chapter 8 applications of perimeter , area and volume

235

8.1 Pythagoras’ theoremPythagoras’ theorem links the sides of a right-angled triangle. In a right-angled triangle the side opposite the right angle is called the hypotenuse. The hypotenuse is always the longest side.

Pythagoras’ theorem

Pythagoras’ theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

(Hypotenuse)2 = (side)2 + (other side)2

a

b

h

h2 = a2 + b2

8.1

Hypotenuse(longest side)

C H A P T E R

8Applications of perimeter,

area and volume

Syllabus topic — MM2 Applications of perimeter, area and volume

Find unknown sides using Pythagoras’ theorem

Calculate the perimeter of simple fi gures

Calculate the area of composite shapes

Calculate the area from a fi eld diagram

Calculate the volume of prisms and cylinders

Relate capacity to volume

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© The Powers Family Trust 2013 Cambridge University Press

236 Preliminary Mathematics General

Pythagoras’ theorem is used to fi nd a missing side of a right-angled triangle if two of the sides are given. It can also be used to prove that a triangle is right angled.

Example 1 Finding the length of the hypotenuse

Find the length of the hypotenuse, correct to two decimal places.

Solution1 Write Pythagoras’ theorem.2 Substitute the length of the sides.3 Calculate the value of h2.4 Take the square root to find h.5 Express answer correct to two decimal

places.

h2 = a2 + b2

= 92 + 52

h = 9 52 2+ = 10.30 cm

Example 2 Finding the length of a shorter side

A rectangle has a breadth of 5 mm and a diagonal measuring 12 mm. What is the length of the rectangle, correct to one decimal place?

Solution1 Draw a triangle and label the information.2 Mark the unknown length as x.

3 Write Pythagoras’ theorem.4 Substitute the length of the sides.5 Make x2 the subject.6 Take the square root to find x.7 Express answer to correct one decimal place.8 Write the answer in words.

12 mm

x mm

5 mm

h2 = a2 + b2

122 = x2 + 52

x2 = 122 − 52

x = 12 52 2−

= 10.9 mmThe length of the rectangle is 10.9 mm.

9 cm

h cm5 cm

12 mm5 mm



237Chapter 8 — Applications of perimeter, area and volume

Exercise 8A 1 Find the length of the hypotenuse, correct to one decimal place.

a

8 cm

h cm6 cm

b 5 cm

h cm12 cm

c

10 mm

h mm24 mm

d

10 mm

h mm 20 mm

e

4.2 cm

h cm2.5 cm

f 54 cm

h cm63.2 cm

2 Find the value of x, correct to two decimal places.a

12 cm

x cm

15 cm b 15 cm

x cm 21 cm

c

6 mm

x mm13 mm

d

24 mm 32 mm

x mm

e 2.3 cm

x cm

4.8 cm

f

14.1 cmx cm

9.5 cm




Development

3 Calculate the length of the side marked with the pronumeral. (Answer to the nearest millimetre.)a

42 mm

y mm

35 mm

b

30 mm

a mm

63 mm

c 16 mm

x mm 28 mm

d 10 mm

d mm

27 mm

e

52 mmb mm

33 mm f

8 mmm mm

12 mm

4 Find, correct to one decimal place, the length of the diagonal of a rectangle with dimensions 7.5 metres by 5.0 metres.

5 Find the value of the pronumerals, correct to two decimal places.a

90 cm 72 cm

x cm

b

4 cm

7 cm y x cm

6 cm

6 Calculate the length of x, correct to one decimal place.a

4 cm

7 cm

y cm

x cm6 cm

b

x cm18 cm

14 cm

25 cm

7.5 m

5.0 m




8.2 PerimeterPerimeter is the total length of the outside edges of a shape. It is the distance of the boundary.

Perimeter formulasName Shape Perimeter

Trianglec

a bP = a + b + c

Quadrilateral c

a b

dP = a + b + c + d

Square s P = 4s

Rectangleb

l

P = 2(l + b)

Circler Circumference

C = 2πrC = πd

Example 3 Finding the perimeter of a rectangle

Find the perimeter of the following rectangle.

Solution1 The shape is a rectangle, so use the

formula P = 2(l + b).2 Substitute the values for l and b (l = 8

and b = 3).3 Evaluate. 4 Write the answer in words.

P = 2(l + b)

= 2 × (8 + 3)

= 22 mPerimeter of the rectangle is 22 m.

8.2

3 m

8 m




Example 4 Finding the perimeter of a triangle

Find the perimeter of the following triangle. Answer correct to one decimal place.

Solution1 Find the length of the hypotenuse or h.

2 Write Pythagoras’ theorem.

3 Substitute the length of the sides.4 Evaluate the value of h.5 Add the lengths of sides to find the

perimeter.6 Express answer correct to 1 decimal place.7 Write the answer in words.

h2 = 4.22 + 3.72

h 4.2 3.7 5.6 cm2 2= + ≈

P = 3.7 + 4.2 + 5.6 = 13.5 cm

Perimeter of the triangle is 13.5 cm.

Example 5 Finding the circumference of a circle

Find the perimeter of a circle with a radius of 9 mm. Answer correct to two decimal places.

Solution1 The shape is a circle, use the formula

C = 2π r.2 Substitute the value for r (r = 9).3 Evaluate.4 Write the answer in words.

C r== × ×=

2

2 9

56 55

ππ

. mm

Perimeter of the circle is 56.55 mm.

Example 6 Finding the perimeter of a semicircle

Find the perimeter of a semicircle with a diameter of 4 m. Answer correct to two decimal places.

Solution

1 The shape is a semicircle, use the formula C = π d ÷ 2.

2 Substitute the value for d (d = 4) to find the curved distance.

3 Evaluate.

4 Add the curved distance to the diameter.5 Evaluate.6 Write the answer in words.

Cd

2

π=

4

2

π= ×

6.28 m=P = 6.28 + 4 = 10.28 mPerimeter is 10.28 m.

3.7 cm

4.2 cm

9 mm

4 m




Perimeter of composite shapesA composite shape is made up of two or more plane shapes. The perimeter of a composite shape is calculated by adding the distances that make up the boundary of the shape.

Perimeter of a composite shape

1 Sides with the same markings are of equal length.2 Unknown side lengths of some sides are determined by using the given

lengths of the other sides. 3 Pythagoras’ theorem is used to fi nd unknown side lengths involving a right

triangle.4 Distances that are part of a circle are found using C = 2π r.5 Add the distances that make up the boundary of the shape to calculate the

perimeter.

Example 7 Finding the perimeter of composite shapes

Find the perimeter of each of these shapes.a 2 cm

8 cm

5 cm

6 cm

b

5 m

5 m

Solution1 Find the unknown side lengths using the

measurements given in the question.2 Add the lengths of all the edges to find

the perimeter.3 Evaluate.4 Write the answer in words.5 Use the formula C = 2π r ÷ 4 for the

curved distance.6 Substitute the value for r (r = 5).7 Evaluate.8 Add the curved distance to other edges.9 Evaluate.10 Write the answer in words.

a 2 cm

8 cm

5 cm

6 cm 8 − 2 = 6 cm6 + 5 = 11 cm

P = + + + + + =2 6 6 5 8 11 38 cm Perimeter is 38 cm.

b Cr= = × × =2

4

2 5

47 85

π π. m

P = + + + +=

7 85 5 5 5 5

27 85

.

. m

Perimeter is 27.85 m.




Exercise 8B 1 Find the perimeter of each quadrilateral. Answer correct to one decimal place.

a

5.4 cm

b

13.4 m

7.2 m

c

20 m9.5 m

2 Find the perimeter of a square with a side length of 12.3 m. Answer correct to one decimal place.

3 Find the perimeter of each right triangle. Answer correct to one decimal place.a

8 m15 m

b

5 cm2 cm

c

8.5 mm

7 mm

4 Find the perimeter of a right triangle with a base of 10.25 cm and a height of 15.15 cm. Answer correct to two decimal places.

5 Find the perimeter of each circle. Answer correct to one decimal place.a

3 m

b

2 cm

c

14 mm

6 What is the circumference of each circle? Answer correct to one decimal place.a Radius of 4 cm b Radius of 19 mc Radius of 34 mm d Diameter of 50 mme Diameter of 22 m f Diameter of 6 cm




Development

7 Find the perimeter of each semicircle. Answer correct to one decimal place.a

5 m

b 2.1 m

c

16 m

8 Find the perimeter of each shape. Answer correct to two decimal places.a

7 m

b

3 mm

c

5 cm

9 Find the perimeter of each composite shape. Answer to the nearest whole number.a

4 m8 m

6 m

10 m

b

4 m

2 m2 m

2 m

1 m

c

10 cm

8 cm

d 1 m

6 m3 m

6 m

e

4 m 5 m

f 4 m

4 m

10 An annulus consists of two circles with the same centre. Find the perimeter of an annulus if the inner diameter is 3 cm and the outer diameter is 6 cm. Answer correct to the nearest centimetre.

11 A rectangle ABCD has length AB = 12 cm and a width of BC = 6 cm. a Find the value of x.b Calculate the perimeter of quadrilateral AECF.

(Answer correct to two decimal places).

6 cm3 cm

B

x cm

C

A

DF

E




8.3 AreaThe area of a shape is the amount of surface enclosed by the boundaries of the shape. It is measured by counting the number of squares that fi t inside the shape. When calculating area, the answer will be in square units.

100 mm2 = 1 cm2 10 000 cm2 = 1 m2

10 000 m2 = 1 ha 1 000 000 m2 = 1 km2

1 ha = 10 000 m2

To calculate the area of the most common shapes, we use a formula. These formulas are listed below.

Area formulas Name Shape Area

Triangleh

b

A bA bhA b=A b1A b1A b2

A b2

A b

Square s A = s2

Rectangle

l

bA = lb

Parallelogram

b

hA = bh

Trapezium

b

h

a

A a b hA a= +A aA a= +A a1A a1A aA a= +A a1A a= +A a2

A a2

A aA a= +A a2

A a= +A a( )A a( )A a b h( )b h= +( )= +A a= +A a( )A a= +A a

Rhombus xy

A xA xyA xyA xA x=A x1A x1A x2

A x2

A x

Circler

A = π r2

8.3




Example 8 Finding the area of a triangle

Find the area of the following triangle.

Solution

1 The shape is a triangle, so use the formula A bA bhA b=A b1A b1A b2

A b2

A b .

2 Substitute the values for b and h (b = 8.1 and h = 5.5).3 Evaluate.4 Write the answer using the correct units.

A bA bhA b=A b

= ×= × ×

=

1A b

1A b

21

28 1 5 5

22 275 2

. .×. .×8 1. .8 1 5 5. .5 5

. m275. m275

Example 9 Finding the area of a trapezium

Find the area of the following shape.

Solution1 The shape is a trapezium, so use the formula

A a b hA a= +A aA a= +A a1

A a1

A a2

( )A a( )A a b h( )b h= +( )= +A a= +A a( )A a= +A a .

2 Substitute the values for a, b and h.3 Evaluate.4 Write the answer using the correct units.

A a b hA a= +A aA a= +A a

= += +

=

1A a

1A a

21

22 9= +2 9= + 5 1 3

12 2

( )A a( )A a b h( )b h= +( )= +A a= +A a( )A a= +A a

( .= +( .= +2 9( .2 9= +2 9= +( .= +2 9= + . )5 1. )5 1

cm

The area of the shape is 12 cm2.

Example 10 Finding the area of a parallelogram

Find the area of the following shape.

Solution1 The shape is a trapezium, so use the formula

A = bh.2 Substitute the values for b and h (b = 6.5 and

h = 4).3 Evaluate.4 Write the answer using the correct units.

A bhA b=A b= ×=

6 5= ×6 5= × 4

26 2

6 5.6 5= ×6 5= ×.= ×6 5= ×mm

The area of the shape is 26 mm2.

8.1 m

5.5 m

5.1 cm

2.9 cm

3 cm

6.5 mm

4 mm




Example 11 Finding the area of a circle

Find the area of a circle with a radius of 5 metres. Give your answer correct to one decimal place.

Solution1 The shape is a circle, so use the formula

A = π r2.

2 Substitute the value for r (r = 5).

3 Evaluate correct to one decimal place.

4 Write the answer using the correct units.

A rA r=A r

= ×=

πA rπA r

π= ×π= ×

2

2

2

5

78 5. m5. m5

The area of the circle is 78.5 m2.

Area of composite shapesA composite shape is made up of two or more plane shapes. The area of a composite shape is calculated by adding or subtracting the areas of simple shapes.

Area of composite shapes

• Composite shapes are made up of more than one simple shape.• Area of composite shapes can be found by adding or subtracting the

areas of simple shapes.

Example 12 Finding the area of a composite shape

Find the area of the composite shape. Answer correct to one decimal place.

Solution1 Divide the shape into a rectangle and a

semicircle.2 Use the formula A = lb for the rectangle.3 Substitute and evaluate.

4 Use the formula A r= 1

22π for the semicircle.

5 Substitute and evaluate.

6 Add the area of the rectangle to the semicircle.7 Evaluate.8 Write the answer using the correct units.

A lb== ×

=

12 10

120 2cm

A r=

= × ×

≈

1

21

25

39 3

2

2

2

π

π

. cm

A = 120 + 39.3 = 159.3 cm2

The area of the shape is 159.3 cm2.

5 m

12 cm

10 cm




Exercise 8C 1 Find the area of each triangle. Answer correct to one decimal place.

a

23 m

13 m

b

2 cm4 cm

c

6 mm

6.5 mm

d

19 m

7.6 m

e

15.5 mm

13 mmf

9.5 m8.5 m

2 Find the area of each shape. Answer correct to one decimal place.

a

6.1 cm

b

11.2 m

6.4 m

c

22 m

9 m

d

7 mm

10 mm

e 4 m

7 m

f

4 cm

6.7 cm

3.8 cm

3 Find the area of a triangle with a base of 8.25 cm and a height of 10.15 cm. Answer correct to the nearest square centimetre.

4 Find the area of a square with a side length of 105.1 m. Answer correct to the nearest square metre.

5 Find the area of a circle with a radius of 7 cm. Answer correct to the nearest square centimetre.




Development


a

5 km

2.3 km

b

9.1 cm

9.1 cm

c

Diagonals are4.4 mm and 6.8 mm

d

4.8 cm4.1 cm

5.7 cm

9.8 cm

e

6 mm

12.5 mm

8 mm

f

17 m

15 m

8 m

7 Jasmine is planning to build a circular pond. The radius of the pond is 1.5 m. What is the area of the pond, correct to the nearest square metre?

8 A 25 m swimming pool increases in depth from 1.3 m at the shallow end to 2.6 m at the deep end. Calculate the area of the side wall of the pool. Answer correct to the nearest square metre.

9 Philip wants to tile a rectangular area measuring 2.2 m by 3 m in his backyard. The tiles he wishes to use are 50 cm by 50 cm. How many tiles will he need? Answer correct to the nearest whole number.

10 Find the area of each composite shape.

a

32 cm30 cm

b

8 cm

6 cm 3 cm

4 cm c

20 cm

15 cm

4 cm

1.3 m2.6 m

25 m




11 An annulus consists of two circles with the same centre. Find the area of an annulus if the inner diameter is 6 cm and the outer diameter is 10 cm. Answer correct to the nearest square centimetre.

12 A metal parallelogram has two identical squares removed from its shape. The two squares have a side length of 2 cm. Find the shaded area. Answer correct to the nearest square centimetre.

13 A lawn is to be laid around a rectangular garden bed.a What is the amount of lawn required? b Find the cost of the new lawn if the required turf

costs $20 per square metre.

14 The cross-section of an ice-cream cone is shown opposite.a What is the radius of the semicircle?b What is the height of the triangle?c Calculate the area of the region. Answer

correct to one decimal place.

15 What is the area of a quadrant if it has a radius of 8 mm? Answer correct to two decimal places.

16 Decking for a house consists of a square and a triangle. The square has a side length of 8 metres and the triangle is isosceles.a Use Pythagoras’ theorem to find the value of x.b Calculate the area of the shaded region.

17 A metal worker cut circles with a diameter of 2 cm from a rectangular sheet of tin 4 cm by 8 cm.a What is the area of the rectangular sheet?b How many circles can be cut from the rectangular sheet?c What is the area of the remaining metal after the circles have been removed from the

rectangular sheet? Answer correct to two decimal places.

6 cm 10 cm

5 cm

10 cm

7 cm

20 m

25 m

13 m

19 m

10 cm

16 cm 14 cm

8 m

8 m

x m

x m




8.4 Field diagramsField diagrams are used to calculate the area of irregularly shaped blocks of land. Measuring and recording the data in a field diagram is called surveying. One type of survey is called the traverse or offset survey.

Traverse or offset surveyThis type of survey involves measuring distances along a suitable diagonal or traverse. The perpendicular distances from the traverse to the vertices of the shape are called the offsets. When conducting a traverse survey, measurements are taken of the traverse and the offsets. These measurements are recorded in a field book entry.

Traverse

Offset

Offset

The field book entry records the distances along the traverse between two vertical lines. The distances along the offsets are recorded on either side of these measurements.

Field Book Entry

D163

C 65 11075 28 B0

A

Field Diagram

A

B

C

D

65

28

75

35

53

The measurements along the traverse are the distances starting from the bottom. For example, the distance 110 to the offset at point C is the distance from point A. This results in a distance of 35 between offset B and C.

Conducting a traverse survey

1 Choose a suitable diagonal or traverse.2 Place a tape measure along the traverse.3 Starting from the bottom, measure the offsets using a taut string.4 Record the measurements in a field book entry.




Example 13 Calculating area from a field diagram

A fi eld diagram from a traverse survey is shown opposite. Measurements are in metres.a Find the area of the quadrilateral ABCD. Answer

correct to one decimal place.b Find the distance AB. Answer correct to two decimal

places.

Solution1 Divide the quadrilateral into two triangles –

∆ADB and ∆ADC.

2 Find the area of each triangle using the formula A bA bhA b=A b1A b1A b

2A b

2A b .

3 Substitute the values for b and h.4 Evaluate.5 Express answer correct to one decimal place

and with the correct units.

6 Add the area of the two triangles.

a Area of quadrilateral ABCD For ∆ADB, b = 163 and h = 28

A bA bhA b=A b

= ×= × ×

=

1A b

1A b

21

2163 28

2282 2m

For ∆ADC, b = 163 and h = 65

A bA bhA b=A b

= ×= × ×

=

1A b

1A b

21

2163 65

5297 5 2. m5. m5

Total area = 2282 + 5297.5 = 7579.5 m2

7 Write Pythagoras’ theorem by substituting the length of the sides.

8 Take the square root to find AB.9 Evaluate.

10 Express answer correct to two decimal places.

b Distance AB

AB

AB

75 28

75 2880.056 230 2480.06 m

2 2752 275 2

2 2282 228

= +75= +752 2= +2 2752 275= +752 275

= += +75= +752 2= +2 2

==

Distance AB is 80.06 m.

A

B

C

D

65

28

75

35

53




Exercise 8D 1 Find the area of the following fields using each field diagram. Units are in metres.

a

35

6030

60

b

2850

5020

c

80

80

4550

d

A

B

C

D

60

45

50

40

50

e

A

B

C

D

60

20

48

110

18f

A

B

C

D

25

32

24

12

26

2 Find the area of the following fields using each field book entry. Units are in metres.

a B60

C 50 250

A

b D7550 20 B

C 15 400

A

c D150

C 70 11075 30 B0

A

d E140100 32 D

C 55 9030 32 B0

A

e E90

C 36 8075 10 D

B 36 200

A

f E4030 10 D

C 40 2010 10 B0

A




Development

3 The diagram on the right shows a block of land that has been surveyed. All measurements are in metres.a Find the area of the quadrilateral ABCD. Answer

correct to one decimal place.b What is the length of AB? Answer correct to the

nearest metre.

4 The diagram below shows a block of land that has been surveyed. All measurements are in metres.

A

B

DC

E

67

11

46

5444

15F

G

a Find the area of the triangle ABF. Answer correct to one decimal place.b Find the area of the triangle ACE. Answer correct to one decimal place.c Find the area of the triangle DGE. Answer correct to one decimal place.d Find the area of the trapezium BFGD. Answer correct to one decimal place.e What is the total area of the block of land? Answer correct to one decimal place.f Find the distance AB. Answer correct to two decimal places.

5 The field book entry below shows a block of land that has been surveyed. All measurements are in metres.

E7158 43 D

C 23 3425 14 B0

A

a Find the area of ABDEC. Answer correct to one decimal place.b What is the length of AC? Answer correct to the nearest metre.

A

B

C

D

25

14

13

32

35




8.5 Volume of prisms and cylindersVolume is the amount of space occupied by a three-dimensional object. It is measured by counting the number of cubes that fi t inside the solid. When calculating volume, the answer will be in cubic units.

1000 mm3 = 1 cm3

1 000 000 cm3 = 1 m3

1 000 000 000 m3 = 1 km3

To calculate the volume of the most common solids, we use a formula. Some of these formulas are listed below. The volume of a prism is found by using its cross-sectional area. Prisms are three-dimensional objects that have a uniform cross-section along their entire length.

Volume formulas

Name Solid Volume

Cube

ss

s

V = Ah = (s2) × s = s3

Rectangular prism

l

b

h

V = Ah = lb × h = lbh

Triangular prism

A = 12 bh

h

V Ah

bh h

V A=V A

=

×1

2

Cylinder h

r

V = Ah = (πr2) × h = πr2h

8.5




Example 14 Finding the volume of a right prism

A rectangular prism has a length of 8 cm, a breadth of 2 cm and a height of 4 cm. Find the volume of this rectangular prism. Answer in cubic centimetres.

Solution1 Use the volume formula for a right prism V = Ah.2 Determine the shape of the base and the formula

to calculate the area of the base A = lb.3 Substitute the values into the formula.4 Evaluate.5 Give the answer to the correct units.

V = Ah

= lbh = 8 × 2 × 4

= 64 cm3

Example 15 Finding the volume of a cylinder

A cylinder has a radius of 8 mm and a height of 12 mm. Find the volume of the cylinder. Answer correct to two decimal places in cubic millimetres.

Solution1 Use the volume formula for a cylinder V = πr2h.2 Substitute the r = 8 and h = 12 into the formula.3 Evaluate.4 Write the answer correct to two decimal places and

with the correct units.

V = πr2h = π × 82 × 12 = 2412.743 158 mm3

≈ 2412.74 mm3

2 cm

4 cm

8 cm

8 mm

12 mm




Exercise 8E 1 Find the volume of the following prisms where A is the area of the base.

a

12 mA = 8 m2

b

40 m

A = 110 m2

c

16 m

A = 7 m2

2 What is the volume of a rectangular prism with a base area of 15 mm2 and a height of 11 mm?

3 Find the volume of a triangular prism with a height of 15 m and a base area of 50 m2.

4 Find the volume of the following solids. Answer to the nearest whole number.

a

9 m4 m

3 m

b 18 cm

18 cm

18 cm

c

8 mm

6 mm

10 mm

10 mm

d

15 m6 m

10 m

e

6 mm

14 mm f

20 mm

7 mm

5 What is the volume of a rectangular prism with dimensions 4.5 cm by 6.5 cm by 10.5 cm? Answer correct to one decimal place.

6 A closed cylindrical plastic container is 20 cm high and its circular end surfaces each have a radius of 5 cm. What is its volume correct to two decimal places?




Development

7 A water tank is in the shape of a closed cylinder with a radius of 10 m and height of 8 m.a What is the area of the top circular face of the water

tank? Answer correct to one decimal place.b Determine the volume of the water tank. Answer

correct to one decimal place.

8 A hollow container is in the shape of a rectangular prism as shown.

a What is the volume of the container if it was solid?b What is the area of the base?c What is volume of the hollow container?

9 A step is shown opposite.a What is the area of the base?b Determine the volume of the step.

10 What is the volume, correct to one decimal place, of a cylindrical paint tin with a height of 30 cm and a diameter of 25 cm?

11 Find the volume of an equilateral triangular prism with a height 10 cm and a side length of 3 cm. Answer correct to three decimal places.

12 A vase with a volume of 200 cm3 is packed into the cardboard box shown below. The space around the vase is filled with foam to protect from breaking. The parcel is sealed and posted.

6 cm

10 cm8 cm

a What is the volume of the foam?b What is the area of cardboard on the surface of the box?

8 m

10 m

10 m

10 m6 m 2 m

6 m

10 m

2 m

12 m5 m

1 m




8.6 CapacityThe capacity of a container is the amount of liquid it can hold. Some solids have both a volume and a capacity. For example, a can of soft drink is a cylinder that has a volume (V = π r2h) and a capacity (360 mL). The base unit for capacity is a litre (L). Three commonly used units for capacity are a megalitre, (ML), kilolitre (kL) and millilitre (mL).

Capacity

1 ML = 1000 kL1 ML = 1 000 000 L1 kL = 1000 L1 L = 1000 mL

1 cm3 = 1 mL 1 cm3 = 0.001 L1000 cm3 = 1 L

1 m3 = 1 000 000 cm3

1 m3 = 1 000 000 mL1 m3 = 1000 L1 m3 = 1 kL

Example 16 Finding the capacity

The container shown opposite is fi lled with water. a Find the volume of the container in cubic centimetres.b Find the capacity of the container in litres.

Solution1 Use the volume formula for a right prism

V = Ah.2 Determine the shape of the base and the

formula to calculate the area of the base A = lb.3 Substitute the values into the formula.4 Evaluate.5 Give answer to the correct units.

a V = Ah

= lbh = 70 × 40 × 30

= 84 000 cm3

6 To change cm3 to L multiply by 0.001. (1 cm3 = 0.001 L)7 Alternative method is to convert to mL. (1 cm3 = 1 mL)

b Capacity = 84 000 × 0.001 L = 84 L Capacity = 84 000 × 1 mL = 84 000 mL = 84 L

70 cm40 cm

30 cm




Exercise 8F 1 A can of soft drink has a capacity of 375 mL. How many cans of soft drink would it take

to fill a 1.2 L bottle? How much would remain?

2 A medicine bottle has a capacity of 0.3 L.a What is the capacity in millilitres?b How many tablespoons (15 mL) does

the bottle contain?c How many teaspoons (5 mL) does the

bottle contain?d The correct dosage is 10 mL, 3 times a

day. How many doses does the bottle contain?

3 Complete the following.a 4 3cm mL= b 2000 3cm L=c 70 3cm mL= d 34000 3cm L=

e 900 3cm mL= f 500 3cm L=

g 43 3m km k3m k3 Lm k=m k h 30 3m Lm L3m L3m L=m L

i 103 3m km k3m k3 Lm k=m k j 7 m L3 =

k 5 3m km k3m k3 Lm k=m k l 8 m mL3 =

4 What is the capacity of a rectangular prism whose base area is 20 cm2 and height is 10 cm? Answer correct to the nearest millilitre.

5 Find the capacity of a triangular prism with a height of 18 m and a base area of 40 m2. Answer in litres, correct to two significant figures.

6 Find the capacity of a rectangular pyramid whose base area is 12 cm2 and height is 15 cm. Answer correct to the nearest millilitre.

7 Find the capacity of a cylindrical plastic container 16 cm high and with circular end surfaces of radius 8 cm. Answer correct to the nearest litre.




Development

8 Find the capacity of the following solids in millilitres, correct to two decimal places.

a

8 m3 m

4 m

b

4 mm

3 mm

6 mm

5 mm

c 18 mm

10 mm

d

3 m

9 m

10 m

e A = 21 m2

7 m

f

6 m

10 m

9 Find the capacity of a cube whose side length is 75 mm. Answer in millilitres, correct to two decimal places.

10 A water tank is the shape of a cylinder with a radius of 2 m and height of 2.5 m.a What is the area of the top circular face of the

water tank? Answer correct to one decimal place.

b Determine the volume of the water tank in cubic metres. Answer correct to one decimal place.

c What is the capacity of the tank, to the nearest kilolitre?

11 A swimming pool is the shape of a rectangular prism as shown opposite. The swimming pool is filled 25 cm from the top.a What is the volume of water in cubic metres?b How much water does the swimming pool

contain, to the nearest kilolitre?

2.5 m

2 m

15 m10 m

1.7 m

14.4



261Chapter 8 — Applications of perimeter, area and volumeReview

Pythagoras’ theorem Pythagoras’ theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

h a b2 2 2= +

Perimeter Perimeter is the total length of the outside edges of a shape. It is the distance of the boundary.Rectangle P = 2(l + b)Circle C = 2π r or C = π d

Area Triangle A bA bhA b=A b1A b1A b2

A b2

A b Square A = s2

Rectangle A = lb Parallelogram A = bh

Trapezium A a b hA a= +A aA a= +A a1A a1A aA a= +A a1A a= +A a2

A a2

A aA a= +A a2

A a= +A a( )A a( )A a b h( )b h= +( )= +A a= +A a( )A a= +A a Rhombus A xA xyA xyA xA x=A x1A x1A x2

A x2

A x

Circle A = πr 2

Area of composite shapes • Composite shapes are made up of more than one simple shape.• Area of composite shapes can be found by adding or subtracting

the areas of simple shapes.

Field diagrams Field diagrams are used to calculate the area of irregularly shaped blocks of land. A traverse survey measures distances along a suitable diagonal or traverse.

Volume of prisms and

cylinders

Cube V = Ah = (s2) × s = s3

Rectangular prism V = Ah = lb × h = lbh

Triangular prism V Ah bh h= =V A= =V Ah b= =h b( )h b( )h bh b( )h bh h( )h hh h×h h( )1( )h b( )h b1h b( )h b( )2( )h b( )h b2

h b( )h b

Cylinder V = Ah = (πr2) × h = πr2h

Capacity The amount of liquid a container can hold. Base unit is a litre.

Chapter summary – Applications of perimeter, area and volume Study guide 8



262 Preliminary Mathematics GeneralRe

view 1 What is the length of the hypotenuse?

A 400 cm B 20 cmC 28 cm D 7.46 cm

2 What is the perimeter of a quadrant with a radius of 5 mm?A 3.9 mm B 7.9 mm C 13.9 mm D 17.9 mm

3 What is the perimeter of the composite shape?A 25 m B 29 mC 30 m D 32 m

4 What is the area of the composite shape in question 3?A 25 m2 B 51 m2

C 63 m2 D 75 m2

5 What is the area of a triangle with a base of 5 m and a perpendicular height of 8 m?A 13 m2 B 20 m2

C 40 m2 D 80 m2

6 What is the area of the trapezium?A 42 cm2 B 63 cm2

C 96 cm2 D 126 cm2

7 What is the area of ABCD using the field book entry?A 40 B 450C 600 D 1200

8 What is the volume of a cylinder with a height of 6 cm and radius of 2 cm?A 75 cm3 B 113 cm3 C 302 cm3 D 452 cm3

9 A cubic water tank has a side length of 6 m. What is the capacity of the tank?A 36 kL B 216 kL C 360 kL D 216 000 kL

12 cm16 cm

4 m

9 m

5 m

7 m

7 cm8 cm

12 cm

6 cm

A

C 10

4030100

D

20 B

Sample HSC – Objective-response questions



263Chapter 8 — Applications of perimeter, area and volumeReview

1 Find the value of x, correct to two decimal places.a

28 m

21 m

x m b

14 mm 21 mm

x mm c

47 cm27 cm

x cm

2 Calculate the length of x, correct to the nearest millimetre.

3 Find the perimeter of each shape. Answer correct to one decimal place.a

10 cm5 cm

b

6.7 cm

c

11 cm

d 12 m e

6 mm

f 7 m

9 m

11 m

4 m


a

9 m

6 m

b

10 m

3.5 m

c

11.2 cm

d

8.4 cm

5.6 cm

e

4 mm

7 mm

f 2 m

3 m

18 mm

6 mm

10 mm

x

Sample HSC – Short-answer questions



Chapter summary – Earning Money

Revi

ew264 Preliminary Mathematics General

5 Mahendra is planning to tile a rectangular area measuring 3.6 m by 3.2 m in his backyard. The tiles he wishes to use are 40 cm by 40 cm. a What is the area of the rectangle?b What is the area (in m2) of the tiles?c How many tiles will he need?

6 The diagram on the right shows a block of land that has been surveyed. All measurements are in metres.a Find the area of the quadrilateral PQRS. Answer correct to

one decimal place.b What is the length of PQ? Answer correct to the nearest metre.

7 Find the volume of the following solids.

a 11 mm

11 mm

11 mm

b

10 m

3.2 m4.5 m

c

4 mm

3 mm

4 mm

5 mm

8 A closed cylindrical container is 16 mm long and its end surfaces have a radius of 8 mm.a What is the area of the circular ends? Answer

correct to the nearest square millimetre.b What is the volume of the container? Answer

correct to the nearest cubic millimetre.

9 Find the capacity of a triangular prism with a height of 50 cm and a base area of 120 cm2. Answer in litres.

Challenge questions 8

P

Q

R

S

46

52

40

18

58

16 mm

8 mm



chapter 8 applications of perimeter , area and volume

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