 60o Q R 50 m In right angled triangle PQR h PQ h Fig.8.10 or 3 = tan 60° = = RQ 50m 50m h = 50 3m = 50x1.732m = 86.6 m height of the building = 86.6 m o
o
Example 13: The length of a string between a kite and a point on a ground is 90 m. If the 15 string makes an angle θ with the level of the ground such that tanθ = how high is the kite? 8 Solution: 8, 15, 17 form the sides of a right angled triangle A
tanθ =
15 ⇒ 8
17
15 8
h 15 sinθ = = 90m 17 h = 90 ×
90 m h
θ
θ C
B
Fig.8.11
15 m = 79.41 m 17
187
Example 14: A person stands at a distance of 40 m from a building and observes the top and the bottom of a flag pole on the building at angles of elevation 60° and 45°. Find the height of the building and the height of the flag pole. D Solution: Let hm (= BC) be the height of the building h1 m (= CD) h1 be the height of the flag pole and A is the person. C In the right angled triangle ABC h BC h 60 We have tan 45o = = 45 AB 40m A B 40 m h l = or h = 40 m 40 m Fig.8.12 o
o
Thus the height of the building is 40 m. To find the height of the flag pole CD, let us first find out the length BD in the right angled triangle ABD. BD BD or 3= Now tan 60o = 40m AB BD = 40 3 m = 40 × 1.732 m = 69.28 m Hence the height of the flag pole h1 = CD = BD – BC = 69.28 m – 40 m = 29.28 m
Example 15: From the top of a light house, the angles of depression of two ships on either sides of the light house are observed as 30° and 45°. If the height of the light house is 200 metres, find the distance between the ships. Solution: Let AB be the light house = 200 m and EF be the horizontal level through A. C and D be the two ships. Since the angles of depression of two ships are 30° and 45° respectively. A
E
∠EAC = 30°
30o
∴∠ACB = 30°
∠FAD = 45° ∴∠ADB = 45° In the right angled triangle ACB. tan 30° = BC
AB BC
or
1
=
3
200m
200m BC
= 200 √3 m
= 200 × 1.732 m = 346.4 m In the right angled triangle ADB tan 45° =
AB
or 1
=
F o
45
45o
o
30
C
B
Fig.8.13
200m
BD BD BD = 200 m Distance between the ships = CD = BC + BD = 346.4 m + 200 m = 546.4 m
188
D
Example 16: A tree stands vertically on the bank of a river. From a point on the other bank directly opposite to the tree, the angle of elevation of the top of the tree is 60°. From a point 40 m behind this point on the same bank, the angle of elevation of the top of the tree is 30°. Find the height of the tree and the width of the river. A Solution: Let h (=AB) be a height of the tree. Let C and D be the two points on the other river bank opposite to the tree, so that BC is the width of the river. Let BC = d h At D, the angle of elevation of the top of the tree is 30° That is ∠BDA = 30° 60 30 At C, the angle of elevation of the top of the tree is 60° D B d 40 m C That is ∠BCA = 60° Fig.8.14 In a right angled triangle ABC AB h or 3 = tan 60° = BC d o
o
h = d 3 In a right angled triangle ABD AB 1 or tan 30° = BD 3 d + 40m h = 3 From equations (1) and (2) we get d + 40m d 3 = 3
(1) =
h d + 40m
or
3 h = d + 40m (2)
d 3 × 3 = d + 40 m; 3d = d + 40 m 2d = 40 m; d = 20 m Thus the width of the river is 20 m The height of the tree h
= d
3 = 20 3 m
= 20 × 1.732 m = 34.64 m
Example 17: The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower. Solution Let AB be the hill = 100 m and CD be the tower At A, the angle of depression of D is 30° That is ∠EAD = 30° At A, the angle of depression of C is 60° That is ∠EAC = 60° Let DF be the perpendicular from D to AB Then ∠ADF = 30°, ∠ACB = 60° In a right angled triangle ABC 189
E 60
D
30
o
30
o
A
o
F 100 m
o
60
C
B
Fig.8.15
tan 60°
=
BC
=
In rectangle CBFD,
AB BC 100
3
3 =
or
100m BC
or
BC 3 = 100 m
m ∴ DF =
CB = DF
In a right angled triangle ADF AF tan 30° = or
1
DF
100
3
=
100 3
AF 100 m 3
m
or
AF 3 =
100 3
m
100 m = 33.33m (accuracy to 2 decimal places) 3 3x 3 BF = AB – AF = 100 m – 33.33 m = 66.67 m The height of the tower is CD = BF [since CBFD is a rectangle] = 66.67 m =
AF
m=
Example 18: The top of a tower was observed from the top and the bottom of a building of height 20 m at angles of elevation 45° and 60°. Find the height of the tower. Solution : Let AB be the tower CD be the building of height 20 m. Let DE be perpendicular to AB from D. At C, the angle of elevation of A is 60°. That is ∠BCA = 60° At D, the angle of elevation of A is 45°. That is ∠EDA = 45° BE = CD = 20 m (since BEDC is a rectangle) A Let AE be x in metres, then AB is = x +20 m x In a right angled triangle ABC 45 x + 20m AB E D or tan 60° = 3 = BC BC 20 m 20 m x + 20m 60 BC 3 = x + 20m or BC = 3 C B x + 20m ED = BC = Fig.8.16 3 In a right angled triangle AED x x + 20m AE or 1 = or = x tan 45° = x + 20m ED 3 3 o
o
3x = x
=
x + 20 m or 20
3 x–x = 20 m or x( 3 –1)= 20 m
3 −1
190
x
= =
20( 3 + 1) m ( 3 − 1)( 3 + 1)
(Rationalising the denominator)
20( 3 + 1) m 2 2 ( 3) − 1
20( 3 + 1)
20( 3 + 1) m 2 3 −1 = 10(1.732+1)m = 10(2.732) m = 27.32 m ∴ Height of the tower AB = x + 20 m = 27.32 m + 20 m = 47.32 m =
m
=
Example 19: A light house was observed from two points in a line with it, but on opposite sides of it. The distance between the point is 120 m. If the angle of elevation are 30° and 45°, A find the height of the light house. Solution: Let AB = h be the height of the light house. Let C and D be the two points in a line with it lying on opposite sides of the light house. h At C, the angle of elevation of A is 45° That is ∠BCA = 45° 30 45 At D, the angle of elevation of A is 30° xm (120-x)m C D B That is ∠BDA = 30° Given the distance between two points is 120 m Fig.8.17 That is CD = 120 m Let BC be x in metres then BD= 120 m–x. In a right angled triangle ABC AB h tan 45° = or 1 = BC x h = x (1) In a right angled triangle ABD h AB 1 = tan 30° = or or h 3 = 120m − x 120m − x BD 3 o
o
=
h
120m − x
(2)
3 From equation (1) and (2) we get 120m − x x = ; 3
3 x +x = 120 m; x =
=
120m 3 +1
3 x = 120 m–x;
x( 3 +1) = 120 m =
120( 3 − 1) ( 3 + 1)( 3 − 1)
m
120( 3 − 1) m ( 3 ) 2 − 12
191
=
120( 3 − 1)
3 −1
m
=
120( 3 − 1)
2
m
= 60 (1.732–1) m = 60 x 0.732 m = 43.92 m ∴ Height of the light house = 43.92 m Example 20: An aeroplane at an altitude of 2500 metres observes the angles of depression of opposite points on the two banks of a river to be 41° 20′ and 52°10′. Find the width of the river. Solution: Let A be the position of the aeroplane at an instant and D be the point vertically below A on the ground. Then DA = 2500 m. Let B and C be the two opposite points on the two banks of a river such that B, C and D all line in a straight line. Then ∠DBA = 41°20′ ∠DCA = 52°10′. Let CD = x. Let BC = y be the width of the river. In a right angled triangle CDA A CD x = Cot 52°10′ = AD 2500m x = 2500 cot 52°10′ m (1) 2500 m In a right angled triangle BDA BD x+y cot 41°20' = = AD 2500m y B D C(2) x ∴ x + y = 2500 cot 41°20′ m (2) – (1) ⇒ x+y–x = 2500 cot 41°20′–2500 cot 52°10′m Fig.8.18 y = 2500 (cot 41°20′–cot 52°10′)m = 2500 (tan 48o 40′ – tan 37°50′)m = 2500 (1.1370 – 0.7766) = 2500 x 0.3604 m y = 901 m
Example 21: The angle of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it are complementary. Prove that the height of the tower is ab . Solution: Let CD = h be the height of the tower and A and B be the two points such that CA = a; CB = b. If ∠CAD = θ. Then ∠CBD = 90° – θ (since they are complementary angles) D In a right angled triangle ACD CD h tan θ = = AC a h h (1) ∴ tanθ = a In a right angled triangle BCD a CD h B C A tan(90° − θ) = or cot θ =
BC
tan θ (1) and (2) ⇒
h a
= =
b h b h
b
b
(2) or h2 = ab or h =
192
ab
Fig.8.19
Exercise 8.2 1.
The angle of elevation of the top of a tower at a distance of 50 m from its foot on horizontal plane is found to be 60°. Find the height of the tower.
2.
A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60°. Find how far the ladder is from the foot of the wall.
3.
From the top of a tower 30 metres height a person observes the base of a tree at an angle of depression measuring 30°. Find the distance between the tree and the tower.
4.
A surveyor wants to determine the height of a light house. He measures the angle at A and finds that tan A = ¾. What is the height of the light house if A is 40 m from the base?
5.
What is the elevation of the sun when the length of the shadow of a pole is 3 times the height of the pole?
6.
A man standing on top of a multistoreyed building 45 m high is looking at two advertising pillars on the same side whose angles of depression are 30° and 45°. What is the distance between the two pillars?
7.
Two men are on the opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 150 m, find the distance between the men.
8.
The angle of elevation of a multistoreyed building from a point on the road changes from 30° to 60° as one walks 120 m along the road towards the building, find the height of the building.
9.
A flag staff stands on the top of 6 m high tower. From a point on the ground the angle of elevation of the top of the flag staff is 60° and from the same point the angle of elevation of the top of the tower is 45°. Find the height of the flag staff.
10.
The angles of depression of the top and the bottom of a 12 m tall building from the top of a tower are 45° and 60° respectively. Find the height of the tower.
11.
A man standing on the deck of a ship which is 20 m above the water level, observes the angle of elevation of the top of a hill 60° and the angle of the depression of the base of the hill is 30° calculate the distance of the hill from the ship and the height of the hill.
12.
From the top of a hill 240 metres height the angles of the depression of the top and the bottom of a tower are 30° and 45°. Find the height of the tower.
13.
Two persons standing 20 m apart observed the top of a tree in between them at angles of elevation of 30° and 45°. Find the height of the tree.
14.
Two poles of equal heights are standing opposite to each other on either side of road, which is 100 metres wide from a point between them on the road, the angles of elevation of their tops are 30° and 45°. Find the height of the poles. 193
ANSWERS Exercise 8.1 1)
a) 0.3746
b) 0.8192
c) 1.4826
d) 0.4446
e) 0.9078
g) 0.7540
h) 0.7911
i) 0.6876
j) 0.7743
2)
a) 10°
b) 25°
c) 81°
d) 55°24′
3)
a) 1.0247
b) 1.8802
c) 3.6765
4)
1.566 cm
5) 221.1 cm2 6) 57.15 cm 7) 18.10 cm2 8) 38.637 cm 9) 1/2
e) 65° 42′
f) 1.8967
f) 57° 37′
Exercise 8.2 1)
86.6 m
2) 3.464
3) 51.96 m
6)
32.94 m
7) 409.8 m
8) 103.92 m 9) 4.392
12)
101.44 m 13) 7.32 m
194
4) 30 m
5) 18°25′ 10) 28.392 m 11) 34.64 m
9. PRACTICAL GEOMETRY 9.0
INTRODUCTION
Geometry has practical applications in many fields. For example, architects and carpenters must understand the properties of geometric objects to construct stable and attractive buildings. Navigators of aeroplanes, ships and spacecrafts rely on geometric ideas to chart and follow correct course. Designers, engineers, metal workers and photographers also use geometric principles in their career. Ancient Egyptians developed geometric ideas that could be used to reestablish land boundaries after the annual flooding of the Nile river. They also used geometry in building the pyramids. Babylonians used geometry in measurements needed for building and surveying. In Indian Geometry, Brahmagupta (598AD-665 AD) made significant contribution. He was particularly interested in finding the triangles with rational sides and rational areas; and quadrilaterals with rational sides that can be inscribed in a circle called cyclic quadrilateral. One of his theorem expresses the relationship between the sides and the diagonals of a cyclic quadrilateral. He gave the area of a cyclic quadrilateral as
( s − a )( s − b)( s − c)( s − d ) where
2s = a + b + c + d = sum of the sides. In Theoretical Geometry, we do not construct exact geometrical figures but we draw rough sketches of the figures to give support to our logical reasoning. Here, we do not need any geometrical instruments to draw the diagrams. Drawing geometrical figures using geometrical instruments is called the Practical geometry. We are in need of much skill while doing practical geometry. For example, drawing a tangent at a point on a circle can be done easily by protractor but in drawing the same using the compass we need much skill. In our earlier classes, we have learnt the following : (i) Construction of an equilateral triangle. (ii) Construction of a right angled triangle. (iii) Construction of concentric circles. (iv) Construction of circumcircle and incircle of a triangle. (v) To locate the orthocentre of a triangle. In the present chapter we shall learn how to construct the following : (i) To construct cyclic quadrilaterals. (ii) To construct a triangle with the given altitude. (iii) To construct a triangle with the given median. (iv) To draw a tangent to a circle from a given point on the circle. (v) To draw two tangents to a circle from a given point outside the circle. Note: All figures given in this chapter are not drawn to scale.
195
9.1 CONSTRUCTION OF CYCLIC QUADRILATERAL A quadrilateral formed by joining any four points on a circle is called a cyclic quadrilateral. In other words, if all the four vertices of a quadrilateral lie on a circle then it is called a cyclic quadrilateral.
R S
The opposite angles of a cyclic quadrilateral are supplementary. In the Fig. 9.1 note that the vertices P, Q, R and S of a cyclic quadrilateral lie on the circumference of a circle. m∠P + m ∠R = 180°, m∠Q + m∠S Type I
Q
P
= 180°.
Fig. 9.1
Given three sides and one diagonal
Construct a cyclic quadrilateral ABCD, whose sides AB, BC and AD and the diagonal AC are given. C
Rough diagram C
D
D
O B
A Fig. 9.2
B
A
Fig. 9.3
Construction :
• Draw a line segment AB • With A as centre and AC as radius draw an arc. • With B as centre and BC as radius draw another arc to cut the previous arc at C. • • • • •
Join BC and AC. Draw the perpendicular bisectors of AB and BC cutting each other at O. With O as centre and OA or OB or OC as radius draw a circle. With A as centre and AD as radius draw an arc cutting the circle at D. Join CD and DA. Now ABCD is the required cyclic quadrilateral.
196
Example 1 : Construct a cyclic quadrilateral ABCD given AB=6 cm, BC = 8 cm, AC = 8.5 cm and AD = 5 cm. Rough diagram
C
C
O
A
8 cm
8. 5
5 cm
8 cm
8.5 cm
5 cm
D
cm
D
6 cm
B
Fig. 9.4
A
B
6 cm
Fig. 9.5
Construction : • Draw a line segment AB = 6 cm • With A as centre and radius 8.5 cm draw an arc. • With B as centre and radius 8 cm draw another arc to cut the previous arc at C. • Join BC and AC • Draw the perpendicular bisectors of AB and BC cutting each other at `O'. • With O as centre and OA or OB or OC as radius draw a circle. • With A as centre and 5 cm as radius draw an arc cutting the circle at D. • Join CD and DA. • Now ABCD is the required cyclic quadrilateral. Type II
Given two sides and two diagonals Construct a cyclic quadrilateral ABCD given two sides AB and BC and the two diagonals AC and BD. Rough diagram
D
D
C
C
O
A
B
B
A Fig. 9.6
Fig. 9.7
197
Construction : • Draw the line segment AB. • With A as centre and AC as radius draw an arc.
• • • • • • •
With B as centre and BC as radius draw an arc cutting the previous arc at C. Join BC and AC. Draw the perpendicular bisectors of AB and BC cutting each other at `O'. With `O' as centre and OA or OB or OC as radius draw a circle. With B as centre and BD as radius draw an arc cutting the circle at D. Join AD and CD. ABCD is the required cyclic quadrilateral.
Example 2 : Construct a cyclic quadrilateral ABCD given AB = 4 cm, BC = 2.6 cm, AC = 5.5 cm, BD = 5.8 cm. Rough diagram D
D
A
cm
5.5
A
B
4 cm
cm
4 cm
2.6 c m
2.6 c m
5.5
C
cm
cm
O
5.8
8 5.
C
B
Fig. 9.8
Fig. 9.9
Construction : • Draw a line segment AB = 4 cm. • With A as centre and radius 5.5 cm draw an arc. • With B as centre and radius 2.6 cm draw another arc cutting the previous arc at C. • Join BC and AC. • Draw the perpendicular bisectors of AB and BC cutting each other at 'O'. • With O as centre and OA or OB or OC as radius draw a circle. • With B as centre and 5.8 cm radius draw an arc cutting the circle at D. • Join AD and BD. • Now ABCD is the required cyclic quadrilateral. Type III
Given two sides and two angles Construct a cyclic quadrilateral ABCD whose sides AB and BC and two angles ∠BAC and ∠ACD are given. 198
y
Rough diagram
D
C
φ
x
D
C
φ
O
θ B
A
θ A
Fig. 9.10
B
Fig. 9.11
Construction : • Draw a line segment AB. • Through A draw AX such that ∠BAX = θ. • With B as centre and BC as radius draw an arc cutting AX at C. • Join AC and BC. • Draw the perpendicular bisectors of AB and BC cutting each other at 'O'. • With 'O' as centre and OA or OB or OC as radius draw a circle. • Through C draw CY such that ∠ACY = ∠ACD = φ. • Let CY cut the circle at D. • Join AD. • ABCD is the required cyclic quadrilateral. Example 3: Construct a cyclic quadrilateral ABCD given AB = 5 cm, BC = 4 cm, ∠BAC = 35° and ∠ACD = 70° y D
Rough diagram C
70o
x
D
4 cm
O 35o A
C
o
4 cm
70
5 cm
o
35 A
B
5 cm Fig. 9.12
Fig. 9.13
199
B
Construction : • Draw a line segment AB = 5 cm. • Through A draw AX such that ∠BAX = 35°. • With B as centre and 4 cm radius draw an arc cutting AX at C. • Join AC and BC. • Draw the perpendicular bisectors of AB and BC cutting each other at O. • With O as centre and OA or OB or OC as radius draw a circle. • Through C draw CY such that ∠ACY = 70°. • Let CY cut the circle at D. • Join AD. • ABCD is the required cyclic quadrilateral. Type IV Given two sides, one diagonal and one angle Construct a cyclic quadrilateral ABCD whose sides AB and AD, the diagonal AC and ∠BAC are given. Rough diagram
D C
x
C
D
O A
θ A
θ
B
Fig. 9.14
B
Fig. 9.15
Construction : • Draw a line segment AB. • Through A draw AX such that ∠BAX = ∠BAC = θ. • With A as centre and AC as radius draw an arc cutting AX at C. • Join BC. • Draw the perpendicular bisectors of AB and BC. Let them intersect at 'O'. • With O as centre and OA or OB or OC as radius draw a circle. • With A as centre and AD as radius draw an arc cutting the circle at D. • Join CD and AD. • ABCD is the required cyclic quadrilateral. Example 4 : Construct a cyclic quadrilateral ABCD given AB = 7.5 cm, AC = 10 cm, ∠BAC = 30° and AD = 6.5 cm. 200
Rough diagram
D C
X
6.5 cm
6.5 cm
O 30 7.5 cm ο
A
C
D
cm 10
A
B
c 10
m
30o 7.5 cm
B
Fig. 9.16 Fig. 9.17
Construction : • Draw a line segment AB=7.5 cm. • Through A draw AX such that ∠BAX = 30°. • With A as centre and 10 cm as radius draw an arc on AX. Let it cut AX at C. • Join BC and AC. • Draw the perpendicular bisectors of AB and BC. Let them intersect at 'O'. • With O as centre and OA or OB or OC as radius draw a circle. • With A as centre and 6.5 cm as radius draw an arc cutting the circle at D. • Join CD and AD. • ABCD is the required cyclic quadrilateral. Exercise 9.1 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
Construct a cyclic quadrilateral ABCD with AB = 8 cm, BC = 7 cm, AC = 6 cm and AD = 4 cm. Construct a cyclic quadrilateral EFGH given EF = 7 cm, EH = 6 cm, FH = 10 cm and FG = 6.4 cm. Construct a cyclic quadrilateral PQRS given PQ = 4.5 cm, QR = 5.5 cm, PR = 6.5 cm and PS = 4 cm. Construct a cyclic quadrilateral EFGH given EF = 7 cm, EG = 7.5 cm, EH = 6 cm and GH = 5.5 cm. Construct a cyclic quadrilateral ABCD given AB = 7 cm, BC = 5 cm, AC = 6 cm and BD = 6.5 cm. Construct a cyclic quadrilateral EFGH given EF = 6 cm, FG = 5.3 cm, EG = 8.2 cm and FH = 7.8 cm. Construct a cyclic quadrilateral PQRS given PQ = 8 cm, QR = 5.9 cm, PR = 8.6 cm and RS = 7.9 cm Construct a cyclic quadrilateral ABCD given AB = 5.8 cm, AD = 2.6 cm, AC = 6.7 cm and BD = 6.8 cm. Construct a cyclic quadrilateral ABCD given AB = 7.2 cm, ∠ABD = 45°, ∠BAD = 100° and BC = 4 cm. Construct a cyclic quadrilateral PQRS given PQ = 8 cm, QR = 7 cm, ∠QPR = 40° and ∠PRS = 60°. Construct a cyclic quadrilateral EFGH given EF = 5 cm, FG = 4.8 cm, ∠EFG = 85° and ∠FEH = 72°. 201
12. 13. a) b)
9.2
Construct a cyclic quadrilateral PQRS given PQ = 7 cm, QR = 8.4 cm, ∠PQR = 90° and ∠SRP = 56°. Construct a cyclic quadrilateral ABCD with the given measurements. AB = 7 cm, m∠A = 100°, BD = 10 cm, CD = 5 cm. AB = 8.5 cm, BC = 7.5 cm, AC = 10 cm, m∠ACD = 35°.
CONSTRUCTION OF TRIANGLES
D
Construct a segment of a circle on a given line segment E containing given angle θ. O Given : Let AB be a given line segment and θ be a given angle. Required : To describe on AB, a segment of a circle containing B A θ the angle θ. Construction : Fig. 9.18 • At A make ∠BAC = θ. • Draw AD ⊥ AC. • Draw the perpendicular bisector of AB meeting AD at 'O'. • With O as centre and OA as radius draw the circle ABE. AEB is the required segment containing the angle θ. D Example 5: Construct a segment of a circle on AB = 5 cm containing an angle 30°. E Construction: • Draw a line segment AB = 5 cm. O • Draw AC such that ∠BAC = 30°. • Draw AD ⊥ AC. 5 cm B A 30 • Draw the perpendicular bisector of AB meeting AD at 'O'. Fig. 9.19 • With O as centre and OA as radius draw the circle ABE. C • AEB is the required segment containing the angle 30°. K Type I Construct a triangle given the base, the vertical F angle and the altitude on the base J C H Given : Let AB be the given base, θ be the given vertical angle, θ l be the altitude on the base. O Required : To construct a triangle ABC with base AB and vertical angle θ such that the altitude from C on AB is of B G A θ length l. Construction : • Draw the base AB. Fig. 9.20 • At A make ∠BAE = θ. • Draw AF⊥AE. • Draw the perpendicular bisector GO of AB meeting E AF at O. • With O as centre and OA as radius draw the circle ABK. o
202
• The bigger segment AKB contains the angle θ. • On the perpendicular bisector GO, mark a point H such that GH = l. • Draw a line parallel to AB through H meeting the segment AKB at C and J. • Join AC and BC. • ABC is the required triangle. Proof : In ΔABC, AB is the base. Since ∠ACB is an angle in the alternate segment, it is equal to θ (i.e) ∠ACB = θ. Altitude from C to AB = the distance between the parallels AB and CJ = GH = l. Thus ABC is the required triangle. Example 6 : Construct a triangle ABC such that BC = 7 cm, m∠A = 60° and altitude from A to BC is 4.5 cm. K
F
A F
60
O G 7 cm
o
4.5 cm
4.5 cm 60o
A
H
60o
B
Rough diagram
A'
C
B
7 cm
C
Fig. 9.21 Fig. 9.22
E
Construction : • Draw a line segment BC = 7 cm. • At B make ∠CBE = 60°. • Draw BF ⊥ BE. • Draw the perpendicular bisector of BC meeting BF at 'O'. • With O as centre and radius equal to OB or OC draw a circle BCK. • Then the segment BKC contains the angle 60°. • On the perpendicular bisector GO mark a point 'H' such that GH = 4.5 cm. • Through H draw AHA′ parallel to BC. C • Join BA, CA. O θ l • ABC is the required triangle. Type II Construct a triangle, given the base, the vertical A θ D angle and the median on the base. Given : Let AB be the given base, θ be the given vertical angle, l be the median through C Required : To construct a triangle ABC with base AB, ∠C = θ and the median from C to the base AB equal to l.
K E J
B
Fig. 9.23
R
Construction : • Draw the base AB. 203
•
Draw AR such that ∠BAR = θ.
•
Draw AE ⊥ AR.
•
Draw the perpendicular bisector DO of AB meeting AE at 'O'.
•
With O as centre and OA as radius draw the circle ABK.
•
The bigger segment AKB contains the angle θ.
•
With D as centre and median length l as radius draw an arc meeting the bigger segment AKB at C and J.
•
Join AC and BC.
•
ABC is the required triangle.
•
AB is the base of the ΔABC.
•
Since ∠ACB is the angle in the alternate segment, it is equal to θ (i.e) ∠ACB = θ.
•
D is the mid point of AB. ∴CD is the median through C. Hence ABC is the required triangle.
Proof :
Example 7 : Construct a ΔABC given the base BC = 5 cm, m∠BAC = 45° and median AD = 4 cm. Rough diagram
K F
A o
A
1
4c
45o
m
o
45
4 cm
45
B
A
O
D 5 cm
C
B Fig. 9.25
D 5 cm
C
Fig. 9.24
E
Construction : • Draw a line segment BC = 5 cm. • At B make ∠CBE = 45°. • Draw BF ⊥ BE. • Draw the perpendicular bisector of BC meeting BC at D. • The perpendicular bisector meets BF at 'O'. • With 'O' as centre and OB or OC as radius draw the circle BCK. • The segment BKC contains the vertical angle 45°. • With D as centre and radius 4 cm draw an arc on the circle to cut the circle at A and A′. • Join BA and CA. • ABC is the required triangle. 204
Exercise 9.2 1. Construct a segment of a circle on a) CD = 6 cm containing an angle of 60°. b) EF = 4 cm containing an angle of 100°. c) PQ = 6.5 cm containing an angle of 135°. 2. Construct a triangle ABC in which base BC = 4 cm, ∠A = 60 ° and altitude through the vertex is equal to 3 cm. 3. Construct a triangle ABC such that AB = 6 cm, m∠C = 40° and altitude from C to AB is of length 4 cm. 4. Construct a triangle PQR such that QR = 5.1 cm m∠P=60° and altitude from P to QR is 3.2 cm. 5. Construct a ΔABC in which BC = 4.5 cm, m∠A = 60° and median AD through A is 3.2 cm. 6. Construct a ΔABC having base AB = 4.4 cm, vertical angle ∠C = 60° and median through vertex equal to 2.7 cm.
9.3
CONSTRUCTION OF TANGENTS
TO CIRCLES
Draw a tangent to a circle at a given point on it. Construction : • Draw a circle of any radius with O as its centre. • Let P be any point on the circle. Join OP. • With P as centre draw an arc cutting OP at M. • With M as centre and with the same radius draw an arc cutting the previous arc at N. Again with N as centre and with the same radius draw an arc cutting the first arc at L. • Draw the bisector PT of the angle ∠NPL. • ∠OPT = 90° and PT is a tangent at P. Example 8 : Draw a circle with centre 'O' and radius 3 cm. Take a point P on it. Draw a tangent to the circle at the point P.
T
L N o
O
90 M
Fig. 9.26
T
Rough diagram T L
N
O O
3 cm M
3 cm
P
Fig. 9.28
Construction : • Draw a circle of radius 3 cm with 'O' as its centre. • Let P be any point on the circle. Join OP. • With P as centre draw an arc cutting OP at M.
205
P
Fig. 9.27
P
• • •
With M as centre and with the same radius draw an arc cutting the former arc at N. Again with N as centre and with the same radius draw an arc cutting the first arc at L. Draw the bisector PT of the angle ∠NPL. ∠OPT = 90° and PT is a tangent at P.
Draw two tangents to a given circle from a point outside the circle Given : Let O be the centre and r be the radius of the given circle. Let P be any point outside the circle. A
r
O
P
r
G
B Fig. 9.29
Required : To construct two tangents to the given circle through the point P outside the circle. Construction : • Draw a circle of radius 'r' with O as centre. • Mark a point P outside the circle and join OP. • Draw the perpendicular bisector of OP. Let it meets OP at G. • With G as centre and GO as radius draw a circle cutting the given circle at A and B. • Join PA and PB. • PA and PB are the tangents from P to the given circle. Proof : Join OA and OB. Since OP is a diameter and angle in a semicircle is 90°, ∠OAP = ∠OBP = 90°. 1) OA = OB is the radius to the given circle and ∠OBP = ∠OAP = 90°. 2) Thus AP and BP are the two tangents from P to the given circle. Example 9 : Draw a circle of diameter 10 cm. Take a point P at a distance of 13 cm from the centre. Draw two tangents from P to the circle and measure the length of the tangents. A
5 cm
G
13 cm
A
P
5c m
O
Rough diagram
cm
O
cm 12
13 cm
m 5c
5c m
12
B
B Fig. 9.30
Fig. 9.31
206
P
Construction : • Draw a circle of radius 5 cm with centre 'O'. • Mark a point P at a distance of 13 cm from 'O'. Join OP. • Draw the perpendicular bisector of OP. Let it meet OP at G. • With G as centre and OG or GP as radius draw a circle cutting the given circle at A and B. • Join PA and PB • PA and PB are the required tangents. PA = PB = 12 cm is got by measurement. Verification: OP = 13 cm, OA = 5 cm, ΔOPA is a right angled triangle ∴ PA2 = OP2 - OA2 = 132 - 52 = 169 - 25 = 144 ∴ PA = 12 cm Exercise 9.3 1. Draw a circle with centre O and radius 2 cm. Take a point P on it. Draw a tangent to the circle at point P. 2. Draw a circle of radius 3.4 cm. Take a point P on it. Draw tangent to the circle at point P. 3. Draw a circle with centre O and radius 2.5 cm. Take a point P outside the circle at a distance of 5.8 cm from its centre. Draw two tangents to the circle from the point P. Verify the length of the tangents by algebraic calculation. 4. Take a point P at a distance of 7 cm from the centre of a circle of radius 3 cm and from P draw two tangents PA and PB to the circle. Measure the length of each tangent. 5. Draw two tangents to a circle whose diameter is 6 cm from a point P, at a distance of 5 cm from the centre. Measure the length of these tangents. 6. Using ruler and compass draw two tangents to a circle whose diameter is 6 cm from a point P at a distance of 9 cm from its centre.
ANSWERS Exercise 9.3 (4) 6.3 cm
(5) 4 cm
207
10. STATISTICS 10.0 INTRODUCTION The word ‘Statistics’ has been derived from either Latin word ‘Status’ or Italian word ‘Statista’ or German word ‘Statistik’ or French word ‘Statistique’ each of which means a political state. By the word statistics we mean the “numerical statements as well as statistical methodology”. Now a days statistics is used for solving or analyzing the problems of the state. It supplies essential information for developmental activities in all the departments. Statistics is also very widely used in all the disciplines. Sir. Ronald A. Fisher (1890–1962) who is called the “Father of Statistics” contributed many useful calculations from the statistical data. He applied statistics to several fields such as Psychology, Genetics and Education.
10.1
DISPERSION
In earlier classes we have studied that collection of data is represented in different forms. From the data we have learnt to calculate the measures of the central tendency like mean, median and mode. These central measures do not give us all the details about the distribution. Further descriptions of the data called measures of dispersion are necessary. According to A.L. Bowley, “Dispersion is the measure of the variation of the individual item”. That is the dispersion is used to indicate the extent to which the data is spread. Measures of dispersion are many but we are restricting ourselves to study range, standard deviation and coefficient of variation for the individual series only. Range Range is the simplest measure of dispersion. It is defined as the difference between the largest and the smallest values in the series. Range = L – S,
L = largest value, S = smallest value L−S Coefficient of range = L+S Example 1 : Find the range of the data 27,28,34,36,39,59. Also find the coefficient of range. Solution : Largest value L = 59; Smallest value S = 27 Range = L – S = 59–27 = 32 L − S 59 − 27 32 Coefficient of Range = = = = 0.372 L + S 59 + 27 86 Example 2: The weights of seven persons in kg are 46, 49.5, 52.5, 38, 45, 79.5, 84.5. Find the range and the coefficient of range. Solution : Largest value L = 84.5; Smallest value S = 38 208
Range = L – S = 84.5 – 38 = 46.5 kg L − S 84.5 − 38 46.5 = 0.379 = Coefficient of Range = = L + S 84.5 + 38 122.5
Example 3: The largest value of a data is 98. If the range of the data is 73, find the smallest value of the data. Solution : Largest value L = 98; Range = 73 Range = L – S or 73 = 98 – S or S = 98 – 73 = 25 ∴The smallest value = 25. Example 4: The least score of a cricket player of the school team is 5 runs in a series of ten matches. If his range of scores is 87, find his highest score in the series. Solution : Least Score S = 5 ; Range = 87 Range = L – S or 87 = L – 5 or L = 87 + 5 = 92 ∴The highest score = 92 runs. Standard Deviation
Standard Deviation is defined as the positive square root of the mean of the squared deviations of the data from the mean. It is denoted by σ . n
∑ (x
σ=
Standard Deviation
i =1
i
− x) 2
n
(or) σ =
n
∑ d i2 / n i =1
where d i = x i − x
Variance: Variance is defined as the square of the Standard Deviation (S.D) and is denoted by σ2. n
Variance σ2 =
∑ (x i =1
i
− x) 2
=
n
n
∑d i =1
2 i
where d i = x i − x
/n
The following methods are used for calculating standard deviation • Direct method Assumed mean method • Actual mean method Step deviation method Direct Method: By definition, the standard deviation is calculated by the formula
σ=
Σ(x − x) 2 n 2
Here
Σ(x − x) 2 Σ(x 2 − 2xx + x ) Σx 2 Σx Σ x = = − 2x + n n n n n
Σx 2 nx − 2x × x + = n n
Therefore
σ=
2
2
2 2 2 Σx 2 Σx 2 ⎛ Σx ⎞ Σx 2 −x = −⎜ − 2x + x = = ⎟ n n n ⎝ n ⎠
Σx 2 ⎛ Σx ⎞ − ⎜ ⎟ n ⎝ n ⎠
2
209
2
This formula is used to find S.D. in direct method
• Find Σx and Σx2 • Substitute the values of Σx, Σx2 and the number of data n in the formula Σx 2 ⎛ Σx ⎞ − ⎜ ⎟ n ⎝ n ⎠
σ=
2
Example 5: Calculate the standard deviation for the data 14, 22, 9, 15, 20, 17, 12, 11 Solution x 14 22 9 15 20 17 12 11
x2 196 484 81 225 400 289 144 121
Σx = 120
Σx = 1940
⎛ Σx ⎞ − ⎜ ⎟ ⎝ n ⎠
2
1940 ⎛ 120 ⎞ − ⎜ ⎟ 8 ⎝ 8 ⎠
2
= =
242.5 − 225
σ=
Σx n
2
= 17.5 σ = 4.18
2
Actual Mean Method: The formula used in this method is σ =
(or)
σ=
Σd 2 n
4.18 4 17.5000 16 81 150 81 828 6900 6624 276
Σ( x − x ) 2 n
where d = x − x
•
Σx n Calculate the deviation d = x − x for each value of the series and find Σd2
•
Substitute Σd2, n in the formula and calculate σ =
•
Calculate the mean x =
Σd 2 n
Example 6: The marks obtained by 10 students in a class test out of 100 marks are 62, 49, 71, 75, 33, 41, 100, 88, 50, 31. Calculate the standard deviation of the marks. 600 Σx 62 + 49 + 71 + 75 + 33 + 41 + 100 + 88 + 50 + 31 Solution: = = 60 x = = 10 10 n
210
x 62 49 71 75 33 41 100 88 50 31
d = x − x = x − 60 2 –11 11 15 –27 –19 40 28 –10 –29 Σd = 0
d2 4 121 121 225 729 361 1600 784 100 841 Σd2 = 4886
Σd 2 n
σ=
4886 10 = 488.6 = 22.10 =
Therefore, standard deviation is 22.10 Assumed Mean Method: When the data is large or the mean is not an integer we use assumed mean method to calculate the standard deviation. • Choose one of the items nearer to the middle value in the data say A as the assumed mean. • Calculate deviation d = x–A for each value of the series and find Σd, Σd2 • Substitute Σd, Σd2 and n in the formula and calculate the standard deviation
Σd 2
The formula for standard deviation is σ =
n
(
− Σd
n
)
2
where d = x–A
Example 7: The following are the bowling rate per over of a player in 12 cricket matches : 6.5, 5.0, 5.2, 5.3, 5.5, 5.0, 4.7, 4.5, 6.3, 3.0, 4.0, 9.0. Find the standard deviation. Take the assumed mean A=5 Bowling rate per over 6.5 5.0 5.2 5.3 5.5 5.0 4.7 4.5 6.3 3.0 4.0 9.0
d = x–A = x–5
d2
1.5 0 0.2 0.3 0.5 0 –0.3 –0.5 1.3 –2.0 –1.0 4.0 Σd = 4.0
2.25 0 0.04 0.09 0.25 0 0.09 0.25 1.69 4.00 1.00 16.00 Σd2 = 25.66
Σd = 4, Σd2 = 25.66, n = 12 σ=
Σd 2 ⎛ Σd ⎞ −⎜ ⎟ n ⎝ n ⎠
2
2
=
25.66 ⎛ 4 ⎞ −⎜ ⎟ 12 ⎝ 12 ⎠
=
2.1383 − 0.1089
=
2.0294 = 1.42
Step Deviation Method • Choose one of the nearest middle values in the series as the assumed mean A • Calculate d=x–A for each value of the series
211
x−A where c is the common factor of all ‘d’ c • Find Σd′, Σd′2 • Substitute the values of Σd′, Σd′2, c and n in the formula and calculate the standard deviation.
• Find d ' =
Σd '2 ⎛ Σd ' ⎞ −⎜ ⎟ ×c n ⎝ n ⎠ 2
The formula for standard deviation is σ =
where d ' =
x−A c
Example 8 : The following are the income (in Rs.) of ten employees of a firm 600, 620, 640, 620, 680, 670, 680, 640, 700, 650. Calculate the standard deviation of the income. Solution: Take assumed mean A = 640; Common factor c = 10 Income (in Rs.) x 600 620 640 620 680 670 680 640 700 650
d = x–640 d ' =
x − 640 10
–4 –2 0 –2 4 3 4 0 6 1 Σd′ = 10
–40 –20 0 –20 40 30 40 0 60 10
d2 16 4 0 4 16 9 16 0 36 1 2 Σd′ = 102
n = 10, Σd′ = 10, Σd′2 = 102 2 Σd '2 ⎛ Σd ' ⎞ σ = −⎜ ⎟ ×c n ⎝ n ⎠ 2 102 ⎛ 10 ⎞ = − ⎜ ⎟ × 10 10 ⎝ 10 ⎠
= 10.2 − 1 × 10 = = 3.033 × 10 = Rs.30.33
σ
9.2 × 10
Example 9: Find standard deviation for the following values : 30,80, 60, 70, 20, 40, 50 Solution: i) Direct method x x2 2 30 900 Σx 2 ⎛ Σx ⎞ σ = − ⎜ ⎟ 80 6400 n ⎝ n ⎠ 60 3600 2 70 4900 20300 ⎛ 350 ⎞ = − ⎜ ⎟ 20 400 7 ⎝ 7 ⎠ 40 1600 400 = 50 2500 2 20 σ = Σx = 350 Σx =20300 ii) Actual mean method x
=
Σx n
=
30 + 80 + 60 + 70 + 20 + 40 + 50 7 212
=
350 7
=
50
x 30 80 60 70 20 40 50
d = x – 50 –20 30 10 20 –30 –10 0 Σd = 0
d2 400 900 100 400 900 100 0 2 Σd = 2800
σ
= =
σ
= =
Σd 2 n 2800 7 400 20
iii) Assumed mean method: Take assumed mean A = 40 x 30 80 60 70 20 40 50
d = x – 40 –10 40 20 30 –20 0 10 Σd = 70
d2 100 1600 400 900 400 0 100 Σd2 = 3500
iv) Step deviation method: Take A = 60, c = 10 x - 60 d' = d2 x d = x – 60 10 30 80 60 70 20 40 50
–30 20 0 10 –40 –20 –10
–3 2 0 1 –4 –2 –1 Σd′ = –7
9 4 0 1 16 4 1 Σd′2 = 35
σ
σ
σ =
2
3500 ⎛ 70 ⎞ −⎜ ⎟ 7 ⎝7⎠
2
n
= = = =
⎛ Σd ⎞ ⎟ ⎝ n ⎠
Σd 2
=
−⎜
500 − 10 2 400 20
Σd '2 ⎛ Σd ' ⎞ −⎜ ⎟ n ⎝ n ⎠
=
35 ⎛ −7 ⎞ −⎜ ⎟ 7 ⎝ 7 ⎠
=
5 − 1 × 10
=
4 × 10
2
×c
2
× 10
σ = 20
In all the four methods we get the same standard deviation σ = 20. Example 10: A school has the strength of 1600 students. The no.of absentees to the school on 10 different days are as follows : 26,34,22,36,26,41,48,22,36,26. Calculate the variance & SD. Solution: Take the assumed mean A = 30 Variance Marks x d = x–A = x – 30 d2 2 26 –4 16 Σd 2 ⎛ Σd ⎞ 2 σ = −⎜ ⎟ 34 4 16 n ⎝ n ⎠ 22 –8 64 2 36 6 36 709 ⎛ 17 ⎞ 26 –4 16 = −⎜ ⎟ 10 ⎝ 10 ⎠ 41 11 121 48 22 36 26
18 –8 6 –4 Σd = 17
324 64 36 16 Σd2 = 709
213
= 70.9 – 2.89 = 68.01 S.D σ = 68.01 = 8.24
Example 11: Find the standard deviation of the following data 5,10, 15, 20, 25. Add 3 to each item and find the new standard deviation. Solution: Here c = 5 and take assumed mean A = 15
x 5 10 15 20 25
x-15 d' = 5 –2 –1 0 1 2 Σd′ = 0
Σd '2 ⎛ Σd ' ⎞ −⎜ ⎟ n ⎝ n ⎠
σ =
2
d′
4 1 0 1 4 2 Σd′ = 10
=
10 −0 ×5 5
=
2 ×5
2
×c
= 5 2
When we add 3 to each value we get 8, 13, 18, 23, 28 x -18 d' = x d′2 σ 5 8 –2 4 13 –1 1 18 0 0 23 1 1 28 2 4 2 Σd′ = 0 Σd′ = 10
=
Σd '2 ⎛ Σd ' ⎞ −⎜ ⎟ n ⎝ n ⎠
=
10 −0 ×5 5
=
2 ×5
2
×c
= 5 2
From the above example we conclude: The standard deviation of a series remains unchanged when each value is added or subtracted by the same quantity. Example 12: The marks of 5 students scored out of 50 are 20, 25, 30, 35, 40. Find the standard deviation of the marks. When we convert the marks to 100 find the new standard deviation. Solution : Take the assumed mean A = 30, c = 5
x 20 25 30 35 40
x - 30 d' = 5 –2 –1 0 1 2 Σd′ = 0
d′2
σ =
4 1 0 1 4 Σd′2 = 10
Σd '2 ⎛ Σd ' ⎞ −⎜ ⎟ n ⎝ n ⎠
=
10 −0 ×5 5
=
2 ×5
2
×c
= 5 2
To convert the marks out of 100, it is enough to multiply the marks by 2. Then the new marks are 40, 50, 60, 70, 80. Here c = 10 and take A = 60 214
x 40 50 60 75 80
x - 60 10
d′2
–2 –1 0 1 2
4 1 0 1 4
Σd′ = 0
Σd′2 = 10
d' =
σ =
Σd '2 ⎛ Σd ' ⎞ −⎜ ⎟ n ⎝ n ⎠
=
10 − 0 × 10 5
=
2 × 10
2
×c
= 10 2
From the above example we conclude the following result: The standard deviation of a series gets multiplied or divided by the quantity k if each value is multiplied or divided by k. Example 13: The standard deviation of 10 values is 4. If each value is increased by 3, find the standard deviation and the variance of the new set of values. Solution: S.D. of given 10 values = 4; Increment in each value = 3
S.D. is unchanged by the increments in the values. ∴ σ = 4; Variance σ2 = 42 = 16 Example 14: The variance of 5 values is 9. If each value is doubled then find the standard deviation of new values. Solution: Variance of 5 values = 9; or S.D. of 5 values = 3 When each value is doubled, S.D. is also doubled
∴ S.D. of 5 new values = 3 x 2 = 6 Example 15: Find the standard deviation of the first n natural numbers. Solution: Mean x =
σ=
1 + 2 + 3 + ... + n n(n + 1) n + 1 and Σx 2 = = = n 2n 2
Σx 2 ⎛ Σx ⎞ −⎜ ⎟ n ⎝ n ⎠
n ( n + 1)( 2n + 1)
6
2
=
(n + 1)(2n + 1) ⎛ n + 1 ⎞2 −⎜ ⎟ = 6 ⎝ 2 ⎠
=
(n + 1) ⎡4n + 2 − 3n − 3⎤ ⎥= ⎢ 6 2 ⎦ ⎣
(n − 1) ⎡2n + 1 n + 1⎤ ⎢ 3 − 2 ⎥ 2 ⎣ ⎦ (n + 1) (n − 1) = (n2 − 1)/12 12
Example 16: The mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the items and also the sum of the squares of all the items.
215
⎯ ∑x x = n , ∑x = n x = 100 × 48 = 4800
⎯
Solution:
2
⎯ ∑x2 ∑x2 σ2 = n − ( x ) ⇒ 102 = 100 − 482
∑x2 = 100(482 + 102) = 100(2304 + 100) = (240400) ∴ ∑x = 4800, ∑x2 = 240400
(
⎯ 2
Example 17: Given that ∑x = 99, n = 9 ∑(x – 10)2 = 79. Find ∑x2 and ∑ x − x Solution: n = 9, ∑x = 99 ;
∑x 99 x = n = 9 = 11
⎯
Σ(x – 10)2 = 79 ; Σ(x2 – 20x + 100) = 79 or
)
Σx2 – (20 × 99) + 900 = 79
or
Σx2 – 20 Σx + (100 × 9) = 79
or
Σx2 = 79 + 1980 – 900 = 1159
Σ(x – x )2 = Σ(x – 11)2 = Σ(x2 – 22x + 121) = Σx2 – 22Σx + (121 × 9) Σ(x – x )2 = 1159 – (22 × 99) + 1089 = 70 ∴
Σx2 = 1159; Σ(x – x )2 = 70
Note: Sometimes when we calculate mean x and standard deviation σ, some values of the data may be taken incorrectly in the calculations. After the calculations, when we detect the mistake, we can calculate the correct values of the mean and the standard deviation without repeating the whole procedure. Example 18: The mean and standard deviation of a set of 100 observations were worked as 40 and 5 respectively. By mistake 40 was entered as 50. Calculate the correct mean and standard deviation. ⎯ ⎯ ∑x Solution: Mean x = n , ∑x = n x In correct Σx = 100 × 40 = 4000. Correct Σx = 4000 – 50 + 40 = 3990 correct ∑ x 3990 ∴ Correct mean = = 100 = 39.90 n 2
Variance
Correct
∑x2 ⎛∑x⎞ σ = n − ⎜ n ⎟ ; Incorrect σ = 5, x = 40 ⎝ ⎠ ∑x2 25 = 100 − 1600 ; Incorrect Σx2 = 2500 + 160000 = 162500 2
Σ x2
= 162500 – 502 + 402 = 162500 – 2500 + 1600 = 161600
216
Correct variance =
σ2
correct ∑x2 161600 − (correct mean)2 = 100 − (39.9)2 = 1616 − 1592.01 n
= 23.99;
Correct S.D
σ=
23.99 = 4.89
Example 19:
The mean and standard deviation of 100 items are found to be 40 and 10. At the time of calculations two items were wrongly taken as 30 and 70 instead of 3 and 27. Find the correct mean and standard deviation. ⎯ ⎯ ∑x Solution: Mean x = n , ∑x = n x ; Incorrect Σx = 100 × 40 = 4000 Correct
Σx = 4000 – 70 – 30 + 3 + 27 = 3930 ;
Correct mean x =
3930 = 39.3 100
2 ⎯ ∑x2 ∑x2 2 2 2 ( ) − x or 10 = n 100 − 40 ⇒ ∑x = 10000 + 160000 Incorrect Σx2 = 170000 Correct Σx2 = 170000 – 302 – 702 + 32 + 272 = 170000 – 900 – 4900 + 9 + 729 Correct Σx2 = 164938 ⎯ 2 correct ∑x2 − (correct x) Correct σ2 = n 164938 = 100 − (39.3)2 = 1649.38 − 1544.49 = 104.89
Incorrect σ2
=
⇒
σ
= 104.89 = 10.24
∴ Correct mean = 39.3 ; Correct standard deviation = 10.24.
Coefficient of Variation According to Karl Pearson “Coefficient of variation is the percentage variation in mean, standard deviation being considered as the total variation in the mean”.
Coefficient of variation is defined as follows: σ C.V. = ⎯ × 100% where σ is the S.D, x is the mean of the given data. It is also called x relative standard deviation. If we wish to compare the variability of two or more series, we can use the coefficient of variation. The series of data for which the coefficient of variation is large indicates that the group is more variable (less stable/less uniform/less consistent). If the coefficient of variation is small it indicates that the group is less variable (more stable / more uniform / more consistent. Example 20: Find the coefficient of variation of the following data: 16, 13, 17, 21, 18 217
Solution: Mean x =
16 +13 +17 + 21 +18 85 = = 17 5 5 2
x d = x – 17 16 –1 13 –4 17 0 21 4 18 1 Σd = 0
∑d 34 S.D. σ = 5 = 6.8 n = σ = 2.61 σ 2.61 C.V. = ⎯ × 100 = 17 × 100 x C.V. = 15.35%
d2 1 16 0 16 1 2 Σd = 34
Note: Of the three values x , σ, C.V. if two are given then the third can be calculated using the C.V. formula. Example 21: The coefficient of variation of a series is 69% and its standard deviation is 15.6. Find the arithmetic mean of the series. ⎯ σ σ Solution: Coefficient of variation C.V. = − × 100 ⇒ x = C.V × 100 x ⎯ σ 15.6 Arithmetic mean of the series x = C.V × 100 = 69 × 100 = 22.6 Example 22: The standard deviation and the mean of 20 values are 21.2 and 36.6. Find the coefficient of variation. σ 21.2 Solution: σ = 21.2, x = 36.6; C.V. = − × 100 = 35.5 × 100 = 57.92% x Example 23: A group of 100 candidates have their average height 163.8 cm with coefficient of variation 3.2 %.What is the standard deviation of their height? −
σ x 163.8 Solution: C.V. = − × 100% ; σ = 100 × C.V = 100 × 3.2 = 5.24 x Example 24: A factory has two sections with 50 and 60 employees respectively. Their average weekly wages are Rs.386 and Rs.475. The standard deviations are 9 and 10. (i). Which section has a larger wage bill? (ii) Which section has greater variability in wages? Solution : i) Wage bill for section A = 50 x 386 = Rs.19,300. Wage bill for section B = 60 x 475 = Rs.28,500 ⇒ Section B has larger wage bill. 9 10 ii) C.V. for section A = 386 × 100 − 2.33% ; C.V. for section B = 475 × 100 = 2.11% Section B is more consistent ⇒ There is greater variability in the wages of section A. Example 25 : The mean of the runs scored by three batsman A, B and C in the same series of 10 innings are 58,48 and 14 respectively. The S.D. of their runs are 15, 12 and 2 respectively who is the most consistent batsman?
218
Solution:
Player Mean S.D. A 58 15 B 48 12 C 14 2 12 15 C.V. for A = 58 × 100 = 25.86% ; C.V. for B = 48 × 100 = 25.0% ; 2 C.V. for C = 14 × 100 = 14.29% Since C.V. for C is the least ⇒ C is the most consistent batsman. Exercise 10.1 1. The score of a batsman are 38, 70, 48, 34, 42, 56. Find the standard deviation. 2. Calculate the standard deviation of the marks secured by a student of Std X in the terminal exam as given: 87, 79, 65, 63 80. 3. There are five post bags which are weighed as 11,14, 15, 17, 18 (kgs). Calculate the standard deviation and variance. 4. Calculate the standard deviation and variance of the heights (in mm) of 5 men as given: 1593, 1640, 1614, 1633, 1586. 5. Find the standard deviation of the following data a+5, a+3, a+10, a+2, a+10 (Hint : First take the data as 5, 3, 10, 2 10). 6. The variance of 5 scores in 16. If each one of them is divided by 2, find the standard deviation and variance of the new scores. 7. Mean of 200 items is 80 and their standard deviation is 10. Find the sum of the items and also the sum of the squares of all the items. 8. The sum of 10 observations is 60; Σ (x – x )2 = 36. Find Σx2 and Σ(x – 5)2. 9. The mean and standard deviation of 18 observations are 7 and 4 respectively. But on comparison with original data it was found that the value 12 was incorrectly taken as 21. Calculate the correct mean and standard deviation. 10. The mean and standard deviation of 200 items are found to be 60 and 20 respectively. If at the time of calculation two items were wrongly taken as 3 and 67 instead of 13 and 17, find the correct mean and standard deviation. 11. The pocket expenses (in Rs.) of 10 students are 30,50, 60, 50, 80, 50, 40, 80, 90, 70. Calculate the coefficient of variation. 12. The C.V. of two series are 75% and 90%. Their S.D. are 15 and 18 respectively. Find their arithmetic means. 13. The coefficient of variations of two series are 60% and 80%. The arithmetic means are 33.3 and 20 respectively. What are their S.Ds? 14. The following are the details of weekly wages paid to a worker in two branches A and B of a firm:
219
A B No of workers 160 150 Average wage Rs. 369 Rs. 427 Variance 100 121 Find the branch in which there is greater variability in wages. 15. Information regarding the price movements of the shares of three companies is given below. Company Average Price (in Rs.) S.D. (in Rs) A 18.00 5.40 B 22.50 4.50 C 24.00 6.00 Which company’s share is more stable in price?
10.2 PROBABILITY The probability theory has its origin in the games of chance pertaining to gambling. Jerome Cardon an Italian mathematician wrote a book on ‘Games of chance’ published in 1663. French mathematicians Blaise Pascal and Pierre de Fermat developed a systematic procedure for probability theory. A Swiss mathematician James Bernoulli made an extensive study on probability. Abraham de Moivre (1667 – 1754) and Thomas Bayes (1702 – 1761) made significant contribution to this theory. In 19th century, Pierre Simon de Laplace made an extensive research and published his book on ‘Theory of Analytical Probability’ in 1812 which became the first book in theory of probability. Consider the following statements: Our cricket team is likely to win the worldcup. 1. Meterological officials predict that the rainfall will be normal during the monsoon. 2. The prices of essential commodities are likely to be stable. 3. In all these cases, when we arrive at a decision we face uncertainity. The possibility (chance) for happening is called probability. The following terms are used in studying the theory of probability. Random experiment: An operation which produces an outcome is known as experiment. When an experiment is conducted repeatedly under the same conditions the results can not be unique but may be one of the various possible outcomes. Such an experiment is called a random experiment. In a random experiment, we cannot predict the outcome.
Tossing a coin is a random experiment. When we toss a coin either head or tail may turn up. Some more examples of random experiment: 1. Rolling a die. 2. Drawing a card from a pack of cards. Taking out a ball from a bag containing balls of different colours. 3. Trial : Performing a random experiment is called a trial. Sample space : The set of all possible outcomes of a random experiment is called a sample space and is denoted by S. 220
When we roll a die, the possible outcomes are 1, 2, 3, 4, 5, 6 . Sample space is S = { 1, 2, 3, 4, 5, 6} Event: Any possible outcome or combination of outcomes is called an event. That is every subset of the sample space S is called an event. Events are usually denoted by A, B, C, D, E, F. When a coin is tossed, getting a head or tail is an event. S = { H, T}, A = {H}, B = {T}. Here events A and B are subsets of the sample space S. Equally likely events: Two or more events are said to be equally likely if each one of them has an equal chance of occurring. In tossing a coin, getting a head and getting a tail are equally likely events. Mutually exclusive events: Two or more events are said to be mutually exclusive when the S occurrence of anyone event excludes the occurrence of the other event. Mutually exclusive events can not occur simultaneously. In throwing a die, let E be the event of getting E F an odd number and F be the event of getting an even 2 number. 1 E = {1, 3, 5} 5 6 F = {2, 4, 6} 3 4 The number that turns up cannot be odd and even simultaneously. Therefore events E and F are mutually exclusive. Fig.10.1 Exhaustive events: If two or more events together constitute the sample space S then these events are said to be exhaustive events. In throwing a die, the events of getting an odd number and the event of getting an even number together form the sample space. So they are exhaustive events. In an experiment of tossing three coins, consider the following events. A: exactly one head appears, B: exactly two heads appear C: exactly three heads appear D : atleast two tails appear S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} A = {HTT, THT, TTH} B = {HHT, HTH, THH} C = {HHH} D = { TTH, THT, HTT, TTT} S The events A, B, C and D together form the sample space S. That is, S = A∪B∪C∪D. Therefore A, B, C and D are called C B exhaustive events. Note : The events B, C and D are mutually exclusive and exhaustive.
A
Complementary events: Let E be an event of a random experiment and S the sample space. All the other outcomes which are not in E belong to the subset S–E. The event S – E is called the complement of E. It is denoted by E . Note that E and E are mutually exclusive and exhaustive events.
D
Fig.10.2
S 1
221
3
2 6
4 5
In throwing a die, let E be an event of getting a ‘multiple of 3’ E = { 3, 6} E = { 1, 2, 4, 5} Sure Event: Since S ⊆ S, S itself is an event and S is called sure or certain event. Fig.10.3
In tossing two coins simultaneously, let E be an event of getting less than 3 tails. E = { HH, HT, TH, TT} = S Therefore E is the sure event. Impossible Event : Let F be an event of getting more than two heads in tossing two coins simultaneously.. F = { } = ϕ. So F is an impossible event. ⇒ A sample space is a sure event. An empty set ϕ is an impossible event. Favourable outcomes: The outcomes corresponding to the desired event are called the favourable outcomes.
In rolling a die there are six outcomes. Let E be an event of getting an even number. Then the outcomes 2, 4, 6 are favourable to the event E. Probability of an event: If a sample space contains n outcomes, m of which are favourable to an event E, then the probability of an event E, denoted by P (E), is defined as the ratio of m to n. Number of favourable outcomes for E n(E) m m P(E) = = or P(E) = = Total number of outcomes n n(S) n
In tossing a coin, let E be the event of getting a head and F be the event of getting a tail S = { H, T}, E = {H}, F = { T } n(E) 1 n(F) 1 = ; Probability E, F are given by P(E) = P(F) = = n(S) 2 n(S) 2 Note : (1) The probability of sure event is 1. That is P(S) = 1. (2) The probability of impossible event is 0. That is P(ϕ) = 0. (3) The probability of any event E is between 0 and 1. That is 0 ≤ P(E) ≤ 1 (4) The probability of non-occurrence of an event E, denoted by E , is given by −
P(E) =
Number of outcomes not favourable for E n − m m = n = 1− n Total number of outcomes −
P(E) = 1 − P(E) Note : Let us try to understand the meaning of A ∪ B and A ∩ B in the context of probability theory. A ∪ B means ‘event A or the event B or both’. A ∩ B means ‘event A and event B’
222
Thus by writing P (A ∪ B) we mean ‘Probability that event A or event B occurs or both events occur’. Again by P(A ∩ B) we mean ‘Probability that event A and event B occur’. Example 26 : A fair die is rolled. Find the probability of getting i) 3 on the face of the die iii) a number greater than 1 on the die. ii) an odd number on the face of the die. iv) prime factors of 6 on the die. Solution: In rolling a die, the sample space S ={ 1, 2, 3, 4, 5, 6} : n (S) = 6. (i) Let A be an event of getting 3 n(A) 1 A = { 3 }, n (A) = 1 = ∴ P(A) = n(S) 6 (ii) Let B be an event of getting an odd number n(B) 3 1 P(B) = = = B = { 1, 3, 5}, n (B) = 3 ∴ n(S) 6 2 (iii) Let C be an event of getting a number greater than 1 n(C) 5 P(C) = = ∴ C = { 2, 3, 4, 5, 6}, n (C) = 5 n(S) 6 (i) Let D be an event by getting prime factors of 6 n(D) 2 1 ∴ D = { 2, 3}, n (D) = 2 P(D) = = = n(S) 6 3 Example 27: One card is drawn at random from a shuffled pack of 52 cards. What is the probability that it will be (i) a spade (ii) a black card (iii) a king (iv) any card except queen. Solution : The total number of cards n (S)= 52. (i) Let A be the event of drawing spade. There are 13 spade cards, n(A) = 13. n(A) 13 1 P(A) = = = n(S) 52 4 (ii) Let B be the event of drawing a black card. There are 26 black cards, n (B) = 26. n(B) 26 1 P(B) = = = n(S) 52 2 (iii) Let C be the event of drawing a king card. There are 4 king cards, n(C) = 4 n(C) 4 1 P(C) = = = n(S) 52 13 (iv) Let D be the event of drawing a card which is not a queen. There are 4 queen cards. Therefore n(D) = 52–4 = 48. n(D) 48 12 P(D) = = = n(S) 52 13 Example 28: Two coins are tossed simultaneously. What is the probability of getting two heads?. Solution: In tossing two coins the sample space S = {HH, HT, TH, TT}, n(S) = 4. Let A denote the event of getting two heads A = {HH}, n(A) = 1. n(A) 1 = Therefore P(A) = n(S) 4
223
Example 29: An integer is chosen at random from 1 to 50. Find the probability that the number is divisible by 5. Solution : Sample space S = { 1, 2, 3, …. 50}, n(S) = 50. Let A denote the event of getting a number divisible by 5. So, A = { 5, 10, 15, 20, 25, 30, 35, 40, 45, 50}, n(A) = 10. n(A) 10 1 P(A) = = = n(S) 50 5 Example 30: There are 5 items defective in the sample of 25 items. Find the probability that an item chosen at random from the sample space is not defective. Solution : Total number of items n(S) = 25. Number of defective items = 5. Number of items which are not defective = 25 – 5 = 20. n(A) 20 4 Let A be the event of selecting an item which is not defective. P(A) = = = n(S) 25 5 Example 31 : If three coins are tossed, then what is the chance of getting exactly one head? Solution : Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Let A be the event of getting exactly one head. A = { HTT, THT, TTH}, n(A) =3. n(A) 3 P(A) = = ∴ n(S) 8 Example 32: Four coins are tossed simultaneously. What is the probability of getting exactly one tail?. Solution: In tossing four coins simultaneously the sample space. S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, TTHH, THHT, HTTT, THTT, TTHT, TTTH, TTTT} Let A be the event of getting exactly one tail. A = { THHH, HTHH, HHTH, HHHT}, n(A) = 4, n(S) = 16. n(A) 4 1 = = Therefore P(A) = n(S) 16 4 Example 33: What is the probability that (a) a leap year selected at random will contain 53 Sundays. (b) a non-leap year selected at random will contain 53 Sundays. Solution: (a) Number of days in a leap year = 366 days = 52 weeks + 2 days. 52 weeks contain 52 Sundays. 2 The chances of 53rd Sunday falling on one of the remaining two days period = . 7 2 Therefore the probability of 53 Sundays in a leap year . 7 (b) Number of days in a non-leap year = 365 days = 52 weeks + 1 day. There are 52 Sundays and the remaining one day may be any one of the seven days. 1 Therefore the probability of 53 Sundays in a non-leap year = . 7 224
Example 34: Two dice are thrown. What is the probability of getting (i) the same number on both dice (ii) a total of face numbers 12 Solution : The sample space in throwing two dice is given by S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. i) Let A be the event of getting the same number on the dice A = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }; n(A) = 6, n(S) = 36 n(A) 6 1 Therefore P(A) = = = n(S) 36 6 ii) Let B be the event of getting the total of face numbers 12. n(B) 1 = B = {(6,6)}, n(B) = 1. Therefore P(B) = n(S) 36 Example 35: Three dice are rolled once. What is the chance that the sum of the face numbers on the three dice is greater than 15? Solution: When three dice are rolled, the sample space S = {(1,1,1), (1,1,2), (1,1,3) ...(6,6,6)}. S contains 6 × 6 × 6 = 216 outcomes. Let A be the event of getting the sum of face numbers greater than 15.
A = { (4,6,6), (6,4,6), (6,6,4), (5,5,6), (5,6,5), (6,5,5), (5,6,6), (6,5,6), (6,6,5), (6,6,6)}. n(S) = 216, n(A) = 10. n(A) 10 5 . Therefore P(A) = = = n(S) 216 108 Example 36: A point is chosen at random inside a circle of radius 2. What is the probability that this point is nearer to the centre than to the circle. Solution: Radius of the circle is 2. Consider a concentric circle of radius 1. If the chosen point P is inside the smaller circle then it is nearer to the O 1 centre O than to the circle. Probability that the point is nearer to the centre than to the circle P 2 π× 1 1 Area of the circle with radius 1 2 = Area of the circle with radius 2 = 2 = 4 π×2 Fig.10.4 S
Addition theorem on probability A
If A and B are any two events associated with a sample space S, then P (A∪B) = P (A) + P(B) – P(A∩B) or P (A or B) = P(A) + P(B) – P(A and B) Proof: Let the events A and B of a sample space S be represented as sets in the diagram. From set theory, we know n(A∪B) = n(A) + n(B) – n(A∩B) Divide both sides by n(S)
225
B A
B
Fig.10.5
n(A ∪ B) n(A) n(B) n(A ∩ B) ⇒ P(A∪B) = P(A) + P(B) – P(A∩B) n(S) = n(S) + n(S) − n(S) Corollary : If A and B are mutually exclusive events then A ∩ B = ϕ. Therefore P(A∩B) = 0. The addition theorem on probability for mutually exclusive events is P(A∪B) = P(A) + P(B) Extension of Addition theorem on probability for three events
If A, B and C are any three events associated with a sample space S, then P(A∪B∪C) = P(A) + P(B) + P(C) – P(A∩B) – P(B∩C) – P(A∩C) + P (A∩B∩C) and if A, B and C are mutually exclusive, then P(A∪B∪C) = P(A) + P(B) + P(C). Example 37: Two dice are thrown, what is the probability of getting ‘a multiple of 3 on the face of the first die or the total of the numbers on the faces is 7’. Solution : Sample space in throwing two dice is S = { (1,1) , (1,2), (1,3), …….(6,6)} Let A denote the event of getting a multiple of 3 on the first die. A = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} Let B denote the event of getting the total of 7. B = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} A∩B = {(3,4), (6,1)}. n(S) = 36, n(A) = 12, n(B) =6, n(A∩B) = 2 n(A) 12 1 P(A) = = = n(S) 36 3 n(B) 6 1 P(B) = = = n(S) 36 6 n(A ∩ B) 2 1 P(A ∩ B) = n(S) = 36 = 18
Required probability is given by P(A∪B) = P(A) + P(B) – P(A∩B) 1 1 1 8 4 + = = = 3 6 18 18 9 Example 38: A card is drawn from the pack of 52 cards. Find the probability of getting a spade or heart. Solution : Total number of cards n(S) =52. Let A be the event of drawing a spade and B the event of drawing a heart n(A) = 13, n(B) = 13. 13 1 13 1 P(A) = = , P(B) = = 52 4 52 4 A∩B denote the event of drawing a card which is spade and heart. But there is no such card in the pack. That is A∩B = ϕ, n(A∩B) = 0. 1 1 1 A, B are mutually exclusive ⇒ P(A ∪ B) = P(A) + P(B) = 4 + 4 = 2 Example 39: Two dice are thrown. If A is the set outcomes with odd sum of the face numbers of the dice, B the set of outcomes with even sum, C the set of outcomes with sum 11 and D the set of outcomes with sum 4 then find the following. i) P(A), P(B) and P(A∪B); ii) P(C), P(D) and P(C∪D); iii) P(A∩C) and P(B∩D) 226
Solution : The following table shows all possible outcomes of throwing of two dice. Sum of the face numbers on the dice are entered in the table.
+ 1 2 3 4 5 6 A B C
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
= {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)} = {(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)} = { (5,6), (6,5)}; D {(1,3), (2,2), (3,1)}
n(S) = 36, n(A) = 18, n(B) = 18, n(C) = 2 , n(D) = 3 i)
P(A) =
18 1 18 1 = , P(B) = = 36 2 36 2
1 1 A and B are mutually exclusive events ⇒ P(A ∪ B) = P(A) + P(B) = 2 + 2 = 1 2 1 3 1 P(C) = ii) = , P(D) = = 36 18 36 12 1 1 2+3 5 C and D are mutually exclusive events ⇒ P(C ∪ D) + P(C) + P(D) , 18 , 12 = 36 = 36 iii)
A ∩C = {(5,6), (6,5)}, n(A ∩ C) = 2
B∩D = {(1,3), (2,2), (3,1)}, n(B∩D) = 3
2 1 3 1 P(A ∩ C) = 36 = 18 ; P(B ∩ D) = 36 = 12 Example 40: A die is thrown twice what is the probability that atleast one of the two throws comes up with the number 4? Solution: Throwing a die twice is same as throwing two dice once. Therefore n(S) = 36.
Let A be the event of getting 4 in the first throw and B be the event of getting 4 in the second throw. A = { (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) }; B = {(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)} A∩B = {(4,4)}
227
6 6 1 P(A) = 36 ; P(B) = 36 ; P(A ∩ B) = 36 6 11 11 6 y addition theorem P(A∪B) = P(A) + P(B) – P(A∩B) = 36 + 36 − 36 = 36 Example 41 : The probability that a student passes a mathematics test is 2/3, the probability that he passes both mathematics and science tests is 14/45. The probability that he passes atleast one test is 4/5. What is the probability that he passes the science test. 2 14 4 Solution: Given values are P(M) = 3 , P(M ∩ S) = 45 and P(M ∪ S) = 5 4 2 14 P(M∪S) = P(M) + P(S) – P(M∩S) or = + P(S) − 5 3 45 4 P(S) = . 9 Independent events: In a random experiment two events are called independent events of the occurrence or non-occurrence of one event does not influence the occurrence or nonoccurrence of the other event. If two events A and B are independent, then P(A∩B) = P(A) × P(B). That is P(A and B) = P(A) P(B) Let us consider the following random experiment, First a coin is tossed and then a die is rolled. Let A be the event : the coin shows head and B be the event: the die shows an even number A and B are independent events. Sample space S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} A = { H1, H2, H3, H4, H5, H6} B = [H2, H4, H6, T2, T4, T6}; A ∩ B = { H2, H4, H6}
P(A) =
6 1 = ; 12 2
P(B) =
3 1 1 1 1 6 1 = ; P(A ∩ B) = 12 = 4 ; P(A) P(B) = 2 × 2 = 4 12 2
Therefore P(A∩B) = P(A) P(B) Example 42 : Two friends A and B appeared for an interview for two vacancies. The 1 1 probability of A’s selection is and that of B’s selection is . Find the probability that 3 2 i) both of them will be selected ii) only one of them will be selected iii) none of them will be selected. 1 1 Solution: P(A) = Probability of selecting A = 3 P(B) = Probability of selecting B = 2 1 2 P(A) = Probability of not selecting A = 1 − 3 = 3 1 1 P(B) = Probability of not selecting B = 1 − 2 = 2 Selection or non-selection of any one of the candidate is not affecting the selection of the other. Therefore A and B are independent events.
228
i)
1 1 1 Probability of selecting both A and B = P(A∩B) = P(A) P(B) = 3 × 2 = 6
ii)
Probability of selecting anyone of them = P(Selecting A and not selecting B) + P(not selecting A and selecting B)
(
−
) (
−
)
= P A ∩ B + P A ∩ B = P(A) P(B) + P(A) P(B) 1 1 2 1 1 2 3 1 =3×2+3×2=6+6= = 6 2 − − − − 2 1 1 Probability of not selecting both A and B = P A ∩ B = P A P B = 3 × 2 = 3 Example 43 : A bag contains 5 green and 7 red colour balls. Two colour balls are taken out one after the other, from the bag. What is the probability of one is green and the other is red? Solution: There are 5 green and 7 red balls. Probability of taking two balls, 1 green and 1 red - Probability of taking a green ball first and a red ball next + probability of taking a red all first and a green ball next. That is (taking 1 green, 1 red balls) = P (G ∩R) + P (R∩G) 5 7 P(taking a green ball first) = ; P (taking a red ball second) = 12 11 7 35 5 ∴ P (G ∩R) = P (G) P(R) = × 11 = 132 Since G and R are independent events 12 7 5 P(taking a red ball first) = ; P (taking a green ball second) = 12 11 7 5 35 ∴ P (R ∩G) = P(R) P(G) = 12 × 11 = 132 since R and G are independent events. Probability of taking two balls, 1 green and 1 red ball = P (G ∩R) + P (R ∩G) 35 35 70 35 = = + = 132 132 132 66 Exercise 10.2
(
iii)
1. 2. 3. 4. 5. 6.
) () ()
A die is rolled once. Find the probability of getting i) an even number ii) a number less than 5 iii) a number multiple of 3 In a simultaneous toss of two coins, find the probability that i) two heads turn up ii) exactly one head turns up A card is drawn from a pack of 52 cards find the probability that i) it is a red card ii) it is an ace. An integer is chosen from 1 to 50. Find the probability that it is a multiple of? An integer is chosen from 10 to 30. Find the probability that it is a prime number. Find the probability of choosing a perfect cube from the numbers 1 to 300. (Hint : perfect cubes are 1, 8, 27...)
229
7. 8.
9. 10. 11. 12. 13. 14.
15.
There are 5 dozen oranges in a box. Four oranges are rotten. Find the probability of choosing a good orange. A coin is tossed three times. Find the probability of getting i) head and tail alternatively iii) atleast two heads ii) exactly two heads iv) no head In a simultaneous loss of four coins, what is the probability of getting i) exactly 2 heads ii) more than 2 heads iii) less than 2 heads iv) atleast one head Three dice are thrown. Find the probability that each of the three dice shows the same number on its face. From a set of 17 cards numbered 1,2,3... 17 one card is draw at random. Find the chance that the number is a multiple of 3 or 7. A card is drawn from a pack of 52 cards. Find the probability that it is either red card or king card. Two dice are rolled once. Find the probability of getting an even number on the second die or the total of face numbers 10. 4 2 3 The probability that A, B and C can solve a problem are , and respectively. If A 5 3 7 8 2 and B can solve with probability , B and C can solve with probability , A and C 15 7 12 8 can solve with probability all the three can solve the problem with probability , 35 35 what is the probability that the problem gets solved. Two persons X and Y appeared in an interview for two vacancies in an office. The chance 1 1 for X’s selection is and the chance for Y’s selection is . Find the chance that (i) both 5 7 of them are selected (ii) only one of them is selected. (iii) none of them is selected. ANSWERS
Exercise 10.1 1) 12 .11 5) 3.4 9) 6.5, 2.5 13) 19.98, 16
2) 9.26 6) 2, 4 10) 59.8, 20.09 14) Company A
3) 2.45 kg, 6 kg 7) 16000, 1300000, 11) 30.67 % 15) Company B
Exercise 10.2 1 2 1 1 1 1 1. (i) (ii) (iii) 2. (i) (ii) 3. (i) 2 3 3 4 2 2 14 1 3 1 1 3 7. 8. (i) (ii) (iii) (iv) 9.(i) 15 4 8 2 8 8 7 7 19 101 11. 12. 13. 14. 15. (i) 17 13 36 105
230
4) 21.25 mm, 451.76 mm 8) 396, 46 12) 20, 20
1 7 2 4. 5. 13 50 7 5 5 15 (ii) (iii) (iv) 16 16 16 1 2 24 (ii) (iii) 35 7 35 (ii)
6.
1 50
10.
1 36
11. GRAPHS 11.0 INTRODUCTION In several fields we come across variables assuming real values. The different variables that govern a particular situation may be connected by an equation. Using the equation connected by two variables x and y we can get a value for y for each given value of x and obtain a set of ordered pair (x, y) of real numbers. All these ordered pairs (x,y) can be plotted as points in the Cartesian plane where the horizontal line represents x-axis and the vertical line represents y-axis. These points now define the graph of the equation. The graph shows the nature of relationship between the variables x and y. In Class IX we have seen some linear graphs. In this chapter we are going to draw quadratic graphs (parabolas) and some special graphs. Using these graphs we are going to solve quadratic equations. Graphs drawn attractively are eye-catching. Graphs are a good visual aid. But graphs do not give accurate solution of the variables as are obtained by calculations. However if the curves are drawn plotting more number of points then the graph will be very smooth ensuring solution of better accuracy.
11.1 QUADRATIC
GRAPHS
If two variables x and y are connected by an equation of the form y = ax2 + bx + c then it is called a quadratic polynomial. We have already seen in Algebra that the equation ax2 + bx + c = 0 is a quadratic equation. This is the reason why we call y = ax2 + bx + c as a quadratic graph. For each value of x the equation y = ax2 + b x + c gives a value of y and we obtain an ordered pair (x, y) of real numbers. The set of all such ordered pairs define the graph of y = ax2 + bx + c called the quadratic graph. To draw the quadratic graph y = ax2 + bx + c We now proceed to draw quadratic graphs of different nature. The following procedure is used in drawing quadratic graphs.
• By substituting selected values for x in the equation y = ax2 + bx + c, we get corresponding values for y and then we form a table. • Draw x-axis and y-axis on the graph sheet and choose a suitable scale on the co-ordinates axes. The scale for both the axes are chosen depending on the values of the co-ordinates obtained. • Plot the points in the Cartesian plane of the graph sheet. • Join these points by a smooth curve, then we get the required quadratic graph of y = ax2 + bx + c. Note: All figures given in this chapter are not to scale. 231
Example 1: Draw the graph of y = x2 and use this to find the square root of 15. Solution: Draw x-axis and y-axis on the graph sheet. Mark the scales on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Assign values for x = –4 to 4 and we get the corresponding y values tabulated as follows: x y = x2
–4 16
–3 9
–2 4
–1 1
0 0
1 1
2 4
3 9
4 16 x-axis : 1 cm = 1 unit y-axis : 1 cm = 2 units
y
Plot the points (–4,16), (–3,9), (–2,4), (–1, 1), (0,0), (1,1), (2,4), (3,9), (4,16) on the graph sheet and join the points by smooth curve. This curve is called the parabola y = x2 (Fig.11.1).
(-4,16) 2
y=x
(-3,9)
Note that it is symmetrical about y-axis. In the graph, draw a horizontal dotted line through y = 15 to intersect the curve at a point and draw a vertical dotted lines through that point which meets the x-axis at x = ± 3.9. Therefore the square root of 15 is ±3.9 approximately.
(-2,4)
20 18 15 16 14 12 10 8 6 4 2
(-1,1) -7 -6 -5 -4 -3 -2 -1
x’
(4,16)
(3,9) (2,4)
(1,1) o 1 2 3 4 5 6 7 -2 -4 3.9 -6 -8
x
y’
Fig.11.1 Example 2: Draw the graph of y = –x2 Solution: Draw x-axis and y-axis on the graph sheet. Mark the scales as on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Assign values from –4 to 4 for x to get the corresponding values for y. We form a table as follows: x –4 y = –x2 –16
–3 –9
–2 –4
–1 –1
0 0
1 –1
2 –4
3 –9
4 –16
y
Plot the points (–4, –16), (–3, –9), (–2, –4), (–1, –1), (0,0), (1, –1), (2, –4), (3, –9), (4, –16) on the graph sheet and join these points by a smooth curve. We get the required parabola y = –x2 (Fig.11.2).
x’
-7 -6 -5 -4 -3 -2 (-2,-4) (-3,-9)
y=- 2 x
(-4,-16)
-2 -4 -6 -8 -10 -12 -14 -16 -18
x-axis : 1 cm = 1 unit y-axis : 1 cm = 2 units
6 4 2 (0,0) o 2 3 4 5 6 7 (2,-4) (3,-9)
(4,-16)
y’
Fig.11.2 232
x
Example 3: Draw the graph of y = x2 – 4 Solution: Mark the scales on the graph sheet on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Assigning values x = –4 to 4 we calculate the values for y and prepare the following table : x x2 –4 y = x2 – 4
–4 16 –4 12
–3 9 –4 5
–2 4 –4 0
–1 1 –4 –3
0 0 –4 –4
1 1 –4 –3
2 4 –4 0
3 9 –4 5
x-axis : 1 cm = 1 unit y-axis : 1 cm = 2 units
y
(-4,12)
16 14 12 10 8 6 4 2
2
y=x
-4
We plot the points (–4,12), (–3,5), (–2,0) (–1,–3), (0, –4), (1, –3), (2,0), (3,5), (4,12) on the graph sheet. Join these points by a smooth curve we get the required graph of the parabola y = x2 – 4 (Fig.11.3).
4 16 –4 12
(-3,5)
(4,12)
(3,5)
(2,0) (-2,0) -7 -6 -5 -4 -3 -2 -1 o 1 2 3 4 5 6 7 -2 (1,-3) (-1,-3) -4 (0,-4) -6
x’
x
y’
Fig.11.3 Example 4: Draw the graph of y = x2 – 2x – 3. Solution: Draw x-axis and y-axis on the graph sheet and mark the scales on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Assign values x = –4 to 5 and calculate the corresponding y values to form a table as follows: x x2 –2x –3 y = x2 – 2x – 3
–4 16 8 –3 21
–3 9 6 –3 12
–2 4 4 –3 5
–1 1 2 –3 0
0 0 0 –3 –3
1 1 –2 –3 –4
2 4 –4 –3 –3
3 9 –6 –3 0
4 16 –8 –3 5
x-axis : 1 cm = 1 unit y-axis : 1 cm = 2 units
y
(-4,21)
2
y=x
We plot the points (–4,21), (–3,12), (–2,5), (–1,0), (0, –3), (1, –4), (2, –3), (3,0), (4,5) and (5,12) on the graph sheet and join these points by a smooth curve. We get a required graph of the parabola y = x2 – 2x – 3 (Fig.11.4).
- 2x -
(-3,12)
3
(-2,5) x’
5 25 –10 –3 12
24 22 20 18 16 14 12 10 8 6 4 2
(4,5)
(3,0) (-1,0) -7 -6 -5 -4 -3 -2 -1 o 1 2 3 4 5 6 7 -2 (2,-3) (0,-3) -4 (1,-4) -6 y’
Fig.11.4 233
(5,12)
x
Example 5: Draw the graph of y = 2x2 – 5x + 2 Solution: Draw x-axis and y-axis on the graph sheet. Mark the scales on x-axis 1 cm = 1 unit and on y-axis 1 cm = 5 units. Assign values from x = –4 to 4 to calculate the values for y. We form a table as follows: x 2x2 –5x +2 y = 2x2 – 5x + 2
–4 32 20 2 54
–3 18 15 2 35
–2 8 10 2 20
–1 2 5 2 9
0 0 0 2 2
We plot the points (–4,54), (–3,35), (–2,20), (–1,9), (0,2), (1, –1), (2,0), (3,5), (4,14) on the graph sheet. Join these points by smooth arcs to get the required graph of the parabola y = 2x2 – 5x + 2 (Fig.11.5).
1 2 –5 2 –1
2 8 –10 2 0 y
2 x+ x -5
(-4,54)
y =2 2
To solve the quadratic ax2 + bx + c = 0 graphically
(-3,35)
equation
(-2,20)
While studying Algebra we have solved the quadratic equation ax2 + bx + c = 0 by algebraic method. Now we are going to solve this equation by graphical method. The quadratic equation has two distinct real roots or two equal real roots or no real roots.
3 18 –15 2 5
x’
4 32 –20 2 14
x-axis : 1 cm = 1 unit y-axis : 1 cm = 5 units
55 50 45 40 35 30 20 15 10 5 (2,0)
(4,14) (-1,9) (3,5) (0,2) -7 -6 -5 -4 -3 -2 -1 o 1 2 3 4 5 6 7 -5 (1,-1) -10 -15
x
y’
Fig.11.5 In this section while we solve the quadratic equations by graphical method we adopt two methods. In the first method, we draw the graph of the given equation and get the points of intersection of the curve with x-axis to determine the roots. In the second method we split the quadratic equation into two equations representing a parabola and a straight line. The x co-ordinates of the points of intersection of the parabola and the straight line will give us the roots of the given quadratic equation. Example 6: Solve graphically the equation x2 + 3x – 4 = 0 Solution: First let us draw the graph of y = x2 + 3x – 4 by the usual method. Let us form a table as follows: x x2 +3x –4 y = x2 + 3x – 4
–5 –4 25 16 –15 –12 –4 –4 6 0
–3 9 –9 –4 –4
–2 4 –6 –4 –6
–1 1 –3 –4 –6
0 0 0 –4 –4
1 1 3 –4 0
2 4 6 –4 6
3 9 9 –4 14
4 16 12 –4 24
Choose on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Plot the above points and join these points by a smooth curve. We get the graph of y = x2 + 3x – 4 (Fig.11.6). Now by solving the equations, we get 234
y = 0 = – y =
x2 + 3x – 4 x2 + 3x – 4 – – + 0
x axis : 1 cm = 1 unit y axis : 1 cm = 2 units
y
2
y=x
24 22 20 18 16 14 12 10 8 6 4 2
+ 3x 4
which is the equation of x-axis. The points of intersection of the parabola with x-axis are (–4,0) and (1,0). Therefore the solution set is {–4,1}.
(-5,6) (-4,0) -7 -6 -5 -4 -3 -2 -1
x’
∴ The roots of the equation x2 + 3x – 4 = 0 are –4 and 1.
(-3,-4) (-2,-6) (-1,-6)
(4,24)
(3,14)
(2,6)
(1,0) o 1 2 3 4 5 6 7
x
-2 -4 (0,-4) -6 y’
Fig.11.6 Example 7: Solve graphically 2x2 – x – 3 = 0 Solution: First form a table for the parabola y = 2x2 – x – 3 x 2x2 –x –3 y = 2x2 – x – 3
–4 32 4 –3 33
–3 18 3 –3 18
–2 8 2 –3 7
–1 2 1 –3 0
0 0 0 –3 –3
1 2 –1 –3 –2
2 8 –2 –3 3
3 18 –3 –3 12
y
Choose the scales on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Mark the above points and join them by smooth arcs. We get the graph of the parabola y = 2x2 – x – 3 (Fig.11.7). Solving the equation of the parabola and the given equation we get
(-4,33) 2 -3 y = 2x - x
y = 0 = – y =
(-3,18)
2x2 – x – 3 2x2 – x – 3 – + + 0
This is the equation of x-axis. The points of intersection of the curve with x-axis are (–1,0) and (3/2, 0), (or) (1.5, 0). The x-co-ordinates of these points are the roots of the given equation. ∴The solution set is {–1,1.5}
4 32 –4 –3 25
(-2,7)
(-1,0) -7 -6 -5 -4 -3 -2 -1
x’
32 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2
x axis : 1 cm = 1 unit y axis : 1 cm = 2 units
(4,25)
(3,12)
(2,3)
o 1 2 3 4 5 6 7 -2 (1,-2) (0,-3)-4 -6 y’
Fig.11.7
235
x
Example 8: Solve graphically (2x + 1) (3 – x) = 0. Solution: First form a table for the parabola y = (2x + 1) (3 – x) = 6x – 2x2 + 3 – x = –2x2 + 5x + 3 x –2x2 +5x +3 y = –2x2 + 5x + 3
–3 –2 –18 –8 –15 –10 3 3 –30 –15
–1 0 1 2 –2 0 –2 –8 –5 0 5 10 3 3 3 3 –4 3 6 5
3 4 5 –18 –32 –50 15 20 25 3 3 3 0 –9 –22
x axis : 1 cm = 1 unit y axis : 1 cm = 2 units
y
Choose the scale on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Plot the above points on the graph sheet and join these points. We get the graph of the parabola y = –2x2 + 5x + 3 (Fig.11.8).
x’
x
2 5x + 3 y = -2 x +
10 8 (1,6) (2,5) 6 4 (0,3) 2 (3,0) -7 -6 -5 -4 -3 -2 -1 o 1 2 3 4 5 6 7 -2 (-1,-4) -4 -6 -8 (4,-9) -10 -12 -14 (-2,-15) -16 -18 -20 -22 (5,-22) -24 -26 -28 (-3,-30) -30 -32
By solving y = –2x2 + 5x + 3 and 0 = –2x2 + 5x + 3 we get y = 0, the x-axis. The x-co-ordinates of the points of intersection of the parabola with x-axis give the solution of the given equation.
y’
∴ The solution set is {–0.5, 3}
Fig.11.8 2
Example 9: Draw the graph of y = x + 3x and solve the equation x2 + 3x = 0 Solution: First we form the following table for the parabola y = x2 + 3x y
–3 9 –9 0
–2 4 –6 –2
–1 1 –3 –2
0 0 0 0
1 2 1 4 3 6 4 10
3 9 9 18
4 16 12 28
Choose the scale on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units and plot the above points on the graph sheet. By joining these points we get the parabola y = x2 + 3x (Fig.11.9). Solving the equation y = x2 + 3x and x2 + 3x = 0 we get y = 0, the x-axis. The roots of the required equation are the x-co-ordinates of the points of intersection of the parabola with x-axis. ∴ The solution set is {–3, 0}.
x axis : 1 cm = 1 unit y axis : 1 cm = 2 units
30 28 (4,28) 26 24 22 20 18 (3,18) 16 14 12 10 (2,10) 8 6 4 (1,4) 2 y = x2 + 3x
x –5 –4 x2 25 16 3x –15 –12 y = x2 + 3x 10 4
(-5,10)
(-4,4) x’
(-3,0) -7 -6 -5 -4 -3 -2 -1 o 1 2 3 4 5 6 7 (-2,-2) (-1,-2) -2 -4 -6 y’
Fig.11.9 236
x
Example 10: Draw the graph of y = x2 – x – 8 and hence solve the equation x2 – 2x – 15 = 0. Solution: First we form a table for y = x2 – x – 8 –4 16 4 –8 12
–3 9 3 –8 4
–2 4 2 –8 –2
–1 1 1 –8 –6
0 0 0 –8 –8
1 1 –1 –8 –8
2 4 –2 –8 –6
3 9 –3 –8 –2
4 16 –4 –8 4
5 25 –5 –8 12
(-4,12)
Choose the scale on x-axis 1 cm = 1 unit and on y-axis 1 cm = 1 unit. Mark these points on the graph sheet and join these points. We get the graph of y = x2 – x – 8. Solving the equations we get, y = 0 = – y =
x2 x2 – x
(-1,6) (-2,5)
(-3,4)
– x – 8 – 2x – 15 + + + 7
x’
-7 -6 -5 -4 -3 -2 -1 (-2,-2)
which is a straight line. Again form the table for the straight line y = x + 7 and draw the straight line. x –2 –1 0 y=x+7 5 6 7
1 8
(-1,-6)
2 9
(0,-8)
The x-co-ordinates of the points of intersection of y = x2 – x – 8 and y = x + 7 give the solution set. ∴ The solution set is {–3, 5}. Note that the two roots are real and distinct. Example 11: Draw the graph of y = x2 – 2x + 9 and use it to solve the equation x2 – 2x + 1 = 0. Solution: First form a table for y = x2 – 2x + 9. –1 1 2 9 12
0 1 2 3 0 1 4 9 0 –2 –4 –6 9 9 9 9 9 8 9 12
(-3,24)
4 16 –8 9 17
Choose the scale on x-axis 1 cm = 1 unit and on y-axis 1 cm = 2 units. Mark the above points on the graph sheet and join these points to get the graph of the parabola y = x2 – 2x + 9. Solving we get, y = x2 – 2x + 9 0 = x2 – 2x + 1 – – + – y =
(-4,33)
9
–2 4 4 9 17
y
- 2x +
–3 9 6 9 24
x
Fig.11.10
2
–4 16 8 9 33
13 12 7 (5,12) + x 11 = y 10 (2,9) 9 8 (1,8) 7(0,7) 6 5 4 (4,4) 3 2 1 o 1 2 3 4 5 6 7 -1 -2 (3,-2) -3 -4 -5 -6 (2,-6) -7 -8 (1,-8) y’
y=x
x x2 –2x +9 y = x2 – 2x +9
x axis : 1 cm = 1 unit y axis : 1 cm = 1 unit
y
2 8 y=x -x -
x x2 –x –8 y = x2 – x – 8
36 34 32 30 28 26 24 22 20
x axis : 1 cm = 1 unit y axis : 1 cm = 2 units
(4,17) 16 14 12 (-1,12) (3,12) 10 y=8 (0,9) (2,9) 8 (1,8) 6 4 2
(-2,17)
x’
-7 -6 -5 -4 -3 -2 -1
o 1 2 3 4 5 6 7 -2 -4 -6 y’
8
Fig.11.11 236
x
which is a straight line parallel to x-axis. This straight line intersects the curve at two coincident points (1,8), (1,8). It touches the curve at the point (1,8). The x co-ordinates of these points 1 and 1 are the roots of the given equation, x2 – 2x + 1 = 0. Therefore the solution set is {1,1}. Note that the roots are real and equal. Example 12: Draw the graph of y = x2 + x – 12 and hence solve the equation x2 + 2x + 2 = 0. Solution: First we form a table for y = x2 + x – 12. x
–5 –4 –3 –2 –1
0
1
2
3
4
25 16
1
0
1
4
9
16
+x
–5 –4 –3 –2 –1
0
1
2
3
4
–12
–12 –12 –12 –12 –12 –12 –12 –12 –12 –12
x
2
2
y = x + x – 12
8
0
9
4
–6 –10 –12 –12 –10 –6
0
8
Choose the scale on x-axis 1 cm = 1 unit and on y-axis 1cm = 1 unit. Mark the above points on the graph sheet and join them. Then we get the graph of y = x2 + x – 12. By solving y = x2 + x – 12 and x2 + 2x + 2 = 0 we get the equation of straight line y = – x – 14. y = 0 = – y=
x2 + x – 12 x2 + 2x + 2 – – – – x – 14
Form a table for y = – x – 14 and draw the graph of it. x axis: 1 cm = 1 unit y axis: 1 cm = 1 unit
y
–2 –1 0 1 2 –12 –13 –14 –15 –16
(-5,8)
x’
Since the straight line does not intersect the parabola we are unable to find the x–co-ordinates of the points of intersection. Therefore the roots of the given equation x2 + 2x + 2 = 0 are not real.
8 7 6 5 4 3 2 1
(4,8) y = x2 + x 12
x y
(3,0) (-4,0) -7 -6 -5 -4 -3 -2 -1 o 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 -6 (-3,-6) (2,-6) -7 -8 -9 (-2,-10) -10 (1,-10) -11 2) ,0 -12 (-2,-12) ( 1 (-1,-13) -13 -14) , -14 (0 5) ,-1 -15 y (1 -16) = , -16 - x (2 -1 y’ 4
Fig.11.12 237
x
Exercise 11.1 1.
2.
3. 4. 5. 6. 7.
Draw the graph of the following equations. 1 (a) y = x 2 (b) y =2x2+ x (c) y = x2–x–12 (d) y = (2 + x) (x – 4) (e) y = 2x2 – x + 3 2 Solve graphically the equations. (a) x2 – 9 = 0 (b) x2 – 4x = 0 (c) x2 + 4x – 12 = 0 (d) x2 – 5x + 6 = 0 (e) 2x2 – x – 3 = 0 (f) (2x – 1) (3–x) = 0 (g) 2x2 – x – 6 = 0 (h) (2x + 1) (x – 3) = 0 2 Draw the graph of y = x – 4x + 3 and hence solve the equation x2 – 4x + 3 = 0. Draw the graph of y = x2 – 2x – 8 and hence use it to solve x2 – 2x – 8 = 0. Draw the graph of y = x2 – 3x and use it to solve x2 – 3x – 4 = 0. Draw the graph of y = 2x2 + x – 6 and hence find the roots of 2x2 + x – 10 = 0, Draw the graph of y = x2 – 4x + 6 and use the graph to show that the equation x2 + 3x + 5 = 0 does not have a real root.
11.2 SOME SPECIAL GRAPHS This section demonstrates how to draw two types of graphs one of direct variation and the other of indirect variation. The first one is well known as a straight line while the second one is called a Rectangular Hyperbola. Finally we will find the approximate area under a curve by the rule of trapezoidal approximation. Example 13: The following table gives the cost and the number of pens bought.
Number of Pens (x) Cost in Rs. (y)
2 4 5 6 8 10 20 40 50 60 80 100
For the above table find what type of variation involved and draw the suitable graph.
y 100 90 80 70 60 50 40 30 20 10
x-axis : 1 cm = 1 unit y axis : 1 cm = 10 units
(10,100)
Number of Years (x) Value of the machine in Rs.(y)
238
y=
10
x
Solution: From the above table we observe (8,80) that as x increases y also increases. Therefore it is in direct variation. ∴ we get y α x, ∴ y = Kx where K is a (6,60) constant of proportionality. (5,50) y (4,40) Since = K , and from the table we find x 20 40 50 60 (2,20) = = = 2 4 5 6 80 100 O 1 2 3 4 5 6 7 8 9 10 = = = 10 = K 8 10 ∴ we get K = 10. The relation y = 10 x is a straight line as exhibited in the graph. Fig.11.13 Example 14: For the following table draw the suitable graph. 1 2 4 8 10 1000 500 250 125 100
x
Estimate the value of the machine for the 5th year. Solution: From the table we observe that as x increases y decreases. This type of variation is called indirect variation. That is y α 1/x or xy = K where K is a constant of proportionality. Also from the table we find that K = 1 × 1000 = 2 × 500 = 4 × 250 = 8 × 125 = 10 × 100 = 1000
x axis: 1 cm = 1 unit y axis: 1 cm = 100 units
The relation is xy = 1000 is a rectangular hyperbola as exhibited in the graph (Fig.11.15). From the graph we see that the estimated value of the machine for 5th year is Rs.200.
Fig.11.14 Example 15: It is estimated that the water tank can be filled in 72 hours if 4 pumps are used. Find graphically how many pumps would be needed to complete the work in 48 hours. Solution: Let x denote the number of pumps and y denote the number of hours. As the number of pumps increases the number of hours y (time) will be reduced. Therefore they are in indirect variation. The relation is y α 1/x. ∴ xy = k where k is the constant of proportionality. Given y = 72 when x = 4. ∴ k = 72 × 4 = 288. ∴ The relation becomes xy = 288. 288 Let us now form the table for curve y = and the draw the graph (Fig.11.16). x x axis: 1 cm = 1 unit y axis: 1 cm = 10 units
x y
3 96
4 72
6 48
8 36
9 12 32 24
From the graph we observe that to we need (x = ) 6 pumps to fill the tank in (y = ) 48 hours.
Fig.11.15 239
Exercise 11.2 1.
Draw the graphs for the following tables and identify the variation and also find the constant of proportionality. (a)
2. 3. 4.
x y
2 8
3 5 8 15 12 20 32 60
(b) x y
2 4 5 8 10 50 25 20 12.5 10
A bus travels with a speed of 30 km per hour. Give the distance time formula and draw the graph of it. Use it to find the distance when it travels 3½ hours journey. Draw the graph of xy = 12, x, y > 0. Use the graph to find y when x = 5 and x when y = 8. A train travels from Chennai to Arkonam for a distance nearly 100 kms. Give the speed time formula for it and draw the graph. Using this find the number of hours to travel with a speed of 16 km per hour.
ANSWERS Exercise 11.1 (2) (a) –3, 3 (b) 0, 4 (c) –6, 2 (d) 2, 3 (e) –1, 1.5 (f) 0.5, 3 (g) –1.5, 2 (h) –0.5, 3. (3) 1,3 (4) –2, 4, (5) –1, 4 (6) –2.5, 2 (8) no real roots.
Exercise 11.2 (1) (a) k = 4 (b) k = 100 (2) 105 km (3) y = 2.4, x = 1.5 (4) 6 hrs 15 min
240
