chapter 12 - geometry

8/14/2019 Chapter 12 - Geometry

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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 calledalternateangles.

Angles 5 and 3 are calledalternateangles.

Angles 2 and 6 are calledcorrespondingangles.Angles 1 and 5 are calledcorrespondingangles.

Angles 4 and 8 are calledcorrespondingangles.

Angles 3 and 7 are calledcorrespondingangles.

Angles 3 and 6 are calledcointeriorangles.

Angles 4 and 5 are calledcointeriorangles.

Angles 1 and 3 are calledvertically opposite anglesand 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.

1 234

5 67

8

l2

l1

l3

Linesl1and l2are cut by a transversal l3.

When linesl1andl2are parallel,correspondinganglesare of equal magnitude, alternateangles are of equal

magnitude andcointeriorangles are supplementary.1 2

34

5 6

78 l2

l1

l3

Converse results also hold:

If corresponding angles are equal, then

l1is parallel tol2.

If alternate angles are equal, thenl1is parallel tol2.

If cointerior angles are supplementary, thenl1is parallel tol2.

362ISBN: 9781107655904 Peter Jones, Michael Evans, Kay Lipson 2012 Cambridge University Press

Photocopying is restricted under law and this material must not be transferred to another party


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Chapter 12 Geometry 363

Example 1 Parallel line properties

Find the values of the pronumerals.

65e

cd

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= 20Also 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

1Anglesdandbare:

A alternate B cointerior C corresponding

D supplementary E vertically opposite

ISBN: 9781107655904 Peter Jones, Michael Evans, Kay Lipson 2012 Cambridge University Press



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364 Essential Further Mathematics Module 2 Geometry and trigonometry

2Anglesdandaare:



3Anglescandhare:

A alternate B cointerior C correspondingD supplementary E vertically opposite

4Anglesbandfare:



5Anglescandeare:



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

ax

z

y70

b

xz

y

40

c(2z)

y 80

d

xz

y50

e

xy

120

f

(x + 40)

(2x 40)

12.2 Properties of triangles: a reviewa,bandcare the magnitudes of the interiorangles of the triangleABC.

B

A C

a c d

b

dis the magnitude of an exteriorangle atC.

Thesumof the magnitudes of theinterior anglesof

a triangle is equal to180:a + b + c= 180.b + a= d. The magnitude of an exterior angleisequal to thesumof the magnitudes of thetwo

opposite interior angles.

A triangle is said to beequilateralif all its sides

are of the same length:AB =BC= CA.Theanglesof anequilateral triangleare all of

magnitude60.

A C

B

60

60 60




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Thebisectorof each of the angles of an equilateral

trianglemeets the opposite side atright anglesand

passes through themidpointof that side.

A C

B

O

A triangle is said to beisoscelesif it has two sidesof 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= 360A triangle is said to be aright-angled triangleif it

has one angle of magnitude 90. A

f

B e

d

C

Example 4 Angle sum of a triangle


A

B

Cx20

22

y

Solution

20 + 22 + x = 180 (sum angles = 180) 42 + x = 180 or x= 138

138

+ y

= 180

(sum angles

=180)

y= 42Or, to find x:

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




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Exercise 12B

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

a

5080

x y

A C

B b

P R

Q

x

30

x

c

A

B

Cy

x

70 30

d

X

Y

Z

x

50

y

e

A C

B

z

y w x40

f

a

bc

4075

g

AEy

y xB C

D

12.3 Properties of regular polygons: a review

Equilateral triangle Square Regular pentagon Regular hexagon Regular octagon

Aregular polygonhas allsides of equal lengthand allangles of equal magnitude.

Apolygonwith n sidescan be divided into n triangles. The first three polygons below are

regular polygons.

O O O

The anglesum of the interior anglesof ann-sidedconvex polygonis given by the

formula:

S= [180(n 2)]= (180n 360)

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

vertex to another vertex lies inside the polygon.

The magnitude of each of the interior anglesof ann-sided polygon is given by:

x= (180n 360)

n




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Exercise 12C

1Name each of the following regular polygons.

a b c d e

2ABCDis a square.BDandACare diagonals that meet atO.

a Find the size of each of the angles atO.

b What type of triangle is:

i BAD? ii AOB?

O

A

B C

D

3ABCDEis a regular pentagon.

a Find the value of:i x ii y A

B

C

DE

O

y

x

b Find the sum of the interior angles of the regular

pentagonABCDE.

4ABCDEFis a regular hexagon.

Find the value of:

a x b y O

C

B

AF

E

D

x

y

5State the sum of the interior angles of:

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

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

does the polygon have?

7A circle is circumscribed about a hexagonABCDEF.

a Find the area of the circle ifOA= 2 cm.b Find the area of the shaded region.

OE

F

A

B

C

D

8The diagram shows a tessellation of regular hexagons

and equilateral triangles.

State the values ofaandband use these to explain

the existence of the tessellation.

ab

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




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Solution

1 Using Pythagoras theorem, write down an

expression forxin terms of the two other

sides of the right-angled triangle. Solve forx.

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.

Example 9 Pythagoras theorem

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 foryin terms of the two other

sides of the right-angled triangle. Solve fory.

5.62+ y2= 8.62 (Pythagoras)y2= 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.

Example 10 Pythagoras theorem

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. Letxbe the length of the

short side.

130 m

100 m

xm

Let x m be the length of the shorter

side.


expression forxin terms of the two other sides

of the right-angled triangle. Solve forx.

x 2+ 1002= 1302 (Pythagoras)x 2= 1302 1002= 6900x=

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

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

a

10 cm

6 cm

b

11 cm

5 cm

c10 cm 3 cm

d9 cm

7 cm

e 33 cm

44 cm

A

C Bf

15 cm12 cm

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

a

xcm

3.2 cm

4.8 cm

b

xcm6.2 cm

2.8 cm c

xcm

9.8 cm

5.2 cm

d

4 cm

3 cm3.5 cm

x cm

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


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

4In triangleVWX, there is a right angle atX.VX= 2.4 cm andXW= 4.6 cm. FindVW.

5FindAD, the height of the triangle.

20 cm

32 cm 32 cm

DC B

A

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

reach?

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

8TriangleABCis isosceles. Find the length ofCB. A

8 14

BC




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9In a circle of centreO, a chordABis of length 4 cm.

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

of the chord fromO.

xcm

A B

O

10Find the value ofx.

18 cm

xcm

xcm

11How high is the kite above the ground?

90 m

170m

X

Y

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

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

a 8

2 cm b 8 cm

14Find the length ofAB.

A

D

B

C20 cm

8 cm12 cm

15ABCDis a square of side length 2 cm.

IfCA = CE, find the length ofDE.AB

C D E

16The midpoints of a square of side length 2 cm

are joined to form a new square. Find the area

of the new square.




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12.5 Similar figures

Similarity

In this section we informally define two objects to be similarif 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 toR1is: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 35A'

B'

C'

corresponding sides are in the same ratio

AB

AB= B

C

BC= A

C

AC= k

kis the scale factor.




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

45 45A

B

C A'

B'

C'

ABAB

= ACAC

If triangleABCis similar to triangleXYZthis 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 toX,Bcorresponds toY,Ccorresponds toZ.

Example 11 Similar triangles

Find the value of length of sideACin ABC,

correct to two decimal places.

A Cx cm

A' C' 3.013 cm

3 cm5 cm 20

B

B'

206.25 cm

3.75 cm

Solution

1 TriangleABCis 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

AC= AB

AB .

Use this fact to write down an expression

involvingx. Solve forx.

x

3.013= 5

6.25

x= 56.25

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 sideABin ABC.

A3 cm C

2.5 cm

Y

X

xcm

B6cm

Solution

1 TriangleABCis similar to triangleAXY

(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

involvingx. Solve forx. Note that

ifA B=

x , thenA X=

x+

6.

x

x+ 6=3

5.5 5.5x= 3(x+ 6)

= 3x+ 18

3 Write down your answer.

2.5x

=18 or x

=

18

2.5=7.2

The length of side AB is 7.2 cm.

Exercise 12E

1Find the value ofxfor each of the following pairs of similar triangles.

a A

B C

4 cm 5 cm82

56

A'

B' C'

9 cm xcm

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 cmxcm

Y

d

A

B

C

D

E

x cm12 cm

13 cm

14 cm 10 cm

e

A

C

B10 cm

xcm

R

Q P

6 cm

8 cm

f

A

C

B D

E

6 cm

xcm

2 cm

4 cm

g A

B C

12 cm 16 cm

8 cmxcmP Q

h B

A C

D

xcm

2 cm

3 cm 2 cmE

i A

B C

QP

2 cm

6 cm 8 cm

x cm

j

10 cm

1.5cm

2 cmQ

C

B

P

A

x cm




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2Given thatAD = 14,ED = 12,BC= 15 andEB = 4,findAC,AEandAB.

1412

15

4A

DC

E B

3A tree casts a shadow of 33 m and at the same time a stick 30 cm long casts a shadow24 cm long. How high is the tree?

33 m

0.24 m

0.3 m

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

cable as shown. An ant crawls along the cable starting atA.

How high is the ant when it is 15 m fromA?

40m

20 m

A

5A 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?

6A man stands atAand looks at pointYacross the river.

He gets a friend to place a stone at Xso thatA,XandY

are collinear. He then measuresAB,BXandXCto be

15 m, 30 m and 45 m, respectively. FindCY, the

distance across the river.

river

30 m

15 m45 m

Y

C X

A

B

7Find the height,hm, 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.

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

and 12 cm deep. Findx, wherexcm is the width of the

widest rectangular box of height 8 cm that can be placed

on a stair under the plank. 20 cm

xcm 12 cmpla

nk

8 cm

9The sloping edge of a technical drawing table is 1 mfrom front to back. Calculate the height above the ground

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

80 cm

1m

30cm

A

92 cm




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10Two 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 ofx.x m

0.8 m

(1.3 x) m

1.5 m

11A 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.

12A 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.

13A 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= r2= 32

= 28.27 cm2




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2 Multiply by the height. Volume = 28.2743 Make sure that accuracy is given to the

correct number of decimal places.

= 113.10 cm3 (correct to twodecimal places)

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

Solid Formula

Cylinder (radiusrcm, heighthcm)

V= r2hhcm

r cm

Cube (all edgesxcm)

V= x3

Rectangular prism (lengthlcm,V= lwhwidthwcm, heighthcm)

lcm

h cm

wcm

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= 12bhl

Volume of a pyramid

x

h

The formula for finding the volume of a right pyramid

can be stated as:

Volume of pyramid= 13base areaperpendicular height

For the square pyramid shown:

V= 13x2 h

The termrightin 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= 13A h

= 13 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= 13

x 2 h

= 13 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 = 13base area heightV= 1

3r2h

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

V= 43r3 S= 4r2

whereris 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= 43

r3

= 43 43

= 268.08 cm3

(two d.p.)

Volume of cone = 13

r2h

= 13 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 = 4r2= 4 102

= 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 formulaV= r2hto find thevolume of the cylinder.

The volume of the cylinder

Vcyl.= 52 5 = 392.699 . . . mm33 Find the volume of hemisphere noting

that the volume of a hemisphere is

V= 12

43r3

= 23r3.

The volume of the hemisphere

Vhem.= 12 4

3 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. Thenetof 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 + 2r l= 2r2 + 2rl

2r

A = 2rlA = r2A = r2

l

rr

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




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Solid Formula

Cylinder (radiusrcm, heighthcm)

S= 2r2 + 2rhhcm

r cm

Cube (all edgesxcm)

S= 6x2

Rectangular prism (lengthlcm,

S= 2(lw+ lh + wh)widthwcm, heighthcm)

lcm

h cm

wcm

Triangular prism

h cm

b cm l 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 cm23 Determine the area of one of the

isosceles triangles.

The area of one triangle

= 12 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+475= 100+ 300= 400 cm2




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Exercise 12F

1Find 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 cm2height = 6.2 cm

d

radius 2.3 cm and length 4.8 cm

2Each 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 m3Find 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 cmc hemisphere of diameter 3.8 mm d cone with diameter 15 mm and height 10 mm

4The 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.

5The 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 youranswer correct to the nearest cm3.

6Find the total surface areas of shapesaandbof question1. Give answers correct to the

nearest cm2.

7For the triangular prism shown, find

2 m

3 m

0.5 m

a the volume in m3

b thetotalsurface area in m2, correct to one decimal place




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

2.55 m2.5 ma Determine the volume enclosed by the tent, correct to thenearest m3.

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

correct to the nearest m2.

9Find, 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

10Find:

a the surface area

b the volume

of the object shown. 10 m5 m

3 m4 m

2 m

11The 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

12Find:

a the surface

b the volume

of the solid shown opposite.

10 cm

7 cm

4 cm

4 cm

13The 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 2radius circle 1

= 43

4 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 2length rectangle 1

= 63= 2

Ratio of areas = 246= 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 2height triangle 1

= 93= 3

Ratio of areas = 546= 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

ofkscales the area by a factor ofk2.

Scaling areas

If twoshapes are similarand thescale factor is k, then theareaof 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 factork. k= 2510= 2.5

2 Write down the area of the small triangle. Area of small triangle= 40 cm23 Area of larger triangle = k2 area of

smaller triangle.

Area of larger triangle = 2.5240= 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 factork. Note we are

scaling down.

k= 1260= 0.2

2 Write down the area of the larger heart. Area of larger heart= 100 cm23 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

= 36cm3

Scale factor= k= radius sphere 2radius sphere 1

= 43

Ratio of volumes =256

3

36= 256

108

= 6427

= 43

3 = k3Volume = 4

3 43

=256

3

cm

3




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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 2side length 1

= 42= 2

Ratio of volumes =64

8= 8

= (2)3 = k3 Volume = 4 4 4

= 64 cm3

Similar cylinders

1 cm

2 cm

3 cm

6 cm

Volume = 12 2= 2cm3

Scale factor= k= radius 2radius 1

= 31= 3

Ratio of volumes = 54

2= 27 = (3)3 = k3

Volume = 32 6= 54cm3

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

ofkscales the volume by a factor ofk3.

Scaling volumes

If twosolids are similarand thescale factor is k, then thevolumeof the similar

solid= k3 volume of the original solid.

Example 22 Similar solids

1.5 cm 6 cm

volume =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 factork. Note that we

are scaling down.

k= 1.56= 0.25

2 Write down the volume of the larger cuboid. Volume larger cuboid = 120 cm33 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 D

O

C

V

9 cm

4 cm

Pyramid 1 Pyramid 2

Solution

1 Determine the scale factor,k. Use the base

measurements.

k= 54= 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= 11.25Substitute the appropriate values and evaluate.

4 Write down your answer.

Volume

The height of Pyramid 2 is 11.25 cm.

5 Volume Pyramid 2 = k3 volume Pyramid 1.Substitute the appropriate values and evaluate.

Volume 1 = 48 cm3Volume2= 1.253 48= 93.75

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

Exercise 12G

1Two 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 factorkfor scaling up.

b The area of the smaller hexagon is 15 cm2.

Determine the area of the larger hexagon.

15 cm22.4 cm

7.2 cm

2TriangleABCis similar to triangleXYZ.

The length scale factork= 2. The areaof triangleABCis 6 cm2. Find the area

of triangleXYZ.A C

B

X Z

Y

26 cm2

3The two rectangles are similar. The area

of rectangleABCDis 20 cm2. Find the

area of rectangleA B CD.3 cm 5 cm20 cm2

A

B C

D

B'

A' D'

C'

4The two shapes shown are similar. The length

scale factor for scaling down is 23 . The areaof the shape to the right is 30 cm2. What is the

area of the shape to the left?

30 cm2




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5The 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

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

of the smaller sphere. The volume of the smaller sphere is 24 mm 3. 2.5rr

a Write down the length scale factorkfor scaling up.

b Determine the volume of the larger sphere in mm3.

7The 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

8The 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 factorkfor scaling up.

b What is the length scale factorkfor 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

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

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

10TrianglesXBYandABCare similar.

The area of triangleXBYis 1.8 cm2.

Determine the area of triangleABC.

A

B

C

X Y

7 cm

3 cm

11TrianglesXQYandPQRare similar.

The area of trianglePQRis 7.5 cm2.

a Determinek, the length scale factor for scaling down.

b Determine the area of triangleXQY.

RP

YX

Q

3 cm

2 cm




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Key ideas and chapter summary

Alternate, corresponding, Angles 4 and 6 are examples ofalternateangles.

Angles 2 and 6 are examples ofcorrespondingangles.

Angles 3 and 6 are examples ofcointeriorangles.

Angles 1 and 3 are examples

ofvertically oppositeangles

and are of equal magnitude.

1 234

5 67

8

l1

l2

l3

cointerior and vertically

opposite angles

Angles associated with parallel When linesl1andl2are parallelcorrespondingangles

of equal magnitude,alternateangles are of equal

magnitude andcointeriorangles are supplementary.

AlternateAlternate

Corresponding

Corresponding

Converse results also hold:

If corresponding angles are equal, thenl1is parallel tol2.

If alternate angles are equal, thenl1is parallel tol2.

If cointerior angles are supplementary, thenl1is parallel tol2.

lines crossed by a transversal

line

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 beequilateralif

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 beisoscelesif

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 Apolygonis 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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Convex polygon A polygon is said to beconvexif any diagonal lies

inside the polygon.

Regular polygon Aregular polygonhas all sides of equal length and

all angles of equal magnitude.

Sum of the interior angles The angle sum of the interior angles of ann-sided

polygon is given by the formula:S= (180n 360).Pythagoras theorem Pythagoras theoremstates that for a right-angled

triangleABCwith side lengthsa,bandc,

a2 + b2 = c2, wherecis the longest side.Similar figures We informally define two objects to besimilarif 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.

AB

AB= B

C

BC= A

C

AC= k

wherekis the scale factor Two pairs of corresponding sides have the same

ratio and the included angles are equal.

Volumes of solidsCylinder:

V=

r2hhcm

r cm

Cube:

V= x3

x

Rectangular prism:

V= lwh lcm

h cm

wcm

Right-angled triangular prism

V= 12bhl

h cm

b cm l cm

Surface area of solids Cylinder:

S= 2r2 + 2rhhcm

r cm




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Cube:

S= 6x2

x

Rectangular prism:

S= 2 (lw+ lh + wh) lcm

h cm

wcm

Right-angled triangular prism

S= bh + bl+ hl+ l

b2 + h2h 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 ofthe original shape.

k =3

2

k2 =9

4 If two solids are similar and the scale factor

isk, then the volume of the similar

solid= k3 volume of the original solid.

k = 32

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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Multiple-choice questions

Questions 1 to 3 relate to the diagram

S QR

P

120 70

1 AnglePRS=A 20 B 40 C 50 D 60 E 110

2 AngleRPS=A 20 B 40 C 50 D 60 E 110

3 Given thatPSbisects angleQPR, the size of anglePQSis:

A 20 B 35 C 40 D 50 E 60

Questions 4 to 6 relate to the diagram

Linesmandlare parallel and cut by a transversal,q.

m

q

l

125

x

z y

4 The value ofxis:

A 65 B 125 C 62.5 D 60 E 55

5 The value ofyis:

A 65 B 125 C 62.5 D 60 E 55

6 The value ofzis:

A 65 B 125 C 62.5 D 60 E 55

7 The triangleABChas a right angle atA. The length

of sideBC, in cm, is:

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

A B

C

8 cm

6 cm

8 The triangleABChas a right angle atA. The length of

sideBC, to the nearest cm, is:

A 10 B 14 C 9 D 12 E 11

A B

C

9 cm

7 cm

9 TrianglesABCandXYZare similar isosceles triangles.

5 cm

12 cm12 cm

5 cm

3 cm B X Y

Z

C

A

The length ofXYis:

A 4 cm B 5 cm C 4.2 cm

D 8.5 cm E 7.2 cm

10 YZis parallel toYZandYY= 12Y X.

The area of triangleXYZis 60 cm2. The area of triangleX YZis:

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 ofxis:

A 12 B 27 C 2.16

D 20.8 E 13.81

1.2 cm

18 m

1.8 m

xm

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 TrianglesABCandXYZare similar isosceles triangles.

7 cm7 cm

4 cm

10 cm 10 cm

A B

C Z

YX

The length ofXY, 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 2 B 452 2 C 528 2

chapter 12 - geometry

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