CUAU031-EVANS September · PDF fileChapter 12 — Geometry 367 12.4 Pythagoras’...

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C H A P T E R

12MODULE 2

Geometry

What are the properties of parallel lines?

What are the basic properties of triangles?

What are the basic properties of regular polygons?

How do we use and apply similarity and Pythagoras’ theorem in two dimensions?

How do we explore ratios of areas of similar figures?

How do we explore ratios of volumes of similar solids?

12.1 Properties of parallel lines – a reviewAngles 4 and 6 are called alternate angles.

Angles 5 and 3 are called alternate angles.

Angles 2 and 6 are called corresponding angles.




Angles 3 and 6 are called cointerior angles.

Angles 4 and 5 are called cointerior angles.

Angles 1 and 3 are called vertically opposite angles and are of equal magnitude.

Other pairs of vertically opposite angles are 2 and 4, 5 and 7, and 6 and 8.

Angles 1 and 2 are supplementary, i.e. their magnitudes add to 180◦.

1 2

34

5 678

l2

l1

l3

Lines l1 and l2 are cut by a transversal l3.

When lines l1 and l2 are parallel, corresponding angles

are of equal magnitude, alternate angles are of equal

magnitude and cointerior angles are supplementary.1 2

34

5 678 l2

l1

l3

Converse results also hold:

If corresponding angles are equal then

l1 is parallel to l2.

If alternate angles are equal then l1 is parallel to l2.

If cointerior angles are supplementary then l1 is parallel to l2.

360

SAMPLE

Cambridge University Press • Uncorrected Sample pages • 978-0-521-61328-6 • 2008 © Jones, Evans, Lipson TI-Nspire & Casio ClassPad material in collaboration with Brown and McMenamin


Chapter 12 — Geometry 361

Example 1 Parallel line properties

Find the values of the pronumerals.

65°e°c°d°

b° a°

Solution

a = 65 (corresponding)

d = 65 (alternate with a)

b = 115 (cointerior with d)

e = 115 (corresponding with b)

c = 115 (vertically opposite e)

There are lots of ways of finding these values. One sequence of reasoning has been used here.



(2x – 50)°(x + 10)°

Solution

2x − 50 = x + 10 (alternate angles)

∴ 2x − x = 50 + 10

∴ x = 60



(x + 100)°

(y + 60)°(2x + 80)°

Solution

x + 100 = 2x + 80 (alternate angles)

∴ 100 − 80 = 2x − x

∴ x = 20

Also x + 100 + y + 60 = 180 (cointerior)

and x = 20

∴ y + 180 = 180 or y = 0

Exercise 12A

Questions 1 to 5 apply to the following diagram

adc b

l1

ehg f

l3

l2

1 Angles d and b are:

A alternate B cointerior C corresponding

D supplementary E vertically opposite

SAMPLE



362 Essential Further Mathematics – Module 2 Geometry and trigonometry

2 Angles d and a are:



3 Angles c and h are:



4 Angles b and f are:



5 Angles c and e are:



6 Find the values of the pronumerals in each of the following:

ax°

z°

y° 70°

bx°

z°y°

40°

c(2z)°

y° 80°

dx°z°

y° 50°

e

x°y°

120°

f

(x + 40)°

(2x – 40)°

12.2 Properties of triangles–a reviewa◦, b◦ and c◦ are the magnitudes of the interior angles of the triangle ABC.

B

A C

a° c° d°

b°

d◦ is the magnitude of an exterior angle at C.

The sum of the magnitudes of the interior angles of

a triangle is equal to 180◦: a◦ + b◦ + c◦ = 180◦.

b◦ + a◦ = d◦. The magnitude of an exterior angle is

equal to the sum of the magnitudes of the two

opposite interior angles.

A triangle is said to be equilateral if all its sides

are of the same length: AB = BC = CA.

The angles of an equilateral triangle are all of

magnitude 60◦.

A C

B

60°

60° 60°

SAMPLE




The bisector of each of the angles of an equilateral

triangle meets the opposite side at right angles and

passes through the midpoint of that side.

A C

B

O

A triangle is said to be isosceles if it has two sides

of equal length. If a triangle is isosceles, the angles

opposite each of the equal sides are equal.

A C

B

The sum of the magnitudes of the exterior angles

of a triangle is equal to 360◦: e◦ + d◦ + f ◦ = 360◦

A triangle is said to be a right-angled triangle if it

has one angle of magnitude 90◦. A

f °

B e°

d°C

Example 4 Angle sum of a triangle


A

B

Cx°20°

22°

y°

Solution

20◦ + 22◦ + x◦ = 180◦ (sum angles � = 180◦)

∴ 42◦ + x◦ = 180◦ or x = 138

138◦ + y ◦ = 180◦ (sum angles = 180◦)

∴ y = 42

Or, to find x:

Two of the angles sum to 42◦ and therefore the third angle is 138◦. To find y ◦. The two angles

sum to 180◦. Therefore the second is 42◦

Example 5 Angle sum of an isosceles triangle


A C

B

x° x°

100°

Solution

100◦ + 2x ◦ = 80◦ (sum angles � = 180◦)

∴ 2x ◦ = 80◦ or x = 40

Or, observe the two unknown numbers are the same and

must sum to 80◦, therefore each of them has size 40◦.SAMPLE




Exercise 12B

1 Find the values of the pronumerals in each of the following:

a

50°80°

x° y°A C

B b

P R

Q

x°

30°

x°

c

A

B

Cy°

x°

70° 30°

d

X

Y

Zx°

50°

y°

e

A C

Bz°

y° w° x°40°

f

a°

b° c°

40°75°

g

A

Ey°

y° x°B C

D

12.3 Properties of regular polygons—a review

Equilateral triangle Square Regular pentagon Regular hexagon Regular octagon

A regular polygon has all sides of equal length and all angles of equal magnitude.

A polygon with n sides can be divided into n triangles. The first three polygons below are

regular polygons.

O O O

The angle sum of the interior angles of an n-sided convex polygon is given by the

formula:

S = [180(n − 2)]◦ = (180n − 360)◦

The result holds for any convex polygon. Convex means that a line you draw from any

vertex to another vertex lies inside the polygon.

The magnitude of each of the interior angles of an n-sided polygon is given by:

x = (180n − 360)◦

n

SAMPLE




The angle bisectors of a regular polygon meet at a point O.

For a regular polygon, a circle can be drawn with centre O on which all the vertices lie.

O O O

The sum of the angles at O of a regular polygon is 360◦.

The sum of the exterior angles of a regular polygon is 360◦.

Example 6 Angle properties of an octagon

The diagram opposite shows a regular octagon.

Ox°

y°

A

B

C

DE

F

G

H

a Show that x = 45.

b Find the size of angle y.

Solutiona 8x ◦ = 360◦(sum angles at O = 360◦)

∴ x ◦ = 360◦

8= 45◦

∴ x = 45

b y ◦ + y ◦ + 45◦ = 180◦(�OBC isosceles)

∴ 2y ◦ = 135 or y = 67.5

Example 7 Angle sum of an octagon

Find the sum of the interior angles of an 8-sided convex polygon (octagon).

Solution

1 Use the rule for the sum of the interior

angles of an n-sided polygon:

S = (180n − 360)◦

2 In this example, n = 8. Substitute and

evaluate.

3 Write down your answer.

S = (180n − 360)◦

n = 8

∴ S = (180 × 8 − 360)◦ = 1080◦

The sum of the interior angles is 1080◦.SAM

PLE




Exercise 12C

1 Name each of the following regular polygons.

a b c d e

2 ABCD is a square. BD and AC are diagonals which meet at O.

a Find the size of each of the angles at O.

b What type of triangle is triangle:

i BAD ii AOB

O

A

B C

D

3 ABCDE is a regular pentagon.

a Find the value of:

i x ii y A

B

C

DE

Oy°x°

b Find the sum of the interior angles of the regular

pentagon ABCDE.

4 ABCDEF is a regular hexagon.

Find the value of:

a x b y O

C

B

AF

E

D

x°y°

5 State the sum of the interior angles of:

a a 7-sided regular polygon b a hexagon c an octagon

6 The angle sum of a regular polygon is 1260◦. How many sides

does the polygon have?

7 A circle is circumscribed about a hexagon ABCDEF.

a Find the area of the circle if OA = 2 cm.

b Find the area of the shaded region.

O

E

F

A

B

C

D8 The diagram shows a tessellation of regular hexagons

and equilateral triangles.

State the values of a and b and use these to explain

the existence of the tessellation.

a°b°

9 If the magnitude of each angle of a regular polygon is 135◦, how many sides does the

polygon have?

SAMPLE




12.4 Pythagoras’ theoremPythagoras’ theoremPythagoras’ theorem states that for a right-angled

triangle ABC with side lengths a, b and c, as shown

in the diagram, a2 + b2 = c2.A

c

bC

a

B

Pythagoras’ theorem can be illustrated by the diagram shown

here. The sum of the areas of the two smaller squares is

equal to the area of the square on the longest side

(hypotenuse).area = c2 cm2

area = a2 cm2

a cm

b cmc cm area =

b2 cm2

There are many different proofs of Pythagoras’

theorem. A proof due to the 20th President of the

United States, James A. Garfield, is produced using

the following diagram.

aY Z

c

c

b

E

a

X b W

Area of trapezium WXYZ = 12 (a + b)(a + b)

Area of �EYZ + area of �EWX + area of �EWZ

= 12 ab + 1

2 c2 + 12 ab

= ab + 12 c2

∴ a2

2+ ab + b2

2= ab + 1

2c2

∴ a2 + b2 = c2

Pythagorean triadsA triple of natural numbers (a, b, c) is called a

Pythagorean triad if c2 = a2 + b2.

The table presents the first six such ‘primitive’

triples. The adjective ‘primitive’ indicates that the

highest common factor of the three numbers is 1.

a 3 5 7 8 9 11

b 4 12 24 15 40 60

c 5 13 25 17 41 61

Example 8 Pythagoras’ theorem

Find the value, correct to two decimal places, of

the unknown length for the triangle opposite.x cm

5.3 cm

6.1 cm

SAMPLE




Solution

1 Using Pythagoras’ theorem, write down an

expression for x in terms of the two other

sides of the right-angled triangle. Solve for x.

x2 = 5.32 + 6.12 (Pythagoras)

∴ x =√

5.32 + 6.12 = 8.080 . . .

The length is 8.08 cm correct to two

decimal places.

2 Write down your answer for the length correct

to two decimal places.


Find the value, correct to two decimal places, of

the unknown length for the triangle opposite.8.6 cm

5.6 cm

y cmSolution


expression for y in terms of the two other

sides of the right-angled triangle. Solve for y.

5.62 + y 2 = 8.62 (Pythagoras)

∴ y 2 = 8.62 − 5.62

∴ y =√

8.62 − 5.62 = 6.526 . . .

2 Write down your answer for the length correct

to two decimal places.

The length is 6.53 cm correct to two

decimal places.


The diagonal of a soccer ground is 130 m and the long side of the ground measures 100 m.

Find the length of the short side correct to the nearest cm.

Solution

1 Draw a diagram. Let x be the length of the

short side.130 m

100 m

x m

Let x m be the length of the shorter

side.


expression for x in terms of the two other sides

of the right-angled triangle. Solve for x.

x 2 + 1002 = 1302 (Pythagoras)

∴ x 2 = 1302 − 1002 = 6900

∴ x =√

1302 − 1002 = 83.066 . . .

4 Write down your answer correct to the

nearest cm.

Correct to the nearest centimetre, the

length of the short side is 83.07 m.

SAMPLE




Exercise 12D

1 Find the length of the ‘unknown’ side for each of the following:

a

10 cm

6 cmb

11 cm

5 cm

c10 cm 3 cm

d9 cm

7 cm

e 33 cm

44 cm

A

C Bf

15 cm12 cm

2 In each of the following find the value of x correct to two decimal places.

a

x cm

3.2 cm

4.8 cm

b

x cm6.2 cm

2.8 cm cx cm

9.8 cm

5.2 cm

d

4 cm

3 cm3.5 cm

x cm

3 Find the value of x for each of the following (x > 0). Give your answers correct

to 2 decimal places.

a x2 = 62 + 42 b 52 + x2 = 92 c 4.62 + 6.12 = x2

4 In triangle VWX, there is a right angle at X.VX = 2.4 cm and XW = 4.6 cm. Find VW.

5 Find AD, the height of the triangle.

20 cm

32 cm 32 cm

DC B

A

6 An 18 m ladder is 7 m away from the bottom of a vertical wall. How far up the wall does it

reach?

7 Find the length of the diagonal of a rectangle with dimensions 40 m × 9 m.

8 Triangle ABC is isosceles. Find the length of CB. A

814

BC

SAMPLE




9 In a circle of centre O, a chord AB is of length 4 cm.

The radius of the circle is 14 cm. Find the distance

of the chord from O.

x cm

A B

O

10 Find the value of x.

18 cm

x cm

x cm

11 How high is the kite above the ground?

90 m

170 m

X

Y

12 A square has an area of 169 cm2. What is the length of the diagonal?

13 Find the area of a square with a diagonal of length:

a 8√

2 cm b 8 cm

14 Find the length of AB.

A

D

B

C20 cm

8 cm12 cm

15 ABCD is a square of side length 2 cm.

If CA = CE, find the length of DE.

AB

C D E

16 The midpoints of a square of side length 2 cm

are joined to form a new square. Find the area

of the new square.SAMPLE




12.5 Similar figuresSimilarityIn this section we informally define two objects to be similar if they have the same shape

but not the same size.

ExamplesAny two circles are similar to each other. Any two squares are similar to each other.

C1 C2

3 cm 4 cm S2S13 cm 4 cm

It is not true that any rectangle is similar to any other rectangle. For example, rectangle 1 is not

similar to rectangle 2.

4 cm

1 cmR1 1 cm8 cmR2

A rectangle similar to R1 is:8 cm

2 cmR3

So, for two rectangles to be similar, their corresponding sides must be in the same

ratio

(8

2= 4

1

).

Similar trianglesTwo triangles are similar if one of the following conditions holds:

corresponding angles in the triangles are equal

A

100°

45° 35°

B

C

100°

45° 35°A'

B'

C'

corresponding sides are in the same ratio

A′B′

AB= B′C′

BC= A′C′

AC= k

k is the scale factor.

SAMPLE




two pairs of corresponding sides have the same ratio and the included angles are equal

45° 45°A

B

C A'

B'

C'

AB

A′B′ = AC

A′C′

If triangle ABC is similar to triangle XYZ this can be written symbolically as �ABC ∼ �XYZ.

The triangles are named so that angles of equal magnitude hold the same position, i.e. A

corresponds to X, B corresponds to Y, C corresponds to Z.

Example 11 Similar triangles

Find the value of length of side AC in �ABC

correct to two decimal places.

A Cx cm

A' C'3.013 cm

3 cm5 cm 20°

BB'

20°6.25 cm3.75 cm

Solution

1 Triangle ABC is similar to triangle A′B ′C ′: two

pairs of corresponding sides have the same

ratio

(5

3= 6.25

3.75

)and included angles (20◦).

Triangles similar

2 For similar triangles, the ratios of corresponding

sides are equal, for example,AC

A′C ′ = AB

A′B ′ .

Use this fact to write down an expression

involving x. Solve for x.

∴ x

3.013= 5

6.25

∴ x = 5

6.25× 3.013

= 2.4104

3 Write down your answer correct to two decimal

places.

The length of side AC is 2.41 cm,


Example 12 Similar triangles

Find the value of length of side AB in �ABC.

A3 cm C

2.5 cm

Y

X

x cm

B6 cm

Solution

1 Triangle ABC is similar to triangle AXY

(corresponding angles are equal).

Triangles similar

SAMPLE




2 For similar triangles, the ratios of corresponding

sides are equal (for example,AB

AX= AC

AY).

Use this fact to write down an expression

involving x. Solve for x. Note that,

if AB = x then AX = x + 6.

∴ x

x + 6= 3

5.5

∴ 5.5x = 3(x + 6)

= 3x + 18


∴ 2.5x = 18 or x = 18

2.5= 7.2

The length of side AB is 7.2 cm.

Exercise 12E

1 Find the value of x for each of the following pairs of similar triangles.

a A

B C

4 cm 5 cm82°56°

A'

B' C'

9 cm x cm

56°

82°

b

B A

C

18°

135°

10 cm6 cm

X Y

Z18°

135°

5 cmx cm

c

B

C

A8 cmX

12 cm

13 cmx cm

Y

d

A

B

C

D

E

°

° x cm12 cm

13 cm

14 cm 10 cm×

×

e

A

C

B10 cm

x cm

°R

Q P

6 cm

8 cm°

f

° °A

C

B D

E

6 cm

x cm2 cm

4 cm

g A

B C

12 cm 16 cm

8 cmx cmP Q

h B

AC

D

x cm

2 cm

3 cm 2 cmE

i A

B C

QP

2 cm

6 cm 8 cm

x cm

j

10 cm

1.5 cm

2 cmQ C

B

P

A

x cmSAM

PLE




2 Given that AD = 14, ED = 12, BC = 15 and EB = 4,

find AC, AE and AB.14

12

15

4A

DC

E B

3 A tree casts a shadow of 33 m and at the same time a stick 30 cm long casts a shadow

24 cm long. How high is the tree?

33 m°

0.24 m

0.3 m°

4 A 20 metre high neon sign is supported by a 40 m steel

cable as shown. An ant crawls along the cable starting at A.

How high is the ant when it is 15 m from A?

40 m20 m

A

5 A hill has a gradient of 1 in 20, i.e. for every 20 m horizontally there is a 1 m increase in

height. If you go 300 m horizontally, how high up will you be?

6 A man stands at A and looks at point Y across the river.

He gets a friend to place a stone at X so that A, X and Y

are collinear. He then measures AB, BX and XC to be

15 m, 30 m and 45 m respectively. Find CY, the

distance across the river.

river

30 m

15 m45 m

Y

C X

A

B

7 Find the height, h m, of a tree that casts a shadow 32 m long at the same time that a vertical

straight stick 2 m long casts a shadow 6.2 m long.

8 A plank is placed straight up stairs that are 20 cm wide

and 12 cm deep. Find x, where x cm is the width of the

widest rectangular box of height 8 cm that can be placed

on a stair under the plank. 20 cm

x cm 12 cmplank

8 cm

9 The sloping edge of a technical drawing table is 1 m

from front to back. Calculate the height above the ground

of a point A, which is 30 cm from the front edge.

80 cm

1 m

30 cm

A

92 cmSAMPLE




10 Two similar rods 1.3 m long have to be hinged together

to support a table 1.5 m wide. The rods have been fixed

to the floor 0.8 m apart. Find the position of the hinge

by finding the value of x.x m

0.8 m

(1.3 – x) m

1.5 m

11 A man whose eye is 1.7 m from the ground when standing 3.5 m in front of a wall 3 m high

can just see the top of a tower that is 100 m away from the wall. Find the height of the tower.

12 A man is 8 m up a 10 m ladder, the top of which leans against a vertical wall and touches it

at a height of 9 m above the ground. Find the height of the man above the ground.

13 A spotlight is at a height of 0.6 m above ground level.

A vertical post 1.1 m high stands 3 m away, and 5 m

further away there is a vertical wall. How high up the

wall does the shadow reach?vertical post

1.1 mspotlight

0.6 m

wall

12.6 Volumes and surface areasVolume of a prismA prism is a solid which has a constant cross-section. Examples are cubes, cylinders,

rectangular prisms and triangular prisms

The volume of a prism can be found by using its cross-sectional area.

volume = area of cross-section × height (or length)

V = A × h

Answers will be in cubic units, i.e. mm3, cm3, m3 etc.

Example 13 Volume of a cylinder

Find the volume of this cylinder which has radius 3 cm

and height 4 cm correct to two decimal places.3 cm

4 cmSolution

1 Find the cross-sectional area of the prism. Area of cross-section = πr 2 = π × 32

= 28.27 cm 2

SAMPLE




2 Multiply by the height Volume = 28.27 × 4

3 Make sure that accuracy is given to the

correct number of decimal places

= 113.10 cm 3 (correct to two

decimal places)

The formulas for determining the volumes of some ‘standard’ prisms are given here.

Solid Formula

Cylinder (radius r cm, height h cm)

V = �r2h

h cm

r cm

Cube (all edges x cm)

V = x3

Rectangular prism (length l cm,

V = lwhwidth w cm, height h cm)

l cm

h cm

w cm

Triangular prism The triangular prism shown has a right-angled

triangle base but the following formula holds

for all triangular prismsh cm

b cm l cm

V = 12 bhl

Volume of a pyramid

x

h

The formula for finding the volume of a right pyramid

can be stated as:

Volume of pyramid = 13 × base area

× perpendicular height

For the square pyramid shown:

V = 13 x2 h

The term right in this context means that the apex of the pyramid is directly over the centre of

the base.

Example 14 Volume of a pyramid

Find the volume of this hexagonal pyramid with a base area of 40 cm2

and a height of 20 cm. Give the answer correct to one decimal place.

Solution

V = 1

3× A × h

= 1

3× 40 × 20

= 266.7 cm3 (correct to one decimal place)

20 cm

SAMPLE




Example 15 Volume of a pyramid

Find the volume of this square pyramid with a square base

with each edge 10 cm and a height of 27 cm.

Solution

V = 1

3x 2 h

= 1

3× 10 × 10 × 27

= 900 cm3

27 cm

10 cm10 cm

Volume of a coneThe formula for finding the volume of a cone can be stated as:

Volume of cone = 13 × base area × height

V = 13 �r2h

Volume of a sphereThe formula for the volume of a sphere is:

V = 43 �r3

where r is the radius of the sphere.

r

Example 16 Volume of a sphere

Find the volume of this sphere.

Solution

Volume of sphere = 4

3πr 3

= 4

3× π × 43

= 268.08 cm3 (2 decimal places)

4 cm

Composite shapesUsing the shapes above, new shapes can be made. The volumes of these can be found by

summing the volumes of the component solids.SAMPLE




Example 17 Volume of a composite shape

A hemisphere is placed on top of a cylinder to form a capsule.

The radius of both the hemisphere and the cylinder is 5 mm.

The height of the cylinder is also 5 mm. What is the volume of the

composite solid in cubic millimeters, correct to two decimal places?

Solution

1 Use the formula V = �r2h to find the

volume of the cylinder

The volume of the cylinder

= π × 52 × 5 = 125π mm3

2 Find the volume of hemisphere using

the formula that the volume of a

hemisphere = 12

(43 �r3

) = 23 �r3.

The volume of the hemisphere

= 2

3π × 53 = 250

3π mm3

3 Add the two together. Therefore the volume of the composite =125π + 250

3π = 625 π

3= 654.498 . . .

The volume of the composite = 654.50 mm3

(correct to two decimal places)4 Write down your answer.

Surface area of three-dimensional shapesThe surface area of a solid can be found by calculating and totalling the area of each of its

surfaces. The net of the cylinder in the diagram demonstrates how this can be done.

l

r

Total area = 2�r2 + 2�rl

= 2�r (r + l)

2πr

A = 2πrlA = πr2A = πr2

l

rr

Here are some more formulas for the surface areas of some solids. The derivation of these is

left to the reader.SAMPLE




Solid Formula

Cylinder (radius r cm, height h cm)

S = 2�r2 + 2�rhh cm

r cm

Cube (all edges x cm)

S = 6x2

Rectangular prism (length l cm,

S = 2(lw + lh + wh)width w cm, height h cm)

l cm

h cm

w cm

Triangular prism

h cm

b cm l cm

S = bh + bl + hl + l√

b2 + h2

Example 18 Surface area of a right square pyramid

Find the surface of the right square pyramid shown if the

square base has each edge 10 cm in length and the isosceles

triangles each have height 15 cm.

Solution

1 Draw the net of the pyramid.

2 First determine the area of the square. Area of the square = 102 = 100 cm2

3 Determine the area of one of the

isosceles triangles.

The area of one triangle

= 1

2× 10 × 15 = 75 cm2

4 Find the sum of the areas of the four

triangles and add to the area of the

square.

The surface area of the solid

= 100 + 4 × 75 = 100 + 300

= 400 cm2

SAMPLE




Exercise 12F

1 Find the volume in cm3 of each of the following shapes, correct to two decimal places.

a

radius 6.3 cm and height 2.1 cm

b

dimensions 2.1 cm, 8.3 cm and 12.2 cmc

area of cross section = 2.8 cm2

height = 6.2 cm

d

radius 2.3 cm and length 4.8 cm

2 Find, correct to two decimal places, the surface area and

volume of the solid shown given that the cross section is

a right angled isosceles triangle. 12 cm

4 cm

4 cm

3 The box shown has dimensions length: 13 cm,

width: 4 cm and height 3 cm.

a Find the surface area of the box in square centimetres (cm2).

b Find the volume of the box in cubic centimetres (cm3).

4 Find the volumes, to two decimal places, of spheres with:

a radius = 4 mm b diameter = 23 cm c radius = 3.8 m d diameter = 15 cm

5 Find the volume, to two decimal places, of hemispheres with:

a radius = 12 cm b diameter = 32 mm c radius = 16 mm d radius = 15 cm

6 Calculate the volume of a right pyramid with a rectangular base 18 m by 15 m. The vertex

of the pyramid is 20 m perpendicularly above the centre of the base.

7 Each side of the square base of one of the great Egyptian pyramids is 275 m long. Calculate

the volume of the pyramid, to the nearest cubic metre, if it has a perpendicular height of

175 m.

8 Find the surface area and volume of the right square pyramid shown.

12 cm

10 cmA

D

V

B

XOC

The length of each edge of the square base is 10 cm and the

height of the pyramid is 12 cm.

SAMPLE




9 The diagram shows a capsule, which consists of two hemispheres,

each of radius 2 cm, and a cylinder length 5 cm and radius 2 cm.

The surface area of a sphere is given by the formula S = 4�r2 and

the surface area of the curved section of a cylinder is given by the

formula S = 2�rh.

Find the surface area and volume of the capsule. Give your answers


5 cm

2 cm

10 Find:

a the surface area

b the volume

of the object shown. 10 m5 m

3 m4 m

2 m

11 The diagram opposite shows a right pyramid on a cube.

Each edge of the cube is 14 cm.

The height of the pyramid is 24 cm.

Find:

a the volume of the solid

b the surface area of the solid

14 cm

24 cm

12 Find:

a the surface

b the volume

of the solid shown opposite.

10 cm

7 cm

4 cm

4 cm

13 The solid opposite consists of a half cylinder on

a rectangular prism. Find, correct to two decimal

places:

a the surface area

b the volume

5 cm

10 cm

20 cm

SAMPLE




12.7 Areas, volumes and similarityAreasSome examples of similar shapes and the ratio of their areas are considered in the following.

Similar circles

3 cm

Area = � × 32

Scale factor = k = radius circle 2

radius circle 1= 4

34 cm

Ratio of areas = � × 42

� × 32= 42

32=

(4

3

)2

= k2

Area = � × 42

Similar rectangles

3 cm

2 cm6 cm

4 cm

Area = 3 × 2

= 6 cm2

Scale factor = k = length rectangle 2

length rectangle 1= 6

3= 2

Ratio of areas = 24

6= 4 = (2)2 = k2 Area = 6 × 4

= 24 cm2

Similar triangles

5 cm

4 cm

3 cm 15 cm

12 cm

9 cm

Area = 12 × 4 × 3

= 6 cm2

Scale factor = k = height triangle 2

height triangle 1= 9

3= 3

Ratio of areas = 54

6= 9 = (3)2 = k2

Area = 12 × 12 × 9

= 54 cm2

A similar pattern emerges for other shapes. Scaling the linear dimension of a shape by a factor

of k scales the area by a factor of k2.

Scaling areasIf two shapes are similar and the scale factor is k, then the area of the similar

shape = k2× area of the original shape.

Example 19 Using area scale factors with similarity

10 cm 25 cm

40 cm2

The two triangles shown are similar.

The base of the smaller triangle has a length of 10 cm.

Its area is 40 cm2.

The base of the larger triangle has a length of 25 cm.

Determine its area.

SAMPLE




Solution

1 Determine the scale factor k. k = 25

10= 2.5

2 Write down the area of the small triangle. Area of small triangle = 40 cm2

3 Area of larger triangle = k2× area of

smaller triangle.

∴ Area of larger triangle = 2.52 × 40

= 250

Substitute the appropriate values and

evaluate.

The area of the larger triangle is 250 cm2.4 Write down your answer.

Example 20 Scale factors and area

12 cm 60 cm

Area = 100 cm2

The two hearts shown are similar shapes.

The width of the larger heart is 60 cm.

Its area is 100 cm2.

The width of the smaller heart is 12 cm.

Determine its area.

Solution

1 Determine the scale factor k. Note we are

scaling down.

k = 12

60= 0.2

2 Write down the area of the larger heart. Area of larger heart = 100 cm2

3 Area of smaller heart = k2× area of

larger heart.

∴ Area of smaller heart = 0.22 × 100

= 4

Substitute the appropriate values and

evaluate.

4 Write down your answer. The area of the smaller heart is 4 cm2.

VolumesTwo solids are considered to be similar if they have the same shape and the ratio of their

corresponding linear dimensions is equal.

Some examples of similar volume and the ratio of their areas are considered in the

following.

Similar spheres

3 cm

4 cm

Volume = 43 � × 33

= 36� cm3

Scale factor = k = radius sphere 2

radius sphere 1= 4

3

Ratio of volumes =256

3�

36�= 256

108

= 64

27=

(4

3

)3

= k3

Volume = 43 � × 43

= 256

3� cm3

SAMPLE




Similar cubes

2 cm

2 cm

2 cm

4 cm

4 cm

4 cmVolume = 2 × 2 × 2

= 8 cm3

Scale factor = k = side length 2

side length 1= 4

2= 2

Ratio of volumes = 64

8= 8

= (2)3 = k3Volume = 4 × 4 × 4

= 64 cm3

Similar cylinders

1 cm

2 cm

3 cm

6 cm

Volume = � × 12 × 2

= 2� cm3

Scale factor = k = radius 2

radius 1= 3

1= 3

Ratio of volumes = 54�

2�= 27 = (3)3 = k3

Volume = � × 32 × 6

= 54� cm3

A similar pattern emerges for other solids. Scaling the linear dimension of a solid by a factor

of k scales the volume by a factor of k3.

Scaling volumesIf two solids are similar and the scale factor is k, then the volume of the similar

solid = k3× volume of the original solid.

Example 21 Similar solids

1.5 cm 6 cmvolume = 120 cm3

The two cuboids shown are similar solids.

The height of the larger cuboid is 6 cm.

Its volume is 120 cm3.

The height of the smaller cuboid is 1.5 cm.

Determine its volume.

Solution

1 Determine the scale factor k. Note we are

scaling down.k = 1.5

6= 0.25

2 Write down the volume of the larger cuboid. Volume larger cuboid = 120 cm3

3 Volume smaller cuboid = k3× volume

larger cuboid.

Volume smaller cuboid = 0.253 × 120

= 1.875

Substitute the appropriate values and evaluate.

4 Write down your answer. The volume of the smaller cuboid is

1.875 cm3.

SAMPLE




Example 22 Similar solids

The two square pyramids shown are similar

with a base dimensions 4 and 5 cm respectively.

The height of the first pyramid is 9 cm and

its volume is 48 cm3. Find the height and

volume of the second pyramid.

B'

O'A' D'

C'

V'

5 cmB

A DO

C

V

9 cm

4 cm

Pyramid 1 Pyramid 2

Solution

1 Determine the scale factor k. Use the base

measurements.

k = 5

4= 1.25

Height

2 Write down the height of Pyramid 1. Height 1 = 9 cm

3 Height Pyramid 2 = k× height Pyramid 1. ∴ Height 2 = 1.25 × 9 = 11.25



Volume The height of Pyramid 2 is 11.25 cm.

5 Volume Pyramid 2 = k3× volume Pyramid 1


Volume 1 = 48 cm3

∴ Volume 2 = 1.253 × 48 = 93.75

6 Write down your answer. The volume of Pyramid 2 is 93.75 cm3.

Exercise 12G

1 Triangle ABC is similar to triangle XYZ.

The length scale factor k = 1.2. The area

of triangle ABC is 6 cm2. Find the area

of triangle XYZ.A C

B

X Z

Y×1.2

2 The two rectangles are similar. The area

of rectangle ABCD is 20 cm2. Find the

area of rectangle A′B ′C ′D′.3 cm 5 cm

A

B C

D

B'

A' D'

C'

3 The two shapes shown are similar. The length

scale factor is 32 . The area of the shape to the

right is 30 cm2. What is the area of the shape

to the left?

4 Triangle ABC is similar to triangle XYZ.

XY

AB= YZ

BC= ZX

CA= 2.1.

The area of triangle XYZ is 20 cm2. Find the area of triangle ABC.

SAMPLE




5 Triangles ABC and A′B ′C ′ are equilateral triangles.

2 cma cm

2 cma cma cm

2 cm 2 cm

B

AF

C A'

B'

C'F'

a Find the length of BF.

b Find a.

c Find the ratioarea of triangle A′B ′C ′

area of triangle ABC.

6 The areas of two similar triangles are 16 and 25. What is the ratio of a pair of

corresponding sides?

7 The areas of two similar triangles are 144 and 81. If the base of the larger triangle is 30,

what is the corresponding base of the smaller triangle?

8 These two rectangular prisms are similar. The length scale

factor is 1.8. The volume of the first solid is 20 cm3.

What is the volume of the second solid?

9 Two cones are similar. The ratio of volumes is 8 : 125. Find the ratio of the:

i heights ii lengths of sloping edges iii areas of bases.

10 A cone has water poured into it as shown. Find the ratio

of the volume of empty space in the cone to volume

of water.

5 cm

10 cm

11 Consider two similar cuboids that have edges where lengths are in the ratio 1 : 4.

a Find the ratio of the surface area of the two cuboids.

b Find the ratio of the volumes.

12 An inverted right circular cone of capacity 100 m3 is filled with

water to half its depth. Find the volume of water.

13 The ratio of the radii of two spheres is 2 : 5. Find the ratio of:

i the surface areas ii the volumes

14 Two right circular cones are as shown. Find:

10 cm

30 cm

15 cm45 cm

a the ratio of the heights of the cones

b the ratio of the surface areas

c the ratio of the volumes

15 The ratio of the volumes of two cubes is 1 : 27. Find:

a the ratio of the surface areas of the cubes

b the ratio of the edges of the cubes

SAMPLE



Review


Key ideas and chapter summary

Alternate, corresponding, Angles 4 and 6 are examples of alternate angles.

Angles 2 and 6 are examples of corresponding angles.

Angles 3 and 6 are examples of cointerior angles.

Angles 1 and 3 are examples

of vertically opposite angles

and are of equal magnitude.

1 234

5 678

l1

l2

l3

cointerior and verticallyopposite angles

Angles associated with parallel When lines l1 and l2 are parallel corresponding angles

are of equal magnitude, alternate angles are of equal

magnitude and cointerior angles are supplementary.

AlternateAlternate

Corresponding

Corresponding

Converse results also hold:

If corresponding angles are equal then l1 is parallel to l2.

If alternate angles are equal then l1 is parallel to l2.

If cointerior angles are supplementary then l1 is parallel to l2.

lines crossed by a transversalline

Angle sum of triangle The sum of the magnitudes of the interior angles of a

triangle is equal to 180◦: a◦ + b◦ + c◦ = 180◦.

Equilateral triangle A triangle is said to be equilateral if

all of its sides are of the same length.

The angles of an equilateral triangle

are all of magnitude 60◦.

10 cm 10 cm

10 cm

Isosceles triangle A triangle is said to be isosceles if

it has two sides of equal length.

If a triangle is isosceles the angles

opposite each of the equal sides are equal.

5 cm 5 cm

Polygon A polygon is a closed geometric shape with sides which are

segments of straight lines. Examples are:3 sides: Triangle

4 sides: Quadrilateral

5 sides: Pentagon

6 sides: Hexagon

SAMPLE



Rev

iew


Convex polygon A polygon is said to be convex if any diagonal lies

inside the polygon.

Regular polygon A regular polygon has all sides of equal length and

all angles of equal magnitude.

Sum of the interior angles The angle sum of the interior angles of an n-sided

polygon is given by the formula: S = (180n − 360)◦.

Pythagoras’ theorem Pythagoras’ theorem states that for a right-angled

triangle ABC with side lengths a, b and c,

a2 + b2 = c2, where c is the longest side.

Similar figures We informally define two objects to be similar if they

have the same shape but not the same size.

Conditions for similarity of triangles � Corresponding angles in the triangles are equal.� Corresponding sides are in the same ratio.

A′B ′

AB= B ′C ′

BC= A′C ′

AC= k

where k is the scale factor� Two pairs of corresponding sides have the same

ratio and the included angles are equal.

Volumes of solidsCylinder:

V = �r2hh cm

r cm

Cube:

V = x3

x

Rectangular prism:

V = lwh l cm

h cm

w cm

Right-angled triangular prism

V = 12 bhl

h cm

b cm l cm

Surface area of solids Cylinder:

S = 2�r2 + 2�rhh cm

r cmSAMPLE



Review


Cube:

S = 6x2

x

Rectangular prism:

S = 2 (lw + lh + wh) l cm

h cm

w cm

Right-angled triangular prism

S = bh + bl + hl + l√

b2 + h2

h cm

b cm l cm

Scaling, areas and volumes � If two shapes are similar and the scale factor is k,

then the area of the similar shape = k2× area of

the original shape.

k = 32

k2 = 94

� If two solids are similar and the scale factor

is k, then the volume of the similar

solid = k3× volume of the original solid.

k = 32

k3 = 278

Skills check

Having completed this chapter you should be able to:

apply the properties of parallel lines and triangles and regular polygons to find the

size of an angle given suitable information

find the size of each interior angle of a regular polygon with a given number of sides

use the definition of objects such as triangles, quadrilaterals, squares, pentagons,

hexagons, equilateral triangles, isosceles triangles to determine angles

recognise when two objects are similar

determine unknown lengths and angles through use of similar triangles

find surface areas and volumes of solids

use Pythagoras’ theorem to find unknown lengths in right-angled triangles

use similarity of two- and three-dimensional shapes to determine areas and volumes

SAMPLE



Rev

iew


Multiple-choice questions

Questions 1 to 3 relate to the diagram

S QR

P

120° 70°

1 Angle PRS =A 20◦ B 40◦ C 50◦ D 60◦ E 110◦

2 Angle RPS =A 20◦ B 40◦ C 50◦ D 60◦ E 110◦

3 Given that PS bisects angle QPR, the size of angle PQS is:

A 20◦ B 35◦ C 40◦ D 50◦ E 60◦

Questions 4 to 6 relate to the diagram

Lines m and l are parallel and cut by a transversal q.

m

q

l

125°

x°z° y°

4 The value of x is:

A 65 B 125 C 62.5 D 60 E 55

5 The value of y is:

A 65 B 125 C 62.5 D 60 E 55

6 The value of z is:

A 65 B 125 C 62.5 D 60 E 55

7 The triangle ABC has a right angle at A. The length

of side BC in cm is:

A 10 B 14 C 9 D 9. 8 E 11

A B

C

8 cm

6 cm

8 The triangle ABC has a right angle at A. The length of

side BC to the nearest cm is:

A 10 B 14 C 9 D 12 E 11

A B

C

9 cm

7 cm

9 Triangles ABC and XYZ are similar isosceles triangles.

5 cm

12 cm12 cm

5 cm

3 cm B X Y

Z

C

A

The length of XY is:

A 4 cm B 5 cm C 4.2 cm

D 8.5 cm E 7.2 cm

10 YZ is parallel to Y ′Z ′ and Y ′Y = 12 Y X .

The area of triangle XYZ is 60cm2. The area of triangle XY ′Z ′ is:

A 20 cm2 B 30 cm2 C 15 cm2

D20

3cm2 E

80

3cm2

Z

X

Y

Z'Y'

SAMPLE



Review


11 The value of x is:

A 12 B 27 C 2.16

D 20.8 E 13.81

1.2 cm

18 m

1.8 m

x m

12 A regular convex polygon has 12 sides. The magnitude of each of its interior angles

is:

A 30◦ B 45◦ C 60◦ D 150◦ E 120◦

13 Triangles ABC and XYZ are similar isosceles triangles.

7 cm7 cm

4 cm

10 cm 10 cm

A B

CZ

YX

The length of XY correct to one decimal place is:

A 4.8 cm B 5.7 cm C 4.2 cm

D 8.5 cm E 8.2 cm

14 Two similar cylinders are shown. The ratio of the volume

of the smaller cylinder to the larger cylinder is:

10 cm

40 cm

60 cm15 cmA 1:4 B 1:16 C 1:64

D 15:60 E 1:3

15 Each interior angle of a regular polygon measures 135◦. The number of sides the

polygon has is:

A 4 B 6 C 8 D 10 E 7

16 Each side length of a square is 10 cm. The length of the

diagonal is:

A 10 B 5√

2 C 10√

2 D 8 E 1.4

A B

CD

10 cm

SAMPLE


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