Year 10 Surface Area & Volume 3 -...

Year 10 Mathematics Surface Area and Volume Practice Test 3

Name__________________________

1 Find the area of each shape

a) b) c)

d) e) f) 2 Find the volume of the following solids

3 Find the volume of this square pyramid 4 Find the volume of this cone to 1 decimal place a) b)

a)

b)

232 Surface Area and Volume (Chapter 8) Syllabus reference MS5.2.2

B. VOLUME OF PYRAMIDS AND CONES

From Investigation 1 the volume of a pyramid is given by:

Exercise 8B

Find the volume of these square pyramids.a b c

h hh

conetriangular-based pyramidsquare-based pyramid

Volume = (area of base × height)

V = Ah

13--

13--

Example 1

Find the volume of the following square pyramid.

Volume = × area of base × height

= × 10 × 10 × 12 cm3

= 400 cm3

12 cm

10 cm

13---

13---

1

15 cm

8 cm

20 cm

15 cm

0.8 m

1.3 m

Example 2

Find the volume of this cone.

Volume = × area of base × height

= × πr2 × h

= × π × 52 × 8

= 209.4 cm2

8 cm

5 cm

13---

13---

13---


Find the volume of the following solids.a b c

d e f

g h i

j k

Example 4

Find the volume of the following solids.a b

a Volume b Volume= length × breadth × height = area of base × height= lbh = πr2 × h= 7.5 cm × 6 cm × 4 cm = π × 5 × 5 × 10= 180 cm3 = 250π

= 785.4 cm3

Volume =cross-sectional area times perpendicular

height.7.5 cm

6 cm

4 cm10 cm10 cm

6

8 m

7 m

2 m

6 cm 6 cm

12 cm 20 cm

16 cm

1 cm6 cm

7 cm

12 cm

7 cm

3 cm

8 cm

5 cm

13 cm

14 cm

15 cm20 cm

12 cm

11 cm

20 cm25 cm

13 cm

13 cm

20 cm

8 cm


Find the volume of the following solids.a b c

d e f

g h i

j k

Example 4

Find the volume of the following solids.a b

a Volume b Volume= length × breadth × height = area of base × height= lbh = πr2 × h= 7.5 cm × 6 cm × 4 cm = π × 5 × 5 × 10= 180 cm3 = 250π

= 785.4 cm3

Volume =cross-sectional area times perpendicular

height.7.5 cm

6 cm

4 cm10 cm10 cm

6

8 m

7 m

2 m

6 cm 6 cm

12 cm 20 cm

16 cm

1 cm6 cm

7 cm

12 cm

7 cm

3 cm

8 cm

5 cm

13 cm

14 cm

15 cm20 cm

12 cm

11 cm

20 cm25 cm

13 cm

13 cm

20 cm

8 cm

229Surface Area and Volume (Chapter 8) Syllabus reference MS5.2.2

Find the area of the following plane shapes.

a b c

d e f

Find the area of the following.a a rectangle of length 3.8 cm, breadth 1.9 cmb a triangle with base 15 cm, height 7.8 cmc a parallelogram with base 12 cm, perpendicular height 15 cmd a trapezium with parallel sides 15 cm and 18 cm, 12 cm aparte a rhombus with diagonals 4 cm and 6.4 cmf a circle of radius 15.3 cm

Find the shaded areas.a b c d

a A = lb b A = bh c A = bh

= 6.8 × 3.6 cm2 = × 8 × 3 m2 = 15 × 6 cm2

= 24.48 cm2 = 12 m2 = 90 cm2

d A = h(a + b) e A = xy f A = πr2

= (5)(10 + 17) m2 = × 6 × 7 cm2 = π × 6 × 6 cm2

= 67.5 m2 = 21 cm2 = 113.1 cm2

12---

12---

12--- 1

2---

12--- 1

2---

3

15 m

8 m 3 cm

4 cm

5 cm 8 cm

7 m

11 m

6 m

8 cm

6 cm 7 cm

4

5

20 cm

12cm8 cm

2 cm

5 cm

4 cm

5 cm

10 cm

9 cm 13 cm

12 cm5 cm



a b c

d e f




= 6.8 × 3.6 cm2 = × 8 × 3 m2 = 15 × 6 cm2

= 24.48 cm2 = 12 m2 = 90 cm2

d A = h(a + b) e A = xy f A = πr2

= (5)(10 + 17) m2 = × 6 × 7 cm2 = π × 6 × 6 cm2

= 67.5 m2 = 21 cm2 = 113.1 cm2

12---

12---

12--- 1

2---

12--- 1

2---

3

15 m

8 m 3 cm

4 cm

5 cm 8 cm

7 m

11 m

6 m

8 cm

6 cm 7 cm

4

5

20 cm

12cm8 cm

2 cm

5 cm

4 cm

5 cm

10 cm

9 cm 13 cm

12 cm5 cm



a b c

d e f




= 6.8 × 3.6 cm2 = × 8 × 3 m2 = 15 × 6 cm2

= 24.48 cm2 = 12 m2 = 90 cm2

d A = h(a + b) e A = xy f A = πr2

= (5)(10 + 17) m2 = × 6 × 7 cm2 = π × 6 × 6 cm2

= 67.5 m2 = 21 cm2 = 113.1 cm2

12---

12---

12--- 1

2---

12--- 1

2---

3

15 m

8 m 3 cm

4 cm

5 cm 8 cm

7 m

11 m

6 m

8 cm

6 cm 7 cm

4

5

20 cm

12cm8 cm

2 cm

5 cm

4 cm

5 cm

10 cm

9 cm 13 cm

12 cm5 cm



a b c

d e f




= 6.8 × 3.6 cm2 = × 8 × 3 m2 = 15 × 6 cm2

= 24.48 cm2 = 12 m2 = 90 cm2

d A = h(a + b) e A = xy f A = πr2

= (5)(10 + 17) m2 = × 6 × 7 cm2 = π × 6 × 6 cm2

= 67.5 m2 = 21 cm2 = 113.1 cm2

12---

12---

12--- 1

2---

12--- 1

2---

3

15 m

8 m 3 cm

4 cm

5 cm 8 cm

7 m

11 m

6 m

8 cm

6 cm 7 cm

4

5

20 cm

12cm8 cm

2 cm

5 cm

4 cm

5 cm

10 cm

9 cm 13 cm

12 cm5 cm



a b c

d e f




= 6.8 × 3.6 cm2 = × 8 × 3 m2 = 15 × 6 cm2

= 24.48 cm2 = 12 m2 = 90 cm2

d A = h(a + b) e A = xy f A = πr2

= (5)(10 + 17) m2 = × 6 × 7 cm2 = π × 6 × 6 cm2

= 67.5 m2 = 21 cm2 = 113.1 cm2

12---

12---

12--- 1

2---

12--- 1

2---

3

15 m

8 m 3 cm

4 cm

5 cm 8 cm

7 m

11 m

6 m

8 cm

6 cm 7 cm

4

5

20 cm

12cm8 cm

2 cm

5 cm

4 cm

5 cm

10 cm

9 cm 13 cm

12 cm5 cm



a b c

d e f




= 6.8 × 3.6 cm2 = × 8 × 3 m2 = 15 × 6 cm2

= 24.48 cm2 = 12 m2 = 90 cm2

d A = h(a + b) e A = xy f A = πr2

= (5)(10 + 17) m2 = × 6 × 7 cm2 = π × 6 × 6 cm2

= 67.5 m2 = 21 cm2 = 113.1 cm2

12---

12---

12--- 1

2---

12--- 1

2---

3

15 m

8 m 3 cm

4 cm

5 cm 8 cm

7 m

11 m

6 m

8 cm

6 cm 7 cm

4

5

20 cm

12cm8 cm

2 cm

5 cm

4 cm

5 cm

10 cm

9 cm 13 cm

12 cm5 cm


Find the volume of these cones.a b c

Find the volume of these pyramids.

a b c

2

75 mm

80 mm

1.7 m

0.62 m

11 cm

7 cm

Example 3

Find the volume of this triangular pyramid.

Base is a triangle so A = bh

= × 5 × 7

= 17.5 cm2

∴ Volume = × area of base × height

= × 17.5 × 8 cm3

= 46.7 cm3

8 cm

5 cm

7 cm

12---

5 cm

7 cm

12---

13---

13---

3

11 cm

8 cm

7 cm

0.9 m

1.5 m

1.4 m

65 mm

32 mm80 mm


Extension

Find the volume of these solids.

a b c

Find the volume of the following composite solids to 1 decimal place.

a b c

Example 4


The measurement for perpendicular height is missing,but can be calculated using Pythagoras’ theorem.

h2 + 62 = 102

∴ h2 = 100 – 36= 64

∴ h = 8 (as h > 0)


= × (π × 62) × 8 cm3

= 96π cm3

! 301.6 cm3

10 cm

6 cm

6

h 10

13---

13---

4

13 cmx cm

x cm

6 cm

7 cm

5 cm 5 cm

5 cmx cm

3 cm

5

5 cm

12 cm

3 cm

8 cm

5 cm

15 c

m

12 c

m

9 cm

6 cm

10 cm

5 Find the volume of these pyramids

a) b)

6 A square pyramid with volume 380 cm3 has height 8 cm. Find the side length of the square base.

7 Find the volume of a sphere of diameter 10 cm.

8 A sphere has volume 140 cm3. Find the radius to 1 decimal place.

9 Find the volume of these solids

a)

b)

c)

10 Find the surface area of these closed cylinders

11 Find the surface area of this open cylinder 12 Find the height of an open cylinder of radius 8 cm with curved surface area of

1000 cm2

a)

b)


Find the volume of these cones.a b c

Find the volume of these pyramids.

a b c

2

75 mm

80 mm

1.7 m

0.62 m

11 cm

7 cm

Example 3

Find the volume of this triangular pyramid.

Base is a triangle so A = bh

= × 5 × 7

= 17.5 cm2


= × 17.5 × 8 cm3

= 46.7 cm3

8 cm

5 cm

7 cm

12---

5 cm

7 cm

12---

13---

13---

3

11 cm

8 cm

7 cm

0.9 m

1.5 m

1.4 m

65 mm

32 mm80 mm


Extension

Find the volume of these solids.

a b c

Find the volume of the following composite solids to 1 decimal place.

a b c

Example 4


The measurement for perpendicular height is missing,but can be calculated using Pythagoras’ theorem.

h2 + 62 = 102

∴ h2 = 100 – 36= 64

∴ h = 8 (as h > 0)


= × (π × 62) × 8 cm3

= 96π cm3

! 301.6 cm3

10 cm

6 cm

6

h 10

13---

13---

4

13 cmx cm

x cm

6 cm

7 cm

5 cm 5 cm

5 cmx cm

3 cm

5

5 cm

12 cm

3 cm

8 cm

5 cm

15 c

m

12 c

m

9 cm

6 cm

10 cm


A rectangular swimming pool 6 m by 5 m by 2 m deep, costs $0.50 per cubic metre per month to maintain. What is the cost of maintaining the pool over a year.

Concrete costs $150 per cubic metre. What will it cost to concrete a driveway 20 m long and 3 m wide to a depth of 12 cm?

A rectangular tank is a pyramid with internal dimensions 4 m by 4 m by 2 m. Find its volume.

A conical tank has base area 8 m2 and height 3 m. Find its volume.

A lake has an average depth of 2 m and surface area of 35 ha. Find its volume in m2.

A building is in the shape of a pentagonal prism with a pentagonal pyramid on top. The base area is 140 m2 and the prism and pyramid each have perpendicular height of 8 m. Find the volume of air contained.

Extension

Edge AD of a square-based pyramid is 4 cmlong. The base has sides 2 cm. Find:a the length of BD, using Pythagorasb the altitude AF, using FD, AD and Pythagorasc the volume of the pyramid.

Find the formula for the volume, V, of:a a cone of base radius r and height 2rb a hemispherical bowl of radius rc a cylinder of radius r and height 3r

Find the volumes of the following figures.

a b c

How many cubic metres of concrete would be needed to lay a slab 3000 mm by 4000 mm and 60 mm thick?

What measurements would you need to make to find the volume of a building that is a hexagonal prism with a hexagonal pyramid on top?

A swimming pool has its length doubled, the depth tripled and the width halved. How does the volume of the new pool compare with that of the old pool?

An ice-cream cone in the shape of a cone has the height halved and the radius tripled. How does the volume of the new cone compare with that of the old cone?

12

13

14

15

16

17

18

2 cm

4 cm

A

B C

D

E

F

19

20

2 m

1.4 m

4.2 m 6 m7 cm

9 cm

5 cm

3 cm

1 cm

21

22

23

24








Extension




a b c





12

13

14

15

16

17

18

2 cm

4 cm

A

B C

D

E

F

19

20

2 m

1.4 m

4.2 m 6 m7 cm

9 cm

5 cm

3 cm

1 cm

21

22

23

24








Extension




a b c





12

13

14

15

16

17

18

2 cm

4 cm

A

B C

D

E

F

19

20

2 m

1.4 m

4.2 m 6 m7 cm

9 cm

5 cm

3 cm

1 cm

21

22

23

24


E. SURFACE AREA OF RIGHT CYLINDERSThe formula for the surface area of a cylinder can be developed by ‘cutting’ the cylinder and laying it out flat. The net then gives a formula for the surface area.

A = 2(area of circle) + area of rectangle

= 2 × πr2 + 2πr × h

= 2πr2 + 2πrh

The surface area of a closed cylinder is:

A = 2πrh + 2πr2

The surface area of an open cylinder is:

A = 2πrh

Exercise 8E

Find the surface area of these closed cylinders.

a b c

h

cut

h

r

r

2πr

The curved part is made from arectangle of length 2πr and breadth h.

Remember: Areaof a circle is πr2.

Example 1

Find the surface area of this closed cylinder.

Surface area = 2πr2 + 2πrh= 2π × 62 + 2π × 6 × 15 cm2

! 791.7 cm2

6 m

15 cm

1

☞5 cm

12 cm

3 m

1.4 m112 mm

4 mm


d e f

Find the surface area of the following open cylinders.

a b c

Solve the following problems.a Determine the volume of paint required to cover the outside of a cylindrical container

12 m long with diameter 10 m if each litre of paint covers 15 square metres.b Which has the greater surface area: a cylinder of length 15 cm and radius 8 cm, or a

cylinder of length 8 cm and radius 10 cm?

Find the surface area of:a an open can with radius of 4 cm and 15 cm highb an open-ended pipe of 10 cm radius and 5 m longc

Determine the cost of painting the exterior walls and top of a cylindrical wheat silo that is 40 m high and 20 m in diameter, given that each litre of paint costs $7.25 and covers 8 m2.

18 cm

28 cm

30 cm

10 cm8 m

1.3 m

Example 2

Find the surface area of this open cylinder.

Surface area = 2πrh= 2 × π × 8 × 23= 1156.1 cm2

8 cm

23 cm

2

8 m

13 cm

3 cm20 cm

20 cm

25 cm

3

4

14 cm 20 cm

5


d e f

Find the surface area of the following open cylinders.

a b c

Solve the following problems.a Determine the volume of paint required to cover the outside of a cylindrical container

12 m long with diameter 10 m if each litre of paint covers 15 square metres.b Which has the greater surface area: a cylinder of length 15 cm and radius 8 cm, or a

cylinder of length 8 cm and radius 10 cm?

Find the surface area of:a an open can with radius of 4 cm and 15 cm highb an open-ended pipe of 10 cm radius and 5 m longc

Determine the cost of painting the exterior walls and top of a cylindrical wheat silo that is 40 m high and 20 m in diameter, given that each litre of paint costs $7.25 and covers 8 m2.

18 cm

28 cm

30 cm

10 cm8 m

1.3 m

Example 2

Find the surface area of this open cylinder.

Surface area = 2πrh= 2 × π × 8 × 23= 1156.1 cm2

8 cm

23 cm

2

8 m

13 cm

3 cm20 cm

20 cm

25 cm

3

4

14 cm 20 cm

5

ANSWERS 1 a) 120 m2 b) 6 cm2 c) 40 cm2 d) 54 m2 e) 24 cm2 f) 153.9 cm2 2 a) 112 m3 b) 2261.9 cm3 3 320 cm3 4 a) 564.4 cm3 b) 37.7 cm3 5 a) 102.7 cm3 a) 28 cm3

6 Side length = 11.93733 cm

7 4188.79 cm3

8 3.22 cm

9 a) 68.04 m3 b) 488.2 cm3 c) 22 cm

10 a) 534.1 cm3 b) 2092.3 cm3 11 552.9 cm3 12 19.8 cm

Year 10 Surface Area & Volume 3 -...

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