Ch 12 Geometry _Maths

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9780521740517c12.xml CUAU031-EVANS October 22, 2011 14:15

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 67

8

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 6

78 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.

362

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Uncorrected sample pages • Cambridge University Press • © Jones, Evans, Lipson 2011 • 978-1-107-65590-4 • Ph 03 8671 1400

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Chapter 12 — Geometry 363

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

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364 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°

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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


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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

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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◦.SAMPLE


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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 that meet at O.

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

b What type of triangle is:

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?

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12.4 Pythagoras’ theorem

Pythagoras’ theorem

Pythagoras’ 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

= 12ab + 1

2c2 + 1

2ab

= ab + 12c2

∴ 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

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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


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 . . .

3 Write down your answer, correct to the

nearest cm.

Correct to the nearest centimetre, the

length of the short side is 83.07 m.

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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


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

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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


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12.5 Similar figures

Similarity

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

but not the same size.

Examples

Any 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 cm

8 cm

R2

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.

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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 cm

3.75 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,

correct to two decimal places.

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

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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 cm

6 cmX Y

Z18°

135°

5 cm

x 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 cmSAMPLE


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

find AC, AE and AB.

1412

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


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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 m

spotlight

0.6 m

wall

12.6 Volumes and surface areasVolume of a prismA prism is a solid that 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 cm2

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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 cm3 (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 cml cm

V = 12bhl

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 = 13x2 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

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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 and surface area of a sphereThe formulas for the volume and the surface area of a sphere are:

V = 43�r3 S = 4�r2

where r is the radius of the sphere.

r

Example 16 Volume of a sphere and a cone

Find the volume of a sphere with radius 4 cm and a cone with radius 4 cm and height 10 cm.

Solution

Volume of sphere = 4

3πr 3

= 4

3× π × 43

= 268.08 cm3 (two d.p.)

Volume of cone = 1

3πr 2h

= 1

3× π × 42 × 10

= 167.55 cm3 (two d.p.)

Example 17 Surface area of a sphere

Find the surface area of a sphere with radius 10 cm.

Solution

Surface area of sphere = 4πr 2

= 4π × 102

= 1256.64 cm2 (2 d.p.)

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

summing the volumes of the component shapes.

Example 18 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 The composite shape is made up from

a cylindrical base plus a hemispherical top.

The volume of the composite shape is the

volume of the cylinder plus the volume

of the hemisphere (half a sphere)

= +

2 Use the formula V = �r2h to find the

volume of the cylinder.

The volume of the cylinder

Vcyl. = π × 52 × 5 = 392.699 . . . mm3

3 Find the volume of hemisphere noting

that the volume of a hemisphere is

V = 12

(43�r3

) = 23�r3.

The volume of the hemisphere

Vhem. = 1

2× 4

3π × 53 = 261.799 . . . mm3

4 Add the two together. The volume of the composite = 654.50 mm3

(correct to two decimal places)5 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

The surface area of the cylinder

= area of ends + area of curved surface

= area of two circles + area of rectangle

= 2 × �r2 + 2�r × l = 2�r2 + 2�rl

2πr

A = 2πrlA = πr2A = πr2

l

rr

The formulas for the surface areas of some common three-dimensional shapes follow.

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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 cml cm

S = bh + bl + hl + l√

b2 + h2

Example 19 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

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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 cm

c

area of cross section = 2.8 cm2

height = 6.2 cm

d

radius 2.3 cm and length 4.8 cm

2 Each side of the square base of one of the great Egyptian

pyramids is 275 m long. It has a perpendicular height of 175 m.

Calculate the volume of this pyramid, correct to the nearest

cubic metre.

175 m

275 m275 m3 Find the volume, correct to one decimal place, of a:

a sphere with radius 1.5 m b cone with radius 6 cm and height 15 cm

c hemisphere of diameter 3.8 mm d cone with diameter 15 mm and height 10 mm

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

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

Find the volume of the capsule correct to the nearest cm3.

5 The diagram shows a composite shape made from a cylinder and

two cones. Both the cylinder and the two cones have a radius of

12 cm. The length of the cylinder is 8 cm and height of the cones

is 10 cm. Find the volume of the composite shape. Give your

answer correct to the nearest cm3.

6 Find the total surface areas of shapes a and b of question 1. Give answers correct to the

nearest cm2.

7 For the triangular prism shown, find

2 m

3 m0.5 m

a the volume in m3

b the total surface area in m2, correct to one decimal place

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8 A hemispherical dome tent has a diameter of 2.5 m, as shown.

2.52.5 m2.5 ma Determine the volume enclosed by the tent, correct to the

nearest m3.

b Determine the total surface area of the tent (including its floor),

correct to the nearest m2.

9 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

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

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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 areas

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

shape = k2 × area of the original shape.

Example 20 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.

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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 21 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

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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 volumes

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

solid = k3 × volume of the original solid.

Example 22 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 that 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.

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Example 23 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

∴ Volume2 = 1.253 × 48 = 93.75

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

Exercise 12G

1 Two regular hexagons are shown.

The side length of the smaller hexagon is 2.4 cm.

The side length of the larger hexagon is 7.2 cm.

a Determine the length scale factor k for scaling up.

b The area of the smaller hexagon is 15 cm2.

Determine the area of the larger hexagon.

15 cm2 2.4 cm7.2 cm

2 Triangle ABC is similar to triangle XYZ.

The length scale factor k = 2. The area

of triangle ABC is 6 cm2. Find the area

of triangle XYZ.A C

B

X Z

Y× 2

6 cm2

3 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 cm20 cm2

A

B C

D

B'

A' D'

C'

4 The two shapes shown are similar. The length

scale factor for scaling down is 23. The area

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

area of the shape to the left?

30 cm2

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5 The octagonal prisms are similar.

The height of the smaller prism is 2 m. The height of the

larger prism is 3 m.

The surface area of the smaller prism is 18 m2.

Determine the surface area of the larger prism in m2.

2 m3 m

6 The radius of the larger sphere is 2.5 times the radius

of the smaller sphere. The volume of the smaller sphere is 24 mm3. 2.5rr

a Write down the length scale factor k for scaling up.

b Determine the volume of the larger sphere in mm3.

7 The two rectangular prisms are similar.

The length of the smaller prism is 10 cm. The length of

the larger prism is 15 cm.

The volume of the larger prism is 3375 cm3.

Determine the volume of the smaller prism in cm3.

10 cm

15 cm

3375 cm3

8 The two cones shown are similar. The smaller cone has a diameter

of 10 cm. The larger cone has a diameter of 30 cm.

a Determine the length scale factor k for scaling up.

b What is the length scale factor k for scaling down?

c The height of the larger cone is 45 cm. Determine

the height of the smaller cone.

d The surface area of the smaller cone is 326.9 cm2. Determine

the surface area of the larger cone correct to the nearest cm2.

e The volume of the smaller cone is 392.7 cm3. What is the volume of the larger cone,

correct to the nearest cm3?

10 cm

30 cm

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

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

10 Triangles XBY and ABC are similar.

The area of triangle XBY is 1.8 cm2.

Determine the area of triangle ABC.A

B

C

X Y

7 cm

3 cm

11 Triangles XQY and PQR are similar.

The area of triangle PQR is 7.5 cm2.

a Determine k, the length scale factor for scaling down.

b Determine the area of triangle XQY.

RP

YX

Q

3 cm

2 cm

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wChapter 12 — Geometry 389

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 67

8

l1

l2

l3

cointerior and verticallyopposite angles

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

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 that are

segments of straight lines. Examples are:

3 sides: Triangle

4 sides: Quadrilateral

5 sides: Pentagon

6 sides: Hexagon

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w390 Essential Further Mathematics – Module 2 Geometry and trigonometry

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 = 12bhl

h cm

b cml cm

Surface area of solids Cylinder:

S = 2�r2 + 2�rhh cm

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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 cml 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 = 3

2

k2 = 9

4� 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 = 3

2

k3 = 27

8

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.

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w392 Essential Further Mathematics – Module 2 Geometry and trigonometry

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 = 12Y X .

The area of triangle XYZ is 60 cm2. 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'

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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 The rectangular prism shown has a volume of 12.8 cm3.

A second rectangular prism is made with half the length,

four times the height and double the width.

The volume of the second prism (in cm3) is:width

height

length

A 6.4 B 12.8 C 51.2

D 102.4 E 204.8

15 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

16 To the nearest mm2, the surface area of a sphere of radius

of radius 8 mm is:

A 202 mm2 B 268 mm2 C 804 mm2

D 808 mm2 E 2145 mm2

17 An equilateral triangle of side length 7 cm is cut from a

circular sheet of metal of diameter 20 cm.

The area of the resulting shape (in cm2) is closest to:

A 21 B 293 C 314

D 335 E 921

7 cm

20 cm

18 The diagram shows a composite shape that consists of a

hemisphere of radius 6 cm placed on top of a cylinder of

height 8 cm and radius 6 cm.

The total surface area of the composite shape (including

the base) is closest to:

A 302 cm2 B 452 cm2 C 528 cm2

D 641 cm2 E 754 cm2

SAMPLE


Ch 12 Geometry _Maths

Documents

similar pattern

angled triangular

ab bc ac

sided convex

triangular

diagram opposite

total surface

angled triangle