Answers and Explanations 1 11 21 31 41 51 61 71
1. c
c a c c a b b c
2 12 22 32 42 52 62 72
c d b d b c b b
3 13 23 33 43 53 63 73
a a a c d c b a
4 14 24 34 44 54 64 74
d b a c b c c a
5 15 25 35 45 55 65 75
d a d d a a d b
Let Let the the radi radius us of the the o out uter er cir circl cle e be be x = OQ
6 16 26 36 46 56 66 76
a d b a c c a d
5. d
Sinc Since e CD > DE, DE, opti option on (b) (b) can canno nott be tthe he ans answe werr. Similarly, since AB > AF, Option (c) cannot be the answer. We are not sure about the positions of points B and F. Hence, (a) cannot be the answer.
6. a
The grad gradie ient nt of the the lin line e AD AD is is – 1. 1. Coordinates of B are ( – 1, – 1, 0).
Hence, perimeter of the circle = 2πx But OQ = BC = x (diagonals of the square BQCO) Perimeter of ABCD = 4x Hence, ratio = 2. c
2 πx 4x
=
π 2
.
So {L1(13) + 2 π } > 4 and hence
L 2 (17 )
c d d b a c b d
8 18 28 38 48 58 68
a a c c c a d
9 19 29 39 49 59 69
d c b c b b d
10 20 30 40 50 60 70
b b d b a d b
A
Foll Follow owin ing g rule rule shou should ld be be used used in in this this case case:: The The perimeter of any polygon circumscribed about a circle is always greater than the circumference of the circle and the perimeter of any polygon inscribed in a circle is always less than the circumference of the circle. Since, the circles is of radius 1, its circumference will be 2 π . Hence, L1(13) > 2 π and L2(17) < π .
{L1(13) + 2π}
7 17 27 37 47 57 67 77
x+y=1
B
(0, 0)
D
( – 1, 0)
will C
be greater than 2.
(0 , – 1)
Equation of line BC is x + y = – 1. 1. 3. a 5 00
7. c
0 1 0
Let the the are area of s se ector tor S1 be x units. Then the area of the corresponding sectors shall be 2x, 4x, 8x,16x, 32x and 64x. Since every successive sector has an angle that is twice the previous one, the total area
5 00
then shall be 127x units. This is
2
2
Area of a
2
4. d
= 1016x units. Hence, angle of sector S 1 is
= 2,500 sq m
∆ is maximum when it is an isosceles ∆.
So perpendicular sides should be of length
100 2
.
We hav have e no nott be been en giv given en the the dis distan tances ces betwe between en any any two points.
Geometry and Mensuration - Actual CAT Problems ‘99-’05
of the total area of
the circle. Hence, the total area of the circle will be 127x × 8
Area of shaded region
= 1 × 1 0 0 × 100
1 8
8. a
π 1016
.
We kn know th that (a (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac = 3ab + 3bc + 3ac Now assume values of a, b, c and substitute in this equation to check the options. Short cut: (a – (a – b) b)2 + (b – (b – c) c)2 + (c – (c – a) a)2 = 0. Hence, a = b = c.
Page 1
9. d
12. a E
(2
x
–
2x )
x
x
C
x
β G
γ
α
A
D
F
B
2m E 2α
C G
A
α 2α 2α α B
α
Let the length of the edge cut at each corner be x m. Since the resulting figure is a regular octagon,
α
3α
3α 2α F
α
+
2
= 2 – 2x
=2 ⇒ x=
2)
2 2
+1
A 3 km N
r
O
E
9 km
9 km
S
P
∆APS and ∆AOC are similar triangles. Where OC = r
∴
r r+3
9
= 81
+
(2r
+
3 )2
Now use the options. Hence, the diameter is 9 km.
C
Let Let BC BC = y and and AB = x. x. Then area of ∆CEF = Area(∆CEB) – CEB) – Area( Area(∆CFB) =
D F
E
In order to reach E from A, it can walk clockwise as well as anticlockwise. In all cases, it will have to take odd number of jumps from one vertex to another. But the sum will be even. In simple case, if n = 4, then an = 2. For a2n – 1 = 7 (odd), we cannot reach the point E.
Page 2
2 x (1
W
14. 14. a
11. d
+ x 2 = 2 – 2x ⇒ x
C
B
G
⇒ 13. b
10. a
H
x2
D
Let ∠EAD = α. Then ∠AFG = α and also ∠ACB = α. Therefore, ∠CBD = 2α (exterior angle to ∆ABC). Also ∠CDB = 2α (since CB = CD). Further, ∠FGC = 2 α (exterior angle to ∆AFG). Since GF = EF, ∠FEG = 2α. Now ∠DCE = ∠DEC = β (say). Then ∠DEF = β – 2 – 2α. Note that ∠DCB = 180 – 180 – ((α + β). Therefore, in ∆DCB, 180 – 180 – ((α + β) + 2α + 2α = 180 or β = 3α. Further ∠EFD = ∠EDF = γ (say). Then ∠EDC = γ – 2 – 2α. If CD and EF meet at P, then ∠FPD = 180 – 180 – 5 5α (because β = 3 α). Now in ∆PFD, 180 – 5α + γ + 2α = 180 or γ = 3α. Therefore, in ∆EFD, α + 2 γ = 180 or α + 6 α = 180 or α = 26 or approximately 25.
A
∴
xy 1 2x 1 x . . y – . . y = 6 2 3 2 3
Area of ABCD = xy ∴ Ratio of area of ∆CEF and area of ABCD is
xy 1 : xy = 6 6
Work with options options.. Length Length of of wire wire must must be a multipl multiple e of 6 and 8. Number of poles should be one more than the multiple.
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
15. d Since area of ∆ABC = 80 =
1 AB × CD 2
80 × 2 = 8 . In ∆ACD; AD = 20 Hence DB = 20 – 20 – 6 6 = 14.
102 – 8 – 82 = 6
So CD =
Wall Ladder x
8m
142
So CB = 18. c
2m Ground Let the length of the ladder be x feet. We have 82 + y2 = x2 and (y + 2) = x Hence, 64 + (x – (x – 2) 2)2 = x2 – 4x + 4 = x2 ⇒ 64 + x2 – 4x ⇒ 68 = 4x ⇒ x = 17
25
=
260 unit
B x
y y
x
C
F
Here ∠ACE=180 – ACE=180 – 2x 2x , ∠BCF = 180 – 180 – 2y 2y and x + y + 40° 40 ° = 180° (In ∆DEF) So x + y = 140 140° So ∠ACB= 180°– 180°– ∠ACE – ACE – ∠BCF = 180° 180° – (180 – (180° ° – 2x) – 2x) – – (180 (180° ° – 2y) – 2y) = 2(x + y) y) – 180 – 180° ° = 2 × 140 140 – – 180 180 = 100° 100°
C
25
40
196 + 64
A
20 24
=
40 °
E D
82
D
y
16. d
+
20
19. b A
B
32
z
CE =
2
2
25 – 20 – 20 = 15
(Since DBC is isosceles triangle.) Assume ABCD is a quadrilateral where AB = 32 m, AD = 24 m, DC = 25 m, CB = 25 m
x
So area of
Area of
∆ ADB =
x
= 384 sq. m
We can find the value of x, using the answer choices given in the question. We put (a), (b), (c) and (d) individually in the figure and find out the consistency of the figure. Only (b), i.e. 11 is consistent with the figure.
2
Hence area of ABCD = 384 + 300 = 684 sq. m
xm
20. c
C
x+4
x – 3
∆ BCD = 2 × 1 × 15 × 20 = 300 sq. m
17. a
3
y
and ∠DAB is right angle. Then DB = 40 m because ∆ADB is a right-angled triangle and DBC is an isosceles triangle.
1 × 32 × 24 2
–
xm
10
20
Path 60
A
D 20
B
Let’ Let’s assume AB be the longest side of 20 unit and another side AC is 10 unit. Here CD
⊥ AB.
Let width of the path be x metres. Then area of the path = 516 sq. m 2x) – 60 60 × 20 = 516 ⇒ (60 + 2x)(20 + 2x) – – 1200 = 516 ⇒ 1200 + 120x + 40x + 4x 2 – 1200 2 160x – 516 516 = 0 ⇒ x2 + 40x – 40x – 129 129 = 0 ⇒ 4x + 160x – Using the answer choices, we get x = 3.
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
Page 3
21. b 24. d
A 3 0° y
B
D
B
E
C
3
D
C
Area of ∆ABE = 7 cm2 Area of ∆ABEF = 14 cm 2 Area of ∆ABCD = 14 × 4 = 56 cm2
Let BC = x and AD = y. As per bisector theorem, Hence, BD =
⇒ 2×4× y×
BD DC
=
4 AB = 3 AC 25. b
4x 3x ; DC = 7 7 ( 4)2
In ∆ABD, cos30° = 3 2
(1, 1)
16x 2 49 2× 4× y
+ y2 –
(0, 0)
16x 2 3 y = 16 + y2 – 49
3y = 9 + y2 – 3y
1 (2) (1) = 1 2 Note: Answer should be independent of a and area of the triangle does not have square root.
Hence, area =
... (i)
9x 2 49
9x 2 49 2×3× y
9 + y 2 –
26. 26. d
... (ii)
Now (i) × 9 – 16 – 16 × (ii), we get 36 3 y – 4 – 48 3 y = 9 y2 – 16y – 16y 2
⇒ y = 12
3
7
27. c
22. a C 15
20
A
(2, 0)
Let a = 0
1 6 x2 = 16 + y – 49 2
Similarly, from ∆ADC, cos30° =
⇒3
F
3 0°
4
⇒4
A
1 ⇒ Diagonal = 5 2 Distance saved = 3 – 5 ≈ 0.75 ≠ Half the larger side. Hence, incorrect. 3 ⇒ Diagonal = 5 4 Distance saved = (4 + 3) – 3) – 5 5 = 2 = Half the larger side. Chec Check k cho choic ices es,, e.g e.g..
Area = 40 × 20 = 800 If 3 rounds are done, area = 34 × 14 = 476 ⇒ Area > 3 rounds If 4 rounds ⇒ Area left = 32 × 12 = 347 Hence, area should be slightly less than 4 rounds.
28. b
B
B
25
20
15
Let the chord = x cm
∴
1 1 x (15 × 20 ) = × 25 × 2 2 2
P
⇒ x = 24 cm
A
2 5 – x
Total area rea = 14 × 14 = 196 m2 Grazed area =
π × r 2 × 4 4
(15)2 – x – x 2 =
πr 2
= 22 × 7(r = 7)
= 154 m2 Ungrazed area is less than (196 – 154) – 154) = 42 m2, for which there is only one option.
Page 4
C
D
x
23. a
Q M
= (20)2 – (25 – (25 – – x) x) 2
⇒x=9 ⇒ BD = 12 Area of
∆ABD = 1 × 12 × 9 = 54 2
s=
1 (15 + 12 + 9 ) = 18 2
r1 =
Area s
⇒
r1 = 3
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
∆BCD =
Area of
s
= =
IR =
PM = r1 + r2 = 7 cm QM = r2 – r – r1 = 1 cm 32. c
50 cm
If KL = 1, 1, the then n IG IG = 1 and and FI = 2
θ
Thus,
r r hr , one circumference of OR = hr hr. 10 7.5
and one length of the chord road =
Hence, tan θ =
30. 30. c
20π, 30π and 15 5 kmph. Thus time taken to travel one circumference of
=4
r2
⇒
Hence, PQ = 29. 29. d
The corresponding speeds are
× 12 = 96
1 (1 6 + 2 0 + 12 ) = 2 4 2
Area s In ∆PQM, r2
1 × 16 2
2 1
Sum Sum of the length length of the the cho chord rd road roads s = 4r 5 and the length of OR = 4π r.
5:π
Thus the required ratio =
=2
none of 30, 45 and 60° 60°.
Area Area of of quad quadri rila late tera rall ABCD ABCD =
Area of quadrilateral DEFG
=
33. c
34. 34. d
The The tota totall time time take taken n=
r 20
+
r 15
Since r = 15, total time taken =
The surfa surface ce area area of a spher sphere e is propor proportio tional nal to to the square of the radius. Thus,
SB SA
r ∴B rA
2 1
=
=
4 1
(S. A. of B is 300% higher than A)
VB VA
2 .5 C
A
Or, VA is
=
N2
36. b
W2
3 2
=
7r 60
7 hr. = 105 min. 4
2 .5
B
r
NOTE: You will realize that such a circle is not possible (if r = 3.125 how can CE be 5). However we need to check data sufficiency and not data consistency. Since we are able to find the value of r uniquely using second statement the answer is (a).
N1
W1
=
We can get the answer using the second statement only. Let the radius be r. AC = CB = 2.5 and using statement B, CE = 5, thus OC = (r – (r – 5). 5). Using Pythagoras theorem, (r – (r – 5) 5)2 + (2.5)2 = r 2 We get r = 3.125
7 8 × 100 87.5%
For questions 32 to 34:
r 15
O
8 1
7 th less than B i.e. 8
+
E
35. a
The volume of a sphere is proportional to the cube of the radius. Thus,
r 30
(i.e. 90 min.) Thus, r = 15 km. The radius of OR = 2r = 30 kms
1 (2x + 4 x ) × 4 x = 12 x 2
1 (5 x + 2x ) × 2x = 7 x 2
The The total total time time taken taken by the the rou route te giv given en =
Hence, ratio = 12 : 7 31. 31. d
r hr 15
E2
The question question tells tells us that the area area of of triangl triangle e DEF DEF will will
E1
be
th the area of triangle ABC. Thus by knowing 4 either of the statements, we get the area of the triangle DEF.
S2 S1
If the radius of the inner ring road is r, then the radius of the outer ring road will be 2r (since the circumference is double). The length of IR = 2 π r, that of OR = 4π r and that of the chord roads are r 5 (Pythagoras theorem)
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
1
37. c
In th this is kind kind of of pol polygon ygon,, the number number of convex convex angle angles s will always be exactly 4 more than the number of concave angles. NOTE : The number of vertices have to be even. Hence the number of concave and convex corners should add up to an even number. This is true only for the answer choice (c).
Page 5
38. c
A
=
r B
Area Area that that can can be graz grazed ed Area Area of the the fiel field d
26 36
The fraction that cannot be grazed =
10
36 = 28% (approx.)
C
2r
40. a
Since the area of the outer circle is 4 times the area of the inner circle, the radius of the outer circle should be 2 times that of the inner circle. Since AB and AC are the tangents to the inner circle, they should be equal. Also, BC should be a tangent to inner circle. In other words, triangle ABC should be equilateral. The area of the outer circle is 12. Hence the area of inner circle is 3 or the radius is
3
π
It is very clear, clear, that that a regular regular hexag hexagon on can can be d divi ivided ded into six equilateral triangles. And triangle AOF is half of an equilateral triangle. Hence the required ratio = 1 : 12
P
41. b
6 0°
. The area of
equilateral triangle = 3 3 r2, where r is the inradius. Hence the answer is 39. b
A
9 3
π
Given ∠APB = 60° 60° and AB = b.
If the the radius radius of the the field field is is r, r, then then the tota totall area area of tthe he
b 2
∴ PQ = ×
πr 2
field =
. 2 The radius of the semi-circles with centre's P and
R=
r 2
∴ πr 2
Hence, their total area =
2
42. d
r + x . 2
r 2 2
=
Q
Solving this, we get x
=
3
B
P A 6
.
Thus the area of the circle with centre S =
The total area that can be grazed =
=
10 D
8
r + x 2 2 r
3b2 4
C
and RS =
Applying Pythagoras theorem, we get (r – (r – x) x)2 +
+ h2 =
∴ 2h2 = b2
4 Let the radius if the circle with centre S be x.
r
b2 4
3
b , h and PQ form a right angle triangle. 2
Next,
.
Thus, OS = (r – (r – x), x), OR =
B
Q
πr 2 9
Triangle ABC is a right angled triangle. .
1 1 πr 2 + 4 9
13 πr 2
Thus
1 1 × BC × AB = × BD × AC 2 2
Or, 6 × 8 = BD × 10. Thus BD = 4.8. Therefore, BP = BQ = 4.8. So, AP = AB – AB – BP BP = 6 – 6 – 4.8 4.8 = 1.2 and CQ = BC – BC – BQ BQ = 8 – 8 – 4.8 4.8 = 3.2. Thus, AP : CQ = 1.2 : 3.2 = 3 : 8
36 Thus the fraction of the field that can be grazed
Page 6
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
43. b
Using Using the the Basic Basic Propo Proporti rtiona onalility ty Theor Theorem em,, PQ CD
and
=
AB PQ
=
BD QD
(
= (2r +
BQ . BD
= 20° 20° (external angle of
C1C2
A 0 1
10 × + 2) × 10
3 10
r
= B1B2 + B2B3 + B3B1
∆BOC )
3
×
(2 + 2 3 )r , 3 × ( 2 + 2 3 ) r , ( 2 + 2 3 ) r 12 0 ( 3 + 1) 40 ( 3 + 1)
40
∆AOC )
3
i.e.
40 3
i.e.
C
20
r
–
C2C3 C C , 3 1 40 120 3 +1 3
,
3
20
45. c
3
(
= 20 (opposite equal sides)
r – 20
=(
40 3
= 10° 10 ° (opposite equal sides)
= 30° 30° (external angle of Thus k = 3
10
)× 3
r
Now time taken by for each distance are
If y = 10°,
∠BOC ∠OBA ∠OAB ∠AOD
3r
3 10
∴ B will be at B1.
AB BQ = Multiplying the two we get, = 3 : 1. CD QD Thus CD : PQ = BD : BQ = 4 : 3 = 1 : 0.75 44. a
+ 20 ) ×
= 10 3
+ 1)
× 2r,
r,
3 20
(
3 40
× 2r,
)
(1 + 3 ) r 60
(1 + 3 ) r
r,
60
We can observe that time taken for C 1C2 and C2C3
r
B
3 3 3 r+ r= r , which is same as time 20 20 10 taken by A. Therefore, Therefore, C will be at C3.
combined is
Let the radius be r. Thus by Pythagoras’ Pythagoras ’ theorem for
∆ABC we have (r – (r – 10) 10)2 + (r – (r – 20) 20)2 = r 2
i.e. r2 – 60r – 60r + 500 = 0. Thus r = 10 or 50. It would be 10, if the corner of the rectangle had been lying on the inner circumference. But as per the given diagram, the radius of the circle should be 50 cm.
48. b
In simil similar ar triangl triangles, es, ratio ratio of Area Area = R Ratio atio of square squares s of corresponding sides. Hence, A and C reach A3 and C3 respectively.
49. a
The whole whole height height h will will be divided divided into into n equal equal parts parts.. Therefore, spacing between two consecutive turns
For questions 46 to 48: A1A2 = 2r, B1B2 = 2r + r 3 , C 1C2
=
h n
.
= 2r + 2r 3 50. 50. b
Hence, a = 3 × 2r
The four four faces faces throu through gh which which strin string g is passi passing ng can can be shown as
b = 3 × (2r + r 3 )
(
c = 3 × 2r + 2r 3
46. 46. a
)
Diffe Differen rence ce betw between een (1) (1) and and (2) is is 3 3r and that between (2) and (3) is 3 3r . Hence, (1) is the correct choice.
47. c
Time tak taken by by A =
2r 20
+
2r 30
+
2r 15
2r × 9 = = 60
Therefore, B and C will also travel for time
(
Now speed of B = 10 3
+ 20 )
Therefore, the distance covered
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
3 10
3 10 r.
n 4
n
n 4
n
n
n 4 n 4
n
Therefore, length of string in each face
=
n2
n 2 + 4
=
n2
+
r
n2 16
=
17n 4
Therefore, length of string through four faces =
17n 4
×4 =
17n
Page 7
51. c
As h/n h/n = number number of turns turns = 1 (as given given). ). Hence Hence h = n. n.
Area of of ci c ircle Area of rec ta tan gl gle
52. c
PQ || AC
∴
CQ
r2
AP
=
QB
=
PB
4
PD DB
As
=
PD
4
∴ PD = ∴
7
d2 4 lb
4 3
4
∴
3 PB
∴
AP 4 PB 7
=
PD
=
QB
=
DB
AP
CQ
7
=
4
7
×
= ×
∴
AP
3
l2
l2
+ b2 b
b
l
C
96 °
4 3 4
=
… (i)
3
∆ AEB : ∆ CBD
AE
=
CB AE AD
AE AD
∴
x
3
=
lb l
1
=
4lb
∴
∴
3
+ b2
Now
53. c
1
=
4lb
4
∴
3
d2
∴ +
E
1
=
PB
4 3 =7:3
AD DC BC
=
DC
b
=
l AE
We have to find
AD
18 0 – 2y A
y
x
y
D
1
∠A + ∠B = 96 i.e. x + y = 96 … (i) Also x + (180 – (180 – 2y) 2y) + 96 = 180° 180 ° x – 2y – 2y + 96 = 0 x – 2y – 2y = – = – 96 96 Solving (i) and (ii), y = 64° 64° and x = 32° 32°
A
E
I
1 + x2
.
=
C
=
4 3
3 x2
+
3 x2
∴x =
3
=
x
∴
B
l
4
+x=
3
b D
x
… (ii)
∴ ∠DBC = y = 64o
b
=x l Therefore, from (i), we get
B
∴ ∴
, i.e.
b
Let
Using exterior angle theorem
54. a
3
3
QD || PC
∴
lb
π
=
1
=
lb
π r2
=
= 4x
− 4x +
−( −4) ±
3
=0
16 − 4
( 3)
3
2 3
4 ± 1 6 − 12 2 3 4±2 2 3
BD = 2r
Page 8
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
=
6 2 3
=
OR
1
1
56. c A+
=
+ L∞
2 A + L∞ 2
2P ( 2
1 3
(
a
, i.e. 1: 3 .
Q
1−
=
1 2
=
2A
P 2
(
2
+ 1)
2
+ 1)
a
− 1)
=
=
∆
ACE is equilateral triangle with side
(
2 2+ 2
×
3
Area of hexagon =
Area as
∆ACE =
Therefore, ratio =
2A
2×2
D
E
P P
3 a
3
I t ’ s standard property among circle, square and triangle, for a given parameter, area of circle is the highest and area of the triangle is least whereas area of the square is in-between, i.e. c > s > t.
=
3 0°
a
From options, the answer is
P+
C
A
2 3
3
a
a
a
2
OR
55. c
B
58. b
a2 × 6
4 3 4
3 a.
( 3a 3a )2
1 2
1 59. 59. d
2A
2 × 4a
(
2
+ 1)
2 × a2
The The req requi uire red d ans answe werr is is 34 34 × 0.65 × 0.65 = 14.365 Because we get two similar triangles and area is proportional to square of its side.
60. b
x
)
a 2
57. a
o
∠BAC = ∠ACT + ∠ATC = 50 + 30 = 80 And ∠ACT = ∠ABC (Angle in alternate So
segment)
∠ABC = 50o
x /2
∠BCA = 180 − (∠ ABC + ∠ BAC)
= 180 − (50 + 80) = 50o Since
In original rectangle ratio =
∠BOA = 2∠BCA = 2 × 50 = 100o In Smaller rectangle ratio =
Alternative Method: x 2
=
2 x 2
⇒x=2
Join OC
Given
∠OCT = 90° (TC is tangent to OC) – 50° = 40° ∠OCA = 90° – 5 ∠OAC = 40° (OA = OC being the radius) ∠BAC = 50° + 30° = 80° – 40° = 40° = ∠OBA (OA = OB being the ∠OAB = 80° – 4
Area of smaller rectangle =
radius)
∠BOA = 180° – ( – ( ∠OBA + ∠OAB) = 100°
61. b
2 x 2
2
x sq. units × 2 = x = 2 2 sq 2
OP PR 4 = = OQ QS 3 OP = 28 OQ = 21 PQ = OP – OP – OQ OQ = 7 PQ OQ
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
x 2
=
7 1 = 21 3
Page 9
62. b
PR + QS = PQ = 7
=
PR QS
=
4 3
a 3
a 3
⇒ QS = 3 63. c
SO = OQ2 – QS – QS2
=
212 – 3 – 32
=
2 4 × 18
= 12
a3
Circum radius for equilateral triangle
3
=
side 3
64. d Therefore
66. b
Circle C
r 4
C2
r 8
C3
r 16
M
M
⇒
=a
C
2
2
A
E
D
F
8
1 1 × AB × BD = × AD × BE 2 2 2 82
BE =
Area Area of unsh unshad aded ed port portio ion n of C either A re a o f C
= 1−
3
B
Radius r
C1
a 3
Area Area of shad shaded ed port portio ion n A re a o f C
− 22 = 8 × BE 60 4
A E = 22
=
15 2
−
15 2
2
=
4−
15 4
=
1 2
1 + 1 = 7 2 2
r 2 r 2 π + + … 4 8 = 1− 2 πr
BC = EF = 8 − 67. d 1
1 1 = 1 − 2 + 2 + … = 1 − 16 1 4 8 1− 4
=
11 12
O 2 A
65. a
C D
O B
DF, DF, AG and CE are are body body diagonal diagonals s of cube. cube. Let the side of cube = a Therefore body diagonal is a 3
Page 10
Let the radius of smaller circle = r
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
∴ O′B = r 2 ∴ OB = O′B + O′D + OD =r
M
71. b C
+r+2
2
Also OB = 2 2
⇒r
2
+r+2= 2
⇒ r = 6 – 4 – 4
2
A
2 D
68. d
B
E
O
H
L F
G 6 5°
N
A
C
O
2 DL = DH + HL
In ∆ABC,
∠B = 90°
(Angles in semicircle)
[AC ∠ACE = ∠CED [A
Therefore
1
DL = DH +
∠ABE = 90 – 6 – 65 = 25° ∠ABE = ∠ACE ( angle subtended by same arc
Therefore
Also
1
HL = OE =
D
E
Also AE)
B
2 OB = AO = radius = 1.5 DO2
ED]
= OL2 + DL2
3 2 = 1 2 + DH + 1 2 2 2 2
∠CED = 25°
69. b B
A
9 0°
1 cm C 6 0° 6 0° 9 0° 6 0°
P
D
90 3 60
× 2π × 1 =
Also AP = QD =
2
+ 1+
π 2
72. a
π
D
2
π
9 6
α 12
30
30
AB 70
10
C
– (2θ + α) = 180 – (θ + α ) ∠ACB = θ + 180 – (
So here we can say that triangle BCD and triangle ABC will be similar. Hence from the property of similarity
30
12 AC
30
= =
12 9
Hence AB = 16
12
Hence AC = 8 6 9 Hence AD = 7 AC = 8
70 10 30 70
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
θ
B
= 1+ π
70
30
2
A
Here
70. c
1
Q
2 So the minimum distance = AP + PQ + QD =
π
2 –
Hence option (b)
Drawn figure since it have not to be within distance of 1 cm so it will go along APQD. AP =
2 1 DH = ⇒ DH + = 2 ⇒ DH 2
SADC = 8 + 7 + 6 = 21 SBDC = 27 Hence r
=
21 27
=
7 9
Page 11
73. a
75. d
P
A
B
r
r
Q
R 3 0°
r
r R
a
x
x S
A covers 2r + 2r + 4r + 4r = 12 r B covers 2πr + 2π r
1 20 °
4 πr
Here cos30° =
a
SB
2r
SA
Here the side of equilateral triangle is r 3 From the diagram cos120° =
=
12r SA
SB – S – S A
a=r 3
2
r
r
x2
⇒ SB =
× 100 =
= 4πr dis ttaan ce
π 3
SA
– 3 π – 3 3
× 100 = 4.72%
Hence Option (d)
– a2 + x 2 – a 2x 2
76. d
2
a = 3x x=r
(
Hence the circumference will be 2r 1 + 3
)
D 12 C 4 B 16 20 A 12 20 O
Hence answer is (a). 74. b
Let the recta rectangle ngle has m and and n tile tiles s along along its its lengt length h and breadth respectively. The number of white tiles W = 2m + 2(n – 2(n – 2) 2) = 2 (m + n – n – 2) 2) And the number of Red tiles = R = mn – 2 – 2 (m + n – n – 2) 2) Given W = R ⇒ 4 (m + n – n – 2) 2) = mn ⇒ mn – mn – 4m 4m – – 4n 4n = – 8 – 8 ⇒ (m – (m – 4) 4) (n – (n – 4) 4) = 8 ⇒ m – 4 – 4 = 8 or 4 ⇒ m = 12 or 8 ∴ 12 suits the options.
Page 12
OB2 = OA2 – AB – AB2 = 202 – 16 – 162 = 144 OB = 12 OD2 = 202 – 12 – 122 = 400 – 400 – 144 144 = 256 OD = 16 BD = 4 Only one option contains 4 hence other will be 28. Hence option (d)
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
Geometry and Mensuration - Actual CAT Problems ‘99-’ 99-’05
Page 13
