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MAT0511/004
CONTENTS TOPIC 1
GEOMETRY
Outcomes
1 1
Section 1.1
Lines and Angles
Section 1.2
Polygons
26
Section 1.3
Circles
63
3
Topic Summary
66
Checklist
69
TOPIC 2
P E RI M E T E R, AR E A AN D VOLUME
Outcomes
72 72
Section 2.1
Measurements of Perimeter and Area
73
Section 2.2
Surface Area and Volume of Three Dimensional Objects
87
Topic Summary
105
Checklist
107
ANSWERS
109
REFERENCES
127
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GEOMETRY OUTCOMES After studying this topic you should be able to do the following. SECTION 1.1: Lines and angles
Create patterns by translating an object in a given direction, over a given distance.
Create patterns by reflecting an object, for example in a vertical or horizontal line. Recognise objects that have reflection symmetry.
Recognise objects that have . Create by rorotational symmetry tating an object through a given angle between 0 ◦ and 360◦patterns .
Define the following concepts: line, line segment, ray, angle.
Recognise lines that are coincident, parallel, perpendicular .
Classify angles according to measure or relationship with other angles.
Recognise the relationship between parallel lines cut by a transversal line and the resulting corresponding, alternate and co–interior angles .
SECTION 1.2: Polygons
Create patterns by means of tessellating polygons.
Draw the diagonals (if they exist) and altitudes of any polygon.
Recognise polygons that are congruent, and identify corresponding sides or angles of congruent polygons.
Recognise polygons that are similar. Show that corresponding sides of similar polygons are in proportion.
Use the property of similarity of polygons to determine the scale factor required when a given object needs to be enlarged or reduced.
2
Classify triangles according to the lengths of their sides or the measure of their angles.
Use the Theorem of Pythagoras.
Know and apply the four sets of conditions that triangles must satisfy in order to be congruent. The conditions are referred to as SSS, SAS, AAS (or ASA) and RHS.
Recognise triangles that are similar. Use the fact that corresponding sides of similar triangles are in proportion to calculate distance or length.
Classify quadrilaterals according to whether opposite sides are parallel, or equal in length.
Use congruency of triangles to prove certain properties of quadrilaterals. – In a kite prove that the diagonals intersect at right angles, and the longer of the diagonals bisects the shorter diagonal. – In a parallelogram prove that the diagonals bisect each other opposite sides have equal length opposite angles have equal measure. – In a rhombus (in addition to the properties of parallelograms) prove that the diagonals bisect each other at right angles the diagonals bisect the vertices. – In a rectangle (in addition to the properties of parallelograms) prove that the diagonals have equal length. – In a square (in addition to the properties of a rhombus) prove that the diagonals have equal length.
SECTION 1.3: Circles
Use the terminology of circles: centre, radius, chord, diameter, arc, semi– circle, tangent, central angle subtended by an arc.
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1.1 LINES AND ANGLES
1.1A WHAT GEOMETRY MEANS Geometry is the branch of mathematics that considers the size and shape of things. From as early as 2 000 BC there are records of geometric activity in various parts of the world. The Babylonians knew how to find the areas of rectangles and some triangles, and volumes of various objects. They were also responsible for dividing the circumference of a circle into 360 equal parts.
The Elements consists of 13 separate books which involve the study of number theory and elementary algebra, as well as geometry. Six of the books deal with much of the plane and solid geometry that has formed part of secondary school mathematics. Transformation geometry
Geometry came into being in response to human activities such as measuring land areas and volumes of harvested grain. Construction of the pyra mids in Egypt required extensive geometric understanding. By the time of Euclid, about 300 BC, the method of making a sequence of deductions from certain initial clearly stated assumptions was a well–established mathematical practice. This way of thinking forms the basis of Euclidean geometry and other branches of mathematics. When we hear the word geometry, most of us think about Euclidean geometry. Euclid was a professor of mathematics at the University of Alexandria (in Egypt) but it seems that he srcinally studied in Athens. His best–known work is Elements which has had a significant influence on scientific thinking. Today we still study aspects of Euclidean geometry at school, although in later mathematics we will encounter non–Euclidean geometries. At school level, transformation geometry is also sometimes studied. Transformation geometry involves three different processes, namely translation, reflection and rotation. We describe each of these processes by means of an example.
1.1.1 Consider the elephant motif below.
If we repeat it several times, horizontally, we obtain the design on the following page.
4
Figure 1.1.1 Translation
The translation distance applies to any point of the motif.
We say that this design has been created by horizontally shifting (or sliding or translating) the srcinal elephant. The distance that the srcinal motif is shifted may vary. For example, in Figure 1.1.2 if we translate the circle motif horizontally by 2 cm, we have a pattern of separate circles; however, if we translate it only a very short distance, we have an overlapping pattern.
Figure 1.1.2
Shapes can be translated in any direction, upwards, downwards, diagonally, etc. Symmetry
Before we consider reflection we need to look at symmetry. In Book 3, th e discussion after Activity 4.1.1 deals with the symmetry of a graph in the y –axis. Look at the butterfly in the figure below.
Figure 1.1.3
Is it possible to draw a line through the middle of the butterfly that divides the butterfly into two ident ical halves? If you fold the figure along a vertical line through the middle of the butterfly you will see that the one half lies directly on top of the other half. We say that the one half of the butterfly is the mirror image of the other half, and that the vertical line is the axis of symmetry of this figure. A figure may have more than one axis of symmetry.
This figure thus has a vertical axis of symmetry. In Figure 1.1.4 we see an object with a horizontal axis of symmetry.
Figure 1.1.4
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Figure 1.1.5 shows an object which we can fold along many possible lines that pass through its centre in such a way that one half lies directly on top of the other half. The object thus has an infinite number of axes of symmetry.
Figure 1.1.5
Reflection symmetry Reflection symmetry is also called line symmetry.
The objects in Figures 1.1.3, 1.1.4 and 1.1.5 are all symmetric about at least one axis of symmetry. One half of the object reflects the other half in this line. For this reason we say that the objects in these figures all have reflection symmetry. We can create patterns by reflecting any given object in a certain line. We may, for example, reflect objects vertically, or horizontally . To grasp more easily what this means, put your pencil down on a piece of paper.
Now pick it up at the point, keeping the blunt end on the paper, and let it lie down flat again pointing the other way.
We can continue moving in the same direction as often as we choose to. In the sketch below we have reflected the pencil three times in a vertical line.
1.1.2 Consider the following motif.
Create a design by reflecting the motif in various ways.
6 (a) Start with a vertical line through the right hand corne r, then reflect the motif five times moving to the right with no spaces between successive motifs. (b) Start with a horizontal line through the bottom corner, then reflect the motif five times moving downwards with no spaces between successive motifs. (c) Reflect the motif once in a vertical line (as in (a)), then reflect the resulting shape once in a horizontal line through the lower corners.
SOLUTION (a)
(b)
(c)
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1.1.3 Consider the motif below. Create a design in the following way. First reflect the motif downwards in a horizontal line through the bottom corner. Then translate the resulting shape horizontally 7 times by intervals of 5 mm.
SOLUTION
Rotational symmetry
We discuss angles in the next study unit.
We can also turn or rotate objects about a central point . If they look exactly the same after the turn as they did before the turn, we say they have rotational symmetry. The amount of rotation is the angle through which the object is rotated, and it is measured in degrees. Earlier (when we discussed pie graphs in Topic 7 of Book 3) we mentioned that angles can be measured in degrees, and that one complete revolution measures 360◦ . Hence, for example, a full circle rotation measures 360◦ , a half circle rotation is a turn through 180 ◦ , and a quarter circle rotation is a turn through 90◦ . When objects have rotational symmetry, the angle of rotation will always be less than 360 ◦ , since any object will look the same after a turn through one complete revolution.
1.1.4 The object below has rotational symmetry of 180◦ since it looks the same
every time we turn it through one half of a revolution about its central point.
Figure 1.1.6
8 Do you see why the angle of rotation of the object in Example 1.1.4 is not 90 ◦ ? If we rotate it 90 ◦ (i.e. through a quarter of a revolution) we obtain the following figure.
This figure is not the same Figure 1.1.6.
1.1.5 Through how many degrees can you rotate the object below to obtain the identical object?
SOLUTION You can turn this 90◦ or 180◦ or 270 ◦ and still have exactly the same object.
1.1.1 (No solution is suggested for (a) and (b)).
(a) Find (in a magazine, newspaper or advertisement) any object with (i) rotational symmetry (ii) reflection symmetry. Use these two items to create an interesting design.
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(b) Look at designs in your enviro nment, such as designs on baskets, table cloths, walls, covers of books. Can you identify the basic motif used to create the design? See whether each motif has reflection or rotational symmetry. If it has rotational symmetry, through how many degree s can the motif be rotated? If it has reflection symmetry, in what line( s) can the motif be reflected? (c) Draw a capital letter E such as the one shown below. Fold along the dotted line. Does it have line symmetry or rotational symmetry?
The object “capital E” has line symmetry, or reflection symmetry. It is symmetrical about the horizontal line shown. It does not have rotational symmetry, since we cannot rotate it through any angle other than 360 ◦ and still have exactly the same object.
1.1B LINES AND ANGLES You have already worked with the number line, which we use to represent the set of real numbers. (See Book 1, Topic 1.) We have also seen (in Book 3, Topic 1) how this association (between points on a line and real numbers) was extended to ordered pairs of real numbers that represent points in the Cartesian plane. When we look at points, lines and curves in the Cartesian plane, we can describe their behaviour in algebraic terms by using equations.
10 We now look at points and lines separat ely from the Cartesian plane. We are thus no longer considering an algebraic description, but a geometric description, which treats these entities as physical objects that we can see and measure. There is obviously a big difference between a “point” on a number line and a “point” on a page in your book. On the number line ther e are, for example, infinitely many numbers between 0 and 1. A few of them are shown in Figure 1.1.7(a).
0 _81 _41
1 _
1
2
Figure 1.1.7(a)
In the Cartesian plane we can consider the points (1, 0), 12 , 0 ; then 14 , 0 , 18 , 0 , and so on (see Figure 1.1.7(b)), and we soon realise that this process can also continue indefinitely.
( _41 , 0) 1
( _8 , 0 )
1
( _2 , 0 )
( 1, 0)
Figure 1.1.7(b)
However, if we physically draw a line segment, and use even a very fine pencil, we will soon cover the number line with the dots that represent numbers; however there will be many numbers not yet shown, for which there is “no space” on the number line.
A point
We In use the word “poin t” in many different ways. When we look in a Mathematics Dictionary we are even more confused. We read that a point is “an undefined element of geometry”. Euclid called it something that “has position but no non–zero dimensions”. This directly contradicts the Concise Oxford English dictionary’s definition of a point as a “very small mark on a surface”. What is the point of all this? (Yet another use of the word!)
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MAT0511/004
The geometry we will now study
is based on an undefined concept (a point)
has a structure in which the apparent contradictions are overcome through the propositions formulated by Euclid
is consistent with the physical world.
We will not study any of Euclid’s propositions, but it is helpful to remember the role that they play. Terminology
A line
In many sections of the MAT011–K books we have used words such as point, line, line segme nt, angle, and so on. We now give the specifi c mathematical meaning of some of these words.
Definition 1.1.1 A line is a collection of points in a plane. It extends indefinitely in two directions. It has no width.
B
A
The line AB Figure 1.1.8
Note that when we use the word line we mean straight line. We denote the line by means of any two points on the line. Some authors use the notation AB where the double–headed arrow indicates that the line continues indefinitely in both directions.
←→
12
A line segment
Definition 1.1.2 A line segment is part of a line. It has two endpoints. Its length is the distance between the two endpoints.
A
B
The line segment AB Figure 1.1.9
Note that some authors use the notation AB to denote the line segment AB ; they may also use AB to denote the length of the line segment. We use the notation
AB to denote both the line segment and its length. We need the following definition because we use it in the definition of an angle.
A ray
Definition 1.1.3 A ray is part of a line. It has one endpoint. It extends indefinitely in one direction.
B A
The ray AB Figure 1.1.10
−→
Some authors use the notation AB to denote a ray. We denote the ray by first writing down its endpoint (starting point), then any other point on it. For example, if the points C , D or E also lie on the ray shown in Figure 1.1.10, then
−AB →, −AC →, −AD → and −AE → all denote the same ray.
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Since the context usually makes it quite clear whether we are dealing with lines, rays, line segments, or the lengths of line segments, we avoid the sometimes confusing notation
←→ −AB →
for ray
AB
for line segment
AB
for line
AB
for the length of the line segment.
From now on we will use only AB in each case, and we depend on the context to make it clear which meaning is intended.
When we consider the different line segments (or lines or rays) we have three possible situations. The line segments
We can also use the phrase “cut each other” instead of the word “intersect”.
are coincident
intersect in one point (in which case they may be perpendicular)
are parallel to each other.
Figure 1.1.11 illustrates the meanings of these terms.
E
B B D
G C
A
B
A
F
A
D
C
Two coincident
Line segments E F and CD
Line segment
line segments
intersect once, at F . Line segments E F and AB intersect once, at G . EF is perpendicular to AB and we write E F AB. The little block at G indicates the right angle.
linesegment CD . We write AB CD. The arrows denote parallel lines.
⊥
(a)
(b)
AB is parallel to
(c) Figure 1.1.11
14 In Figure 1.1.11(b) we see that the line segments E F and CD intersect the line segment AB in different ways. The difference can be considered in terms of the angles that are made at the points of intersection.
Note that the plural of vertex is vertices.
Definition 1.1.4 An angle is formed by rotating a ray about its fixed endpoint, called the vertex of the angle.
The amount of rotation is the measure of the angle.
This is a mathematical convention, chosen to ensure consistency. We will not consider negative angles in this module.
The direction of rotation determines whether the measurement of the angle is positive or negative. If the ray is ro tated in an anti–clockwise direction, the measurement is positive; if it is rotated in a clockwise direction, the measurement is negative. We can see from Figure 1.1.11(b) that whenever two lines (or line segments or rays) meet at some point such as G or F , they form angles whose vertex is that point.
positive
negative
angle
angle
In the Concise Oxford Dictionary the word “angle” is given as “space between two meeting lines or planes; inclination of two lines to each other; corner; sharp projection” and a few other options. The word angle is deriv ed from the Latin word angulus which means corner.
End position of ray C B2
Starting position of ray
A
vertex
B1
D
ˆ The angle C AD Figure 1.1.12
The angle shown in Figure 1.1.12 is obtained by rotating the ray with endpoint A in an anti–clockwise direction. AB1 is the starting position of the ray, and it is rotated about the endpoint A so that the end position is the ray AB 2 .
15 Notation
MAT0511/004
Consider the angle with vertex A . We denote it by
ˆ (or D AC ˆ , B2 AB ˆ 1 , CAB ˆ 1 etc., using A and any other two points, one CAD on one ray, the other on the other ray; note that A is always written in the middle)
∠CAD
(or ∠DAC, ∠B2 AB1 , etc., using A and any other two points on the rays; once again A is written in the middle)
Aˆ (provided there is no ambiguity about the angle).
In Figure 1.1.12 it is quite clear what we mean when we speak about the angle A, or Aˆ . However, in Figure 1.1.13 we see that it is necessary to be more specific.
S
P
R
Figure 1.1.13
ˆ and In Figure 1.1.13 there are two different angles at the point Q . We have S QP ˆ , and hence if we write just Qˆ it will not be clear what angle we are referring SQR to. Angle Measurement There are also other ways of measuring angles , but we will not discuss those here.
We mentioned previously that angles are measured in degrees. One complete rotation of a ray about its endpoint is a rotation of 360 degrees. We denote this by 360◦ , and we call this one revolution.
360
o
Figure 1.1.14
In our notation we do not always specifically distinguish between the angle itself ˆ in Figure 1.1.13 measures and the measure of the angle. Suppose the angle S QP ◦ 60 . We then usually write ˆ = 60◦ SQP whereas it is more correct to write ˆ = 60◦ . the measure of SQP
16 Degrees can be broken into minutes and seconds. We have 1 degree = 60 minutes (we write 60 ) 1 minute = 60 seconds (we write 60 ). Thus, when we are measuring an angle accurately, we may have a measurement such as 35◦ 20 15 . As in the case of time, where we do not work within a decimal system, we need to remember that 40 , 6◦ means 40◦ and 0, 6 of one degree. We have 0, 6 degree = (0, 6
6 60) minutes 60 minutes = 10 = 36 minutes.
× ×
We can thus write the measurement 40, 6◦ as 40◦ 36 . When we have a diagram in which several angles occur, we often use capital letters to represent the vertices and small letters to indicate the measures of the angles in degrees. See Figure 1.1.15.
A B
e E
a
d cQ
b
C D
Figure 1.1.15
ˆ is a , the measure of From Figure 1.1.15 we understand that the measure of A QB ˆ is b, and so on, where a, b , c , d and e represent specific numbers of degrees. BQC Since one revolution is 360◦ , we know that a + b + c + d + e = 360◦ . Although we recognise that a, b, etc., represent measurements, it is often convenient to refer to angle a , angle b , etc. Although we understand that an angle is defined in terms of the rotation of a ray, we do not usually put arrows on the ends of the line segments to denote rays. It is also clear that angles occur whenever lines or line segments intersect, so we need not restrict ourselves to thinking of angles only in terms of rays .
17 Revolution; straight and right angles
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Angles are classified according to their measures. We have three special names for three specific measurements. 360
360◦
one revolution
180◦
a straight angle
90◦
o
o
180
o
Acute, obtuse and reflex angles
90
a right angle
We have another three names which describ e angles. In these three cases they are classified according to their measures relative to 90 ◦ , 180◦ and 360◦ .
ˆ is an acute angle. If 0◦ < a < 90◦ then A OB B a
O
A
ˆ is an obtuse angle. If 90 ◦ < a < 180◦ then A OB B
a A O
ˆ is a reflex angle. If 180◦ < a < 360◦ then A OB a
O
A
B
Right angles are espec ially important. Many of the structures we depend on everyday make use of right angles. Walls are usually at right angles to floors. Shelves need to be at right angles to the wall, i.e. perpendicular to the wall (and parallel to the floor). Let us consider two lines in the same plane that are not coincident or parallel. When they cut each other they form angles of different measurements at the point of intersection.
18 Note the following.
The lines can intersect in at most one point. See Figure 1.1.16(a).
If two lines in a plane are perpendicular to the same line, they are parallel to each other. See Figure 1.1.16(b), where AB CD since AB PQ and CD PQ.
⊥
⊥
A A
C
D
P
C
Q
B
B
D
(a)
(b) Figure 1.1.16
Adjacent, supplementary, complementary and vertically opposite angles
We have looked at some names given to specific angles, and we now consider some other angle names based on the relationships between two or more angles.
We call two angles adjacent if they have a common vertex and a common ray. C
A
D
B
ˆ and CBD ˆ are adjacent angles with common vertex B and common ABC ray BC.
We call two (or more) angles supplementary if their measures add up to 180 ◦ . A
120
o
B
O 60 C
o
D
ˆ and C OD ˆ are supplementary angles since AOB ˆ + C OD ˆ = 180◦ . We AOB are usually more interested in adjacent supplementary angles, since together they form a straight angle.
19
MAT0511/004
R o
180
P
S
ˆ and RQS ˆ are adjacent supplementary angles, since they are adjaRQP cent (common vertex Q and common ray QR ) and supplementary (since
PQR ˆ and R QS ˆ form the straight angle P QS ˆ ).
We call two (or more) angles complementary angles if their measures add up to 90 ◦ . D E
60
A
o
30
o
C
B
ˆ and E BC ˆ are complementary angles since A BD ˆ + E BC ˆ = 90◦ . ABD We also have adjacent complementary angles, where the measures of two (or more) adjacent angles add up to 90 ◦ . P R
S
ˆ and R QS ˆ are adjacent complementary angles, since they have a comPQR ˆ + RQS ˆ = 90◦ . mon vertex, Q , and a common ray, QR , and P QR
We say that two angles are vertically opposite each other if they are non– adjacent angles formed by two intersecting lines. The measures of two vertically opposite angles are equal. Q
M
O
P
N
ˆ and P ON ˆ are vertically opposite angles and M OQ ˆ = P ON ˆ . M OQ ˆ ˆ Similarly, M OP = QON .
20 Further relationships arise when we consider two parallel lines. Look at Figure 1.1.17, which shows two parallel lines both cut by a line AB .
B
The line AB is sometimes referred to as a transversal
p
1
E
line.
p4
Pp
p
2
C
3
r1 R r2
F
r4
D
r3
A
EC FD; pi and r i (i = 1, 2, 3, 4) represent the measures (in degrees) of the angles shown.
Figure 1.1.17 Corresponding, alternate and co–interior angles We also have: if correspon-
ˆ and BRD ˆ are called corresponding angles. If two parallel lines are BPC cut by another line, the corresponding angles are equal. Thus we have p2 = r2 . Similarly p3 = r3 ; p4 = r4 and p1 = r 1 .
ˆ and PRD ˆ are called alternate angles . If two pa rallel lines are cut E PR by another line, the alternate angles are equal. Thus p4 = r 2 . Similarly
ding angles are equal, then the lines are parallel. The converse of the second statement is also true:
if alternate angles are equal, then the lines are parallel. The converse of the third statement is also true.
p3 = r 1 . ˆ and D RP ˆ are called co–interior angles. If two parallel lines are cut CPR by another line, the co–interior angles are supplementary. Thus we have p3 + r2 = 180◦ . Similarly p4 + r1 = 180◦ .
1.1.6 What is the converse of the statement “If two parallel lines are cut by a transversal line then the co–interior angles so formed are supplementary.”?
SOLUTION If two lines in the same plane are cut by a transversal line and the co–interior angles so formed are supplementary, then the two lines are parallel.
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1.1.2
o
c b 30 d a
A
C
h e g f
B
D
We have AB CD, and b = 30◦ . Find the measurements of all the other angles, and give reasons for your answers.
There are several different ways of finding these angles. For example, e = 30 ◦ since
a = 150◦
a and b are adjacent supplementary angles.
d = 30◦
b and d are vertically opposite angles.
c = 150◦
b and c are adjacent supplementary angles.
e = 30◦
b and e are corresponding angles on parallel lines AB and CD.
f = 150◦
e and f are adjacent supplementary angles.
h = 150◦
h and d are co–interior angles on parallel lines AB and CD.
g = 30◦
g and e are vertically opposite angles.
e and d are alternate angles on parallel lines AB and CD.
22
1.1.3 E
A
G
PC, AB, DE and F G are four lines. DE FG. xi ( i = 1, 2, 3, 4, 5) and y i (i = 1, 2, 3, 4) represent the measures, in degrees, of the angles shown.
x1 Px 2 x3 x5 x4 y1 y2 Q y 4 y3
C
D F
B
Figure 1.1.18
Look at Figure 1.1.18 and complete each of the following.
ˆ is a ........ angle. (a) x2 + x3 + x4 = ........ because A PB ˆ is a ........ angle. (b) x2 = 90◦ . Hence A PC ˆ and D PQ ˆ are ........ angles. Hence x1 ........x4 . (c) APE ˆ are co–interior ˆ and GQP (d) DE FG and hence x5 + y1 = ........ because E PQ angles on the parallel lines DE and F G.
(e) There are two pairs of alternate angles. They are ........ and ........; ........ and ........ . ˆ is a straight angle. Thus (f) x2 = 90 ◦ . Hence x1 + x3 = ........ because DPE ˆ is an ....... angle, because x 1 ........ . We call x 1 and x 3 ........ angles. APE ˆ is a straight (g) Since x1 ........ it follows that x 5 ........ and x 5 ........ because APB ˆ angle. Hence we call E PQ an ........ angle. ˆ are ........ angles on the parallel lines DE and ˆ and PQG (h) x1 = y1 because APE FG.
23
(a) 180◦
MAT0511/004
straight
(b) right (c) vertically opposite
=
(d) 180◦ ˆ ; ˆ and P QG (e) DPQ (f) 90 ◦ (g)
< 90◦
ˆ ˆ and PQF E PQ < 90o
acute > 90◦
complementary
< 180◦
obtuse
(h) corresponding angles
We have seen that two non–coincident lines in a plane are either parallel or cut each other in one point. When more than two line segments are considered they can form many different shapes. The names of the shapes often tell us something about them. We look at some of these shapes in the next two sections.
1.1
1. Reflect the triangle ABC shown below in a vertical line through C to create its mirror image. A
C
B
24 2. Rotate the triangle ABC shown below through an angle of 90 vertex B in
◦ about the
(a) a clockwise direction (b) an anti–clockwise direction. C
A
In case the word “bisect” is unfamiliar: to bisect means to cut in half .
B
ˆ = C PD ˆ , APD ˆ = 90 ◦ , and 3. Consider the following sketch, in which BPC ˆ . PB bisects A PD A
P B
C
D
(a) Identify all the pairs of adjacent angles. ˆ ? (b) What is the measure of C PD ˆ , B PC ˆ and C PD ˆ are complementary angles? (c) Why can we say that APB ˆ and DPC ˆ are (d) Suppose you include an additional point E such that E PD ˆ ? supplementary angles. What will be the measure of E PA 4. Write each of the following angle measurements in terms of degrees, minutes and seconds. (If there are no minutes, or no seconds, write the answer as x ◦ 0 0 where x represents the number of degrees.) (a) 28 , 65◦ (c) 30 , 8◦
(b) 90 (d) 100
◦
, 055
◦
,1
25 5. In the sketch below, AB
MAT0511/004 ˆ . CD and AB ⊥ EF. H G bisects E PD H
E
B
R Q D
A P
C G
F
(a) What are the measures of ˆ (i) E PC ˆ (ii) E PH ˆ ? (iii) C PG (b) Identify four pairs of co–interior angles. (c) Identify the pair of acute alternate angles. ˆ (d) What kind of angle is A QR ? ˆ ? (e) (i) What angle is the complement of F PG ˆ (ii) What angle is the supplement of F PG ? 6. Consider the shapes below.
Show that one of the shapes has rotational symmetry as well as reflection symmetry and that the other shape only has reflection symmetry.
26
1.2 POLYGONS
1.2A TERMINOLOGY In Book 1 we noted that a specific type of algebraic expression consisting of several terms can be described in general as a polynomial. In geometry we have a similar type of classification, in which we call a many– sided figure a polygon. The geometric meaning of polygon is different from the statistical meaning.
A polygon is a closed figure in a plane composed of line segments that only meet at their endpoi nts. The line segments are called the sides of the polygon, and each point where two sides meet is a vertex of the polygon. So, for example, the shape in Figure 1.2.1 represents a polygon; the shape in Figure 1.2.2 does not, because in Figure 1.2.2 the line segments do not only meet at their endpoints. D A
B C
ABCD is a polygon, since the line segments AB, BC, CD and DA meet only at the endpoints A , B, C and D .
Figure 1.2.1 C
A
P D
ABCD is not a polygon since the line segments AD and BC intersect at P, which is not an endpoint of any one of the line segments.
B
Figure 1.2.2 Classification of polygons
Polygons are classified according to the number of angles (or sides) they have.
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Because the sides only meet at the vertices it is clear that any polygon has exactly the same number of angles as the number of sides.
Three sides
Triangle
Four sides
Quadrilateral
Five sides
Pentagon
Six sides
Hexagon
There are also special names for polygons with 7 , 8, 9, 10 and 12 sides. For example, an octagon is an eight–sided polygon . Regular polygons
A polygon is called a regular polygon if all of its sides have the same length and if all of its interior angles have the same measurement. Consider the three special quadrilaterals sketched below.
A rhombus is defined later in Study Unit 1.2C.
Rectangle (not regular)
Rhombus (not regular)
Square (regular)
The rectangle has all four angles equal to right angles, but the sides are not all equal. The rhombus has all sides equal, but not all angles equal. Thus neither of these quadrilaterals is a regular polygon. If, however, the rectangle has equal sides or the rhombus has equal angles we obtain a square, which is a regular polygon. Figure 1.2.3 shows three other regular polygons.
28 A
A
F
A E
B
B
B
C
C
Regular triangle
D
E
C
Regular pentagon
D
Regular hexagon
Figure 1.2.3
Polygons are usually denoted by the vertices, stated in order. Thus we refer to the hexagon in Figure 1.2.3 as the hexagon ABCDEF. Polygons that fit together to create a flat surface with no gaps are said to tessellate. A pattern formed in this way is referre d to as a tessellation. Figure 1.2.4 is a tessellation of regular hexagons.
Figure 1.2.4 Can you think of a regular polygon that will not tessellate? Altitude is sometimes referred to as height.
Note that not all regular polygons will tessellate. Polygons have altitudes and diagonals. A diagonal of a polygon is a line segment whose endpoints are non–adjacent vertices. The altitude from any vertex V to an opposite side is the line segment with endpoint V which is perpendicular to that side. See Example 1.2.1.
1.2.1 Draw diagonals for the quadrilateral ABCD, and draw the altitude from A to CD.
A
B
C
D
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SOLUTION A
B
C
D E
AC and BD are the only two diagonals of quadrilateral ABCD. AE is the altitude from A to CD .
When we compare different polygons, to see whether they are possibly the same shape and size, we look at their corresponding angles, or corresponding sides. Look at the two polygon s shown in Figure 1.2.5 . Can you say what kind they are? (Since they each have fiv e sides, they are pentago ns. The sides are not all the same length, so they are not regular pentagons.)
T
P
E
A
S Not regular Pentagons
B
D
Q R
C
Figure 1.2.5
In polygons ABCDE and PQRST , examples of pairs of corresponding sides are
AE and PT ED and T S DC and SR CB and RQ BA and QP.
30 Similarly, pairs of corresponding angles are
Aˆ and Pˆ Bˆ and Qˆ Cˆ and Rˆ ˆ and Sˆ D Eˆ and Tˆ .
Congruent polygons
If all corresponding sides and corresponding angles of two polygons are equal, we say the polygons are congruent. Thus congruent polygons have exactly the same shape and size . We use the notation to denote congruency.
≡
1.2.1
(a) P A
C
Q
B
R
The two triangles shown above appear to have the same shape and size. List the three pairs of corresponding sides, and the three pairs of corresponding angles, that we can compare to see whether or not the triangles are congruent. (b)
A
B
P
S
Q
R
D
C
ABCD and PQRS are both squares. Why are they not congruent?
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(a) Corresponding sides
AB and QP
(opposite the biggest angle in each triangle)
AC and QR
(opposite the smallest angle in each triangle)
CB and RP
(the remaining sides of the two triangles)
Corresponding angles
Cˆ and Rˆ
(the biggest angles)
Bˆ and Pˆ
(the smallest angles)
Aˆ and Qˆ
(the remaining two angles)
(b) ABCD and PQRS have exactly the same shape , but PQRS is clearly bigger than ABCD, so they do not have the same size. Thus they are not congruent.
Similar polygons
When polygons have the same shape but not the same size, they are called similar polygons. Hence the two squares in Activity 1.2.1(b) are similar, even though they are not congruent. To be certain that they have the same shape , we need to find out whether the measurements of the corresponding angles are equal and the lengths of the corresponding sides are in proportion . In (b) of Activity ˆ = Sˆ = 90◦ ; we 1.2.1 we see that Aˆ = Pˆ = 90◦ , Bˆ = Qˆ = 90◦ , Cˆ = Rˆ = 90◦ , D also have PQ QR RS SP = = = . AB BC CD DA Hence the squares ABCD and PQRS are similar. In a sketch, if we want to show that corresponding sides or corresponding angles have the same measurements, we mark the sides with different numbers of small lines, and the angles with different numbers of arcs, as shown in Figure 1.2.6.
32
Figure 1.2.6
1.2.2 Assume the polygons in the sketch below have the measurements and properties indicated. Note that they are not drawn to scale . P 4 cm
3 cm
S 12 cm
A
D
o
100
9 cm
Q 100o
B
R
C
E
H
o
100
12 cm
F
G
6 cm
(a) Which two polygons are similar, and why? (b) Which pairs of polygons are not similar, and why not?
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The polygons illustrated are all quadrilaterals since they have four sides. (a) Quadrilateral ABCD is similar to quadrilateral PQRS, for the following reasons. Corresponding angles are equal.
Aˆ = Rˆ = 100◦
This is given.
Bˆ = Qˆ = 80◦
We deduce this from the fact that AD BC and hence Aˆ and Bˆ are co–interior angles and thus supplementary. Similarly PQ SR and Qˆ and Rˆ are supplementary.
Cˆ = Pˆ = 100◦
Cˆ + Bˆ = 180◦ since AB Pˆ + Qˆ = 180◦ since PS
ˆ = Sˆ = 80◦ D
ˆ = 180◦ Aˆ + D ˆ ˆ R + S = 180◦
DC. QR. since AB DC. since PS QR.
Corresponding sides are in proportion.
AD Just after Activity 1.2.8 we
RS
SR and QR then PQ = SR and
prove that if PQ
PS
=
PS = QR.
12
=3
4
Since
PQ = 4 cm it also follows that
SR = 4 cm, because otherwise PS and QR would not be parallel.
CD 9 = =3 3 PS BC 12 = =3 4 QP
BC = 12 cm since AD = 12 cm.
9 AB = =3 QR 3
AB = 9 cm since CD = 9 cm. QR = 3 cm since PS = 3 cm.
(b) Quadrilateral ABCD is not similar to quadrilateral EFGH . We realise this as soon as we try to identify corresponding sides and angles. E
H
o
100
12 cm
A
D
o
100
12 cm 9 cm
F
B
C
G
6 cm
34 If we “stand” the quadrilaterals on the longer of their two sides, we see that quadrilateral ABCD “leans” to the right, whereas quadrilateral EFGH “leans” to the left. We also see that the ratio of the longer sides is 12 , i.e. 12 1, but the ratio of the shorter sides is 96 , i.e. 32 . Similarly quadrilateral PQRS is not similar to quadrilateral EFGH . P 4 cm
3 cm
E
H
o
100
S
12 cm
Q 100
o
F
R
G
6 cm
The idea of similarity is important when we need to reproduce a large object on a smaller scale, or enlarge an object. We are all familiar with maps, which are small–scale reproductions of large–scale regions. There may also be times when we need to provide bigger representations of very small objects. Hence we can scale things up or down, depending on the requirements of the situation.
Scale factor
The amount by which an object is enlarged or made smaller is known as the scale factor. Thus, if a figure has doubled in size, we say that the new figure has a scale factor of 2 . If an object is halved in size the new figure has a scale factor of 12 .
1.2.3 A company wants to hang on the wall of one of its function rooms an enlargement of a photograph which is 50 cm across and 70 cm high. The available wall space measures 0, 9 m across and 1 , 5 m high. What scale factor will allo w for maximum enlargement? What will the dimensions of this enlargement be?
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The enlargement will be similar to the srcinal, i.e. the corresponding sides will be in proportion. The srcinal photograph has sides that are in the ratio 5 : 7 (when we consider the ratio of width to height). The available wall space is 0, 9 m wide and 1 , 5 m high.
0,9 m 50 cm 70 cm
1,5 m
Picture
Available wall space
Since the dimensions of the photograph are given in centimetres it is convenient to convert the wall space dimensions to centimetres as well.
The enlargement cannot be more than 90 cm wide or 150 cm high . We first consider whether it is possible for the enlargement to be 90 cm wide. Let us use a scale factor of x . If 50
Alternatively:
then
⇔ ⇔
50 90 50x
x
=
× x = 90
70
x=
x 6300
= = 126
90 9 = . 50 5
Thus, if we use a scale factor of 95 , the width of the enlargement will be 90 cm and its height will be 70 95 cm, i.e. 126 cm. Thus an enlargement with this scale factor will fit into the available wall space. If we consider a scale factor that will make the enlargement 150 cm high, then it will be too wide to fit into the available wall space.
×
The scale factor of 95 gives an enlargement that completely fits the available width. Hence the scale factor of 95 will produce the maximum enlargement, and the enlargement will measure 90 cm across, and 126 cm high.
36
1.2B TRIANGLES In Study Unit 1.2A we introduced triangles as polygons with three sides . The name “tri–angle” suggests that triangles are shapes which have three angles (and hence also three sides). Triangles have various characteristics. Some of these are important when we study (the word, literally translated fromisthe srcinal Greek form, meanstrigonometry triangle measurement). Trigonometry, in turn, an important foundation for other mathematical topics. Consider the triangle in Figure 1.2.7. C
γ a
b
A
β
α c
B
Figure 1.2.7
We use the symbol to denote a triangle. Hence Figure 1.2.7 represents ABC.
α, β and γ are pronounced as alpha, beta and gamma . They are the first three letters of the Greek alphabet.
In Study Unit 1.1B we used a, b and c to represent the measurement of angles. Previously, in Topic 1 of Book 3, we worked according to the convention that a, b and c represent the lengths of the sides opposite the angles A, B and C , respectively. When we do this we often represent the measurements of angles in a triangle by means of letters of the Greek alphabet. In Figure 1.2.7 the measure of Aˆ is denoted by α , the measure of Bˆ is denoted by β and the measure of Cˆ is denoted by γ, where α, β and γ represent numbers of degrees. Triangles can have different shapes, but regardless of their shape, they all have one common characteristic.
THE ANGLE SUM OF A TRIANGLE In any triangle the sum of the measures of the angles is 180◦ .
Thus, in Figure 1.2.7, we have α + β + γ = 180◦ . Like many other statements in geometry, we can prove the statement formally, or we can illustrate it in some way. We now illustrate this statement. Draw any triangle, on a separate piece of paper. Tear off the vertices, and arrange them next to each other, as shown in Figure 1.2.8.
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α X
β
γ
P
Y
Figure 1.2.8
Note that this is not an actual proof.
When we arrange the angles next to one another so that the three vertices meet at the point P, we see that, regardless of the order in which they are arranged, the angles form a straight line. From our classification of angles, we know that ˆ is a straight angle, which measures 180 ◦ . X PY Another fact, which we do not prove here, is the following.
TRIANGLE INEQUALITY The sum of any two sides of a triangle is greater than the third side.
Thus, in Figure 1.2.7, we have
Classification of triangles
a+b
>
c
a+c
>
b
b+c
>
a.
In Study Unit 1.1B angles are classified according to certain characteristics. We may also classify triangles according to various characteristics, such as the lengths of their sides. The lengths of the sides of the triangles influence the sizes of the angles, and vice versa as we see in Figures 1.2.9, 1.2.10 and 1.2.11.
A scalene triangle has no sides of equal length, and hence no angles of equal size. A
C
B
Scalene triangle Figure 1.2.9
The longest side is opposite the biggest angle. Conversely, the biggest angle is opposite the longest side.
38
An isosceles triangle has two equal sides, and hence two equal angles. A
B
C
Isosceles triangle Figure 1.2.10
If AC = AB , then the angles opposite these two sides are also equal, i.e. Bˆ = Cˆ . Conversely, if Bˆ = Cˆ , then the sides opposite these two angles are also equal. Hence AC = AB .
An equilateral triangle has all three sides equal in length, and hence all three angles equal. A o
60
C
B
Equilateral triangle Figure 1.2.11
Since the angles of any triangle add up to 180 ◦ it follows that in equilateral triangles, all angles measure 60 ◦ . If AB = BC = AC then Cˆ = Aˆ = Bˆ = 60◦ . Conversely, if Aˆ = Bˆ = Cˆ (= 60◦ ) then BC = AC = AB .
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We may also classify triangles according to their angles.
An acute angle triangle has all angles acute . A
B
C
Acute angle triangle Figure 1.2.12
It is clear that an acute angle triangle may be scalene, isosceles or equilateral.
An obtuse angle triangle has one obtuse angle. A
C
B
Obtuse angle triangle Figure 1.2.13
Obtuse angle triangles may be scalene or isosceles, but cannot be equilateral.
A right angle triangle (which we usually call a right triangle ) has one right angle. A
C B
Right triangle Figure 1.2.14
A right triangle may be scalen e or isosceles. In a right triangle, the side opposite the right angle is called the hypotenuse.
40
1.2.4 In
ABC, we have AC > BC > AB. A
B
C
(a) Arrange the angles in descending order according to size. (b) How can you classify
ABC?
(a) Since the longest side is AC, the angle opposite AB , i.e. Bˆ , is the biggest. The next longest side is BC , hence the next biggest angle is the angle opposite BC, i.e. Aˆ . Thus in descending order we have
Bˆ , Aˆ , Cˆ .
ABC is a scalene triangle. ABC is an acute angle triangle.
(b) All sides have different lengths, hence All angles are acute, hence
We dealt with right triangles and the Theorem of Pythagoras in Topic 1 of Book 3. You may want to read the relevant parts of that section again, before moving on.
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1.2.5 Consider the right triangle PQR .
P
50
o
R
Indicate which of the following statements are true or false. If possible, correct each false statement in an appropriate way. We use PQ2 (PQ)2 , etc.
to denote
(a) PQ2 = PR2 + RQ2 (b) QR =
±
PQ2
− PR
2
(c) PQ = PR + QR (d) PQ > PR (e) QR > PR
(a)
True
This is a statement of the Theorem of Pythagoras for PQR.
(b)
False QR =
PQ2
− PR
2
The statement is algebraically correct, since 2
2
2
2
2
2
if a = b + c 2then 2b = a c , and hence b = a c . But RQ represents length, which cannot be negative, hence we ignore the negative root.
±√ −
(c) F alse PQ < PR + QR
−
The length of the hypotenuse is less than the combined lengths of the other two sides. See the Triangle Inequality.
42
Congruency We make use of congruency to prove several important mathematical statements, and to derive rules such as the distance formul a.
(d)
True
(e)
False QR > PR
PQ is opposite an angle that measures 90 ◦ . PR is opposite an angle that measures 50 ◦ . QR is opposite an angle that measures 40 ◦ . Since 180 ◦ (50◦ + 90◦ ) = 40◦ we have Pˆ = 40◦ . PR is opposite an angle that measures 50 ◦ .
−
We discussed congruency of polygons in Study Unit 1.2A. Two triangles are congruent if they have exactly the same shape and size , i.e. if corresponding sides are equal in length and if corresponding angles have the same measure. However, we do not need to investigate all pairs of corresponding sides and all pairs of corresponding angles every time that we want to show that two triangles are congruent.
Conditions for congruency
If two triangles have certain properties, then certain other properties will follow. We have four different sets of requirements that must be satisfied in order for triangles to be congruent.
It is clear that if corresponding sides are equal, then the corresponding angles will be equal as well.
(1) Two triangles are congruent if the three sides of one triangle are equal to the three corres ponding sides of the other triangle . We call this the side– side–side (abbreviated SSS) condition for congruency.
We denote equal sides by writing down the vertices of corresponding angles in the correct order. Thus we write
C
Q
A
AB = PQ and not
P
B
AB = QP, since A corresponds to P, and B corresponds to Q .
R
ABC ≡ PQR Figure 1.2.15
We see that AB = PQ , AC = PR , BC = QR. Hence Cˆ = Rˆ , Bˆ = Qˆ , Aˆ = Pˆ . A
B
C
Bˆ is the angle included by the sides AB and BC.
(2) Two triangles are congruent if two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle. We call this th e side–angle–side (abbreviated SAS) condition for congruency.
43
MAT0511/004 W
R
U
S
T
V
RST
≡
WUV
Figure 1.2.16
We see that RS = WU , Sˆ = Uˆ , ST = UV . From this it follows that RT = WV (we can show, by drawing any two such triangles, that under these conditions the third sides of the two triangles are equal). Thus we have the condition SSS and hence RST WUV . Consequently ˆ Rˆ = W and Tˆ = Vˆ
≡
since the angles are opposite equal sides.
When we have AAS we automatically have ASA and
(3) Two triangles are congruent if two angles and one side of one triangle are equal to two corresponding angles and the corresponding side of the other triangle. We call this the angle–side–angle, or angle–angle–side (abbreviated ASA or AAS) condition for congruency.
vice versa: if two pai rs of S
angles are equal the third pair must also be equal.
X
P
R
M
PQR ≡ MX S Figure 1.2.17
ˆ , Qˆ = Xˆ , PQ = MX . We see that Pˆ = M From this it follows that Sˆ = Rˆ (angle sum of a triangle) RP = SM RQ = SX . The last two properties can be shown by drawing any two triangles with the properties shown in Figure 1.2.17. Thus we have the SSS condition and hence PQR MX S.
≡
44
This is a special case of two sides and an angle where we do not necessarily have the angle included between the two sides.
(4) Two right triangles are congruent if the lengths of the hypotenuse and one side of one triangle are respectively equal to the lengths of the hypotenuse and one side of the other triangl e. We call this the right angle– hypotenuse–side (abbreviated RHS) condition for congruency. A
C
M
O
T
P
MPO ≡ TAC Figure 1.2.18
We see that AC = PO , CT = OM , Aˆ = Pˆ = 90◦ . From this it follows that AT = PM (by the Theorem of Pythagoras) and hence we have the SSS condition. Thus MPO TAC. Consequently Cˆ = Oˆ
≡
and
Tˆ = Mˆ since both pairs of angles are opposite equal sides. In Example 1.2.2 we see why ASS is not a condition for congruency.
1.2.2 N
X
Y
Z
U
T
Figure 1.2.19
Figure 1.2.19 shows XY Z and NU T in which none of the angles is a right angle and XY = NU , Y Z = U T , Zˆ = Tˆ .
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Although the lengths of two corresponding sides are equal and one pair of corresponding angles is equal, we see that We use
≡ to denote “is not
XY Z ≡ NU T
congruent to”.
since Yˆ is acute but the corresponding angle, Uˆ , is obtuse. Thus it is necessary that when two sides and one angle of each triangle are involved we must have either RHS or SAS.
1.2.6 (a) Draw any scalene triangle PQR. Can the triangle have a diagona l? Draw the altitude from Q to PR . (b) Draw any equilateral triangle GEF. Draw the altitude from G to EF, so that it meets E F in the point H . Show that GEH GFH .
≡
(a) X
R
P
Q
It is impossible to draw a line from any vertex to a non–adjacent vertex. Hence PQR has no diagonals.
In fact, no triangle has a diagonal. Since a triangle has three sides it can have three altitudes.
QX (b)
PR, hence QX is an altitude of
⊥
PQR.
G
E
F H
46
In GEH and EG = F G
GF H
GH = GH
GEF is equilateral. We write this to show that although GH is one line, it serves as a side for two different triangles.
ˆ = F HG ˆ = 90◦ . E HG
The altitude GH is perpendicular to E F .
Thus GEH
RHS condition is satisfied.
≡ GFH .
From the solution to Activity 1.2.6(b), since
GEH ≡ GFH it follows that
EH = F H and ˆ = F GH ˆ . E GH Thus we see that an altitude of an equilateral triangle bisects the angle at the vertex from which it srcinates and bisects the opposite side of the triangle. Similar triangles
We use ity.
||| to denote similar-
As in the case of polygons in general, we also have similar triangles. We say that two triangles are similar if all three pairs of corresponding angles are equal. In practice this means that we only need to show that two pairs of corresponding angles are equal, because the angles of any triangle add up to 180◦ . S
M
K
P
T
L
Figure 1.2.20
|||
In Figure 1.2.20 we have MPT SLK . We know that when polygons are similar, their corresponding sides are in proportion. This applies to triangles, and thus PT PM MT . = = LK LS SK We can use this relationship to calculate the length of a side of a triangle.
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1.2.3
At a certain time of the day a boy who is 1 , 2 m tall casts a shadow that is 3 , 6 m long. At the same time a tree casts a shadow that is 7 , 2 m long. How high is the tree?
SOLUTION D
In the sketch we have AB representing the boy, BC representing shadow;
the similarly
boy’s
A
DE
represents the tree and EF the tree’s shadow.
1,2 m B
C
E
3,6 m
F
7,2 m
We assume that the boy and the tree both stand upright and make an angle of 90◦ with the horizontal ground. Since the sun is in the same position relative to both the boy and the tree, we can also assume that Cˆ = Fˆ . We thus have two similar triangles, i.e.
ABC ||| DEF. Since the triangles are similar, pairs of corresponding sides are in proportion. Thus DE EF DF = = . AB BC AC We want to find the length of DE . Since we know nothing about DF or AC, we use the equation DE EF = . AB BC We substitute AB = 1, 2 m, BC = 3, 6 m and E F = 7, 2 m into this equation. We then have DE 7, 2 m = 1, 2 m 3, 6 m
48 and thus
DE =
= You may want to revise decimal multiplication and divi-
×
1, 2 7, 2 m 3, 6
×
12 72 m 360
= 2 , 4 m.
sion. See Topic 2 of Book 1.
The tree is thus 2, 4 m high.
1.2C QUADRILATERALS As we have already pointed out, a quadrilateral is a polygon with four sides. The angles in a triangle add up to 180 ◦ . What do you think is the angle sum of any quadrilateral? You may want to try to work this out physic ally. If you draw a few different quadrilaterals, cut off the vertices and rearrange them (as you did in the case of a triangle in Figure 1.2.8) you will see that the angle sum of a quadrilateral is 360◦ . You can also derive this mathematically, using your knowledge of triangles. A quadrilateral can be divided into two triangles. See Figure 1.2.21. A
B
D C
Figure 1.2.21
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In quadrilateral ABCD, ˆ Aˆ + Bˆ + Cˆ + D ˆ +D ˆ + ACB ˆ = C AD ˆ + DCA ˆ +Bˆ + BAC the angles in
the angles in
ACD
ABC
= 180◦ + 180◦ = 360◦ . Because quadrilaterals have four sides, we have two different kinds of quadrilaterals: one kind in which opposite sides (either one pair or both pairs) are parallel, and the other kind in which no sides are parallel. We cannot generalise much about quadrilaterals such at these, because if opposite sides are not parallel, then such quadrilaterals cannot have any common properti es. Figure 1.2.22 shows three different quadrilaterals, in which no opposite sides are parallel. A
A
B C
D
C
B
A
D
C B
Figure 1.2.22
D
50 Kite
The only “interesting” quadrilateral in this category is a kite. A kite is a quadrilat– eral in which pairs of adjacent sides have the same length. See Figure 1.2.23. Q D
R A square and a rhombus are
A
P
C
also kites.
S
B
Figure 1.2.23
1.2.7
When we write “bisects the vertex” we mean “bisects the angle at the vertex”.
Use congruent triangles to show that the diagonals of a kite cut each other at right angles, and that the longer of the two diagonals bisects the shorter one, and bisects the vertices that it joins.
Q
R
T
P
S
In
QPS and QRS
QP = QR PS = RS QS = QS.
Hence
QPS ≡ QRS.
SSS
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Hence ˆ = R QS ˆ PQS and ˆ = R SQ ˆ . PSQ
Hence
QS bisects Qˆ and Sˆ. In
PQT and RQT
QP = QR QT = QT ˆ = R QT ˆ . PQT
Hence
PQT ≡ RQT .
Hence
SAS
QTˆ P = Q Tˆ R. Since
QTˆ R + QTˆ P = 180◦
RTˆ P is a straight angle.
it follows that
QTˆ R = Q Tˆ P = 90◦ . Thus the diagonals are perpendicular to each other. Since PQT RQT it also follows that PT = RT , i.e. the longer diagonal bisects the shorter one.
Trapezoid (Trapesium)
≡
A trapezoid or trapesium is a quadrilateral which has two parallel sides . See Figure 1.2.24. E
D
A F
C
B
Figure 1.2.24
ABCD is a quadrilateral in which DC AB. We call ABCD a trapezoid or trapesium. Either of the parallel sides is called a base of the trapezoid. Any perpendic-
52 ular line such as E F between DC and AB is called an altitude of the trapezoid. A trapezoid in which the non–parallel sides are equal in length is called an isosceles trapezoid. Parallelogram
A parallelogram is a quadrilateral in which opposite sides are parallel . In Figure 1.2.25, we have AB DC and AD BC.
D
C
A
B
Figure 1.2.25
1.2.8 Draw any parallelogram. Draw its diagonals. Measure its sides and its angles. What do you notice about these line segments and angles?
(In this activity (and in others that follow) you are asked to draw parallel lines, perpendicular lines, measure angles, construct certain angles, and so on. We assume you will have learnt to do these constructions at school. If you are not sure what to do, any school geometry book at grade 7 or 8 level will help you. In this module we will not expect you to do accurate constructions.)
R
S
T Q
P
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From even a rough sketch, such as the figure on the previous page, you can see that in parallelogram SRQP
Note that SQ = PR.
opposite sides have equal length, i.e. PQ = RS and QR = SP
opposite angles have equal measurement, i.e. Rˆ = Pˆ and Sˆ = Qˆ
the diagonals bisect each other, i.e. RT = T P and ST = T Q.
Note that these results will be the same regardless of the shape of the parallelogram we draw. In fact, we can once again prove the results, using congruent triangles. Consider parallelogram SRQP.
S 1
2
1
R 2
T
2
1
P
1
2
Q
Figure 1.2.26
In
QSP and SQR Qˆ 1 = Sˆ2 Qˆ 2 = Sˆ1 SQ = SQ .
Alternate angles. Alternate angles. Common to both triangles.
Hence and thus
QSP ≡ SQR
ASA
Rˆ = Pˆ , QP = SR , PS = RQ .
Similarly,
Hence
PSR ≡ RQP. Sˆ = Qˆ .
ASA
54 Since Rˆ = Pˆ and Sˆ = Qˆ , we have shown that the opposite angles of a parallelogram are equal. Since QP = SR and PS = RQ we have shown that the opposite sides of a parallelogram are equal.
Also
SRT ≡ QPT
so that
ASA
ST = QT and
QRT ≡ SPT
and thus If the line segments bisect
ASA
RT = PT .
each other, they intersect at the midpoints of the line seg-
Hence the diagonals bisect each other.
ments.
These properties of parallelograms are interdependent. Each one implies the others, and hence in any quadrilateral, if we have any one of the following conditions the quadrilateral is a parallelogram.
Opposite sides are equal.
Opposite angles are equal.
Each diagonal divides the parallelogram into two congruent triangles.
The diagonals bisect each other.
In the following example we show that one of these conditions implies that the quadrilateral is a parallelogram.
1.2.4 In quadrilateral ABCD we have AB = CD and AD = CB. Show that ABCD is a parallelogram.
SOLUTION We sketch ABCD, and draw the diagonal DB . D
A
C
B
55 In
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ABD and CDB we have BD = DB
Common to both triangles.
AB = CD AD = CB. Thus
ABD ≡ CDB.
SSS
Since the triangles are congruent, all pairs of corresponding angles are equal. ˆ = C DB ˆ . Thus DC AB. (If alternate angles are equal the lines are Hence ABD parallel.)
ˆ = CBD ˆ and hence AD Similarly A DB
BC.
Since opposite sides are parallel, ABCD is a parallelogram.
We have several special parallelograms. Rhombus
If all sides of a parallelogram have equal length, the parallelogram is a rhombus. See Figure 1.2.27.
C
N
M
L
Figure 1.2.27
1.2.9 Draw any rhombus. Draw its diagonals. What do you notice about these diagonals?
56
R
S T
Q
P
The diagonals SQ and PR bisect each other at right angles, and SQ and PR bisect the vertices. Note that SQ = PR .
We can also prove these facts, once again using congruency. ˆ = Q SR ˆ PQS
Alternate angles equal since PQ
But ˆ = PSQ ˆ . PQS
SR.
PS = PQ
Thus ˆ = Q SR ˆ PSQ i.e. ˆ . SQ bisects P SR ˆ , and that PR bisects S RQ ˆ and S PQ ˆ . Similarly it follows that SQ bisects P QR Now, in
PST and RST PS = RS ST = ST ˆ = R ST ˆ . PST
The diagonal
SQ bisects the vertex S .
Hence and thus
PST ≡ RST
SAS
PT = RT i.e. we have shown that the diagonal SQ bisects the diagonal PR . Also
PTˆ S = R Tˆ S and thus
PTˆ S = R Tˆ S = 90◦ .
PTˆ R = 180◦ .
Hence the diagonal SQ is perpendicular to the diagonal PR . In the same way we can show that the diagonal PR is the perpendicular bisector of the diagonal QS .
57 Rectangle and square
MAT0511/004
If one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. If one angle of a rhombus is a right angle, then the rhombus is a square. See Figure 1.2.28.
D
C
A
B
S
R
P
Q
Rectangle
Square Figure 1.2.28
1.2.10 Show why we only have to specify that if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle.
D
C
A
B
ABCD is a parallelogram. Thus AB interior angles are supplementary.
ˆ = 90 ◦ DC, and thus ADC
since the co–
ˆ and A BC ˆ are suppleSimilarly, AD BC and hence the co–interior angles D AB ˆ = 90◦ . mentary. Thus A BC
ˆ We can also show that DCB = 90◦ . A rectangle is a quadrilateral in which each angle measures 90◦ , and thus ABCD is a rectangle.
Similarily we can show that we need only specify that one angle of a rhombus is a right angle for the rhombus to be a square.
58
1.2.11 Draw the diagonals of the rectangle and square given in Figure 1.2.28. Do they bisect each other? Do they intersect at right angles? Do they bisect the vertices?
D
C
A
B
S
R
P
Q
We see that in the rectangle
the diagonals bisect each other
the diagonals do not intersect at right angles
the diagonals do not bisect the vertices
the diagonals have equal length.
In the square
The first three properties follow immediately from the fact that a square is a rhombus.
the diagonals bisect each other
the diagonals are perpendicular to each other
the diagonals bisect the vertices
the diagonals have equal length.
By using congruency we can prove that the diagonals of a rectangle have equal length.
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In rectangle ABCD we have
AB = BA AD = BC Aˆ = Bˆ = 90◦ . Hence
DAB ≡ CBA. Thus
SAS
DB = CA
i.e. the diagonals are equal in length. Similarily we can prove that the diagonals of a square are also equal in length. In any square or rectangle, we can apply the Theorem of Pythagoras to calculate the length of the diagonal.
1.2.12 Calculate the length of DB in the rectangle ABCD sketched below.
A
12 cm
B
5 cm
D
C
By the Theorem of Pythagoras we have
DB2 = AD 2 + AB2 i.e. we have
DB2 = (52 + 122 ) cm2
= (25 + 144) cm2 = 169 cm2 . Thus DB =
√169 cm = 13 cm, since length cannot be negative.
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1.2 1.
(a) Create a design by reflecting a regular octagon with sides 1 cm, in a horizontal line through the lower edge once, and then translating the resulting shape twice by intervals of 3 cm to the right. (b) Create a tessellation using right triangles.
2. A
B
C D
Suppose ABC is an isosceles triangle with AB = AC. Show that the altitude AD bisects BC.
3. A
B
C D
Suppose ABC is an equilateral triangle with sides of length s cm. Show that 3 AD = s cm. 2
√
4. You have seen that the angle sum of a triangle is 180 ◦ , and the angle sum of a quadrilateral is 360 ◦ . Sketch a regular octagon, and calculate the sum of the measures of all the angles. 5.
(a) See Activity 1.2.8. How many pairs of congruent triangles are there after you have drawn both diagonals? (b) If PQRS is any quadrilateral such that SQ divides the quadrilateral into two congruent triangles, namely QRS and SPQ, show that PQRS is a parallelogram.
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6. Consider the trapezoid CDEF sketched below. FG is an altitude of the trapezoid. CG is a quarter of the length of CD , and CF is one third of the length of CD . F
C
E
D G
If the length of CD is x cm, calculate the length of F G, in terms of x . 7. A
D
C
E
B
In the figure we have
AB = DC Bˆ = Cˆ DB Is
⊥
AC .
ECD ≡ ECD? Give reasons for your answer.
8. Sketch a kite in which the pairs of adjacent sides are such that the shorter sides are half the length of the longer sides. (a) Draw the longer of the two diag onals. Show that it cuts the kite into two congruent triangles. (b) Draw the shorter diagonal. Show that the longer diagonal bisects the shorter diagonal.
62 9.
D
A
C
P
In
Q
DPQ and DCA, CA PQ.
Also, DA = AQ, DC = CP. Show that
DPQ ||| DCA, and that CA =
1 PQ. 2
10. Complete the following table. Polygon(regular)
Number of sides
Number of vertices
Number Angle of sum diagonals
Triangle Quadrilateral Pentagon Hexagon Octagon 11. Complete the following statements. (a) The lengths of the diagon als of a rectangle are .......... . (b) The diagonals of a rhombus make angles of ......... with each other at their point of intersection. (c) The diagonals of a ......... and a .......... bisect the vertices as well as each other. (d) (Complete the statement by choosing one of the options suggested.) A parallelogram with diagonals that are almost the same length will “lean” over (further than /less than) a parallelogram in which one diagonal is considerably longer than the other.
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1.3 CIRCLES
1.3A SOME BASIC FACTS ABOUT CIRCLES
The plural of radius is radii.
A circle is a figure in a plane consisting of all points which are the same distance from a fixed point called the centre of the circle. Any line segm ent from the centre to any point on the circle is called a radius of the circle. It should be obvious that all radii of the same circle have the same length.
O
P
Figure 1.3.1
In Figure 1.3.1 the circle has centre O and radius OP . When two or more circles have the same centre they are called concentric circles. Each of the circles in Figure 1.3.2 has the same centre, C .
C
Figure 1.3.2
If we move along a circle from one point to another, we move along an the circle.
arc of
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We pronounce this with a silent “h”, so that it sounds like “kord”.
If we join any two points on a circle by means of a line segment, the line segment is called a chord. A chord which passes through the centre of the circle is called a diameter of the circle. A diameter thus consists of two radii, and hence the length of the diameter is twice the length of the radius. Algebraically we express this as d d = 2r , or r = 2 where d and r represent the lengths of the diameter and radius, respectively. Any diameter divides the circle into two semi–circles. A tangent or tangent line to a circle is any line that touches the circle at exactly one point, i.e. it is a line that that contains exactly one point of the circle.
F C B E O A
D
Figure 1.3.3
In Figure 1.3.3 we have a circle with
centre O
arc AB (also arcs AC, CB, BD, DA, DC, etc.)
chord AB
diameter CD
radius OC (also radius OD ) tangent E F which touches the circle at C .
⊥
In Figure 1.3.3 we note that CD EF . In fact, in any circle, the radius (or diameter) is always perpendicular to the tangent to the circle at the point of contact.
65
MAT0511/004 A
B
P E
C
D
Figure 1.3.4
Consider Figure 1.3.4 above. The arcs AB, BC, CD, DE and EA are equal in length. If we join each of the points A , B , C , D and E on the circle to the centre ˆ is subtended by arc P, then we obtain five central angles. We say that angle B PC BC. By this we mean that we draw the radii PB and PC and thus create an angle ˆ . Equal arcs subtend equal central angles. Thus at the centre, namely B PC ˆ = BPC ˆ = C PD ˆ = DPE ˆ = E PA ˆ . APB You can verify this yourself by drawing any circle, marking off equal arcs and then measuring the central angles subtended by these arcs.
1.3 1. Draw any right triangle ABC with right angle at B and find the midpoint of the hypotenuse, AC. Draw a circle using the midpoin t of the hypotenuse as centre, and the hypotenuse as diameter. (a) Does B lie on the circle? (b) Join A and C to any other point D on the circle, on the opposite side ˆ appear to be? of the circle to B . What does the measure of A DC 2. Use a pair of compasses (if you have one) or draw freehand as accurately as possible, a circle with radius 4 cm, and, using the same centre, a circle with radius 2,5 cm. For the smaller circle, draw two diameters AB and CD perpendicular to each other. Mark the endpoints of the two diameters by means of A and B , C and D . Extend AB in both directions to cut the bigger circle at Z and T . Extend CD in both directions to cut the bigger circle at Q and W . Draw lines parallel to ZABT , through C and through D . Draw lines parallel to QCDW, through A and through B . Write down as many as you can find of the congruent figures that have been created.
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•
Transformation
Translation (shifting) – shifting an object in any given direction (e.g. horizontally or vertically) over any given distance
Symmetry – axis of symmetry: vertical
horizontal
– reflection (or line) symmetry: creating a mirror image of a given object in a vertical or horizontal line – rotational symmetry: object remains identical after rotation through any given angle less than 360◦
• Lines
A line is a collection of points. It extends indefinitely in two directions.
A line segment is part of a line, and has two endpoints. The distance between the two endpoints is its length.
A ray is part of a line. It extends indefinitely in one directio n, and has one endpoint.
Points of intersection – infinitely many (the lines are coincident) – none (the lines are parallel) – only one (the lines may be perpendicular)
• Angles
An angle is formed when a ray is rotated about a fixed endpoint, called the vertex. The amount of rotation is the measure of the angle, expressed in degrees, minutes and seconds.
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Classification of angles according to measure (Suppose the measure of the angle is x , where x represents a certain number of degrees.) – one revolution 360 ◦ – reflex angle 180 ◦ < x < 360◦ – straight angle 180 ◦ – obtuse angle 90 ◦ < x < 180◦ – right angle 90 ◦
– acute angle 0 ◦ < x < 90◦ Relationships between angles – adjacent angles share a common vertex and a common ray – supplementary angles have measures that add up to 180 ◦ – complementary angles have measures that add up to 90 ◦ – vertically opposite angles are equal – when parallel lines are cut by a transversal line, then: corresponding angles are equal alternate angles are equal co–interior angles are supplementary
• Polygons (in general)
Congruent polygons have exactly the same shape and size.
Similar polygons have exactly the same shape but different sizes. The corresponding sides of similar polygons are in proportion.
– Scale factor: the number that represents the amount by which an object is enlarged or reduced so that it is similar to the srcinal.
Polygons have – vertices: the corners of the polygon – diagonals: line segments whos e endpoints are non–adjacent vertices – altitudes: line segments from any ve rtex to an opposite side, perpendicular to that side
Triangles: polygons with three sides
•
Angle sum is 180◦ .
Classification of triangles according to lengths of sides – scalene: no sides with equal length (and hence no equal angles) – isosceles: two equal sides (and hence two equal angles) – equilateral: three equal sides (and hence three equal angles, all measuring 60◦ )
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Classification of triangles according to angle measure – acute angle triangle: all angles acute – obtuse angle triangle: one angle obtuse – right angle triangle: one angle a right angle (the hypotenuse is the side opposite the right angle)
• Congruency of triangles
Conditions for congruency – side, side, side (SSS) – side, angle, side (SAS) – angle, angle, side (AAS or ASA) – right angle, hypotenuse, side (RHS)
• Similar triangles
Similar triangles have equal angles, and corresponding sides in proportion.
• Quadrilaterals:
polygons with four sides
Kite: pairs of adjacent sides have equal length
– the diagonals cut each other at right angles, and the longer of the two diagonals bisects the shorter one
Trapezoid: two sides are parallel
Parallelogram: both pairs of opposite sides are parallel
– opposite sides have equal length – opposite angles have equal measurement – each diagonal divides the parallel ogram into two congruent triangles – the diagonals bisect each other
Rhombus: a parallelogram in which all sides have equal length. In addition to all the properties of parallelograms it is also true that
– the diagonals bisect each other at right angles – the diagonals bisect the angles from which they srcinate (i.e. the vertices).
Rectangle: a parallelogram in which one angle is a right ang le. In addition to all the properties of parallelograms it is also true that
Square: a rhombus in which one ang le is a right angle. In addition to all the properties of a rhombus, it is also true that
– the diagonals have equal length.
– the diagonals have equal length.
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• Circles: a plane figure consisting of all points equidistant from a fixed
point called the centre. Different circles with the same centr e are called concentric circles.
Radius: any line segment from the centre to any point on the circle
Chord: any line segment joining any two points on the circle
Diameter: a chord through the centre, dividing the circle into two semi–circles
Arc: any path along the circle from one point to another. An arc from one endpoint of a diameter to the other endpoint is a semi–circle. An angle at the centre subtended by a chord or arc is a central angle.
Tangent: a line which touches the circle at exactly one point
CHECKLIST Now check that you can do the following. SECTION 1.1
1. Create patterns by transla ting an object in a given direction, over a given distance. Examples 1.1.1, 1.1.3 2. Create patterns by reflecting an object, for example in a vertical or horizontal line. Recognise objects that have reflection symmetry. Examples 1.1.2, 1.1.3; Activity 1.1.1 3. Recognise objects that have rotational symmetry. Create patterns by rotating an object through a given angle between 0 ◦ and 360◦ . Examples 1.1.4, 1.1.5; Activity 1.1.1 4. Define the following concepts: line, line segment, ray, angle. Definitions 1.1.1, 1.1.2, 1.1.3 and 1.1.4 5. Recognise lines that are coincident, parallel, perpendicular. Figures 1.1.11, 1.1.16 6. Classify angles according to measure or relationship with other angles. Pages 17–19; Activity 1.1.3 7. Recognise the relationship between parallel lines cut by a transversal line and the resulting corresponding, alternate and co–interior angles. Page 20; Example 1.1.6; Activities 1.1.2, 1.1.3
70 SECTION 1.2
1. Create patterns by means of tessellating polygons. Figure 1.2.4 2. Draw the diagonals (if they exist) and altitudes of any polygon. Example 1.2.1 3. Recognise polygons that are congruent, and identify corresponding sides or angles of congruent polygons. Activity 1.2.1 4. Recognise polygons that are similar. Show that corresponding sides of similar polygons are in proportion. Activity 1.2.2 5. Use the property of similarity of polygons to determin e the scale factor required when a given object needs to be enlarged or reduced. Activity 1.2.3 6. Classify triangles according to the lengths of their sides or the measures of their angles. Figures 1.2.9, 1.2.10, 1.2.11, 1.2.12, 1.2.13, 1.2.14; Activity 1.2.4 7. Use the Theorem of Pythagoras. Activities 1.2.5, 1.2.12 8. Know and apply the four sets of conditions that triangles must satisfy in order to be congruent. The conditions are referred to as SSS, SAS, AAS (or ASA) and RHS. You do not need to prove that each of these conditions implies congruency. Figures 1.2.15, 1.2.16, 1.2.17, 1.2.18 and related discuss ion; Example 1.2.2; Activity 1.2.6 9. Recognise triangles that are similar. Use the fact that corresponding sides of similar triangles are in proportion to calculate distance or length. Figure 1.2.20 and related discussion; Example 1.2.3 10. Classify quadrilaterals according to whether opposite sides are parallel, or equal in length. Figures 1.2.24, 1.2.25, 1.2.27, 1.2.28 and related discussion. 11. Use congruency of triangles to prove certain properties of quadrilaterals. – In a kite prove that the diagonals intersect at right angles, and the longer of the diagonals bisects the shorter diagonal. Activity 1.2.7
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– In a parallelogram prove that the diagonals bisect each other opposite sides have equal length opposite angles have equal measure. Activity 1.2.8 and the discussion that follows; Example 1.2.4 – In a rhombus (in addition to the properties of parallelograms) prove that the diagonals bisect each other at right angles the diagonals bisect the vertices. Activity 1.2.9 and the discussion that follows – In a rectangle (in addition to the properties of parallelograms) prove that the diagonals have equal length. Activity 1.2.11 and the discussion that follows – In a square (in addition to the properties of a rhombus) prove that the diagonals have equal length. Activity 1.2.11 and the discussion that follows
SECTION 1.3
1. Use the terminology of circles: centre, radius, chord, diameter, arc, semi– circle, tangent, central angle subtended by an arc. Figures 1.3.1, 1.3.2, 1.3.3, 1.3.4 and related discussion
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PERIMETER, AREA AND VOLUME OUTCOMES After studying this topic you should be able to do the following. SECTION 2.1: Permeter and Area (Measurement)
Calculate the perimeters of various polygons.
Calculate the circumference of a circle.
Calculate the horizontal distance covered by a point on a circular object.
Calculate the areas of various regular polygons.
Calculate the area of a circle.
Calculate areas involving a combination of polygons and circles.
SECTION 2.2: Surface Area and Volume of 3-D objects
Draw the net for several familiar three–dimensional objects, i.e. a pyramid, rectangular prism, cube, right circular cylinder.
Calculate the surface area of a pyramid, rectangular prism, cube.
Calculate the surface area of a right circular cylinder and the surface area of a right circular cone.
Calculate the surface area of a sphere.
Calculate the surface area of irregularly shaped objects.
Calculate the volume of familiar three–dimensional objects, i.e. rectangular prism, cube, right circular cylinder, right circular cone, sphere.
Calculate the volume of irregularly shaped objects, or objects that are combinations of other solids.
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2.1 MEASUREMENTS OF PERIMETER AND AREA
2.1A PERIMETER AND AREA OF TWO–DIMENSIONAL OBJECTS Perimeters of polygons
We use the word perimeter to describe the total length of the edges or sides of any polygon. In the case of a polygon that is not regular , we need to measure the lengths of all the sides and add the meas urements. For all poly gons the perimeter is the sum of the lengths of all the sides . In the case of regular polygons, we often have a formula that we can use to calculate the perimeter. In Table 2.1.1 we now state, without proof, the formulas for the perimeters of the polygons we encounter most often (It is important to know these formulas by heart.)
Polygon
Perimeter
Triangle (scalene)
s1 + s2 + s3
The triangle has three sides of different lengths, s 1 , s2 and s 3 units.
Triangle (isosceles)
2 s1 + s2
We denote the length of each of the two equal sides by s 1 units; the length of the third side is s 2 units.
Triangle (equilateral)
3s
The three sides all have the same length, s units.
Quadrilaterals (in general)
s1 + s2 + s3 + s4
The four sides may have different lengths, namely s 1 , s2 , s3 and s 4 units.
Special quadrilaterals Rectangle
2
Square Parallelogram
4
Rhombus
Kite
4 2
2
(l + b)
The length is l units. The breadth is b units.
s
Each of the four sides measures s units. The length of one of the sides is l units; the length of the adjacent side is b units. The four sides all have the same length, s units. Two sides each measure s 1 units and the other two sides each measure s 2 units.
(l + b)
s
(s1 + s2 )
Table 2.1.1
74 From the formulas given in Table 2.1.1 for the perimeter of an equilateral triangle, a square and a rhombus, it is clear that, in general, the perimeter of any regular polygon is ns units, where n is the number of sides, and s is the length of a side.
2.1.1 Calculate the perimeter of each of the following shapes.
6m 5m 7m 2m
1m
Scalene triangle
Rectangle
8 cm 2 cm
3 cm
2 cm 5 cm
2 cm 3 cm 4 cm
Regular pentagon
Regular hexagon
Irregular polygon
SOLUTION Triangle: Rectangle: Pentagon: Hexagon: Irregular polygon:
Perimeter Perimeter Perimeter
= (2 + 6 + 7) m = 2 (5 + 1) m = (5 3) cm
= 15 m. = 12 m. = 15 cm.
× Perimeter = (6 × 2) cm = 12 cm. We deduce the lengths of the remaining edges from the given measurements, and the fact that opposite sides are parallel.
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MAT0511/004 8 cm 2 cm
2 cm 5 cm
7 cm
3 cm 4 cm
13 cm
We then have Perimeter = (7 + 2 + 2 + 8 + 5 + 3 + 4 + 13) cm = 44 cm.
Circumference of a circle We also speak of the “radius” and “diameter” instead of the lengths of the radius and diameter, respectively.
In the case of a circle, the perimeter of a circle is called the circumference of the circle. The formula for the circumfe rence of a circle is πd or 2πr , where d and r are respectively the lengths of the diameter and radius. In this module we first mentioned π in connection with irrational numbers. The Greek mathematicians noticed that in any circle with circumference c and diameter d , the ratio dc is always a constant. They gave this constant the name π, and from this the formula c = πd developed. The fact that the ratio c : d is always constant tells us that the circumference and diameter of a circle are always in direct proportion to each other. Remember that π is irrational. We may approximate it by 3 , 14 or 22 , but it is 7 important to remember that these numbers are not exactly the same as π.
2.1.2 Calculate the circ umference of a circle with radiu s 5 cm. Give your answ er correct to one decimal place.
SOLUTION
Circumference = 2πr
= 2π 5 cm = 10π cm
×
≈
31, 4 cm
Thus the circumference is approximately 31, 4 cm.
The formula for the circumference of a circle is useful to determine the distance covered by a point on a wheel.
76
2.1.3 The diameter of a wheel is 40 cm. How far does a point on the wheel travel when it revolves completely, 200 times? (Assume the wheel moves smoothly and does not slip.) Leave the answer in terms of π.
SOLUTION In one revolution any point on the wheel covers a distance equal to its circumference. (If you are not sure that this is so, experiment with a coin, or saucer, or any small circular object.) Circumference =
d cm
π
= π 40 cm = 40π cm
×
Thus in one revolution a point on the wheel travels 40π cm, and in 200 revolutions it travels ( 200 40π) cm, i.e. 8 000 π cm, i.e. 80 π m.
×
Areas of polygons
For any polygon we can determine “how much space in the plane” it covers. This
We use the word dimensions to denote length and
intuitive sense of taking up space is made more exact in the concept of area. We represent area in terms of square units. Suppose the figure below is a rectangle, with dimensions 10 cm by 6 cm.
breadth.
10 cm
6 cm
Figure 2.1.1
If we divide the rectangle into equal squares with sides of 1 cm, we will have 60 such squares. We say that the area of the rectangle is 60 square centimetres. We denote square centimetres by cm2 , and the area of the rectangle represented in Figure 2.1.1 is thus 60 cm2 . In the following table (Table 2.1.2) we give the formulas for the areas of the polygons we considered earlier. We do not have a general formula for the area of any polygon. However, we can consider irregularly shaped polygons as a combination of rectangles, triangles, etc., and calculate each of these areas separately before combining them to obtain the total area. (Learn to know these formulas.)
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A
B D
a
Area A (in square units)
Triangle (all types)
A = 12 base altitude (We use altitude for perpendicular height.)
Square
×
1 2
ABC = × BC × AD
A = length of side length of side = (length of side)2
×
Area of ABCD = (AB)2 = a2
B
S
R
Rectangle
A = length
× breadth
Area of PQRS = PQ
P
D
A
×
Area of
C
A
Polygon
C
D
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C
E
Parallelogram
A = base
× altitude
Area of ABCD = AB
B
S
R
Rhombus
A = base
T L
Kite
A=
M
K
× DE
× altitude
Area of PQRS = PQ P
× RQ
× ST
1 2
× length of first diagonal × length of second diagonal
Area of KLMN =
1 2
LN
KM
× × N M
Trapezoid L
A=
1 2
× sum of parallel sides × altitude
Area of KLMJ = J
N
K
Table 2.1.2
1 2
(ML + JK )
× MN
78 Area of a circle
We also need to consider the area of a circle. You may know that
Note: Students should know this formula by heart.
Area of a circle = πr2
or π
2
d
Area of a circle =
2
where r represents the radius and d represents the diamete r of the circle. You may be interested to see how the derivation of this formula can be illustrated. The circle with centre C and radius r is divided into 8 congruent sectors. See Figure 2.1.2. P W
Q r C
V
R
S
U T
Figure 2.1.2
We cut up the circle and arrange the sectors as shown in Figure 2.1.3. R
S
T
U
V
r
r
R
P
Figure 2.1.3
W
V
79
Note: It is not necess ary to be able to deduce this formula for examination purposes.
MAT0511/004
The object in Figure 2.1.3 has two “wavy” sides, both denoted by RV . The distance from R to V is half the circumference of the circle, i.e. πr units. The length RR (or V V ) is the same as the radius of the circle, i.e. r units. We can repeat this process, and cut the circle into a large number of much smaller sectors. The smaller we make the sectors, the less curved will be the arcs that make up the long sides of the object similar to that in Figure 2.1.3. Thus we will eventually have an object that is almost identical to a rectangle. The formula for the area of a rectangle is l b. In this case we have l = πr units, and b = r units. Thus, the area of the “circle” (even though it is no longer a circle in shape we have not changed its area) is
×
(l
× b) square units = ( r × r) square units π
i.e. we have area of circle = πr 2 square units.
2.1.4 Calculate the approximate area of a circle with diameter 42 cm. (Let
π
≈
22 .) 7
SOLUTION Area =
r2 22
π
≈
2
× (21) 7
= 1 386 cm 2
cm2
Since d = 42 cm we have r = 21 cm
Thus the area of the circle is approximately 1 386 cm 2 .
2.1.5 Calculate the area of the figure on the following page. The numbers indicate lengths, in centimetres. All angles are 90◦ .
80
P
4
W 5
V
7
U 3
T S 5
R
SOLUTION For convenience we include an additional line segment, namely the line segment joining V and S .
P
4
W 5
V
7
U 3
T
S 5
R
Area of the figure = Area of rectangle PQRW PQ = (5 + 3 + 5) cm, since
all angles are right angles.
= = = =
+ Area of rectangle STUV (PQ PW ) + (ST UT ) ((13 4) + (7 3)) cm2 (52 + 21) cm2
× ×
73 cm 2
Thus the total area of the figure is 73 cm2 .
×
×
81
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2.1.1 A landscape gardener wants to lay paving around a pond in a garden. The area requiring paving is the shaded reg ion shown in Figure 2.1.4 . What is the approximate area, correct to two decimal places, of the section to be paved. (Let π 3, 14.)
≈
1,2 m
5m
12 m
Figure 2.1.4
Paved area = Area of rectangle
=
(12
− Area of circle 2
2
× 5) − (1, 2) m = (60 − (1, 44)) m ≈ (60 − (3, 14) (1, 44)) m = (60 − 4, 5216) m π
2
π
2
2
= 55, 4784 m2
≈
55, 48 m2 (correct to two decimal places)
Thus the approximate area to be paved is 55, 48 m2 .
2.1.6 Find a formula for the area of a regular hexagon in terms of the length of one of its sides.
82
SOLUTION s
T
r
a A
B
Sketch a regular hexagon, with sides of length s cm. Draw the diagonals. The diagonals intersect in only one point , T , and the lengths of the line segments joining the vertices and T are all the same, say r cm. Do you see that the reason for congruency is SSS?
The hexagon thus consists of six congruent triangles. In any one of the triangles, draw an altitude from T to the base. Say its length is a cm. This will be true for each triangle, since the triangles are congruent. Area of hexagon = 6
=
area of
ATB
× × base× height 6 × × s × a cm
= 6
1 2
2
1 2
= 3as cm
cm2
2
We now express a in terms of s . Triangle AT B is isosceles, since AT = BT = r . ◦ Also, ATˆ B = 360 60 ◦ since there are six congruent triangles with common 6 = ˆ = T BA ˆ = 60◦ , and hence vertex T . Thus T AB ATB is an equilateral triangle.
Thus r = s . From Question 3 of Exercise 1.2 we have a = is 3
√ 3 2 s
2
s cm , i.e.
√ 3 3 2
2
2
s cm .
√3 2
s. Thus the area of the hexagon
83
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2.1 1. What is the area of a kite with one diagonal that is one metre long, and the other diagonal half a metre long? 2. The following diagram shows a circular pool with diameter 5 m.
The shaded area represents paving around the pool. The paving is 0 , 5 m wide. What is the area of the paved sec tion? Give your answer correc t to one decimal place. 3. Find the perimeter and area of each of the following figures, correct to 2 decimal places if the answers are not exact. Assume angles drawn to look like right angles are right angles. Assume measurements are in centimetres. (a) A
F
5
E 2
3
D 2
B
C
9
(b) D 6
4
12 6
C 4 6
A 6
2
+
4
2
B
84 (c) 3 3
A
3
9
C
9
B
(d) I 4
4
A 2 J
H 2 G D
5
5
B 2 C
E 2 F
4. What is the area of the shaded sec tion in the follow ing figure? You may assume that AC is a diameter of the circle. Give your answer correct to the nearest square metre. A
B
6m
D
C
5. On a vacant piece of land with area 25 000 m 2 , a sports club plans to build an athletics track, with the shape shown in the following figure.
ym
xm
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In the figure, the two curved sec tions are both semi –circles. In order to accommodate competition races, the lengths of the lanes in which runners will compete must be 400 m. One such lane is shown by means of a dashed line on the sketch. (The starting positions of the runners are staggered so that each runner covers the same distanc e.) The outer boundary of the athletics track is shown here as the solid line. It must measure 420 m, and each of the long straight sides must be twice as long as the diameter of each of the curved sides. (a) What values of x and y satisfy these requirements? Give your answer correct to two decimal places. (b) The club wants to plant trees in the outer sect ion surro unding the track. They have been advised that they should allow 50 m 2 for each tree. What is the maximum number of trees that they can plant? 6. A truck wheel (including the tyre) has a diameter of 140 cm. If the tyre picks up a stone, approximately how far does the stone travel, if the wheel revolves completely, 500 times? (Use π 22 in this question.) 7
≈
7. Consider a circle with diameter d 1 , and area a1 . Double the diameter. Calculate the relationship between the new area (a2 ) and the srcinal area (a1 ). Repeat the process 4 more times and note your results in a table (like the one shown below). How would you describe the relationship between each bigger area and the srcinal area? Diameter
Area a i (i = 1, 2, 3, 4, 5, 6)
d 2d 4d 8d 16d 32d
a1 = a2 = a3 = a4 = a5 = a6 =
Relationship between a i and a 1 for i = 2, 3, 4, 5, 6
8. Calculate the area of trape zoid ABCD which is sketched below. A
D
3 km 2 km
B
7 km
E
C
9. A circle has an area of 64 cm 2 . Find the length of the diameter (correct to one decimal place).
86 10. Suppose a farmer has 500 m of fencing. Will a square field or a circular field fenced with this length of fencing give him more planting space? 11. Show how the formula for the area of a triangle can be derived, if we know the formula for the area of a rectangle. 12. Find the area of each of the following symmetrical figures. Measurements are given in centimetres. If answers are not exact, give them correct to one decimal place. (a)
(b) 3 1
4
9
2 3 2
5
13. Find the area (corr ect to one decimal plac e) of the shaded region in the sketch below. The shaded portion repre sents what is left when a circle with diameter 5 cm is cut from a square with sides 5 cm.
5
87
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2.2 SURFACE AREA AND VOLUME OF THREE– DIMENSIONAL OBJECTS The branch of geometry that deals with the study of figures in space (i.e. three– dimensional figures) whose faces or cross sections are for example polygons or circles, is called solid geometry. This does not mea n that an object has to be solid. For example in solid geometry we may study an empty rectangular box, or a solid wooden cube.
2.2A SOME THREE–DIMENSIONAL OBJECTS Before we consider formulas for calculating volume and surface area we need to know the names and characteristics of various objects. Table 2.2.1 on the next page summarises some of the well–known three–dimensional objects we encounter. Many three–dimensional objects arise from the two–dimensional polygons we have already considered. When we create a three–dimensional object from two–dimensional polygons, we call the two–dimensional polygons the faces of the three–dimensional object. We do not only need to use polygons to create three–dimensional objects. We also have objects whose faces are circles as well as polygons.
It is useful to create a net from which the three–dimensional object can be made. The net of an object is created by theore tically opening out the three– dimensional object so that it becomes a flat surface .
88
NameandDescription The base of a pyramid may be any polygon, not necessarily a square.
Net
Object
Pyramid Square base: Four triangular sides, where the triangles are congruent
and have a common vertex. Rectangular prism Two congruent and parallel rectangular faces (called bases); the other faces are also rectangles, formed by joining corresponding vertices of the bases.
Base
Base
Cube A rectangular prism in which all faces are congruent squares.
Right circular cylinder Two circular faces (called bases), with a rectangle forming the curved sides in such a way that the sides are perpendicular to the circular base.
Table 2.2.1
We can make models of many three–dimensional shapes by drawing the relevant net onto a piece of cardboard or paper, cutting it out and folding it along its edges. We then need to stick the edges together.
89
MAT0511/004
2.2.1 Copy an enlarged version of the net shown in the figure below, onto a piece of flexible cardboard. Cut it out and fold it into a three–dimensional shape.
Figure 2.2.1
SOLUTION Your shape will look like this.
Figure 2.2.2
2.2B SURFACE AREA Nets are two–dimensional representations of three– dimensional objects.
The value of being able to identify the nets from which three–dimensional objects are created is that they enable us to see, in two dimensions, the shape of each of the object’s faces. This in turn helps us to identify the differen t components we need to consider when we calculate the surface area of the object. Surface area is exactly what the term suggests. It is the combined areas of the individual surfaces that are the faces of an object.
In Table 2.2.2 we give three familiar objects, together with their surface areas. You can refer to the nets given in Table 2.2.1 if you are not sure how these areas are determined.
90 Suppose a , b, c, etc. indicate lengths, in centimetres. Then in each case S represents surface area, in square centimetres. Object
Surfacearea S (square centimeters)
Pyramid S = 4(area of triangular face) + 1(area of square base) 1
×
b 2h + (b = 42bh2 + = b = b (2h + b)
h b
× b)
Rectangular prism c b a
S = 2(area of base) + 2(area of long side) + 2(area of short side) = 2 (ab) + 2 (ac) + 2 (bc) = 2 (ab + ac + bc)
Cube S = 6(area of face) = 6a2
a
Table 2.2.2
Do not try to memorise these formulas. You can always calculate surface areas of different objects by finding the area of each of the faces, and then adding the areas.
2.2.1 By open cylinder, we mean a cylinder with a closed base, and open top.
How much tin sheeting will you need to make an open cylinder with a height of 20 cm and a base that has a diameter of 10 cm? Give your answer in square metres, correct to two decimal places.
91
MAT0511/004
We first sketch the object, and the net on which it is based. A cylinder such as this is called a right circular cylinder, because the base is a circle, and the sides are perpendicular to the base.
20 cm
20 cm
5 cm 5 cm
Surface area = area of circle + area of rectangle whose breadth is the circumference of the base of the cylinder, and whose length is the height of the cylinder. Remember that the circumference of a circle is given by 2 πr .
Thus S=
π
(5)2 + 2π(5)
× 20 2
cm2
Diameter = 10 cm and hence radius = 5 cm.
= (25π + 200π) cm = 225π cm2
≈ 706, 86 cm ≈ 707 cm .
2
Using a calculator.
2
To the nearest square centimetre.
× 10−
Do you remember how to convert cm2 to m2 ? If not, see Book 1, Study Unit 5.
You will thus need approximately 707 cm2 of tin sheeting, i.e. 707 i.e. 0 , 0707 m2 , i.e. approximately 0, 07 m 2 of tin sheeting.
4
m2 ,
Right circular cylinder
From Activity 2.2.1 can you deduce the general formula we use to calculate the surface area of an open right circular cylinder? We know that Surface area = area of circular base + area of “rectangle”. Hence, for an open right circular cylinder where the radius of the base is r units and the height is h units, we have surface area S =
π
r2 + 2πrh square units.
See Figure 2.2.3 on the next page.
92 r
h
The surface area of a closed
r
right circular cylinder is (2πr2 + 2πrh) square units.
Figure 2.2.3 Right circular cone
We now consider another object that has a circular base. This is a right circular cone, which is illustrated in Figure 2.2.4.
We call this a right circular cone because it is
symme-
h
trical about a line through the vertex, perpendicular to, and throu gh the centre of,
r
the circle that forms the base.
Figure 2.2.4
Let us now try to draw a net for this object. We need to do this if we want to use paper to make party hats in this shape. If the radius r of the base is 10 cm and the height h is 12 cm then we can use the theorem of Pythagoras to find the length of the slanting side AB . See Figure 2.2.5. A
h = 12 cm r = 10 cm B
Figure 2.2.5
By Pythagoras we find AB =
= =
≈
2
2
h +r √144 √ + 100 cm
244 cm
15, 62 cm.
93
MAT0511/004
If we cut the hat along AB and flatten out the paper, we obtain the following net.
A
h 2 + r 2 cm
B 2 r cm
The paper is now in the shape of a sector of a circle, with radius
Note: It is not necessary to remember these formulas for exam purposes.
√h
2 + r2
√
In general, when the slanting side has length h2 + r 2 units, then the area of this sector is πr h2 + r 2 square units. At this stage you may not know how this formula has been obtained. However, if you continue with mathematics and study trigonometry, you will be able to work out an area such as this.
√
Hence the lateral surface area (i.e. excluding the base) of a right circular cone with radius r and height h is πr
h2 + r2 square units, whereas the surface area h2 + r2
of the closed right circular cone (i.e. including the base) is ( πr2 + πr
square units. Sphere
cm.
We now consider a sphere. A sphere consists of a set of points in space that are all the same distan ce from a fixed point called the cent re. The length of a line segment from any point on the sphere to the centre is the radius of the sphere. A line segment through the centre joining two points on the sphere is the diameter of the sphere.
r
Figure 2.2.6
We cannot draw a net for a sphere and use it to find the surface area. The derivation of the formula for the surface area of a sphere is also beyond the scope of this module. If a sphere has radius r units, then Surface area = 4πr2 square units.
94
2.2.2 How much leather is required for a soccer ball that has a diameter of 21 cm? Use 22 as an approximation for π, and give your answer to the nearest square 7 centimetre.
SOLUTION We assume that the soccer ball has the shape of a sphere. Since the diameter is 21 cm, the radius is 21 cm. Thus 2
Surface area = 4πr2 22 4 7
≈
21 2
21 2
cm2
= 1 386 cm 2 .
Thus the ball requires approximately 1 386 cm2 of leather.
The answer to Example 2.2.2 is not really as simple as the solution makes it appear. A soccer ball is usually made up of a pattern of leather penta gons and hexagons, as shown in Figure 2.2.7.
Figure 2.2.7
Additional leather is needed for the seams, and some of the srcinal flat piece is wasted when the pentagons and hexagons are cut out.
95
MAT0511/004
2.2.2 Calculate the surface area of the container shown in Figure 2.2.8. All corners are right angles.
5 cm 6 cm
3 cm 4 cm 9 cm
Figure 2.2.8
t1 s2 s1
t2 s3
f1 b
For convenience we call the face looking towards us the front (denoted by f1 ); the corresponding side facing away from us is the back (denoted by f2 ). The other vertical faces ( sides) are denoted by s 1 , s2 and s3 ; the horizontal faces on the top are t 1 and t 2 , and the base of the container is denoted by b . The net of the container is sketched on the next page.
96 6 back ( f2 )
4 7 5
9
3
s1
t1
3
9
4
s2
t2
s3
3 4
front ( f1 )
5
base ( b ) 15
Once we have allocated the different dimensions to the different parts of the net it is straightforward to calculate the separate areas, and then to add them to get the surface area of the container. Area of f1 = ((3
2
× 6) + (4 × 15)) cm
= 78 cm2 = Area of f 2
Hence we have the following. Area of f1 and f 2 = 2 Area of base b = 5 Area of s 1 = Area of t 1 = Area of s 2 = Area of t 2 = Area of s 3 = Hence the total surface area is 376 cm2 .
× 78 cm = 156 cm × 15 cm = 75 cm 5 × 7 cm = 35 cm 5 × 6 cm = 30 cm 3 × 5 cm = 15 cm 5 × 9 cm = 45 cm 4 × 5 cm = 20 cm 2 2
2
2
2
2
2
2
2
2
2
2
2
2
97
MAT0511/004
2.2C VOLUME The volume of a solid is the number of unit cubes it contains. Suppose the block shown below has width, height and length all 10 cm. 10 cm 10 cm
10 cm
10 cm
Figure 2.2.9
There are 1 000 cubes, and 1 000 = 10 10 10.
× ×
There are 30 cubes, all measuring 1 cm by 1 cm by 1 cm, and 30 = 5 3 2.
It should be easy for you to see that if we cut up the block along the lines shown here, we will have 1 000 small blocks, or cubes, all congruent to the shaded block in the figure above. Thus we say that the volume of the block is the number of cubes of unit length (unit length in this case means 1 cm). There are thus 1 000 such cubes, each measuring 1 cm across, 1 cm wide and 1 cm in height. Since there are 1 000 cubes, all measuring 1 cm by 1 cm by 1 cm, we say that the volume of the block is 1 000 cubic centimetres. While it is not practical to try to divide all objects into unit cubes, we give all volumes in terms of cubic units, i.e. cubic centimetres, or cubic metres, etc. Hence the volume of the block shown in Figure 2.2.10 is 30 cubic centimetres.
× ×
5 cm
3 cm
2 cm
Figure 2.2.10
98 In the same way that we denote square centimetres (or square metre s, etc.) by cm2 (or m 2 , etc.), we also denote cubic centimetres (or cubic metres , etc.) by cm3 (or m 3 , etc.). It is beyond the scope of this module to deriv e the formulas for volume, so we now state the volume formulas for the more common solids. Object
Volume V (in cubic units)
Rectangular prism V =l h
×w×h × ×
= (l w) h i.e. V = area of base
w
× height
l
Cube V = a3 = a2 a i.e. V = area of base
×
Note:
× height
These formulas should be
a
understood and remembered for exam purposes
Right circular cylinder r
V = πr 2 h i.e. V = area of base
× height
h
r
Right circular cone V= i.e. V =
h
1 2 πr h 3 1 volume 3
×
r
Sphere V= r
4 3 πr 3
Table 2.2.3
of right circular cylinder
99
MAT0511/004
When solids are irregular we need to identify the separate components before trying to calculate the volume.
2.2.3 Calculate the volume of the container given in Figure 2.2.8.
SOLUTION We can consider the container as a combination of two separate rectangular containers. One of them, with volume V1 , has faces s 1 , t 1 , s 2 and parts of f 1 , f 2 and b; the other, with volume V2 , has faces t 2 , s 3 and parts of f1 , f 2 and b . Hence Volume V
= = = =
V1 + V2
(6 5 7) + (5 9 (210 + 180) cm3
× ×
3
× × 4) cm
390 cm 3 .
In Example 2.2.3 we considered the container as a combination of two rectangular containers. In the next activity the container consists of a combination of rectangular and cylindrical containers.
2.2.3 Calculate the volume of the container shown in Figure 2.2.11. It is a rectangular box, with a curved lid that is a cylinder sliced down the middle. Give your answer to the nearest cubic centimetre.
6 cm
8 cm 10 cm
Figure 2.2.11
100
Volume = Volume of box + Volume of half–cylinder 1 = (l b h) + πr 2 h 2 The diameter of the 1 2 = (10 8 6) + π (4) 10 cm3 cylinder is 8 cm, hence 2 its radius is 4 cm.
× ×
× × π
3
×
= (480 + 80 ) cm
≈
731 cm 3
When we consider how much liquid a container holds, we often refer to the volume in millilitres (ml), or litres ( ) instead of cubic centimetres. We have the relationships 1 = 1 000 ml = 1 000 cm 3 i.e. 1 ml = 1 cm3 and 1 = 1 000 cm3 = 10−3 m3 In the next activity we also need to calculate the volume of an irregular container, in this case a swimming pool. We noted in Table 2.2.3 that we can determine the volume of certain objects by calculating area of base
× perpendicular height.
In the next activity, think carefully about the following.
We use the word solid to describe the object whose volume we want to find.
Separating the solid into two or more appropriate solids whose volumes we can find.
Considering what “base” we should use for each of the solids identified.
2.2.4 Calculate the volume of water needed to fill a swimming pool with the dimensions shown in the figure on the next page.
101
MAT0511/004 10 m
6m 1m
2m 5m
We divide the pool into two separate parts, namely the shallow part (with volume V1 ):
6m 1m 5m
and the deeper part (with volume V2 ):
6m
5m
C
B 1m
2m
A
D
Now, volume of pool = V1 + V2 . We draw the deep end of the pool again so that one of the base.
1m
2m 5m 6m
6m 6m 1m
D
A 2m
B
5m
C
sides becomes the
102 The “base” is a trapezoid with parallel sides 1 m and 2 m, and height 5 m (the side that is 5 m long is perpendicular to the sides that are 1 m and 2 m long). The area of a trapezoid is given by A= Do not confuse the height, or altitude, of the base, with the
1 h (a + b) 2
where a and b are the lengths of the parallel sides and h is the perpendicular height, or altitude. We thus have
height of the solid itself.
A=
1 (5) (1 + 2) 2
m2 = 7, 5 m2 .
Thus V2 = Area of base
= (7, 5 6) m3 = 45 m3 .
×
× perpendicular height
We also have V1 = l
×b×h × 6 × 1) m
= (5
3
3
= 30 m . Hence the volume of the pool is V = V1 + V2 = 75 m3 . Since 1 = 10−3 m3 we have 1 m3 = 103 .
Thus in order to fill the pool to the top we need 75 m litres of water.
3
of water, i.e. 75
3
× 10
103
MAT0511/004
2.2 Where necessary, round answers to one decimal place.
1. A circular pipe has a diameter of 3 m. How many litr es (to the near est litre) of oil can fit into a section of pipe that is 50 m long. 2.
1 cm
15 cm
5 cm
The metal block shown above has a cylindrical hole bored through the centre. The open ends have square faces and the remaining four sides are rectangular. How much liquid (to the nearest cubic centimetre) can the block hold at any given time. 3. Consider two cylinders, cylinder C1 with radius r1 cm and height h1 cm; and cylinder C2 with radius r 2 cm and height h 2 cm. If C2 must have the same height as C1 , but contain twice the volume of C1 , how much bigger must r 2 be than r 1 ? 4. Consider the two cubes shown below.
s1
ks1
The sides of the smaller cube measure s1 units. The sides of the bigger cube measure ks1 units, where k is a constant. Find the value of k so that the bigger cube will have double the volume of the smaller cube.
104 5. An office has a water dispenser with paper cups in the shape of cones, with diameter 6 cm and depth 8 cm. How many times must one pape r cup be used to fill an empty kettle up to the 2 mark? 6. A milk carto n contains 500 ml of milk. The carton is dama ged and the contents must be poured into a can that is 8 cm across and 9 cm high. Will all the milk fit into the can ? If not, how mu ch milk (in ml) will be left? 7. 24 cheese wedges are packed into a box, in three layers. The cheese wedges are 0 , 5 cm thick, and 2 , 5 cm along the strai ght edges. Assume no space is left between the wedges, or between the top and bottom layers, and the box. The cardboard used to make the box is 1 mm thick. What is the surface area of the box?
2,5 cm
8. How much wood is was ted if we carve a ball with diamet er 9 cm out of a block of wood in the shape of a cube, with sides 10 cm?
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• Perimeters of polygons
s, s1 , s2 , s3 and s4 , where the
If the lengths of the sides are denoted by
lengths are measured in units such as centimetres, then we have the following formulas.
Triangle:
– scalene triangle: Perimeter = s 1 + s2 + s3 – isosceles triangles: Perimeter = 2s1 + s2 – equilateral triangles: Perimeter = 3s
Quadrilateral: Perimeter = s1 + s2 + s3 + s4
– rectangle and parallelogram: Perimeter = 2 (s1 + s2 ) – square and rhombus: Perimeter = 4s – kite: Perimeter = 2 (s1 + s2 )
Pentagon (regular): Perimeter = 5s
Hexagon (regular): Perimeter 6 s
Irregular figures: Perimeter = sum of the lengths of all the sides
• Circumference of circles
Circumference = 2πr = π d , where r is the radius, d is the diameter, and is an irrational number, at times approximated by 22 or 3 , 14. 7
π
• Areas of polygons Triangle: Area =
Rectangle: Area = length
Square: Area = (length of side)2
Parallelogram and rhombus: Area = base
Note: The formula for the area of an hexagon need not be memorised.
1 2
Kite: Area =
× base × altitude × breadth
1 2
× altitude
× length of first diagonal × length of second diagonal Trapezoid: Area = × sum of parallel sides × altitude Hexagon (regular): Area = 6 × × base × altitude (where base 1 2
1 2
and altitude refer to the base and altitude of any √ one of the six congruent triangles that make up the hexagon) =
3 3 2
2
× ( length of side) .
Irregularly shaped polygons: We can break them up into triangles and quadrilaterals, whose areas are easy to calculate.
Circles
Area =
r2 =
π
π
d 2 2 ,
where r is the radius and d the diameter.
106
• Three–dimensional objects: some examples
Pyramid (square base): has four sides that are congruent triangles with one common vertex.
Rectangular prism : has two congruent and parallel rectangular faces called bases; the other sides are also rectangles formed by joining corresponding vertices of the bases.
Cube: a rectangular prism in which all faces are congruent squares.
Right circular cylinder : a circular top an d base, with a rectangle forming the curved sides, in such a way that the sides are perpendicular to the base.
Right circular cone: an object with vertex directly above the centre of its circular base, symmetrical about the perpendicular line joining the vertex to the circular base.
Sphere: an object in which each point is equi distant from a fixed point called the centre.
• Surface area of three–dimensional objects Pyramid: Surface area = 4× area of triangle + area of square base Rectangular prism: Surface area = 2 × area of base + 2 × area of long side + 2× area of short side Cube: Surface area = 6× area of one face
Right circular cylinder (closed) : Surface area = 2 area of circle + area of rectangle In an open right circular cylinder: Surface area = 1 area of circle + area of rectangle In both cases note that the breadth of the rectangle is equal to the circumference of the circle.
× ×
Note: It is not necessary to memorise the surface area of a right circular cone.
√
Right circular cone: Lateral surface area = πr h2 + r 2 , where r is the radius of the circle forming the base and h is the perpendicular distance from the vertex to the centre of the base. In a closed right circular cone: Surface area = π r h2 + r2 + πr2
√
Sphere: Surface area = 4πr2 , where r is the radius of the sphere. Irregularly–shaped object : We break it up into several shapes whose surface areas we can calculate, and combine the separate areas.
• Volume of three–dimensional objects
Rectangular prism: Volume = area of rectangular base 3
× height
Cube: Volume = (length of side)
Right circular cylinder: Volume = area of circular base
× height
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MAT0511/004
Right circular cone: V = 13 (volume of cylinder with the same circular base and height) = 13 πr2 h, where r is the radius of the circular base and h is the height.
×
4 3 π , 3 r
Sphere: Volume =
Irregularly–shaped object: We break the object into several objects whose volumes we can calculate, and combine the separate volumes.
where r is the radius.
CHECKLIST Now check that you can do the following. SECTION 2.1
1. Calculate the perimeters of various polygons. Table 2.1.1, Example 2.1.1 2. Calculate the circumference of a circle. Example 2.1.2 3. Calculate the horizontal distance covered by a point on a circular object. Example 2.1.3 4. Calculate the areas of various regular polygons. Table 2.1.2 5. Calculate the area of a circle . Example 2.1.4 6. Calculate areas involving a combination of polygons and circles. Examples 2.1.5, 2.1.6; Activity 2.1.1
SECTION 2.2
1. Draw the net for several familiar three–dimensional objects, i.e. a pyramid, rectangular prism, cube, right circular cylinder. Table 2.2.1, Example 2.2.1 2. Calculate the surface area of a pyramid, rectangular prism, cube. Table 2.2.2
108 3. Calculate the surface area of a right circular cylinder and the surface area of a right circular cone. Activity 2.2.1; Figures 2.2.3, 2.2.4, 2.2.5 and the related discussion 4. Calculate the surface area of a sphere. Figure 2.2.6 and the related discussion; Example 2.2.2 5. Calculate the surface area of irregularly shaped objects. Activity 2.2.2 6. Calculate the volume of familiar three–dimensional objects, i.e. rectangular prism, cube, right circular cylinder, right circular cone, sphere. Table 2.2.3 7. Calculate the volume of irregularly shaped objects, or objects that are combinations of other solids. Example 2.2.3; Activities 2.2.3, 2.2.4
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MAT0511/004
ANSWERS TOPIC 1 Exercise 1.1 1. A
A1 C
B
B1
A B C is the mirror image of ABC in the vertical line through C. 1 1
2.
(a) A
C
B
(b) B
C
3.
A
ˆ is adjacent to C PB ˆ . (a) DPC ˆ is adjacent to B PA ˆ . C PB ˆ = 22◦ 30 (b) C PD ˆ + BPC ˆ + CPD ˆ = 90◦ (c) APB ˆ = 112◦ 30 (Since E PA ˆ = 180◦ (d) E PA
−
ˆ + BPC ˆ APB
= 180◦ (45◦ + 22◦ 30 ) = 112◦ 30 .)
−
110 4.
5.
(a) 28 ◦ 39 0
(b) 90
(c) 30◦ 48 0
(c) 100
(a)
◦ 3 18 ◦ 6 0
(i) 90 ◦ (ii) 45◦ (iii) 45◦
(b) Pairs of co–interior angles: ˆ ˆ AQP and C PQ ˆ and D PR ˆ BRP ˆ and D PQ ˆ RQP ˆ and C PR ˆ ARP Note that it is possible to deno te the angles differently: for example ˆ and A QF ˆ refer to the same angle, similarly A RG ˆ and Q RP ˆ refer AQP to the same angle. ˆ and D PR ˆ (c) ARP ˆ is a straight angle. (d) AQR ˆ (e) (i) GPC ˆ or F PR ˆ (ii) GPQ 6. The flower has rotational symmetry. It can be rotated through 90 ◦ , 180 ◦ and 270◦ about its central point without any change taking place. It also has reflection symmetry. It can be reflected in any one of the four dashed lines shown.
The face has reflection symmetry. It can be reflected in a vertical line through the centre of the face.
However, it does not have rotational symmetry, since rotation through any number of degrees results in a different picture. For example, rotation through 180◦ gives the same face, but upside down.
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MAT0511/004
Exercise 1.2 1.
(a)
(b)
2.
RHS: AB = AC BD = BD ˆ = A DC ˆ = 90◦ ADB
ABD ≡ ACD
Hence BD = CD. 3. AB = BC = AC = s since ABC is equilateral. AD bisects BC . (See question 2 above, or the discussion that follows Activity 1.2.6.) Hence BD = CD = 12 s cm. According to the Theorem of Pythagoras
AD2 = AB2
2
− BD
i.e. we have AD2 = s2
= s2
− ( 12 s) − 14 s
2
2
3
= 4 s2 and hence AD =
=
3 2 s 4 3 s. 2
√
112 4.
The octagon consists of eight congruent triangles. The angle sum of each triangle is 180 ◦ . Thus the angle sum of all eight triangles is 8
× 180◦ = 1 440◦.
But the sum of the central angles of the triangles is 360◦ . Hence the angle sum of the octagon is 1 440◦ 5.
− 360◦ = 1 080◦.
(a) There are four pairs of congruent triangles:
(b)
ST P ≡ QT R ST R ≡ QT P SRQ ≡ QPS PSR ≡ RQP P
Q
S
R
QRS ≡ SPQ
ˆ = S QP ˆ . Thus Q SR Hence PQ
RS.
Alternate angles are equal.
ˆ = R QS ˆ . Similarly P SQ Hence PS RQ. Hence PQRSis a parallelogram.
Alternateanglesequal. Opposite sides are parallel.
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MAT0511/004
6. By Pythagoras FG2 + CG2 = FC 2 .
Hence FG2 = FC 2
= =
2 1 3 CD
x
3
= = =
x2
9
2
x
x2 16
m2
4
m2
m2
7x2 m2 144
FG =
8.
2 1 4 CD 2
16x2 9x2 144
and thus
7.
CG2
− − − −− x
12
√
7 m2 .
ABD ≡ ECD. If ABD ≡ ECD then AB = EC . Since EC is the hypotenuse of EBC it follows that EC > DC. Hence EC > AB since DC = AB. Consequently ABD ≡ ECD. (a)
A x B
D 2x
C
In
ABC and ADC AB = AD BC = DC.
AC is common to both.
Thus
ABC ≡ ADC.
SSS
114 (b) A E
B
D
C
BAE and DAE. ABC ≡ ADC it follows that
Consider Since
ˆ = D AE ˆ . BAE Since AB = AD and AE is common to both triangles we have Hence
BAE ≡ DAE.
SAS
BE = E D
i.e. diagonal AC bisects diagonal BD . 9. In
DCA and DPQ
ˆ = DPQ ˆ = DQP ˆ . ˆ and D AC DCA D ˆ is common to both triangles.
Corresponding angles are equal.
Thus
DCA ||| DPQ. Because the triangles are similar, the corresponding sides are in proportion, and thus DC CA DA = = . DP PQ DQ Let DC = x. We then have
i.e. we have
x CA = , 2x PQ CA PQ
=
1 2
and hence CA = 12 PQ.
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MAT0511/004
10. Do you notice a pattern in
Polygon Number Number of Number of Angle (regular) of sides vertices diagonals sum ◦ Triangle 3 3 0 180 ◦ Quadrilateral 4 4 2 360 ◦ Pentagon 5 5 5 540
the angle sum?
Hexagon Octagon 11.
68
68
920
720 1080
(a) the same (b) 90◦ (c) rhombus
square
(d) less than
Exercise 1.3 1. B
C
A D
(a) B lies on the circle (b) We choose D on the circle and draw AD and DC. ˆ = 90◦ . Then A DC
◦◦
116 2. Q R
P K
N
C
J
O
B
D
M
S
T
Z A Y
L X
U
V W
We label all points where the lines described intersect with the outer circle by means of the letters N , P, Q, R, S, T , U , V , W , X , Y and Z . The additional points of intersection of the lines are denoted by J , K , L and M . We see that the figures bounded by line segments AK , KC , and arc AC line segments BJ , JC, and arc BC line segments MB, MD, and arc BD line segments DL , LA, and arc AD . are all congruent. Similarly the figures bounded by arc N Z and line segments N K , KA and AZ arc Y Z and line segments Y L, LA and AZ arc U T and line segments U M , MB and BT arc ST and line segments SJ , JB and BT arc PQ and line segments PK , KC and CQ arc RQ and line segments RJ , JC and CQ arc XW and line segments X L, LD and DW arc V W and line segments V M , MD and DW are all congruent. We also have the congruent figures bounded by arc Y X and line segments Y L and LX arc UV and line segments U M and MV arc RS and line segments RJ and JS arc PN and line segments PK and N K .
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MAT0511/004
We also have the four congruent circle sectors, AOC, BOC, BOD and AOD.
There are many other congruent figures which are formed by combining certain smaller congruent figures. For example, ZX A, TV B, ZPA, and T RB are all congruent.
TOPIC 2 Exercise 2.1 1. A 1 _ 2
B 1m
m
D
E
C
Area of kite = area of
ABC + area of ADC ABC) . ABC ≡ ADC.
= 2 (area of Thus Area of kite = 2
= 2 = 2 = 2 = i.e. the area of the kite is
1 4 1
1 base) altitude 2( 1 AC EB 2 1 (1) 14 m2 2 1 m2 8 2
× × ×
AC bisects BD .
m
m2 .
4
2. Paved area = total area
=
π
2
(3)
− area of pool − (2, 5) m π
2
= (π (9 6, 25)) m2 = 2, 75π m2
≈
−
8, 6 m 2
2
118 3.
(a) 7
A
F 3
E 2
5 G
D 2
B
C
9
Perimeter = (5 + 9 + 2 + 2 + 3 + 7) cm
= 28 cm Total area = area of rectangle AGEF + area of rectangle GBCD
= (21 + 18) cm2 = 39 cm 2 (b) D 6
4
12
C
6
4 6
A 6
2
+
4
B
2
Perimeter = AB + BC + CD + DA
= =
≈
62 + 42 + 6 + 4 + 12 cm
22 +
√
52
cm
(22 + 7, 21) cm
= 29, 21 cm Total area = area of ABD + area of DBC = 12 (12 4) + 12 (6 4) cm2
×
×
= 36 cm 2 (c)
3 3
A
3
C
D 9
9
B
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MAT0511/004
Perimeter = semi–circle AC + BC + BA
= (π(3) + 9 + 9) cm
≈
27, 42 cm
=
2 1 2 π (3)
Total area = area of semi–circle + area of
By Pythagoras, BD = Hence
2
cm
+ 12 (6) (BD)
ABC
√9 − 3 = √72 ≈ 8, 485 cm. 2
Total area
2
9 π + 3 (8, 485) 2
≈ ≈
≈
39, 59 cm2 .
cm2
(14, 137 + 25, 455) cm2
= 39, 592 cm2
(d) I 4
4
A 2 J
H 2 G D
5
5
B 2 C
E 2 F
Perimeter = 34 cm JH = CE since JCEH is a rectangle. Thus
Hence
JH I ≡ CE D.
SSS
total area = area of rectangle ABFG i.e. total area = AG
× AB.
By Pythagoras JH
= = =
2
2
√16JI ++16IHcmcm √ 32 cm.
120 Hence AE
√ 32 + 2 cm √ 4 + 32 cm. √ 5 4 + 32 cm √
= =
Total area =
=
2+
20 + 5 32
≈
2
cm2
(20 + 28, 284) cm2 48, 28 cm 2
≈ 4.
A
B
D
6m
C
ABCD is a parallelogram. Hence AD = BG.
But BC = AB (given), hence ABCD is a rhombus. From the answer to question 1 of Exercise 1.3, we know that since AC is a ˆ and A DC ˆ are both right angles. diameter, A BC Hence ABCD is a square. By Pythagoras AB2 + BC2 = AC2
i.e. 2 (AB)2 = 36 m 2 . Thus AB2 = 18 m
i.e.
AB =
√18 m.
Shaded area = area of circle
=
π
(3)2
= (9π
≈ ≈
−
√ −√area of square 2
− 18) m
10, 27 m 2 10 m 2
18. 18
m2
121 5.
MAT0511/004
(a) We have x = 2y and the perimeter = 420 m. Hence 2π
y 2
+ (2x) = 420.
But x = 2y, and thus 2π Now
y + (2 2
× 2y) = 420.
y 2π 2 + 4y = 420 πy + 4y = 420
⇔ ⇔ ⇔
y (π + 4) = 420 y=
Thus y
Hence x (b)
420 +4
π
.
≈ 58, 81.
≈ 117, 62 and y ≈ 58, 81.
Total area = 25 000 m 2 Track area = area of circle + area of rectangle y 2 = π + xy m2 2
≈ π
2
58, 81 2
+ (117, 62
≈
(2 716, 39 + 6 917, 23) m2
≈
9 634 m 2
× 58, 81)
= 9 633 , 62 m 2
Area available for trees = total area
≈
(25 000
− track area − 9 634) m
= 15 366 m
2
2
Maximum number of trees that can be planted
=
≈ ≈
tree area area required for one tree 15 366 50 307
m2
122 6. Over one revolution of the wheel the stone cov ers a distance equal to the outer circumference of the wheel (i.e. circumference of wheel together with the tyre). Distance covered over one revolution = 2π (70) cm Distance covered over 500 revolutions
= 500
× 2 × 70 cm 22 ≈ 500 × 2 × 7 × 70 cm π
= 220 000 cm = 2 200 m 7. Diameter
Area a i (i = 1, 2, 3, 4, 5, 6)
d
a1 = π
2d
a2 = π
4d
a3 = π
8d
a4 = π
16d
a5 = π
32d
a6 = π
d 2 2
Relationship between ai and a 1 for i = 2, 3, 4, 5, 6
= 4 d2 π
2d 2 2
= πd 2
a2 = 4a1
4d 2 2
= 4 πd 2
a3 = 16a1
= 16πd 2
a4 = 64a1
8d 2 2
16d 2 2
= 64πd 2
a5 = 256a1
32d 2 2
= 256πd 2
a6 = 1 024a1
From the pattern a2 = 4a1 2
a3 = (4) a1 3
a4 = (4) a1 4
a5 = (4) a1 5
a6 = (4) a1
we see that each time the diameter is doubled, the area increases by a factor of 4. 8. Area =
1 2 (AD + BC)
× DE = 10 km
9. Area of circle = πr2 = π
d 2 2
2
123 Thus, if π
d 2 2
MAT0511/004
= 64 cm2 , we have
d
2
2
i.e.
d
2
=
64 π
cm2
√8
=
π
i.e.
cm
√16 ≈ 9, 0 cm.
d=
π
10. If the perimeter of a square is 500 m, each side has leng th 125 m. Hence area of square = 15 625 m2 . If the circumference of a circle is 500 m, its radius is Area of circle =
=
π
250 π
π
π
m.
2
62 500
250
m2
m2
19 894 , 37 m 2
≈
Thus a circular field provides more planting space than a square field. 11. F
A
E
B
D
C
ABC is a triangle, and BCEF , BDAF and DCEA are rectangles.
Now Area of and
ABC = Area of ABD + Area of ADC
Area of FBCE = Area of FBDA + Area of DCEA . Also Area of
ABD =
1 Area 2
of FBDA
Area of
ADC =
1 Area 2
of DCEA .
and
124
Hence Area of
ABC =
= =
12.
1 Area 2 1 BC 2 1 BC 2
of FBCE
× FB × AD.
(a) 19 cm 2 (b) 39 cm 2
13. Shaded area = area of square
=
≈
− − 25
π
5 2 2
area of circle
cm2
5, 4 cm 2
Exercise 2.2 1. The circular pipe has the shape of a cylin der with diameter 3 m. We are interested in a 50 m length of pipe, i.e. we regard 50 m as the height of the cylinder. 2
Volume of cylinder =
π
r h
3 2 50 m 3 2
=
π
≈
353, 42917 m3
Since 1 m = 100 cm = 102 cm we know that 1 m 3 = 1 000 000 cm 3 = 106 cm3 .
Also 1 = 1 000 ml = 1 000 cm 3 .
Hence 1 m 3 = 1 000 = 103 . Hence 353, 42917 m3 = 353, 42917
≈
353 429 .
3
× 10
Thus the pipe contains approximately 353 429 litres of oil.
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MAT0511/004
2. Volume of cylindrical hole =
π
r2 h
Square faces have sides of 5 cm, but there is a space of 1 cm between the hole and each of the edges of the block. Thus the diameter of the hole is 3 cm. The length of the hole is 15 cm. Hence Volume =
3 2 15 2 3
π
≈
cm3
106 cm
3. Volume ofC1 = V1 = π (r1 )2 h1 Volume ofC2 = V2 = π (r2 )2 h2 We must have V2 = 2V1
i.e. π
Since h 2 = h 1 we have
(r2 )2 h2 = 2π (r1 )2 h1 . (r2 )2 = 2 (r1 )2
i.e.
√
r2 =
Thus r 2
2r1 .
√ must be 2 times bigger than r . 1
3
4. Volume of bigger cube = Vb = ( ks1 ) cubic units = k 3 s31 cubic units Volume of smaller cube= Vs = s 31 cubic units If Vb = 2Vs
then k3 s31 = 2s31
i.e. we have k3 = 2
i.e. we have k=
√ 3
2.
3
√
Thus if the constant k is the number 2 then the bigger cube will have 3 sides 2 times bigger than the sides of the smaller cube, and the volume of the bigger cube will be double the volume of the smaller cube.
√
126 5. Volume of one cup
2 1 π 3 r h 2 1 π (3) 3
= =
= 24π cm = 24π ml
× 8 cm
3
3
Now the kettle requires 2 of water, i.e. 2 000 ml of water. 2 000 24π 26, 53
Number of cups required =
≈
Hence we need to use the paper cup 27 times to fill the kettle, 26 times using a full cup and once using just over half a cup. 6. Volume of can =
π
r2h
=
π
(4)2
≈
× 9 cm
452 cm
3
3
= 452 ml The milk will not fit into the can. There will be approximately (500 452) ml, i.e. approximately 48 ml, of milk left. 7.
− Surface area = area of sides (including thickness of cardboard)
+area of top (including thickness of cardboard) +area of bottom (including thickness of cardboard) = (2πr 1, 7) + πr2 + πr2 cm2 =
≈
(2π
× × 2, 6 × 1, 7) + 2
π
(2, 6)2 cm 2
(27, 77 + 42, 77) cm2
= 70, 54 cm2 8. Volume of cube = (10
3
× 10 × 10) cm
= 1 000 cm 3 Volume of ball =
=
Wasted wood
3 4 3 πr 3 4 π 4 5 3 ( , )
≈
381, 7 cm 3
≈
(1 000
3
− 381, 7) cm
= 618, 3 cm
3
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MAT0511/004
REFERENCES 1. Eves, H.: An Introduction to the History of Mathematics Holt, Rinehart and Winston, 1976.
(4th edition),
2. Freeman, R.: How to Learn Maths , National Extension College, 1994. 3. James G. and James R.C.: Mathematics Dictionary (3rd edition), D. van Nostrand Company, Inc., 1968. 4. Poole, B.: Basic Mathematics, Prentice–Hall, 1994. 5. Sykes, J.B.: The Concise Oxford Dictionary of current English (6th edition), Oxford University Press, 1976.