chapter 10 shape and measurement -...

CHAPTER 10 Shape and measurement 379

CHAPTER CONTENTS 10A Pythagoras’ theorem 10B Pythagoras’ theorem in three dimensions 10C Perimeter and area 10D Total surface area (TSA) 10E Volume 10F Capacity 10G Similar figures 10H Similar triangles 10I Symmetry

DIGITAL DOCdoc-960810 Quick Questions

CHAPTER 10

Shape and measurement

10A Pythagoras’ theoremIn a right-angled triangle the longest side (which is opposite to the right-angle) is called the hypotenuse.

For any right-angled triangle with side lengths a, b and hypotenuse c it can be proved that

a2 + b2 = c2

This is called Pythagoras’ theorem.

WORKED EXAMPLE 1

Find the value of the pronumerals to 1 decimal place.

THINK WRITE

1 In ΔABC only BC is known; therefore x should be found fi rst.

2 Use ΔADC to fi nd AC using Pythagoras’ theorem. ΔADC:x2 = 252 + 172

= 625 + 289 = 914 x = 914

≈ 30.2

3 With the added information, fi nd y.Remember that x2 = 914.

ΔABC:

( 914 )2 = y2 + 122

914 = y2 + 144y2 = 914 − 144y2 = 770

AB = 770= 27.748 873 85

4 Round to the required number of decimal places. y ≈ 27.7

hypotenuse

a

b

c

y

x25

17D C

B

A

12

380 Maths Quest 11 Standard General Mathematics

Exercise 10A Pythagoras’ theorem 1 For each of the following triangles, find the length of the unknown side correct to 1 decimal place.

a

9

12a

b 4

10b

c 5

19

c

d

d

6.34.2

e

14.816.2

e f

6.7

f

2 What is the length of a diagonal of:a a square with side lengths of 62 cm (correct to 1 decimal place)b a rectangle with dimensions of 0.8 m and 37 cm? (Give your answer in metres correct to

2 decimal places.)

3 WE1 Find the value of the pronumeral to 1 decimal place in each of the following shapes.a

1.2 m

1.8 m2.5 m

1.4 ma

b

16

32.8

b

4 A long-distance runner wants to complete a 30-km run. She has completed 18 km and is at the marked point. If she took the short cut across the park, would she reach her goal or should she go along the streets? Justify your answer with mathematical evidence.

Home

MarySt

Park crossing

Hampton St

x

y

4.5 km

2 km 5 km5 kmRunner

5 Two of the legs of a triathlon are set in a triangular format.The swim The swim leg has the competitors heading out due north to buoy 1, turning due east to the second buoy and then heading straight back to the starting position.a Draw a diagram of this situation.b Calculate the total length of the swim if the fi rst two legs are 300 m and 400 m respectively.The cycleThe athletes leave the transition area and ride in a southerly direction for 12 km. They then turn due west and ride for 16 km. The last leg has the cyclists heading straight back to the transition area.c Competitor 42’s cycle develops a puncture at the start of the last leg. How far does competitor

42 have to walk to get back to the transition area?

DIGITAL DOCSdoc-9609

SkillSHEET 10.1Pythagoras’ theorem

doc-9610Pythagoras’ theorem


10B Pythagoras’ theorem in three dimensionsThree-dimensional objects can have right-angled triangles within them that can be re-drawn in 2 dimensions, so that Pythagoras’ theorem can be used to fi nd missing measurements.

WORKED EXAMPLE 2

Find the length labelled x. Round your answer to 1 decimal place.

THINK WRITE/DRAW

1 Identify a right-angled triangle that contains x and redraw it in two dimensions.

12

8

A

B C

x

2 Write Pythagoras’ formula. 122 = 82 + x2

3 Solve for x. x2 = 122 − 82

x2 = 144 − 64x2 = 80x = ± 80

4 Given that x is the side length of a triangle then its value is positive.

x = 8.94427.

5 Write the answer, correct to 1 decimal place. The height labelled x is 8.9 units.

WORKED EXAMPLE 3

ABCDE is a square pyramid. Find EH, the height of the pyramid.

THINK WRITE/DRAW

1 (a) The unknown length x (EH) can be found from triangle EHC.

(b) In this triangle, the length of HC is missing as well so it has to be found fi rst. The length of HC is half the length of AC.

(c) In turn, AC can be found from triangle ABC. Redraw this triangle in 2 dimensions and include all measurements. 20

20A

C

B

yH

2 Use Pythagoras’ theorem to fi nd AC. Leave the answer in surd form.

c2 = a2 + b2

y2 = 202 + 202

= 400 + 400= 800

y = 800

= ×400 2

= 20 2

12A

B D

C

F

E

178x

H

BA

D

E

C

45

20


3 Find the length of HC by halving the length of AC. (There is no need to evaluate 20 2, as we will have to square this value in the steps that follow.)

HC = 1

2 AC

= × 20 21

2

= 10 2

4 Redraw the triangle that contains the unknown length x in 2 dimensions (triangle EHC). Write the measurements, including the length of HC found in the previous step. 45

10 2H C

E

x

5 Use Pythagoras’ theorem to fi nd x. (Round your answer to 1 decimal place.)

c2 = a2 + b2

c = 45, a = x, b = 10 2

452 = x2 + (10 2)2

2025 = x2 + 200x2 = 2025 − 200

= 1825

x = 1825x ≈ 42.7

Exercise 10B Pythagoras’ theorem in three dimensions 1 WE2 In each of the following, find the length of the line labelled x. Round your answers to 1 decimal place.

a

G

E F

H

DA

B C

x

20

10

15

b

G

E F

H

BA

C D

x

15

24

8

c

D

A

B C

x 16

12

d

B C

A

28.2

x

62.6

2 WE3 In each of the following, find the length of the line labelled x. Round your answers to 1 decimal place.a

E

G H

F

BA

C D

x

1612

8

b

5

1022

x

c

12

20x

d

9

615

x

3 The diagram at right shows a rectangular metal pencil case. Find:a the length of the longest pen that can be placed fl at on the bottom of

the pencil caseb the length of the longest pen that can be placed in this pencil case.

DIGITAL DOCdoc-9610

Pythagoras’ theorem

20 cm15 cm

10 cm


10C Perimeter and areaThe table below shows the formulas for finding the area and perimeter of some common shapes.

Shape Area Perimeter

Square

L

A = L2 P = 4L

Rectangle L

W

A = L × W P = 2(L + W)

Parallelogram

b

hA = b × hwhere the height measurement must be at a right angle to the base measurement.

P = sum of all sides

Trapezium

b

h

a A = 1

2(a + b)hwhere the height measurement must be at a right angle to the base measurement.


Rhombus x

y –

––

–A =

1

2 x × yP = sum of all sides

Triangle

b

hh

b

h

b

A = 1

2 bhwhere the height measurement must be at a right angle to the base measurement.


Triangle

a

bcA = − − −s s a s b s c( )( )( )

where s = 1

2(a + b + c)(Use when height measurement is unknown.)

P = a + b + c

Circle

rd

A = π r2 C = π d orC = 2π r


WORKED EXAMPLE 4

Calculate a the area and b the perimeter of this shape.

THINK WRITE

a 1 Write the appropriate formula for the area. a A = − − −s s a s b s c( )( )( )

where s = 12 (a + b + c)

2 Identify the values of the pronumerals. a = 7, b = 8, c = 10

3 To fi nd s, substitute a, b and c values into the formula and simplify.

s = 12 (7 + 8 + 10)

= 12 × 25 = 12.5

4 Substitute the values of a, b, c, and s into the formula for the area.

A = − − −12.5(12.5 7)(12.5 8)(12.5 10)

5 Evaluate.(Round the answer to 1 decimal place and include the units.)

A = × × ×12.5 5.5 4.5 2.5

= 773.4375= 27.810 744 33

A = 27.8 m2

b 1 Find the perimeter by adding all the side measurements.

b P = 7 + 8 + 10= 25

2 Write the answer, including the appropriate units. P = 25 m

Composite figuresThe term composite means ‘made up of distinct parts’. Composite fi gures in geometry are fi gures comprising a number of distinct shapes. Depending upon the composite fi gure, to fi nd the overall area or perimeter you may need to add these individual shapes or subtract one from another.

Area of an annulusCircles are said to be concentric if they have the same centre point. The area between the two concentric circles is referred to as an annulus.

Area of annulus = area of larger circle − area of smaller circle= π R2 − π r2

= π (R2 − r2)where R = radius of the large circle

r = radius of the small circle.

WORKED EXAMPLE 5

In one full revolution, the 6-cm-long minute hand of a clock would sweep out a larger circle than the 3-cm-long hour hand. What is the difference in the area they cover to the nearest square centimetre?

THINK WRITE

1 The area required is the annulus. Write the appropriate formula.

A = π (R2 − r2)

2 Identify the value of R (radius of the larger circle) and the value of r (radius of the smaller circle).

R = 6, r = 3

3 Substitute the values of the pronumerals into the formula.

A = π (62 − 32)

7 m 8 m

10 m

A1

A2

Ann

ulus

r

R

TUTORIALeles-1384Worked example 5

3 cm6 cm


4 Evaluate. A = 84.823

5 Write an answer sentence with the value rounded to the nearest square centimetre.

The difference in area covered by the two hands is approximately 85 cm2.

Area of a sector and arc lengthA sector of a circle can be thought of as a wedge- shaped slice of pie.

The area of the sector can be determined by finding the fraction of the whole circle it represents.

Area of sector = 360

2rθ π× ,

where θ is the angle of the sector and r is the radius of the circle.

O

A B

Sector of circle

θ

WORKED EXAMPLE 6

A 10-cm-long minute hand moving from the number 12 to the number 4 position sweeps out a sector. What is the area of this sector?

THINK WRITE

1 Write the formula for the area of the sector. Area of sector = θ π× r

3602

2 Identify the value of the radius. r = 10

3 Calculate the angle of the sector:The angle between consecutive numbers on a clock = 360° ÷ 12

= 30°.From 12 to 4 there are four intervals between the numbers. So to find the angle of a sector, multiply 30° by 4.

θ = 30 × 4= 120

4 Substitute the values of r and θ into the formula and evaluate.

Area of sector = 120

360 × π × 102

= 104.719 755 1

5 Write an answer sentence with the number rounded appropriately and units given.

The minute hand as it rotates through an angle of 120° sweeps through an area of 104.7 cm2.

The curved edge of a sector is called an arc.

Arc length = θ π× d

360or

Arc length = θ π× r

3602

where θ is the angle of a sector,d is the diameter of a circler is the radius of a circle.

4

12 12

3

10 cm

arcθ


WORKED EXAMPLE 7

What distance did the tip end of the 10-cm minute hand travel when it moved from pointing to the number 12 to pointing to the number 4?

THINK WRITE

1 Since the radius is known, write the formula for arc length involving the radius.

Arc length = θ π× r

3602

2 State the value of r. r = 10

3 Find the size of the angle of a sector.(a) The angle between each number on a

clock = 30°.(b) There are 4 intervals between the

numbers; therefore the angle of the sector can be found by multiplying 30° by 4.

θ = 30° × 4= 120°

4 Substitute values of r and θ into the formula and evaluate.

Arc length = π°°

× × ×120

3602 10

= 20.943 951 02

5 Write an answer sentence rounding the value appropriately and writing in the units.

The tip of the minute hand travelled 20.9 cm.

Exercise 10C Perimeter and area 1 WE4 Find: i the area and ii the perimeter of the following shapes, to 2 decimal places.

a

20 mm

14 mm b 140 cm

2 m

122 cm

––

––

c

12 cm

5 cm 9 cm

d

16 cm

4 cm18 cm

e

2.5 m

1.5 m5.5 m

2 Calculate i the area and ii the perimeter of the following shapes. Give your answers to 2 decimal places.

a

30 cm

15 cm

10 cm

13 cm25 cm

b

10 m

c

6 cm 3 MC Examine the diagram at right.

a The circles cover an area of approximately:A 402 cm2 B 201 cm2 C 804 cm2

D 805 cm2 E 603 cm2

b The shaded area is approximately:A 219 cm2 B 421 cm2 C 622 cm2

D 823 cm2 E 220 cm2

c A metal manufacturer is able to cut only four discs from every sheet of metal. What percentage of metal is wasted?A 80% B 21% C 22%D 41% E 61%

4

12 12

3

10 cm

DIGITAL DOCdoc-9611

SkillSHEET 10.2Conversion of units — length

DIGITAL DOCdoc-9612

SkillSHEET 10.3Area and perimeter of

composite shapes

32 cm


4 a A guard dog inside a used car salesyard is tied to a corner post of the fence surrounding the yard. The fence sides meet at a right angle and the dog is on a rope 1.2 m long.

i Draw a diagram of this situation showing the area accessible to the dog. ii To 1 decimal place, how much area does the dog have in which to exercise?b One night the dog is moved to the outside corner of a small rectangular building measuring

2 m × 6 m. The dog’s rope has been lengthened to 3 m. i Draw a diagram of this situation showing the area accessible to the dog. ii Calculate the area available for the dog for exercise. Give your answer to the nearest

square metre.c If the owner ties the dog back to the fence post and he wants the dog to have as much room

to exercise in as it did when it was tied to the building, what length of rope (to the nearest centimetre) does he need to purchase?

Geometry is used extensively in design. Questions 5–10 explore the use of geometrical shapes in the design of country fl ags, company logos, and some commonly used signs.

5 The flag of Japan is a red circle on a white background.Calculate (to 1 decimal place) the radius of the circle, if the area of the circle must be 20% of the total fl ag area.

6 The Commonwealth Bank logo is made up of a yellow square with a black trapezium overlaid at one corner. Calculate, to the nearest whole number, what percentage of the overall design is black.

7 WE5 The collectable plate shown below is 22 cm in diameter and has a golden 0.5-cm-wide ring.

22 cm

1 cm

0.5 cm

Find (to 1 decimal place) the area of the golden ring if its outer edge is 1 cm from the edge of the plate.

8 WE6 A family-size pizza is cut into 8 equal slices. If the diameter of the pizza is 33 cm, find (to the nearest square centimetre) the area of the top part of each slice.

9 WE7 Using the measurements given in the diagram at right, fi nd the length of the bike chain.

10 Early-model vehicles had a single windscreen-wiper blade to remove water from the windscreen. (The bus at right has two single blades of this type.) Using the dimensions given in the diagram below right:a what area (to the nearest

whole number) did the blade cover?

b what percentage (to 1 decimal place) of the windscreen was cleared?

c what distance (to the nearest whole number) does the tip of the blade travel in one full sweep?

3 m

2 m

1 cm3.0 cm 2.

5 cm

10 cm 6 cm150°

160°

42 cm

140°

45 cm

120 cm

60 cm


10D Total surface area (TSA)To find the total surface area (TSA) of a 3-dimensional object, we find the area of all the surfaces, and add them together.

Calculation of the total surface area of these buildings is a complex task.

For some common objects, we have a formula.

Object Net TSA

Cube

L

1 3

2

6

L

L

4 5

TSA = 6L2

Rectangular prism

wl

hw

hh

w

l

l

h

TSA = 2(wh + lw + lh)

Cylinder

hr h

r

r

2πr

TSA = area of 2 circles+ curved surface

= 2π r2 + 2π rh= 2π r(r + h)

Sphere

r

Not shown TSA = 4π r2


Object Net TSA

Cone

S = Slantheighth

r

S

r2 rπ

TSA = area of base (circle)+ area of curved

surface= π r2 + π rS= π r(r + S)

Square-based pyramid

b

hb

h

TSA = area of square+ area of 4 triangles

= b2 + 4 × ⎛⎝⎜

⎞⎠⎟bh1

2

= b2 + 2bh

WORKED EXAMPLE 8

Find the total surface area of the object shown at right.

THINK WRITE

1 Identify the shape. (Shape: rectangular prism)

2 Write the formula for the TSA of a rectangular prism.


3 Allocate a value to the pronumerals. w = 9, h = 17, l = 19

4 Substitute the values of the pronumerals into the formula.

TSA = 2(9 × 17 + 19 × 9 + 19 × 17)

5 Evaluate (brackets fi rst, then multiply by 2). = 2(153 + 171 + 323)= 2 × 647= 1294

6 Write the answer, including units. TSA = 1294 cm2

In some situations you may know the total surface area of an object but be missing a dimension. In this case, you can use the total surface formula to fi nd the unknown dimension.

WORKED EXAMPLE 9

A tennis ball has a surface area of 154 cm2. Will it fi t through a circular hole with a diameter of 6 cm?

THINK WRITE

1 Write the formula for the TSA of a sphere. TSA = 4π r2

2 Allocate the pronumerals a value. TSA = 154 r = ?

3 Substitute known values into the formula. 154 = 4π r2

4 Evaluate. Solving 154 = 4 × π × r2 for r gives r = −3.5007 or r = 3.5007.

5 Given that r is the radius of a tennis ball, its value is positive.

Since r > 0, r = 3.5007.

17 cm

19 cm9 cm


6 The question requires the diameter of the tennis ball so multiply the radius by 2.

d = 2 × r= 2 × 3.5007≈ 7.0

7 Compare the diameter of the hole with the diameter of the ball and write an answer sentence.

The tennis ball will not fi t through the circular hole because its diameter is approximately 7.0 cm, while the hole’s diameter is 6 cm.

Many buildings are composite fi gures made up of prisms and pyramids.

WORKED EXAMPLE 10

The diagram shows the proposed shape for a new container for takeaway Chinese food. Find the TSA of the container.

THINK WRITE

1 Identify the distinct shapes that make up the total object: these are a square-based pyramid and a cube. The base of the pyramid and one face of the cube are not on the surface and therefore their area should not be included.

TSA = 4 triangles + 5 squares.

2 Calculate the area of the triangles. A = 4 × 1

2 × b × h

= 2bhb = 10, h = 10A = 2 × 10 × 10

= 200

10 cm

10 cm



3 Calculate the area of the squares. A = 5L2

L = 10A = 5 × 102

= 500

4 Add the individual TSA together to find the TSA of the whole object.

TSA = 200 + 500= 700 cm2

Exercise 10D Total surface area (TSA) 1 WE8 Find the total surface area of the following objects. Round your answer to 1 decimal place.

a

10 cm

12.2cm

7.5cm

b

32.5 cm

20 cm c

6.2 m

d

8.4 m

10.5 m

e

9.0 cm

16 cm

3 cm f

42 mm

14 mm

2 Match the formulas below with the object.

a

rh

b

wl

h c Sr

(Base not included)Open-ended cone

d h

r

e

r

f

h

b

g

S

r

Base included

h

r

i

blh

i TSA = 2π r2 + 2π rh = 2π r(r + h)

ii TSA = π rS

iii TSA = ( )bh4 1

2iv TSA = π r(r + S)

v TSA = 4π r2 vi TSA = 2(wh + lw + lh)

vii TSA = ( )× bh2 1

2 + (3 × bl) viii TSA =

1

2[2π r(r + h)] + 2rh

= π r(r + h) + 2rh

ix TSA = πr4

2

2

= 2π r2


3 For the following diagrams, find the value of S for the formula:TSA (cone) = π r2 + π rS = π r(r + S)

a

12 cm

6 cm

10.4 cm

b

7 m

9 m

4 For each of the following objects find the value of the pronumeral, rounding your answer to 1 decimal place when required.

a

TSA = 148 m2

4 m

6 mw = ?

b

TSA = 188.5 m2

3 m

h = ?

c

TSA = 100 cm2

s = ?

d = 6 cm

d

b = ?TSA of 1 sphere

= 55.4 cm2

5 WE9 A cylindrical cork is 6 cm high and has a TSA of 33.38 cm2. Could it be used to close a bottle whose neck is 3 cm in diameter? Justify your answer.

6 MC A 60-cm-high cone has a base radius of 32 cm.The TSA could be calculated by using the formula:A π 32(32 + 60) B π 60(32 + 60) C π 32(32 + 68)D π 68(32 + 68) E π 60(32 + 68)

7 MC A spherical candle of TSA = 201 cm2 is to be gift-boxed. The dimensions in centimetres of several differently shaped boxes are given below (length × height × width). The box which will best fit the candle with the least amount of wasted space is:A 6 × 6 × 6 B 4 × 4 × 6 C 8 × 4 × 4D 2 × 8 × 4 E 8 × 8 × 8

8 MC The area covered by the rolling pin shown in one complete turn is:A 2π 5(5 + 30) B 2π 2.5(2.5 + 30)C π 2.5(2.5 + 30) D 2π 2.5 × 30E 2π 5 × 30

9 What area of cardboard needs to be purchased to construct a box 1 m × 1 m × 1 mto contain a television set? Allow an extra 5% of the total surface area to cater for overlaps.

10 a What is the name of the shape (a type of prism) of this chocolate package?

b In your mind, unfold the 3-dimensional shape to its 2-dimensional net. Draw this net, making no allowance for overlap.

c Using the dimensions given, calculate the approximate TSA of the chocolate package. (Round all values to 1 decimal place.)

11 WE10 Certain medicines come in capsules, as shown at right. Find the area of plastic (in square millimetres) needed to produce one such capsule. (Ignore overlap).

12 A commercial bread bin has dimensions as shown in the diagram at right. Find the TSA of the bin.

32 cm

60 cm

30 cm

5 cm

19.5 cm 3.5 cm

10 mm

5 mm

60 cm

90 cm


WorkSHEET 10.1Investigation

doc-9613Cone heads


10E VolumeThe volume of an object is the amount of space that the object occupies.

Volume of a prism = cross-sectional area × height of the prismV = AH

The height is the dimension perpendicular to the cross-sectional area.

Shape Cross-sectional shape Volume

Cylinder

H

r

r

Area = π r2

V = area of a circle × height= π r2 × H

Triangular prism

bHh

b

h

Area = 1

2bh

V = area of a triangle × height=

1

2bh × HNote: Lowercase h represents the height of the triangle.

Rectangular prism

WL

H

W

L

Area = L × W

V = area of a rectangle × height= L × W × H

Cube

L

HL

Area = L2

V = area of a square × height= L2 × H= L2 × L= L3

(since in a cube, H = L)

WORKED EXAMPLE 11

Find the volume of the shape shown correct to 1 decimal place.

THINK WRITE

1 Identify the shape. Triangular prism

2 Write the appropriate formula for the volume. V =

1

2bh × H

3 Allocate values to the pronumerals keeping in mind that b and h are the base and height of a triangular cross-section or base of the prism, while H is the height of the prism.

b = 2.6, h = 2.3, H = 3.2

4 Substitute and evaluate, rounding the answer to 1 decimal place. Include the units.

V = 1

2 × 2.6 × 2.3 × 3.2

= 9.568≈ 9.6 m3

2.6 m

3.2 m2.3

m


Odd-shaped prismsThe object at right is a prism, because horizontal cuts show a uniform cross-section. To calculate the volume of this prism you would need to be given the value for the cross-sectional area and the height.

WORKED EXAMPLE 12

Find the volume of the shape shown at right.

THINK WRITE

1 Write the general formula for the volume of a prism.

V = A × height

2 Allocate the pronumerals a value. A = 32 m2, H = 8.2 m

3 Substitute the values of the pronumerals into the formula and evaluate.

V = 32 × 8.2= 262.4

4 Write the answer including units. V = 262.4 m3

PyramidsA pyramid has a flat base at one end and a point at the other.

Square pyramidCone Rectangular pyramid Triangular pyramid

A pyramid occupies exactly one third of the space taken by a prism of the same base and height.

Volume of a pyramid = 1

3 × area of base × height of object

= 1

3 × A × H

The following table shows the formulas for the volume of some common pyramids.

Shape Flat end (base) shape Volume

Cone

r

H

rV =

1

3 × area of a circle × height

V = 1

3π r2 × H

Square pyramid

H

L

L

V = 1

3 × area of a square × height

V = 1

3L2 × H

Height

Cross-section

8.2 m

Area = 32 m2



Rectangular pyramid

H

LW

W

L

V = 1

3 × area of a rectangle × height

= 1

3L × W × H

Triangular pyramid

H

hb

h

b

V = 1

3 × area of a triangle × height

V = ×⎛⎝⎜

⎞⎠⎟bh H1

31

2

Note: Lowercase h represents the height of the triangle.

SpheresThe volume of a sphere is given by the following formula:

Volume of a sphere = 4

3π r3

where r is the radius of the sphere.

A hemisphere is half of a sphere. Its volume, therefore, is half of the volume of a sphere.

Volume of hemisphere = 1

2 (volume of sphere)

= 1

2 × 4

3π r3

= 2

3π r3

Composite solidsThe volumes of the individual objects need to be found before they are added or subtracted to fi nd the total volume.

WORKED EXAMPLE 13

Find the volume (in m3) of the toy shown, correct to 1 decimal place.

THINK WRITE

1 Identify the components of the shape. Total volume = volume of a cone + volume of a hemisphere

2 Determine the volume of the hemisphere.(a) Write the formula.(b) Allocate values to the pronumerals.

Note that some units are in cm, others in m. As the answer requires cubic metres, change centimetres to metres.

(c) Substitute the value of r into the formula and evaluate.

Volume of hemisphere:

V = 2

3π r 3

r = d ÷ 2= 80 ÷ 2= 40 cm= 0.4 m

V = 2

3π (0.4)3

= 0.134 041 3 m3

r

r

80 cm1.4 m



3 Find the volume of the cone.(a) Write the formula.(b) Allocate values to the pronumerals.

(Height = 1.4 − radius of circle)

Volume of cone:

V = 1

3π r2 × H

r = 0.4 m, H = 1.4 − r= 1.4 − 0.4= 1 m

4 Substitute the values of H and r into the formula and evaluate. V =

1

3π (0.4)2 × 1

= 0.167 551 6 m3

5 Add the individual volumes together to fi nd the total volume of the given shape.

Total volume = 0.134 041 3 + 0.167 551 6= 0.301 592 9

6 Round your answer to 1 decimal place and include units.

V = 0.3 m3

Exercise 10E Volume 1 Match the volume formula with the appropriate 3-dimensional object.

a

Hr

b

A cm2

H c

H

L

H

d

H

h

b

a e

H

r

f

H

bh

i V = 1

2bh × H ii V = 1

3π r2 × H iii = + ×⎡⎣⎢

⎤⎦⎥

V a b h H( )1

2

iv V = π r2 × H v V = 1

3L2 × H vi V = A × H

2 WE 11 Find the volume of each of the following shapes correct to 1 decimal place.

a 6 cm8 cm

10.2cm

b

7.4 cm

c 10 cm

4 cm

d

22.4 cm

12 cm

e

18.5 cm

12.6 cm

DIGITAL DOCdoc-9615

SkillSHEET 10.4Volume


3 WE12 Find the volume of each of the following shapes.

a

8 cmShadedarea = 42 cm2

b

3.5 cm

Shadedarea = 23 cm2

c

Shaded area= 116 mm2

10.5 mm

d

Shaded area = 55 cm2

0.5 m

4 Alexander is ordering a concrete base (in the shape of the trapezoidal prism shown at right) for his favourite garden sculpture. How much will he have to pay if concrete costs $50 per cubic metre and the cost of labour is $45?

5 WE13 Find the volume of each of the following shapes correct to 1 decimal place.a

1.4

m

2 m 1.8 m

60 cm b

113

cm

64 cm 22 cm

c

25 cm

10 cm

15 cm

6 The diagram at right shows 3 tennis balls packed in a cylindrical container. Find:a the volume of each ballb the volume of the cylinderc the volume of space that remains free.

7 A chocolate company wants to make chocolate Christmas ball decorations containing 50 small, candy-coated chocolates to hang on Christmas trees. If each candy-coated chocolate has a volume of approximately 0.8 cm3, what is the diameter of the Christmas ball required to contain them?

8 A large apple takes up approximately 512 cm3 of space. I have 160 apples to pack into one of the following containers. With the aim of having the minimal amount of wasted space, which container would be best for this purpose and why?

a

40 c

m

38 cm60 cm

b

70 cm

38 cm 40 cm

c

34.2 cm

38cm

42 cm

40 cm

9 The flower vase shown below is from a designer shop. Calculate the total volume of the vase, rounding all calculations to 1 decimal place as you go.

18 cm

24 cm

12 cm

DIGITAL DOCdoc-9616SkillSHEET 10.5Conversion of units — volume

1.2 m

60 cm

50 cm

50 cm

7 cm

DIGITAL DOCdoc-9617SkillSHEET 10.6Finding unknown lengths

DIGITAL DOCSdoc-9618InvestigationCone volumedoc-9619Cone volume


10F CapacityThe capacity of a container refers to the amount that it can hold. The capacity or volume of a container is usually measured in cubic units; however, when the volume of a liquid is being discussed it can be referred to in terms of millilitres, litres and kilolitres.

Recall the following facts:1000 millilitres (mL) = 1 litre (L)

1000 litres = 1 kilolitre (kL)

Cubic units are related to the fl uid capacity units as follows:

1 cm3 = 1 mL1000 cm3 = 1 L

1 m3 = 1000 L = 1 kL

WORKED EXAMPLE 14

Convert:a 400 cm3 to mL b 1200 cm3 to mL and to L c 2 kL to m3.

THINK WRITE

a Since 1 cm3 is equivalent to 1 mL, then 400 cm3 is equivalent to 400 mL.

a 400 cm3 = 400 mL

b 1 Each 1 cm3 will hold 1 mL of liquid. Therefore, 1200 cm3 will hold 1200 mL of liquid.

b 1200 cm3 = 1200 mL

2 To change mL to L, divide by 1000 (since there are 1000 mL in 1 L).

= 1.2 L

c One kL is equivalent to 1 m3. Therefore, 2 kL is equivalent to 2 m3.

c 2 kL = 2 m3

To fi nd the capacity of a container in litres, fi nd its volume in cubic units fi rst and then convert.

WORKED EXAMPLE 15

Find the capacity in mL of a rectangular container measuring 10 cm × 12 cm × 14 cm.

THINK WRITE

1 Find the volume of the container in cm3. (Since the container is rectangular, use the formula for the volume of a rectangular prism.)

V = l × w × h= 10 × 12 × 14= 1680 cm3

2 Change cubic centimetres to millilitres. 1680 cm3 = 1680 mL

3 Write the answer in words. The capacity of the given container is 1680 mL.

Exercise 10F Capacity 1 WE 14 Convert the following units as indicated.

a 750 cm3 = _____ mL b 800 cm3 = _____ L

c 2500 cm3 = _____ mL d 40 000 cm3 = _____ L

e 6 m3 = _____ cm3 = _____ mL = _____ L f 12 m3 = _____ L

g 4.2 m3 = _____ kL h 7.5 m3 = _____ kL = _____ L

i 5.2 mL = _____ cm3 j 6 L = _____ cm3

k 20 L = _____ mL = _____ cm3 l 5.3 KL = _____ m3

DIGITAL DOCdoc-9620Capacity


2 WE 15 How many millilitres will a rectangular drink container hold if its dimensions are 11 cm × 4 cm × 15 cm?

3 A hemispherical bowl with a diameter of 30 cm will be used to hold a pre-mixedfruit drink for a party. If you want to fill it to the brim, how many litres of pre-mixed drink can you pour in?

4 One litre of orange and mango juice is packed in a rectangular container of height 172 mm and width 93 mm.Find the length of the container to the nearest mm.

5 A plastic bottle contains 1.25 L of soft drink.a How many cylindrical cups 10 cm high and 7 cm in diameter can be fi lled to

capacity from this bottle?b What is the volume (to the nearest mL) of the drink that remains

in the bottle?

6 A rectangular swimming pool measures 4 m by 3.5 m by 2.5 m.a What is the capacity of the pool in kL?b If the pool is being fi lled at a rate of 14 L per second, how long

will it take for it to be: i 70% full? ii fi lled to capacity?

7 A tub in the shape of a trapezoidal prism has dimensions as shown. If the capacity of the tub is 943.5 L, how deep is it?

10G Similar figuresImagine that you wanted to enlarge ΔABC by a factor of 2. You could place it under a photocopier set at 200%. Or you could draw it yourself.First draw A′B′ 6 cm.Make ∠A′B′C′ equal to 90°.Draw B′C′ 8 cm.Join A′C′.Checking, you will fi nd that A′C′ is 10 cm.

The triangles have identical angles, but the sides in the second triangle are twice as large.Figures like these, which have exactly the same shape are called similar fi gures.We say that ΔA′B′C′ is similar to ΔABC, and write ΔA′B′C′ ~ ΔABCΔABC has been enlarged by a scale factor of 2.Check the area of each fi gure.

ΔABC: = ×

=

Area3 4

26

ΔA′B′C′: = ×

=

Area6 8

224

The area has been increased by a factor of 22.In general, if a fi gure is enlarged by a factor of a, then its area will increase by a factor of a2.

The following example shows how a fi gure can be enlarged from a fi xed point.

WORKED EXAMPLE 16

Enlarge this shape by a scale factor of 2.

30 cm

172 mm

93 mml

2.2 m

1.5 m

d

0.85 m

C′

A′

B′

6 cm

8 cm

C

A

B

5 cm3 cm

4 cm

A B

D C


THINK WRITE/DRAW

1 Select a point inside the shape, say, point O and draw lines from this fi xed point to the shape’s vertices (that is points A, B, C and D).

A B

D

O

C

2 Measure the lengths of OA, OB, OC and OD. Note: The measurements are in cm.

OA = 0.95, OB = 1.05, OC = 1.75,OD = 1.75

3 Multiply each of the measurements by a scale factor of 2.

O′A′ = 0.95 × 2 O′C′ = 1.75 × 2= 1.9 = 3.50

O′B′ = 1.05 × 2 O′D′ = 1.75 × 2= 2.1 = 3.50

4 Extend the lines to the new measurements; that is, continue line OB until it measures 2.1 cm, line OC until it measures 3.5 cm and so on.

A B

C

O

D

B'

C'D'

A'

5 Join the ends of the extended lines to form the similar fi gure. Label the vertices appropriately.

The following example shows how to reduce a fi gure from a fi xed point.

WORKED EXAMPLE 17

Reduce the pentagon labelled ABCDE by a scale factor of 2.

THINK WRITE/DRAW

1 Select a point outside the shape and label it O. Draw lines from this fi xed point O to the shape’s vertices and measure their lengths.Note: The measurements are in cm.

A

B

CD

O

E

OA = 6.8, OB = 7.6, OC = 7.1, OD = 5.8, OE = 5.62 Divide these measurements by

the scale factor of 2. OA′ = 6.8

2 OC′ = 7.1

2 OE′ = 5.6

2

= 3.4 = 3.55 = 2.8

OB′ = 7.6

2 OD′ = 5.8

2

= 3.8 = 2.93 Locate the vertices of a reduced

pentagon by measuring new distances (that is OA′, OB′ and so on) from point O along the existing lines.

A

B

CD

O

EA'

B'

C'D'

E'

4 Join the vertices A′, B′, C′, D′ and E′ to form a reduced fi gure.

A

B

CD

E



Exercise 10G Similar figures1 WE16 Enlarge each of the following shapes by the given scale factor.

a Scale factor = 3

A

C D

BO

b Scale factor = 1.6

A

O

B

CF

E D

2 WE17 Reduce the following shape by the given scale factor.Questions 3 to 7 will help you to investigate the area and volume ratios of similar figures.

3 Which of the following figures are similar?a All rectanglesb All squaresc All pentagonsd All circlese All equilateral trianglesf All isosceles trianglesg All right-angled trianglesh All isosceles right-angled triangles

Mathematically it can be shown that if the side lengths of two similar figures are in the ratio a:b then the areas of the similar figures are in the ratio a2:b2. Squares are always similar figures. For the two squares shown:if the side length ratio = 2:6then the area ratio = 22:62

= 4:36= 1:9.

Therefore, if the area of the smaller square is 4 cm2, then the area of the larger square is:

large square areasmall square area

= scale factor (ratio)

large square area4

= 9

1large square area = 36 cm2

4 Copy the similar rectangles at right into your workbook.a Copy and complete the following statements and calculations.

Smaller:Larger Smaller:Larger

Side length ratio 2: ? or ? :10

Area ratio 22: ? 2 ? 2:102

= 4: ? = ? :100

= ? :25

It does not matter which pair of dimensions you choose; the area ratio will finally work out to be the same.

b If the area of the larger rectangle is 50 cm2, use the area ratio to calculate the area of the smaller rectangle:

area of small rectanglearea of large rectangle

= scale factor

area of small rectangle50

= 4

25area of small rectangle = ? cm2

Check: Area of rectangle = L × WL = 4, W = 2 = 4 × 2

= ? cm2

Scale factor = 2

62

104

252

1


5 a Show that ΔABC ∼ ΔDEF.b What is the ratio of the two triangles’ lengths in simplest form?c What is the ratio of their areas?d If the area of ΔABC is 6 cm2, use the area ratio to show that the area

of ΔDEF is 13.5 cm2.

6 Draw the following cylinders into your workbook.

10 cm6 cm

r = 5 cm

r = 3 cm

a Copy and complete the following statements and calculations.

Using the height dimensions: Using the radius dimensions:

Side length ratio ? :10 or 3: ?

Volume ratio ? :103 or 33: ?

(Evaluate) ? :1000 or 27: ?

(Cancel) ? :125

b If the volume of the larger cylinder is approximately 785.4 cm3, use the volume ratio to calculate the volume of the smaller cylinder to 1 decimal place:

Volume of small cylinderVolume of large cylinder

= scale factor

?Volume of large cylinder

= 27

125

Volume of small cylinder = ? cm3

c Check this value by using the formula: volume = area of cross-section × height.

7 a What is the volume ratio (in simplest form) between the cones shown at right?

b If the volume of the larger cone (to 1 decimal place) is 56.5 cm3, use the volume ratio to calculate the volume of the smaller cone to 1 decimal place.

c Check your answer using the formula for the volume of a cone.

10H Similar trianglesThere are three formal test rules that can be applied to triangles to see if they are similar.

69

A

B C8

34.5

D

E F440°

80°60°

G

H I40°

80°

60°

Test 1 (AAA)Triangles are said to be similar if all the corresponding angles are equal. That is, the three angles in one of the triangles are equal to the three angles in the other triangle.

ΔDEF ∼ ΔGHI, because ∠D = ∠G ∠E = ∠H ∠F = ∠I

Note: If two pairs of corresponding angles are equal, then the 3rd pair is also equal.

7.5

D

EC

A

B F

6

4.5

54

3


WorkSHEET 10.2doc-9622

InvestigationMaking an aeroplane

3 cm

6 cm

2 cm

4 cm

INTERACTIVITYint-0811

Similar triangles


Test 2 (SSS)Triangles are said to be similar if the ratios between the corresponding side lengths are equal.ΔABC ∼ ΔDEF, because the ratio of corresponding side lengths is 2:

= = = =AB

DE

9

4.52,

AC

DF

6

32, and = =

BC

EF

8

42

Test 3 (SAS)Triangles are said to be similar if two of their corresponding sides are in the same ratio, and the angle between these two sides (the included angle) is the same in both triangles.

ΔABC ∼ ΔDEF,

= =EF

BC

24

83 and = =

ED

AB

18

63; ∠B = ∠E

Sometimes it is hard to decide whether the two triangles are similar or not, because they are not orientated the same way. In such cases it is helpful to re-draw the triangles so that the sides and angles that we think might be corresponding are in the same order.

WORKED EXAMPLE 18

Compare each of the trianglesat right with ΔABC and state whether they are similar, or if there is not enough information given for a decision to be made. Justify your answers.a Δ DEFb ΔGHIc Δ JKLd Δ MNO

THINK WRITE/DRAW

a 1 Re-draw ΔDEF so that its angles correspond to those of ΔABC.

a

110°40°30°

30° 40°

110°D

FE

A

B C20

6.6 6

10

2 Compare the corresponding angles and write what you observe.

∠A = ∠D∠B = ∠E∠C = ∠FAll corresponding angles are equal.

3 State whether the triangles are similar and specify the test upon which your conclusion is based.

ΔABC ∼ ΔDEF (AAA).

b 1 Re-draw ΔABC and ΔGHI so that angles/sides correspond.

bH

G I

4.5

7.510

66.6

CB

A

110°110°

40°40°30°

2 Compare the side measurements.=

AC

HI

6

4.5= 1.333 and

=BC

GI

10

7.5= 1.333

18

D

E F2460°

6

A

B C860°

Included angle

110°

110° 110°

30°30°

40°40°

40°

110°

110°

N

A

B

C

E

F DO

M20

6.6

6.6

J

K

L

10

6

10

20

7.5

4.5

G

IH


3 Compare the included angles. ∠C = ∠I = 40°

4 State and justify your conclusion. ΔABC and ΔGHI are similar (SAS).

c 1 Re-draw ΔABC and Δ JKL so the angles/sides correspond.

c

30° 40°

110°A

B C

6.6 6

10

110°

K

J L

6.6

10

2 Look at the side ratios. =

=

BC

JL

10

10

1

and =

=

AB

KJ

6.6

6.6

1

3 Check the included angles, ∠ABC and ∠KJL.

We do not know ∠KJL.

4 State your conclusion. We are unable to show that the triangles are similar.

d 1 Re-draw ΔABC and ΔMNO so the angles/sides correspond.

d

110°

30° 40°110°

M

ON

A

B C20

6.6 6

10

2 None of the three tests (AAA, SSS or SAS) can be performed since we know the measurements of only one side and one angle in the ΔMNO. State this in writing.

We are unable to determine whether the triangles are similar.

Being able to determine the corresponding angles in a triangle is vital to all three tests on similarity. In some cases the actual value of one of the angles is unknown, but by mathematical deduction its size can be found. This requires that you remember your angle properties. Some of these properties are shown in the following table.

Angle description Diagram

1. When two lines intersect, they form vertically opposite angles, which are equal.

a°b°b°

a°

2. When a transversal cuts parallel lines, a number of angles are formed.

Transversal

Parallel lines

3. Alternate angles are on opposite (alternate) sides of the transversal and are always equal. a°

a°b°

b°

4. Corresponding angles are always equal.

a°c° d°

d°b°a°

b°c°

5. Co-interior angles lie on the same side of the transversal and within the parallel lines. They always add up to 180°.

c°b°

a° a° + b° = 180°c° + d° = 180°d°


WORKED EXAMPLE 19

Show that ΔABC ∼ ΔADE.

THINK WRITE/DRAW

1 Draw the triangles separately with angles marked and side measurements shown.

7.2

A E

D

11.25

15

a°

c°b°

4.8

A

B

C10

7.5

b° a°c°

2 State the corresponding pairs of angles which are equal in size. Specify the reason.

∠A = ∠A (shared)∠B = ∠D (corresponding)∠C = ∠E (corresponding)

3 Write your conclusion. ∴ ΔABC ∼ ΔADE using AAA test.

If, in a pair of similar triangles, the lengths of at least one pair of corresponding sides are given, the ratio (scale factor) can be established. It can then be used to fi nd missing lengths in one of the triangles, provided that the corresponding lengths in the other triangle are known.

WORKED EXAMPLE 20

Find the value of the pronumeral in the fi gure at right.

THINK WRITE

1 Establish whether the triangles ABC and ADE are similar by applying the AAA test.

∠A = ∠A (shared)∠B = ∠D (corresponding)∠C = ∠E (corresponding)∴ ΔABC ∼ ΔADE (AAA test)

2 Since the triangles are similar, their corresponding sides are in the same ratio.

=

= x

AE

AC

15

10AD

AB 7.5

3 Solve for x. =

= ×

=

x

x

15

10 7.515 7.5

1011.25

Note: When forming an equation, place the unknown value in the numerator to make calculation steps easier or use a CAS calculator to solve the equation.


4.87.2

A E

BD

C

3.75

10 5

7.5

A EC

BD

x

1015

7.5


Exercise 10H Similar triangles 1 WE18 For each of the following, compare ΔABC and ΔEFD and state whether they are similar, or

there is not enough information given for a decision to be made. Justify your answers.

a 120°

20°

120°

40°

20°

8.25

5.5

6 4

C

E

A

B

D

F

b

A

B CE

F

D

40°

40°115°

25°6 cm

3 cm 3 cm 6 cm

c

A

5.625

3.375

7.8752.5

1.5

3.5

B

C

E

D

F

d 6

6

12

8

A

B

E

F

D

C

40°

110°

30°

2 Calculate the size of the missing angles. Justify your answer.

a

24° a°c°

d°b°

b

68°a°

c°b°

c

32°a°

c° d°e°b°

d 78°

a°

c°

d° e°

b°e

99°

63°

x°

y°

f

58°32°a°

c°b°

d°

e°

3 MC The magnitude of angle a° is:A 30° B 180° − 30° C 60°D 180° − 60° E 70°

4 MC The magnitude of angle a° is:A 64° B 128° C 52°D 154° E 104°

5 WE19 Show that ΔABC is similar to ΔDBE because:a the corresponding angles are equalb the corresponding side ratios are equal (to a value other than 1).

6 In each of the following diagrams, find and re-draw two triangles that are similar. Give reasons for your answer.a A

B

CE D

b A

E

D

B

C

c A

ED

C

B

d

A

D

B

E

C

e A

D

C B

30°a°

64°

a°

6.8 cm

B

A C

ED

6 cm

4.2 cm 3.5 cm

5 cm

4 cm


7 WE20 Find the values of the pronumerals (to 1 decimal place).a 37 cm

31 cm

6.2 cm

9.6 cmx

y *

*

b 9.6

2.412.4

x y

c

5E

C

A

BD

x

18 12

yd A

B

E

DC615

54y

x

e

3.4

m

1.7 m

A

E

B

DC x 4 m

f

2 cm30°

y°

x°

9 cm

3 cm z

A

D E

B C

8 MC The side lengths of triangle ABC are 18 mm, 24 mm and 30 mm. Triangle DEF is similar to triangle ABC and its shortest side is 12 mm. The perimeter of the triangle DEF is:A 48 mm B 52 mm C 64 mmD 72 mm E 108 mm

9 At your 18th birthday party you want to show slides of you and your friends growing up. The diagram at right shows the set-up of the projector lens, slide and projector screen.If: (a) a slide is 5 cm × 5 cm (b) the distance from A to B is 10 cm (c) the projector screen is 1.5 m × 1.5 mhow far horizontally from the screen do you need to place the projector’s lens so the image just covers the entire screen?

10I SymmetryA 2-dimensional shape may be described as having either line symmetry and/or rotational symmetry.

Line symmetryIf it is possible to cut a 2-dimensional shape in such a way that it is divided into two mirror images, then it possesses line symmetry. The line that allows this to occur is called the axis of symmetry.

Shapes can have more than one axis of symmetry:

One axis ofsymmetry

Two axes ofsymmetry Three axes of

symmetry

Infinite numberof axes ofsymmetry

An easy way to check whether the line is an axis of symmetry of a certain shape is to fold the shape along that line. If the two parts coincide (that is, the shape folds onto itself), the line is an axis of symmetry.

AB

SlideLens

Projectorscreen


WORKED EXAMPLE 21

Which of the dotted lines in this fi gure is an axis of symmetry?

THINK WRITE

1 Visualise the triangle as being folded along each dotted line. If the triangle folds onto itself the line is an axis of symmetry.

2 AB is an axis of symmetry. CD and EF are not.

E

F

AB is an axis of symmetry.

Rotational symmetryIf a shape can be rotated about its centre so that sits exactly on itself in less than one complete revolution of 360°, then it is said to have rotational symmetry. The number of times a shape sits on itself in one rotation is called its order of rotational symmetry.

For example, a square possesses rotational symmetry. By marking one corner and rotating the square in a clockwise direction around point × (its centre of rotation), it can be seen that in a turn of 90°, the square appears as it did originally.

Original Rotated onitself 90°

Rotated onitself 180°



=×

With each subsequent turn of 90° the square appears as it originally did. Within the 360° revolution there were four times the square appeared as its original, so a square’s order of rotational symmetry is said to be 4.

An equilateral triangle has an order of rotational symmetry of 3.

=

120°

120°120°

The principles of line and rotational symmetry can also be applied to 3-dimensional objects.

Axes of symmetryAn axis of symmetry is a line about which an object can rotate so that it assumes positions identical with those of the original. Like rotational symmetry in 2-dimensional shapes, a 3-dimensional shape can be described by its order of rotational symmetry. This is the number of times the shape assumes an appearance identical with that in the original position, within one revolution of 360° about its axis of symmetry.

A square-based pyramid has an order of rotational symmetry of 4.In a square-based pyramid there is only one axis of symmetry; however, other

3-dimensional objects can have none or many.

A

B

C

E

D

F

TUTORIALeles-1390Worked example 21A

B

C D


TessellationsThere are many examples in design, art and architecture where shapes are used to cover a surface with patterns which leave no gaps. These patterns are referred to as tessellations.

Tessellations can be divided into three main categories: regular, homogeneous (or semi-regular) and non-homogeneous (irregular).1. In regular tessellations, the pattern is made by only one type of regular polygon.2. In homogeneous tessellations, the pattern is made up of different types of regular polygon.3. In non-homogeneous tessellations, any type of shape can be used.The subject of ‘tessellations’ is also presented in Chapter 6 ‘Geometry in two and three dimensions’ (pages 253–55).

Non-homogeneous (irregular) tessellationsThis category is not restricted to using regular polygons, so any shapes that will fit together to cover a plane surface can be used. The shapes still need to form a definite pattern that is repeated to cover the surface.

Exercise 10I Symmetry 1 WE21 Which of the dotted lines in each of the figures shown is an axis of symmetry?

a A

B

C D

GE

H F

b A

B

G

D

H

E

C F

c A

B

G

D

H

E

C F

d A

B

E

H

F

C

GD

2 Copy the following figures into your workbook and draw in all the possible axes of symmetry.

a b c

d e f

3 a Draw a regular pentagon with sides 1.5 cm in length and internal angles all accurately measuring 108°.

b Draw in all axes of symmetry.The point where these lines intersect in a regular polygon is known as the centre of rotation. Use this point to rotate the pentagon.c Through what angle must the pentagon be rotated before it superimposes itself for the first time?

(It is important to measure the angle very carefully to the nearest degree.)d How many times does the pentagon superimpose itself when turned 360°?e What is its order of rotation?


4 For each of the following shapes, state its order of rotational symmetry.

a b c

d e f

5 Make a square pyramid by copying onto paper or thin cardboard, the net for the square pyramid shown.

Cut around the edge of the net, fold along the dotted lines and then stick the sides together with adhesive tape.a How many axes of symmetry does a square pyramid have?b What is its order of rotational symmetry?

6 Create a table with the following headings.

Vertical symmetry

Horizontal symmetry

Rotational symmetry None

Determine to which column or columns each of the capital letters of the alphabet (shown below) belongs.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z


SummaryPythagoras’ theorem in two dimensions

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

c2 = a2 + b2

Pythagoras’ theorem in three dimensions

Use the following steps when solving 3-dimensional problems.Step 1. Identify the length that needs to be found.Step 2. Identify the triangle that contains the unknown length.Step 3. Re-draw this triangle in two dimensions.If the triangle containing the unknown length has other measurements missing, fi nd these missing values fi rst from other triangles within the fi gure.Unless stated in the problem, do not round calculations until the very last value is calculated and then give an answer with the appropriate number of decimal places.

Perimeter and area Generally:– Unless specifi ed, do not round units until the end of the solution.– If the measurements in question include units, include them in your fi nal answer.– The perimeter is the distance around a closed 2-dimensional shape.– Area refers to the surface enclosed by the boundaries of a 2-dimensional shape.


Square

L

A = L2 P = 4L

Rectangle L

W

A = L × W P = 2(L + W)

Parallelogram

b

h

A = b × hwhere the height measurement must be at a right angle to the base measurement.


Trapezium

b

h

a A = 1

2(a + b)hwhere the height measurement must be at a right angle to the base measurement.


Rhombus x

y –

––

–

A = 1

2 x × y P = sum of all sides

a

b

c

Hypotenuse



Triangle

b

hh

b

h

b

A = 1

2 bhwhere the height measurement must be at a right angle to the base measurement.


Triangle

a

bcA = − − −s s a s b s c( )( )( )

where s = 1

2(a + b + c)(Use when height measurement is unknown.)

P = a + b + c

Circle

rd

A = π r2 C = π d orC = 2π r

An annulus is the area between two concentric circles. Area of annulus: A = π (R2 − r2), where R is the radius of the larger circle and r is the radius of the smaller circle.A sector is formed by two radii of the circle and an arc.

Area of a sector A = θ πr

3602

Arc length of a sector = θ π× r

3602

where θ is the angle of a sector and r is the radius of the circle.To fi nd the area of composite shapes, calculate the area of the individual shapes fi rst.

Total surface area (TSA)

Total surface area is the sum of the surfaces of a 3-dimensional object.

Object Net TSA

Cube

L

1 3

2

6

L

L

4 5

TSA = 6L2

Rectangular prism

wl

hw

hh

w

l

l

h



Object Net TSACylinder

hr h

r

r

2πr

TSA = area of 2 circles+ curved surface

= 2π r2 + 2π rh= 2π r(r + h)

Sphere

r

Not shown TSA = 4π r2

Cone

S = Slantheighth

r

S

r2 rπ

TSA = area of base (circle)+ area of curved

surface

= π r2 + ππ

r

S

2

2 × π S2

= π r2 + π rS= π r(r + S)

Square-based pyramid

b

hb

h

TSA = area of square+ area of 4 triangles

= b2 + 4 × ⎛⎝⎜

⎞⎠⎟bh1

2= b2 + 2bh

Formulas for all types of objects are not possible. For those objects without a formula:1. draw the net of the object 2. work out the different shapes that make up the net3. calculate their individual areas 4. add all the individual parts together.

Volume The volume of an object is the amount of space that the object occupies. It is measured in cubic units.Prisms are 3-dimensional objects with uniform cross-sections and parallel sides. The cross-section is represented by the ends of the prism. The height is the dimension perpendicular to the cross-sectional area.

Volume of a prism = cross-sectional area × height of the prism

Shape Cross-sectional shape VolumeCylinder

H

r

r

Area = π r2

V = area of a circle × height= π r2 × H

Triangular prism

bHh

b

h

Area = 1

2bh

V = area of a triangle × height

= 1

2bh × H



Shape Cross-sectional shape VolumeRectangular prism

WL

H

W

L

Area = L × W

V = area of a rectangle × height= L × W × H

Cube

L

HL

Area = L2

V = area of a square × height= L2 × H= L2 × L= L3

(since in a square, H = L)

Volume of a pyramid = 13 × area of base × perpendicular height


Cone

r

H

r

V = 1

3 × area of a circle × height

V = 1

3π r2 × H

Square pyramid

H

L

L

V = 1

3 × area of a square × height

V = 1

3L2 × H

Rectangular pyramid

H

LW

W

L

V = 1

3 × area of a rectangle × height

= 1

3L × W × H

Triangular pyramid

H

hb

h

b

V = 1

3 × area of a triangle × height

V = ×⎛⎝⎜

⎞⎠⎟bh H1

31

2


Do not confuse the lowercase h in the formula for a triangular prism/pyramid or trapezoidal prism with the uppercase H. Uppercase H = total height of the prism or pyramid, while lowercase h is the height of the triangle, parallelogram of a trapezium.

For a sphere: Volume = 4

3 π r3


Capacity The capacity of a container is the amount that it can hold.To fi nd the fl uid capacity of a container, fi nd its volume in cubic units fi rst and then convert, using the following rules:

1 cm3 = 1 mL, 1 m3 = 1 kL = 1000 L

Similar figures Similar fi gures have the same shape but different sizes. The corresponding angles in each similar fi gure are the same; however, the corresponding side lengths differ by a scale factor.

Similar triangles Triangles are similar if:1. all corresponding angles are equal (AAA), or2. all corresponding sides are in the same ratio (SSS), or3. two sides are in the same ratio and the included angles are equal (SAS).When comparing triangles it is best to have them drawn with the same orientation.When forming an equation to solve for a missing dimension, place the unknown value in the numerator to make the calculation steps easier.

Symmetry The line that divides a 2-dimensional shape into two parts that are mirror images of each other is called the axis of symmetry.A 2-dimensional shape can be classifi ed by the number of lines of symmetry it possesses and the number of times the shape superimposes itself when rotated around a fi xed point, called the order of rotational symmetry.A 3-dimensional shape can be classifi ed by the number of planes of symmetry it possesses and the number of times the shape superimposes itself when rotated around the chosen axis of symmetry (its order of rotational symmetry).


Chapter review 1 Examine the diagram.The value of the pronumeral to 1 decimal place is:

A 6.2 B 6.0C 5.8 D 5.6E 5.4

2 The longest stick that can be placed inside a cube with side length of 2 m has the length (to 2 decimal places) of:A 3.46 m B 3.44 m C 3.50 m D 3.48 m E 3.52 m

3 The perimeter of a rectangle with an area of 56 cm2 and one side of 8 cm is:A 30 cm B 7 cm C 15 cm D 28 cm E 26 cm

4 The area of a one-metre wide gravel path around a lawn 25 m × 15 m is:

A 84 m2 B 80 m2 C 78 m2 D 76 m2 E 82 m2

5 Examine the diagram.The height of the shape shown is:

A 23.5 cm D 20.5 cmB 19.5 cm E 15.5 cmC 25.5 cm

6 The area of the shaded section in this diagram is:

A 12.2 cm2 D 11.0 cm2

B 141.4 cm2 E 1414 mm2

C 1110 mm2

7 The TSA of a rectangular prism is 126 cm2. If the base of the prism is 6 cm × 5 cm, what is its height?

A 2.5 cm B 3 cm C 3.4 cm D 3.5 cm E 4 cm

8 Examine the diagram at right. The TSA of the solid shown is given by:

A 3π m2 B π m3

2

2

C π ⎛⎝

⎞⎠

m5

2

2

D 34π m2 E π ⎛

⎝⎞⎠

m2

2

2

9 Examine the diagram at right. The volume of the shape shown is given by:

A 1

3π m2n + 4

3π m2 B 2

3π m2 + π m2n C π m2n

D 1

3π n(m2 + 2) E 1

3π m2(n + 2m)

10 A cylindrical container of radius r and height h is being filled with sand using a conical container, also of radius r and height h. How many cones of sand are needed to fill the cylinder to capacity?

A π B 3π C 3 D 1

3 E π3

11 A rectangular swimming pool has a capacity of 1000 kL. If the pool is 50 m long and 2 m deep, how wide is it?A 10 m B 20 m C 25 m D 50 m E 100 m

12 If the volume of a container is 1000 mm3, what is its capacity in mL?A 0.1 mL B 1 mL C 10 mL D 100 mL E 1000 mL

13 The volumes of two similar-shaped cylinders are 1000 cm3 and 3375 cm3. In simplest form, what is the ratio of their surface areas?A 10:15 B 4:9 C 4:6 D 2:3 E 9:4

MULTIPLECHOICE

9

5

12

x

8 cm

16 cm

2 cm

7 mm

m

n

m


14 On the architect’s plan, a house block is 7.5 cm wide, while the actual (real-life) width of the block is 60 m. What is the ratio scale of the plan?A 1:80 B 8:1 C 1:8 D 1:800 E 80:1

15 The length of x is:A 4 cm B 16 cmC 10 cm D 8 cmE 6 cm

16 The lengths of x and y, respectively, are:A 12 m and 14 m B 13 m and 16 mC 18 m and 16 m D 16 m and 18 mE 18 m and 14 m

17 A tree casts a shadow of 10 m, while your 0.45-m wooden stick casts a shadow of 100 cm. What is the height of the tree?A 45 m B 4 m C 4.5 m D 5 m E 5.5 m

18 The number of axes of symmetry of the figure shown is:A 2 B 5 C 7D 3 E 9

SHORT ANSWER

1 To allow people access to the gym, the manager has decided to place a ramp over a set of stairs.What length of timber (to the nearest centimetre) does he need to purchase?

Ramp

27 cm

16.5 cm

2 A 2-m surfboard is to be placed in a locker. Will it fit down the side shown by the line AB or will it have to go diagonally across as shown by the line CB?

3 Find (to 1 decimal place): i the perimeterii the areaof the following shapes.a

8.2 cm

220° 35'b

2 m

c 6 cm

15.67 cm

4 To secure the 2.4-m-high poles of a beach volleyball net you need to attach a guy rope to a ring seven-eighths of the way up each pole. The guy ropes are 2.5 m in length.a Draw a diagram of the situation and include all dimensions given.b How far out (to 1 decimal place) from the base of the poles will the guy ropes go?

5 A designer vase has the shape of a truncated, square-based pyramid. The base of the vase is a square with a side length of 15 cm. The area of the square opening is 70.56 cm2. Each of the four sides is a trapezium with slant sides 9 cm long. Find (to the nearest square centimetre) the total surface area of the vase.

6 If the volume of a cone of height 10 cm is 261.8 cm3, show that this volume is increased by a factor of 8 if the dimensions of the cone are doubled.

7 A cylindrical can of drink holds 375 mL of liquid. If the area of the base is 31.25 cm2, find the height of the can.

x cm

68

12

15

1024 12

xy

A C

DB

1.8 m

0.62 m0.62 m


8 The shapes below have either been enlarged or reduced by a scale factor from a fixed point. In both cases, calculate by what scale factor the original will need to be multiplied to produce the image.a

1 cm

1.5 cm

0.4 cm

b

8 cm

6 cm10 cm

5 cm

Note: The image is the shaded area.

9 Two rectangles are similar. The larger rectangle’s dimensions are 24 cm × 12 cm. If the longest side on the smaller rectangle is 16 cm, calculate the other dimension.

10 A circle with a radius of 21 cm has an area of 1385.4 cm2 and another circle has an area of 153.9 cm2. Using the principles of ratios, what is the radius of this circle to the nearest centimetre?

11 Find the length of x in the diagram shown.

12 a Copy the chessboard at right into your workbook and draw in the lines of symmetry.b Rotate the board around its centre of rotation (the point where the lines of

symmetry cross). What is its order of rotation?

EXTENDED RESPONSE

1 Lena and Alex are renovating their bathroom. Inspired by a creative exhibit from the recent Home Show, they decided to use various geometric shapes in their design. The ‘window wall’ is the first section to be renovated. The old window is to be replaced with two new windows as shown below and then the wall is to be tiled.

1.2 m

2.5 mThe window on the right is in the shape of an equilateral triangle, while the one on the left is in the shape of a trapezium. The shorter parallel side of the trapezium is the same length as the sides of the triangular window and the longer parallel side is equal to 2.5 m. Both windows are to be 1.2 m high.Calculate:a the perimeter of each window to the nearest centimetreb the cost of wood (to the nearest dollar) needed to frame the windows, if the wood is sold at

$13.50 per metrec the area of each windowd the total cost of the glass (to the nearest dollar), priced at $45 per m2.The ‘window wall’ is to be tiled with glass tiles in the shape of a rhombus as shown at right.e Will the tiles tessellate? Give reasons for your answer and draw

a small segment of the pattern to illustrate it.f Small spaces between the tiles are to be fi lled with a special

grout. Find the length (to the nearest centimetre) around each tile that needs to be fi lled with grout.

g Find the area of each tile.h If the ‘window wall’ is 4.6 m × 2.7 m, fi nd the total area that needs to be tiled.i The tiles are sold in boxes of 12. Use the results from parts g and h to fi nd the number of boxes that

Lena and Alex need to purchase. Add an extra 10% for breakages to the number of tiles needed.

12 cm

A

B

CD 16 cm

x

20 cm

30 cm


2 After successfully coping with the ‘window wall’ (see Question 1), Alex and Lena decide to improve other parts of their bathroom. First they want to decorate other walls with mosaic features as shown at right.a Find the area of each feature to the nearest cm2.b If Lena wants 6 of each type of feature, fi nd the total

area of the mosaic needed; then add 10% extra for breakages and cutting.

c Mosaic can be purchased in square sheets 30 cm × 30 cm at $22 each. (Each sheet can then be cut as needed, but only whole sheets can be purchased.) Find the total cost of the mosaic features.

Next Alex wants to replace the old mirror with a new one as shown at right.The mirror is in the shape of a major segment, that is, a circle with a section of it cut off straight. (The ‘cut-off’ section will be complemented by small glass shelves.)d If both the radius and the length of the ‘cut-off’ edge

are 50 cm each, fi nd the total area of the mirror (to the nearest cm2).

e Alex wants to have the edges of the mirror bevelled. This can be easily done at the local glass store at a cost of $20 per metre. Find the cost of the bevelling job.

The most daring feature of the new bathroom is the extremely modern looking shower screen. It can be thought of as a section of the curved surface of a cylinder 1.4 m long. The curved edge is in the shape of an arc, subtended by a 120° angle from the

centre of a circle with a radius of 2

3 m.

f Find the length of the curved edge.g Find the total area of the glass in the

shower screen.Bela the builder advises Lena and Alex that since they want a frameless screen, it has to be made from a special toughened 10-mm-thick glass.h What fraction of a cubic metre is the volume of glass in this unusual shower screen?

3 A chocolate manufacturer has brought out a new selection of chocolate shapes. They are filled with a delicious honey/chocolate liquid.

Cone

3

2 Two triangularprisms joined

together

1.5

60°1.5

Sphere1.5

Cube1.5 2.5

Rectangularprism

1.5

1.51.5

1.5

1.5

3Trapezoidal

prism

All measurements given are in cm.a For each chocolate shape, calculate: i the TSA of chocolate required ii the volume of honey/chocolate liquid required (to 1 decimal place).

40 cm

40 cm 40 cm

25 cm

60°

Mirror50 cm

50 cmShelves

1.4 m

Curvededge

120°

Curvededge

2—3

r =


b If a box contains four of each type of chocolate shape, what is the total area of chocolate needed to make one box?

c Chocolate comes in square sheets of 50 cm × 50 cm. How many boxes can be produced per sheet of chocolate?

d How many millilitres of honey/chocolate liquid are required per box?e If the liquid comes in one-litre containers, how many boxes of chocolate can be produced per litre?

4 The scale drawing at right is of a garden shed.a If the real-life width of the shed is 2 m, what is the

scale factor i in ______ cm to ______ m? ii as a ratio ______:______? iii as a fraction?b Use this scale factor to calculate all the real dimensions

indicated by the pronumerals.c If the shed is to be painted, and this includes every side

except the roof (shown at right): i calculate the TSA to be painted ii how many litres of paint are required if one litre

covers 4 m2?d Find the total amount of space available inside the shed. Give the answer

in cubic metres to 1 decimal place.e The longest stick that can fi t inside the shed will go from the corner on

the fl oor, labelled A, to the opposite corner on top of the roof, labelled C. Find the length of such a stick.

5 The diagram below shows a triangular coffee table.

O

a Enlarge the object by a scale factor of 2.The original design of the tabletop is drawn as shown.

b If the longest side of the actual table is 1.6 m, what is the scale ratio?

c Find the actual lengths of the other two sides of the tabletop.The tabletop is made from marble with a glass insert in the middle. The glass insert is also in the shape of the triangle with side lengths 60 cm × 48 cm × 32 cm.d Are the two triangles (that is, the marble top and the glass insert) similar? Justify your answer.e Does the tabletop have: i line symmetry? ii rotational symmetry?f Design a triangular tabletop with three lines of symmetry and an order of rotational symmetry

of 3. Specify the lengths of the sides of your tabletop.g Draw the plan of your design from part f using a scale ratio of 1:20.

3.75 cm2.4 cm

Width3 cm

3 cm

3.75 cm

3 cm

C

A

(a)(b)

(c) (d)

(e)

(f)

Roof section(not painted)

10 cm

8 cm

DIGITAL DOCdoc-9624

Test YourselfChapter 10


ICT activitiesChapter openerDIGITAL DOC

10 Quick Questions doc-9608:(page 379)

10A Pythagoras’ theoremDIGITAL DOCS

SkillSHEET 10.1 doc-9609: (page 380)doc-9610: (page 380)

10B Pythagoras’ theorem in three dimensionsDIGITAL DOC

doc-9610: (page 382)

10C Perimeter and areaTUTORIAL

WE5 eles-1384:(page 384)

DIGITAL DOCSSkillSHEET 10.2 doc-9611:(page 386)SkillSHEET 10.3 doc-9612:

(page 386)

10D Total surface area (TSA)TUTORIAL

WE 10 eles-1385:(page 390)

DIGITAL DOCSInvestigation doc-9613: (page 392)WorkSHEET 10.1 doc-9614:

(page 392)

10E VolumeTUTORIAL

WE 13 eles-1386:(page 395)

DIGITAL DOCSSkillSHEET 10.4 doc-9615: (page 396)SkillSHEET 10.5 doc-9616:

(page 397)

SkillSHEET 10.6 doc-9617:(page 397)Investigation doc-9618: (page 397)doc-9619: (page 397)

10F CapacityDIGITAL DOC

doc-9620: (page 398)

10G Similar figuresTUTORIALS

WE 17 eles-1387:(page 400)

DIGITAL DOCSWorkSHEET 10.2 doc-9621:

(page 402)Investigation doc-9622: (page 402)

10H Similar trianglesINTERACTIVITY

Similar triangle int-0811:(page 402)

TUTORIAL WE19 eles-1389:

(page 405)

10I SymmetryTUTORIAL

WE21 eles-1390: (page 408)

Chapter reviewDIGITAL DOC

Test Yourself doc-9624:(page 420)

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Answers CHAPTER 10

SHAPE AND MEASUREMENT Exercise 10A Pythagoras’ theorem 1 a 15 b 9.2 c 18.3 d 7.6 e 6.6 f 4.7 2 a 87.7 cm b 0.88 m 3 a a = 0.7 m b b = 23.7 4 Across the park: total distance = 28.98 km

Via the streets: total distance = 32.5 kmUsing the streets will enable the runner to complete 30 + km.

5 a Buoy 1 Buoy 2

S/F

b 1200 m c 20 km

Exercise 10B Pythagoras’ theorem in three dimensions 1 a 22.4 b 25.3 c 14.8 d 27.9 2 a 21.5 b 23.1 c 18.1 d 18.5 3 a 25 cm b 26.9 cm

Exercise 10C Perimeter and area 1 a i 97.98 mm2 ii 48 mm b i 1.4 m2 ii 4.88 m c i 168 cm2 ii 54 cm d i 32 cm2 ii 49.34 cm e i 8.75 m2 ii 13.40 m 2 a i 540 cm2 ii 106 cm b i 60.73 m2 ii 45.71 m c i 29.74 cm2 ii 21.42 cm 3 a C b E c B 4 a i

1.2 m

1.2 m ii 1.1 m2

b i

6 m

2 m

3 m

1 m

ii 22 m2

c 5.29 m 5 Radius = 0.6 m or 60 cm 6 33% 7 30.6 cm2

8 107 cm2

9 The chain is 137.4 cm long, so it’s too short.

10 a 2474 cm2

b 34.4% c 110 cm

Exercise 10D Total surface area (TSA) 1 a 577 cm2 b 2670.4 cm2

c 483.1 m2 d 247.0 m2

e 261.2 cm2 f 2463.0 mm2

2 a viii b vi c ii d i e v f iii g iv h ix i vii 3 a 12 cm b 11.4 m 4 a 5.0 m b 7.0 m c 7.6 cm d 8.4 cm 5 No, the diameter of the cork is 1.6 cm,

so it will fall through. 6 C 7 E 8 D 9 6.3 m2

10 a Triangular prism b

c 215.3 cm2

11 235.6 mm2 12 26 482 cm2

Exercise 10E Volume 1 a ii b vi c v d iii e iv f i 2 a 489.6 cm3 b 848.7 cm3 c 502.7 cm3

d 1075.2 cm3 e 768.9 cm3

3 a 336 cm3 b 80.5 cm3

c 1218 mm3 d 2750 cm3

4 $54.00 5 a 6.1 m3 b 123 717.1 cm3

c 2356.2 cm3

6 a 179.59 cm3 b 808.17 cm3 c 269.4 cm3

7 4.2 cm 8 Rectangular prism: 91 200 cm3

Composite prism: 94 004.6 cm3

Trapezium prism: 79 192 cm3

Container a will be best as apples require (512 × 160) 81 920 cm3. Container b has too much space, while c does not have enough.

9 3744 cm3

Exercise 10F Capacity 1 a 750 mL b 0.8 L c 2500 mL d 40 L e 6 000 000 cm3 → 6 000 000 → 6000 L f 12 000 L g 4.2 kL h 7.5 kL → 7500 L i 5.2 cm3

j 6000 cm3 k 20 000 mL = 20 000 cm3

l 5.3 m3

2 660 mL 3 7.1 litres 4 63 mm 5 a 3 b 95 mL 6 a 35 kL b i 29.2 minutes ii 41.7 minutes 7 60 cm

Exercise 10G Similar figures 1 Answers shown in proportion,

but not to size. a

A

A' B'

D'C'

C D

BO

bA

A' B'

C'

D'E'

F'

B

C

DE

FO

2 Answer shown in proportion, but not to size.

3 a Similar b Similar c Not similar d Similar e Similar f Not similar g Not similar h Similar 4 a 2:5 or 4:10 22:52 or 42:102

4:25 or 16:100 4:25 b Area of small rectangle = 8 cm2

5 a Answers will vary. b 3:2 c 9:4 d Answers will vary. 6 a 6:10 or 3:5 63:103 33:53

216:1000 27:125 27:125

b Volume of smaller cylinder

785.4

27

126=

= 169.6 cm3

7 a 8:27 b 16.7 cm3

Exercise 10H Similar triangles 1 a Similar b Similar c Similar d Not enough information 2 a a, b, c, d = 78° b b = 68°; a and c = 112° c c = 32°; a, b, d, e = 74° d a, b, e = 102°; c, d = 78° e y = 36°, x = 63° f a, e = 122°; b = 32°; d, c = 26° 3 A 4 B 5 Answers will vary. 6 a A

x

x

x

CE

B

D EShare same angleCorresponding anglesCorresponding angles

b

x

x x

Share same angleCorresponding anglesCorresponding angles

A

A

ED BC


c

x

x x

Corresponding anglesCorresponding angles

Equal (vertically opposite)angles

A

D

BB EC

d

Corresponding anglesShared angle

A

C E B

A

D

e

Corresponding angles as shown

AA

B DC B

xx

7 a x = 7.4, y = 48 b y = 3.1, x = 3.1 c x = 12.5, y = 19.5 d x = 7.5, y = 10 e x = 4 f x = y = 75°, z = 4 8 A 9 2.9 m

Exercise 10I Symmetry 1 a All, (AB, CD, EF, GH) b AB, CD, EF c AB, CD, EF d None 2 a

3

b

1

c

1

d

2

e

0

f 1 23

4

56

3 a and b

c 72° d 5 times e 5 4 a 2 b 2 c 6 d Infi nite e None f None

5 a Planes of symmetry: 4 Axes of symmetry: 1

b Order of rotational symmetry: 4 6

Verticalsymmetry

Horizontalsymmetry

Rotationalsymmetry None

AHI

MOTUVWXY

BCDEHIKOX

HIOSXZ

FGJLPQRN

CHAPTER REVIEWMULTIPLE CHOICE1 C 2 A 3 A 4 A5 B 6 D 7 B 8 D9 E 10 C 11 A 12 B

13 B 14 D 15 A 16 E17 C 18 C

SHORT ANSWER

1 95 cm 2 The board just fi ts diagonally

(that is, along the line CB). 3 a i P = 48.0 cm ii A = 129.4 cm2

b i P = 12.3 m ii A = 11.5 m2

c i P = 48 cm ii A = 173.7 cm2

4 a Pole

Guy-rope

2.5 m 2.1 m

2.4 m

b 1.4 m

5 TSA = 617 cm2

6 Answers will vary. 7 12 cm 8 a × 2.5 b 1

2× 9 8 cm 10 7 cm 11 9.6 cm 12 a

Lines of symmetry b 2

EXTENDED RESPONSE

1 a Triangular w indow: P = 4.16 m; Trapezoidal window: P = 6.72 m

b $147 c Triangular window: A = 0.834 m2;

Trapezoidal window: A = 2.334 m2

d $143

e Yes

x

xy

x plus y = 180°

Nodey

The sum of angles around each node add to 360º.

f P = 72 cm g A = 300 cm3

h A = 9.252 m2

i 29 boxes 2 a Annulus: A = 766 cm2;

propeller-shaped: A = 838 cm2

b A = 10 586.4 cm2 c $264 d A = 7628 cm2 e $62.36 f 2.42 m g 3.388 m2

h 0.033 88 m3

3 a i

TSA(cm2)

iiVolume(cm3)

IIIIIIIVVVI

13.076 12.897 7.069

13.519.5

17.096

3.12.91.83.45.64.4

b 332.6 cm2

c 7 full boxes d 84.8 mL e 11 full boxes 4 a i 3 cm to 2 m ii 3:200

iii 3200

b a = 2 mb = 2.5 mc = 1.6 md = 2 me = 2 mf = 2.5 m

c i 22.76 m2

ii 5.69 litres d 16.8 m3

e 5.01 m 5 a

b 1 : 16 c 0.96 m d Not similar (does not pass the SSS test) e i No ii No f Equilateral triangle; side length may vary g Answers will vary depending on

answer to part f.

O

chapter 10 shape and measurement -...

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