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CHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON

SOLIDS EXERCISE 124 Page 293

1. Change a volume of 1 200 000 cm3 to cubic metres.

3 6 31m 10 cm= or 3 6 31cm 10 m−= Hence, 1 200 000 3cm = 1 200 000 6 310 m−× = 1.2 3m 2. Change a volume of 5000 3mm to cubic centimetres.

3 3 31cm 10 mm= or 3 3 31mm 10 cm−= Hence, 5000 3mm = 5000 3 310 mm−× = 5 3cm 3. A metal cube has a surface area of 24 2cm . Determine its volume. A cube had 6 sides. Area of each side = 24/6 = 4 2cm Each side is a square hence the length of a side = 4 = 2 cm Volume of cube = 2 × 2 × 2 = 8 3cm 4. A rectangular block of wood has dimensions of 40 mm by 12 mm by 8 mm. Determine (a) its volume in cubic millimetres and (b) its total surface area in square millimetres. (a) Volume of cuboid = l b h× × = 40 × 12 × 8 = 3840 3mm (b) Surface area = 2(bh + hl + lb) = 2(12 × 8 + 8 ×40 + 40 × 12) = 2(96 + 320 + 480) = 2 × 896 = 1792 2mm

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5. Determine the capacity, in litres, of a fish tank measuring 90 cm by 60 cm by 1.8 m, given

1 litre = 1000 3cm . Volume = (90 × 60 × 180) 3cm

Tank capacity = 3

3

90 60 180cm1000cm /litre× × = 972 litre

6. A rectangular block of metal has dimensions of 40 mm by 25 mm by 15 mm. Determine its

volume in 3cm . Find also its mass if the metal has a density of 9 g/cm3.

Volume = length × breadth × width = 40 × 25 × 15 = 15 000 3mm = 15 000 × 310− 3cm = 15 3cm Mass = density × volume = 9 3g/cm × 15 3cm = 135 g 7. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by 2.5 m

(1 litre = 1000 cm3).

Volume = 50 × 40 × 250 3cm

Tank capacity = 3

3

50 40 250cm1000cm /litre× × = 500 litre

8. Determine how many cubic metres of concrete are required for a 120 m long path, 150 mm wide

and 80 mm deep.

Width = 150 mm = 0.15 m and depth = 80 mm = 0.080 m Hence, volume of path = length × breadth × width = 120 × 0.15 × 0.080 = 1.44 3m i.e. concrete required = 31.44m

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9. A cylinder has a diameter 30 mm and height 50 mm. Calculate (a) its volume in cubic centimetres, correct to 1 decimal place, and (b) the total surface area in square centimetres, correct to 1 decimal place. Diameter = 30 mm = 3 cm hence radius, r = 3/2 = 1.5 cm and height, h = 50 mm = 5 cm

(a) Volume = ( )22 1.5 5 11.25r hπ π π= × × = = 35.3 3cm , correct to 1 decimal place

(b) Total surface area = 2πrh + 22 rπ = (2 × π × 1.5 × 5) + (2 × π × 21.5 ) = 15π + 4.5π = 19.5π = 61.3 2cm 10. Find (a) the volume and (b) the total surface area of a right-angled triangular prism of length 80 cm whose triangular end has a base of 12 cm and perpendicular height 5 cm.

(a) Volume of right-angled triangular prism = 12

bhl = 12× 12 × 5 × 80

i.e. Volume = 2400 3cm (a) Total surface area = area of each end + area of three sides

In triangle ABC, 2 2 2AC AB BC= + from which, AC = 2 2 2 25 12AB BC+ = + = 13 cm

Hence, total surface area = 122

bh

+ (AC × 80) + (BC × 80) + (AB × 80)

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= (12 × 5) + (13 × 80) + (12 × 80) + (5 × 80) = 60 + 1040 + 960 + 400 i.e. total surface area = 2460 2cm 11. A steel ingot whose volume is 2 2m is rolled out into a plate which is 30 mm thick and 1.80 m wide. Calculate the length of the plate in metres. Volume of ingot = length × breadth × width i.e. 2 = l × 0.030 × 1.80

from which, length = 2

0.030 1.80× = 37.04 m 12. The volume of a cylinder is 75 3cm . If its height is 9.0 cm, find its radius. Volume of cylinder, V = 2r hπ

i.e. 75 = 2 (9.0)rπ

from which, 2 759.0

rπ

=×

and radius, r = 759.0π ×

= 1.63 cm

13. Calculate the volume of a metal tube whose outside diameter is 8 cm and whose inside diameter

is 6 cm, if the length of the tube is 4 m.

Outer diameter, D = 8 cm and inner diameter, d = 6 cm

Area of cross-section of copper = ( ) ( )2 22 2 8 6

4 4 4 4D d π ππ π

− = −

= 50.2655 – 28.2743 = 21.9912 2cm Hence, volume of metal tube = (cross-sectional area) × length of pipe = 21.9912 × 400 = 8796 cm3

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14. The volume of a cylinder is 400 cm3. If its radius is 5.20 cm, find its height. Determine also its

curved surface area.

Volume of cylinder, V = 2r hπ

i.e. 400 = 2(5.20) hπ

from which, height, 2

400(5.20)

hπ

=×

= 4.709 cm

Curved surface area = 2πrh = (2 × π × 5.20 × 4.709) = 153.9 2cm 15. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by 12 cm. If the length of the

cylinder is to be 60 cm, find its diameter.

Volume of rectangular piece of alloy = 5 × 7 × 12 = 420 3cm Volume of cylinder = 2r hπ

Hence, 420 = 2 (60)rπ from which, 2 420 7(60)

rπ π

= = and radius, r = 7π

= 1.4927 cm

and diameter of cylinder, d = 2r = 2 × 1.4927 = 2.99 cm 16. Find the volume and the total surface area of a regular hexagonal bar of metal of length 3 m if

each side of the hexagon is 6 cm.

A hexagon is shown below.

In triangle 0BC, tan 30° = 3

x from which, x = 3

tan 30° = 5.196 cm

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Hence, area of hexagon = 16 6 5.1962 × ×

= 93.53 2cm

and volume of hexagonal bar = 93.53 × 300 = 28 060 3cm Surface area of bar = [ ] [ ]46 0.06 3 2 93.53 10−× + × in metre units = 1.099 2m 17. A block of lead 1.5 m by 90 cm by 750 mm is hammered out to make a square sheet 15 mm thick. Determine the dimensions of the square sheet, correct to the nearest centimetre. Volume of block of lead = length × breadth × height = 150 × 90 × 75 3cm If length = breadth = x cm and height = 15/10 = 1.5 cm, then ( )2 1.5x = 150 × 90 × 75

from which, 2 150 90 751.5

x × ×= and x = 150 90 75

1.5× × = 821.6 cm = 8.22 m

Hence, dimensions of square sheet are 8.22 m by 8.22 m 18. How long will it take a tap dripping at a rate of 800 3mm /s to fill a three-litre can?

3 litre = 3000 3cm = 3 3 6 33000 10 mm 3 10 mm× = ×

Time to fill can = 6 3

3

3 10 mm800mm /s× = 3750 s = 3750

60 = 62.5 minutes

19. A cylinder is cast from a rectangular piece of alloy 5.20 cm by 6.50 cm by 19.33 cm. If the height of the cylinder is to be 52.0 cm, determine its diameter, correct to the nearest centimetre. Volume of cylinder, V = 2r hπ and volume of rectangular prism = 5.20 × 6.50 ×19.33 3cm

i.e. 5.20 × 6.50 ×19.33 = 2 (52.0)rπ

from which, 2 5.20 6.50 19.3352.0

rπ× ×

=×

and radius, 5.20 6.50 19.3352.0

rπ× ×

=×

= 2.0 cm

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i.e. diameter = 2 × r = 2 × 2.0 = 4 cm 20. How much concrete is required for the construction of the path shown, if the path is 12 cm thick?

Area of path = (8.5 × 2) + ( )21 24π

+ (3.1 × 2) + (2.4 × [2 + 1.2])

= 17 + 3.1416 + 6.2 + 7.68 = 34.022 2m If thickness of path = 12 cm = 0.12 m then

Concrete required = volume of path = 34.022 × 0.12 = 4.08 3m

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EXERCISE 125 Page 296

1. If a cone has a diameter of 80 mm and a perpendicular height of 120 mm, calculate its volume in

cm3 and its curved surface area.

Volume of cone = ( ) ( )221 1 40 1203 3

r hπ π= = 201 061.9 3mm = 201.1 3cm

From diagram below, slant height, l = ( )2 212 4+ = 12.649 cm

Curved surface area = πrl = π(4)(12.649) = 159.0 2cm 2. A square pyramid has a perpendicular height of 4 cm. If a side of the base is 2.4 cm long find the

volume and total surface area of the pyramid.

A sketch of the pyramid is shown below.

Volume of pyramid = ( )( )1 2.4 2.4 4

3× = 7.68 3cm

In the sketch, AB = 4 cm and BC = 2.4/2 = 1.2 cm

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Length AC = ( )2 24 1.2+ = 4.176 cm

Hence, area of a side = ( )( )1 2.4 4.1762

= 5.01 2cm

Total surface area of pyramid = [ ] ( )24 5.01 2.4+ = 25.81 2cm

3. A sphere has a diameter of 6 cm. Determine its volume and surface area.

Volume of sphere = 3

34 4 63 3 2

rπ π =

= 113.1 3cm

Surface are of sphere = 2

2 64 42

rπ π =

= 113.1 2cm

4. If the volume of a sphere is 566 cm3, find its radius.

Volume of sphere = 343

rπ hence, 566 = 343

rπ

from which, 3 3 566 135.122554

rπ

×= =

and radius, r = 3 135.12255 = 5.131 cm 5. A pyramid having a square base has a perpendicular height of 25 cm and a volume of 75 3cm . Determine, in centimetres, the length of each side of the base.

If each side of base = x cm then volume of pyramid = 21 13 3

A h x h× × =

i.e. 75 = 21 (25)3

x and 2 3 7525

x ×= = 9

from which, length of each side of base = 9 = 3 cm 6. A cone has a base diameter of 16 mm and a perpendicular height of 40 mm. Find its volume correct to the nearest cubic millimetre.

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Volume of cone = ( )2

21 1 16 403 3 2

r hπ π =

= 2681 3mm

7. Determine (a) the volume and (b) the surface area of a sphere of radius 40 mm.

(a) Volume of sphere = ( )334 4 403 3

rπ π= = 268 083 3mm or 268.083 3cm

(b) Surface are of sphere = ( )224 4 40rπ π= = 20 106 2mm or 201.06 2cm 8. The volume of a sphere is 325 cm3. Determine its diameter.

Volume of sphere = 3

34 43 3 2

drπ π =

Hence, 325 = 34

3 2dπ

from which, 3 3 325

2 4d

π× =

and 33 325

2 4d

π×

= = 4.265 cm and diameter, d = 2 × 4.265 = 8.53 cm

9. Given the radius of the Earth is 6380 km, calculate, in engineering notation (a) its surface area

in 2km and (b) its volume in 3km .

(a) Surface area of Earth = ( )224 4 6380rπ π= = 6 2512 10 km×

(b) Volume of earth = ( )334 4 63803 3

rπ π= = 12 31.09 10 km× 10. An ingot whose volume is 1.5 3m is to be made into ball bearings whose radii are 8.0 cm. How

many bearings will be produced from the ingot, assuming 5% wastage?

Volume of one ball bearing = ( )334 4 83 3

rπ π= Let number of ball bearings = x

Volume of x bearings = 0.95 × 1.5 × 610 3cm

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Hence, 0.95 × 1.5 × 610 = ( )34 8 [ ]3

xπ

from which, number of bearings, x = 6

3

0.95 1.5 10 34 8π× × ×

× = 664

11. A spherical chemical storage tank has an internal diameter of 5.6 m. Calculate the storage

capacity of the tank, correct to the nearest cubic metre. If 1 litre = 1000 3cm , determine the tank

capacity in litres.

Volume of storage tank = 3

34 4 5.63 3 2

rπ π =

= 91.95 = 92 3m , correct to the nearest cubic metre

Volume of tank = 92 6 310 cm×

If 1 litre = 1000 3cm , then capacity of storage tank = 692 10

1000× litres = 92 000 litres

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EXERCISE 126 Page 300

1. Find the total surface area of a hemisphere of diameter 50 mm.

Total surface area = [ ]2 2 2 2 21 4 2 32

r r r r rπ π π π π+ = + =

= 2503

2π

= 5890 2mm or 58.90 2cm

2. Find (a) the volume and (b) the total surface area of a hemisphere of diameter 6 cm.

Volume of hemisphere = 12

(volume of sphere)

= 23πr3 = 2

3π

36.02

= 56.55 cm3

Total surface area = curved surface area + area of circle

= 12

(surface area of sphere) + πr2

= 12

(4πr 2 ) + πr2 = 2πr2 + πr2 = 3πr2 = 3π26.0

2

= 84.82 cm2 3. Determine the mass of a hemispherical copper container whose external and internal radii are

12 cm and 10 cm, assuming that 1 cm3 of copper weighs 8.9 g.

Volume of hemisphere = [ ]3 3 3 32 2 12 10 cm3 3

rπ π= −

Mass of copper = volume × density = [ ]3 3 3 32 12 10 cm 8.9g/cm3π − × = 13570 g = 13.57 kg

4. A metal plumb bob comprises a hemisphere surmounted by a cone. If the diameter of the

hemisphere and cone are each 4 cm and the total length is 5 cm, find its total volume.

The plumb bob is shown sketched below.

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Volume of bob = ( ) ( ) ( )2 32 31 2 1 22 5 2 2

3 3 3 3r h rπ π π π+ = − +

= 1643

π π+ = 29.32 3cm

5. A marquee is in the form of a cylinder surmounted by a cone. The total height is 6 m and the

cylindrical portion has a height of 3.5 m, with a diameter of 15 m. Calculate the surface area of

material needed to make the marquee, assuming 12% of the material is wasted in the process

The marquee is shown sketched below.

Surface area of material for marquee = 2rl rhπ π+ where l = ( )2 27.5 2.5+ = 7.9057 m Hence, surface area = π(7.5)(7.9057) + 2π(7.5)(3.5) = 186.2735 + 164.9336 = 351.2071 2m If 12% of material is wasted then amount required = 1.12 × 351.2071 = 393.4 2m 6. Determine (a) the volume and (b) the total surface area of the following solids:

(i) a cone of radius 8.0 cm and perpendicular height 10 cm

(ii) a sphere of diameter 7.0 cm

(iii) a hemisphere of radius 3.0 cm

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(iv) a 2.5 cm by 2.5 cm square pyramid of perpendicular height 5.0 cm

(v) a 4.0 cm by 6.0 cm rectangular pyramid of perpendicular height 12.0 cm

(vi) a 4.2 cm by 4.2 cm square pyramid whose sloping edges are each 15.0 cm

(vii) a pyramid having an octagonal base of side 5.0 cm and perpendicular height 20 cm

(i) A sketch of the cone is shown below.

(a) Volume of cone = ( ) ( )221 1 8.0 10

3 3r hπ π= = 670 3cm

(b) Total surface area = 2r rlπ π+ where l = ( )2 210 8.0+ = 12.80625 cm = ( ) ( )( )28.0 8.0 12.80625π π+ = 201.062 + 321.856 = 523 2cm

(ii) (a) Volume of sphere = 34 7.0

3 2π

= 180 3cm

(b) Surface area = 2

2 7.04 42

rπ π =

= 154 2cm

(iii) (a) Volume of hemisphere = ( )332 2 3.03 3

rπ π= = 56.5 3cm

(b) Surface area = ( )22 2 21 (4 ) 3 3 3.02

r r rπ π π π+ = = = 84.8 2cm

(iv) A sketch of the square pyramid is shown below, where AB = 5.0 cm

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(a) Volume of pyramid = ( ) ( )21 2.5 5.03

= 10.4 3cm

(b) In the diagram, AC = ( ) ( )2 2 2 25.0 1.25AB BC+ = + = 5.15388

Surface area = ( )2 12.5 4 2.5 5.153882 + × ×

= 6.25 + 25.7694 = 32.0 2cm

(v) A sketch of the rectangular pyramid is shown below.

(a) Volume of rectangular pyramid = ( )( )1 6.0 4.0 12.03

× = 96.0 3cm

(b) In the diagram, AC = ( )2 212.0 3.0+ = 12.3693 cm and AD = ( )2 212.0 2.0+ = 12.1655 cm

Hence, surface area = ( ) 1 16.0 4.0 2 4.0 12.3696 2 6.0 12.16552 2 × + × × + × ×

= 24 + 49.4784 + 72.993 = 146 2cm (vi) The square pyramid is shown sketched below.

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Diagonal on base = ( )2 24.2 4.2 5.9397+ = cm hence, BC = 1 5.93972× = 2.96985 cm

Hence, perpendicular height, h = ( )2 215.0 2.96985− = 14.703 cm

(a) Volume of pyramid = ( ) ( )21 4.2 14.7033

= 86.5 3cm

(b) AD = ( )2 214.703 2.1+ = 14.8522

Hence, surface area = ( )2 14.2 4 4.2 14.85222 + × ×

= 17.64 + 124.75858

= 142 2cm (vii) A pyramid having an octagonal base is shown sketched below.

One sector is shown in diagram (p) below, where 2.5tan 22.5x

° =

from which, x = 2.5tan 22.5°

= 6.0355 cm

Hence, area of whole base = 18 5.0 6.03552 × ×

= 120.71 2cm

(a) Volume of pyramid = ( )( )1 120.71 203

= 805 3cm

(p) (q) (b) From diagram (q) above, y = ( )2 220 6.0355+ = 20.891 cm

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Total surface area = 120.71 + 18 5.0 20.8912 × ×

= 120.71 + 417.817

= 539 2cm 7. A metal sphere weighing 24 kg is melted down and recast into a solid cone of base radius 8.0 cm.

If the density of the metal is 8000 kg/m3 determine (a) the diameter of the metal sphere and (b) the

perpendicular height of the cone, assuming that 15% of the metal is lost in the process.

Volume of sphere = 3 6 33

mass 24kg 0.003m 0.003 10 cmdensity 8000kg/m

= = = × = 3000 3cm

(a) Volume of sphere = 343

rπ i.e. 3000 = 343

rπ

and radius, r = 33000 3

4π×

= 8.947 cm

Hence, the diameter of the sphere, d = 2r = 2 × 8.947 = 17.9 cm

(b) Volume of cone = 0.85 × 3000 = 2550 3cm = ( )221 1 8.03 3

r h hπ π=

from which, perpendicular height of cone, h = ( )2

2550 38.0π× = 38.0 cm

8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the

side of base is 3.0 cm.

The hexagonal base is shown sketched below.

From the diagram, tan 30° = 1.5

h from which, h = 1.5

tan 30° = 2.598 cm

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Hence, area of hexagonal base = 16 3.0 2.5982 × ×

= 23.3827 2cm

and volume of hexagonal pyramid = ( )( )1 23.3827 16.03

= 125 3cm

9. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere

is 2.5 m and the slant height of the cone is 4.0 m. Determine the volume and surface area of the

buoy.

The buoy is shown in the sketch below.

Height of cone, h = ( )2 24.0 1.25− = 3.80 m

Volume of buoy = ( ) ( ) ( )3 23 22 1 2 11.25 1.25 3.803 3 3 3

r r hπ π π π+ = +

= 4.0906 + 6.2177 = 10.3 3m

Surface area = ( ) ( )( ) ( )221 4 1.25 4.0 2 1.252

rl rπ π π π+ = +

= 5π + 3.125π = 8.125π = 25.5 2m 10. A petrol container is in the form of a central cylindrical portion 5.0 m long with a hemispherical

section surmounted on each end. If the diameters of the hemisphere and cylinder are both 1.2 m

determine the capacity of the tank in litres (1 litre = 1000 cm3).

The petrol container is shown sketched below.

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Volume of container = ( ) ( ) ( )3 23 22 42 0.6 0.6 5.03 3

r r hπ π π π + = +

= 0.288π + 1.8π = 6.55965 3m = 6 36.55965 10 cm×

and tank capacity = 6 3

3

6.56 10 cm1000cm /litre

× = 6560 litres

11. The diagram below shows a metal rod section. Determine its volume and total surface area.

Volume of rod = ( ) ( )221 1( ) 1.0 100 (2.5 2.0 100)2 2

r h l b wπ π+ × × = + × ×

= 50π +500 = 657.1 3cm

Surface area = ( ) ( ) ( ) ( )21 12 2 2 2.50 2.0 2 2.5 100 2.0 1002 2

r h rπ π + + × + × + ×

= π(1.0)(100) + π( )21.0 10 500 200+ + + = 1027 2cm

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12. Find the volume (in 3cm ) of the die-casting shown below. The dimensions are in millimetres.

Volume = ( )21100 60 25 30 502π× × + × ×

= 150 000 + 22 500π = 220 685.835 3mm = 3 3220 685.835 10 cm−× = 220.7 3cm 13. The cross-section of part of a circular ventilation shaft is shown below, ends AB and CD being

open. Calculate (a) the volume of the air, correct to the nearest litre, contained in the part of the

system shown, neglecting the sheet metal thickness, (given 1 litre = 1000 3cm ),

(b) the cross-sectional area of the sheet metal used to make the system, in square metres, and

(c) the cost of the sheet metal if the material costs £11.50 per square metre, assuming that 25%

extra metal is required due to wastage.

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(a) In 3cm , volume of air = ( ) ( ) ( )2 3 2 250 1 4 50 50 80200 150 150

2 4 3 2 2 2π π π π

+ + +

= 125 000π + 5208.33π + 93 750π + 240 000π = 463 958.33π 3cm

= 3

3

463 958.33 cm1000cm /litre

π = 1458 litres, correct to the nearest litre

(b) In 2m , cross-sectional area of the sheet metal

= ( )( ) ( ) ( )( ) ( )( ) ( )2 2 212 0.25 2 4 0.25 2 0.25 1.5 2 0.4 1.5 0.4 0.254

π π π π π + + + + −

= π + 0.0625π + 0.75π + 1.2π + 0.0975π = 3.11π = 9.77035 2m = 9.77 2m correct to 3 significant figures (c) Sheet metal required = 9.77035 × 1.25 2m Cost of sheet metal = 9.77035 × 1.25 × £11.50 = £140.45

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EXERCISE 127 Page 305

1. The radii of the faces of a frustum of a cone are 2.0 cm and 4.0 cm and the thickness of the

frustum is 5.0 cm. Determine its volume and total surface area.

A sketch of a side view of the frustum is shown below.

Volume of frustum = ( )2 213

h R Rr rπ + +

= ( )( ) ( )( )2 21 15.0 4.0 (4.0)(2.0) 2.0 5.0 28.03 3π π+ + = = 147 3cm

From the diagram below, slant length, l = ( )2 25.0 2.0 29+ =

Total surface area = ( ) 2 2l R r r Rπ π π+ + + = ( )( ) ( ) ( )2 229 4.0 2.0 2.0 4.0π π π+ + +

= 32.31π + 4π + 16π = 164 2cm 2. A frustum of a pyramid has square ends, the squares having sides 9.0 cm and 5.0 cm, respectively.

Calculate the volume and total surface area of the frustum if the perpendicular distance between

its ends is 8.0 cm.

A side view of the frustum of the pyramid is shown below.

By similar triangles: CG BHBG AH

= from which, height, CG = ( ) 8.0( ) 2.52.0

BHBGAH

=

= 10.0 cm

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Height of complete pyramid = 10.0 + 8.0 = 18.0 cm

Volume of large pyramid = ( ) ( )21 9.0 18.03

= 486 3cm

Volume of small triangle cut off = ( ) ( )21 5.0 10.03

= 83.33 3cm

Hence, volume of frustum = 486 – 83.33 = 403 3cm A cross-section of the frustum is shown below.

BC = ( )2 28 2+ = 8.246 cm

Area of four trapeziums = ( )( )14 5.0 9.0 8.246 230.8882 + =

2cm

Total surface area of frustum = 2 29.0 5.0 230.888+ + = 337 2cm 3. A cooling tower is in the form of a frustum of a cone. The base has a diameter of 32.0 m, the top

has a diameter of 14.0 m and the vertical height is 24.0 m. Calculate the volume of the tower and

the curved surface area.

A sketch of the cooling tower is shown below.

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Volume of frustum = ( )2 213

h R Rr rπ + +

= ( )( ) ( )2 21 24.0 16.0 (16.0)(7.0) 7.0 8 4173π π+ + = = 10 480 3m

Slant length, l = ( ) ( )( )22 2 224.0 16.0 7.0 25.632mAB BC+ = + − =

Curved surface area = ( )( )( ) 25.632 16.0 7.0 599.54l R rπ π π+ = + = = 1852 2m 4. A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 28.0 cm

and 6.00 cm and the vertical distance between the ends is 30.0 cm, find the area of material

needed to cover the curved surface of the speaker.

A sketch of the loudspeaker diaphragm is shown below.

Slant length, l = ( ) ( )( )22 2 230.0 14.0 3.0 31.953cmAC AB+ = + − =

Curved surface area = πl (R + r) = π(31.953)(14.0 + 3.0) = 1707 2cm

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5. A rectangular prism of metal having dimensions 4.3 cm by 7.2 cm by 12.4 cm is melted down and

recast into a frustum of a square pyramid, 10% of the metal being lost in the process. If the ends of

the frustum are squares of side 3 cm and 8 cm respectively, find the thickness of the frustum.

Volume of frustum of pyramid = 90% of volume of rectangular prism = 0.9(4.3 × 7.2 × 12.4) = 345.514 3cm . A cross-section of the frustum of the square pyramid is shown below (not to scale).

By similar triangles: CG BHBG AH

= from which, height, CG = ( )( ) 1.52.5

BH hBGAH

=

= 0.6 h

Volume of large pyramid = ( ) ( )21 8 0.63

h h+ = 34.133h 3cm

Volume of small triangle cut off = ( ) ( )21 3 0.63

h = 1.8 h 3cm

Hence, 345.514 = 34.133h – 1.8h = 32.333h

Thus, thickness of frustum, h = 345.51432.333

= 10.69 cm

6. Determine the volume and total surface area of a bucket consisting of an inverted frustum of a

cone, of slant height 36.0 cm and end diameters 55.0 cm and 35.0 cm.

A sketch of the bucket is shown below.

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Thickness of frustum, h = ( )2 236.0 (27.5 17.5)− − = 34.5832 cm

Volume of frustum = ( )2 213

h R Rr rπ + +

= ( )( )2 21 34.5832 27.5 (27.5)(17.5) 17.53π + + = 55 910 3cm correct to 4

significant figures Total surface area = ( ) 2l R r rπ π+ + = ( )( ) ( )236.0 27.5 17.5 17.5π π+ + = 1620π + 306.25π = 1926.25π = 6051 2cm 7. A cylindrical tank of diameter 2.0 m and perpendicular height 3.0 m is to be replaced by a tank of

the same capacity but in the form of a frustum of a cone. If the diameters of the ends of the

frustum are 1.0 m and 2.0 m, respectively, determine the vertical height required.

Volume of cylinder = ( ) ( )22 31.0 3.0 3r h mπ π π= = A sketch of the frustum of a cone is shown below.

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Volume of frustum = 3π = ( )2 213

h R Rr rπ + +

= ( )( ) ( )2 21 11.0 (1.0)(0.5) 0.5 1.753 3

h hπ π+ + =

from which, thickness of frustum = vertical height, h = ( )3 9

1 1.751.753

π

π= = 5.14 m

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EXERCISE 128 Page 307

1. Determine the volume and surface area of a frustum of a sphere of diameter 47.85 cm, if the radii

of the ends of the frustum are 14.0 cm and 22.0 cm and the height of the frustum is 10.0 cm.

Volume of frustum of sphere = ( )2 2 21 23 36h h r rπ

+ +

= ( ) ( ) ( )( )2 2210.0

10.0 3 14.0 3 22.06

π+ +

= 11205 3cm = 11 210 3cm correct to 4 significant figures

Surface area of frustum = 2πrh = 2π ( )47.85 10.02

= 1503 2cm

2. Determine the volume (in cm3) and the surface area (in cm2) of a frustum of a sphere if the

diameter of the ends are 80.0 mm and 120.0 mm and the thickness is 30.0 mm.

The frustum is shown shaded in the cross-section sketch below (in cm units).

Volume of frustum of sphere = ( )2 2 21 23 36h h r rπ

+ +

= ( ) 2 2

23.0 8.0 12.03.0 3 36 2 2

π + + in cm units

= ( )9 48 1082π

+ + = 259.2 3cm

Surface area of frustum = 2πrh From the above diagram, 2 2 26r OP= + (1)

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496

2 2 24r OQ= + Now OQ = 3 + OP Hence, ( )22 24 3.0r OP= + + (2) Equating equations (1) and (2) gives: 2 26 OP+ = ( )224 3.0 OP+ + i.e. 236 OP+ = 216 9 6( )OP OP+ + + Thus, 36 = 25 + 6(OP)

from which, OP = 36 25 116 6−

=

From equation (1), 2

2 2 1166

r = +

and radius, r = 2

2 1166

+

= 6.274 cm

Surface area of frustum = 2πrh = 2π(6.274)(3.0) = 118.3 2cm 3. A sphere has a radius of 6.50 cm. Determine its volume and surface area. A frustum of the sphere

is formed by two parallel planes, one through the diameter and the other at a distance h from the

diameter. If the curved surface area of the frustum is to be 15

of the surface area of the sphere, find

the height h and the volume of the frustum.

Volume of sphere = ( )334 4 6.503 3

rπ π= = 1150 3cm

Surface area = ( )224 4 6.50rπ π= = 531 2cm The frustum is shown shaded in the sketch below.

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497

Curved surface area = 2πrh = ( ) 21 531 cm5

in this case

i.e. 2π(6.50)h = ( )1 5315

from which, height, h = ( )( )

1 5315

2 6.50π = 2.60 cm

From the diagram, ( )2 21 6.50 2.60r = − = 5.957 cm

Volume of frustum of sphere = ( )2 2 21 23 36h h r rπ

+ +

= ( ) ( ) ( )( )2 222.60

2.60 3 6.50 3 5.9576

π+ +

= 326.7 3cm 4. A sphere has a diameter of 32.0 mm. Calculate the volume (in cm3) of the frustum of the sphere

contained between two parallel planes distances 12.0 mm and 10.00 mm from the centre and on

opposite sides of it.

A cross-section of the frustum is shown in the sketch below.

From the diagram, ( )2 21 16.0 12.0r = − = 10.583 mm and ( )2 22 16.0 10.0r = − = 12.490 mm

Volume of frustum of sphere = ( )2 2 21 23 36h h r rπ

+ +

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498

= ( ) ( ) ( )( )2 2222.0

22.0 3 10.583 3 12.4906

π+ +

= 14837 3mm = 14.84 3cm 5. A spherical storage tank is filled with liquid to a depth of 30.0 cm. If the inner diameter of the

vessel is 45.0 cm, determine the number of litres of liquid in the container (1 litre = 1000 cm3).

A cross-section of the storage tank is shown sketched below.

Volume of water = 32

3rπ + volume of frustum

From the diagram, ( )2 21 22.50 7.50r = − = 21.21 cm

Volume of frustum of sphere = ( )2 2 21 23 36h h r rπ

+ +

= ( ) ( ) ( )( )2 227.5

7.5 3 22.50 3 21.216

π+ +

= 11 485 3cm

Hence, total volume of water = ( )32 22.503π + 11 845 = 35 341 3cm

Number of litres of water = 3

3

35 341cm1000cm /litre

= 35.34 litres

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EXERCISE 129 Page 309

1. Use the prismoidal rule to find the volume of a frustum of a sphere contained between two parallel

planes on opposite sides of the centre, each of radius 7.0 cm and each 4.0 cm from the centre.

The frustum of the sphere is shown sketched in cross-section below.

Radius, r = ( )2 27.0 4.0+ = 8.062 cm

Using the prismoidal rule, volume of frustum = [ ]1 2 346x A A A+ +

= ( ) ( ) ( )2 2 28.0 7.0 4 8.062 7.06

π π π + +

= 1500 3cm 2. Determine the volume of a cone of perpendicular height 16.0 cm and base diameter 10.0 cm by

using the prismoidal rule.

Using the prismoidal rule: Volume, V = [ ]1 2 346x A A A+ +

Area, 2

21 110.0 25

2A rπ π π = = =

Area,

2

22 25.0 6.252

A rπ π π = = =

Area, ( )223 3 0 0A rπ π= = = and x = 16.0 cm

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Hence, volume of cylinder, V = [ ] [ ]1 2 316.04 25 4(6.25 ) 0

6 6x A A A π π+ + = + +

= 16.06

× 50π = 418.9 3cm

3. A bucket is in the form of a frustum of a cone. The diameter of the base is 28.0 cm and the

diameter of the top is 42.0 cm. If the length is 32.0 cm, determine the capacity of the bucket (in

litres) using the prismoidal rule (1 litre = 1000 cm3).

The bucket is shown in the sketch below.

The radius of the midpoint is 14 212+ = 17.5 cm

Using the prismoidal rule, volume of frustum = [ ]1 2 346x A A A+ +

= ( ) ( ) ( )2 2 232.0 21.0 4 17.5 14.06

π π π + +

= 31 200 3cm

Hence, capacity of bucket = 3

3

31 200cm1000cm /litre

= 31.20 litres

4. Determine the capacity of a water reservoir, in litres, the top being a 30.0 m by 12.0 m rectangle,

the bottom being a 20.0 m by 8.0 m rectangle and the depth being 5.0 m (1 litre = 1000 cm3).

The water reservoir is shown sketched below.

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501

A mid-section will have dimensions of 30 202+

= 25 m by 12 82+

= 10 m

Using the prismoidal rule, volume of frustum = [ ]1 2 346x A A A+ +

= ( ) ( ) ( )5.0 30 12 4 25 10 20 86 × + × + ×

= 1266.7 3m = 61266.7 10× 3cm

Hence, capacity of water reservoir = 6 3

3

1266.7 10 cm1000cm /litre

× = 61.267 10 litre×

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EXERCISE 130 Page 310

1. The diameter of two spherical bearings are in the ratio 2:5. What is the ratio of their volumes?

Diameters are in the ratio 2:5 Hence, ratio of their volumes = 3 32 : 5 i.e. 8:125 2. An engineering component has a mass of 400 g. If each of its dimensions are reduced by 30%,

determine its new mass.

New mass = ( )30.7 400 0.343 400× = × = 137.2 g