# CHAPTER 29 VOLUMES AND SURFACE AREAS OF COMMON SOLIDS

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1 CHAPTER 9 VOLUMES AND SURFACE AREAS OF COMMON EXERCISE 14 Page 9 SOLIDS 1. Change a volume of cm to cubic metres. 1m = 10 cm or 1cm = 10 6m 6 Hence, cm = m = 1. m. Change a volume of 5000 mm to cubic centimetres. 1cm = 10 mm or 1mm = 10 cm Hence, 5000 mm = mm = 5 cm. A metal cube has a surface area of 4 cm. Determine its volume. A cube had 6 sides. Area of each side = 4/6 = 4 cm Each side is a square hence the length of a side = 4 = cm Volume of cube = = 8 cm 4. A rectangular block of wood has dimensions of 40 mm by 1 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 = = 840 mm (b) Surface area = (bh + hl + lb) = ( ) = ( ) = 896 = 179 mm , John Bird

2 5. Determine the capacity, in litres, of a fish tank measuring 90 cm by 60 cm by 1.8 m, given 1 litre = 1000 cm. Volume = ( ) cm Tank capacity = cm 1000 cm /litre = 97 litre 6. A rectangular block of metal has dimensions of 40 mm by 5 mm by 15 mm. Determine its volume in cm. Find also its mass if the metal has a density of 9 g/cm. Volume = length breadth width = = mm Mass = density volume = 9 g/cm 15cm = 15 g = cm = 15 cm 7. Determine the maximum capacity, in litres, of a fish tank measuring 50 cm by 40 cm by.5 m (1 litre = 1000 cm ). Volume = cm Tank capacity = cm 1000 cm /litre = 500 litre 8. Determine how many cubic metres of concrete are required for a 10 m long path, 150 mm wide and 80 mm deep. Width = 150 mm = 0.15 m and depth = 80 mm = m Hence, volume of path = length breadth width = = 1.44 m i.e. concrete required = 1.44 m , John Bird

3 9. A cylinder has a diameter 0 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 = 0 mm = cm hence radius, r = / = 1.5 cm and height, h = 50 mm = 5 cm πrh = π = 11.5π = 5. cm, correct to 1 decimal place (a) Volume = ( ) (b) Total surface area = πrh + π r = ( π 1.5 5) + ( π 1.5 ) = 15π + 4.5π = 19.5π = 61. cm 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 1 cm and perpendicular height 5 cm. (a) Volume of right-angled triangular prism = 1 bhl = i.e. Volume = 400 cm (a) Total surface area = area of each end + area of three sides In triangle ABC, AC = AB + BC from which, AC = AB + BC = = 1 cm Hence, total surface area = 1 bh + (AC 80) + (BC 80) + (AB 80) , John Bird

4 i.e. total surface area = 460 cm = (1 5) + (1 80) + (1 80) + (5 80) = A steel ingot whose volume is m is rolled out into a plate which is 0 mm thick and 1.80 m wide. Calculate the length of the plate in metres. Volume of ingot = length breadth width i.e. = l from which, length = = 7.04 m 1. The volume of a cylinder is 75 cm. If its height is 9.0 cm, find its radius. Volume of cylinder, V = π rh i.e. 75 = π r (9.0) 75 from which, r = π 9.0 and radius, r = 75 π 9.0 = 1.6 cm 1. 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 ( ) ( ) πd πd π 8 π 6 Area of cross-section of copper = = = = cm Hence, volume of metal tube = (cross-sectional area) length of pipe = = 8796 cm , John Bird

5 14. The volume of a cylinder is 400 cm. If its radius is 5.0 cm, find its height. Determine also its curved surface area. Volume of cylinder, V = π rh i.e. 400 = π (5.0) h from which, height, 400 h = π (5.0) = cm Curved surface area = πrh = ( π ) = 15.9 cm 15. A cylinder is cast from a rectangular piece of alloy 5 cm by 7 cm by 1 cm. If the length of the cylinder is to be 60 cm, find its diameter. Volume of rectangular piece of alloy = = 40 cm Volume of cylinder = π rh Hence, 40 = 40 7 π r (60) from which, r = π(60) = π and radius, r = 7 π = cm and diameter of cylinder, d = r = =.99 cm 16. Find the volume and the total surface area of a regular hexagonal bar of metal of length m if each side of the hexagon is 6 cm. A hexagon is shown below. In triangle 0BC, tan 0 = x from which, x = tan 0 = cm , John Bird

6 Hence, area of hexagon = = 9.5 cm and volume of hexagonal bar = = cm Surface area of bar = 6[ 0.06 ] + [ ] in metre units = m 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 = cm If length = breadth = x cm and height = 15/10 = 1.5 cm, then x ( 1.5) = from which, x = and x = = 81.6 cm = 8. m Hence, dimensions of square sheet are 8. m by 8. m 18. How long will it take a tap dripping at a rate of 800 mm /s to fill a three-litre can? litre = 000 cm = mm = 106mm Time to fill can = 10 mm 6 800mm /s = 750 s = = 6.5 minutes 19. A cylinder is cast from a rectangular piece of alloy 5.0 cm by 6.50 cm by 19. cm. If the height of the cylinder is to be 5.0 cm, determine its diameter, correct to the nearest centimetre. Volume of cylinder, V = π rh and volume of rectangular prism = cm i.e = π r (5.0) from which, r = π 5.0 and radius, 47 r = π 5.0 =.0 cm 014, John Bird

7 i.e. diameter = r =.0 = 4 cm 0. How much concrete is required for the construction of the path shown, if the path is 1 cm thick? 1 4 π Area of path = (8.5 ) + ( ) + (.1 ) + (.4 [ + 1.]) = = 4.0 m If thickness of path = 1 cm = 0.1 m then Concrete required = volume of path = = 4.08 m , John Bird

8 EXERCISE 15 Page If a cone has a diameter of 80 mm and a perpendicular height of 10 mm, calculate its volume in cm and its curved surface area. 1 1 = = mm = 01.1 cm Volume of cone = π r h π ( 40) ( 10) From diagram below, slant height, l = ( 1 4) + = cm Curved surface area = πrl = π(4)(1.649) = cm. A square pyramid has a perpendicular height of 4 cm. If a side of the base is.4 cm long find the volume and total surface area of the pyramid. A sketch of the pyramid is shown below = 7.68 cm Volume of pyramid = ( )( ) In the sketch, AB = 4 cm and BC =.4/ = 1. cm , John Bird

9 Length AC = ( 4 1.) + = cm 1 Hence, area of a side = (.4 )( ) = 5.01 cm Total surface area of pyramid = 4[ 5.01] + (.4) = 5.81 cm. A sphere has a diameter of 6 cm. Determine its volume and surface area. Volume of sphere = π π r = = 11.1 cm Surface are of sphere = 4πr = 4π = 11.1 cm 6 4. If the volume of a sphere is 566 cm, find its radius. 4 Volume of sphere = π r hence, 566 = 4 π r 566 from which, r = = π and radius, r = = 5.11 cm 5. A pyramid having a square base has a perpendicular height of 5 cm and a volume of 75 cm. Determine, in centimetres, the length of each side of the base. 1 1 If each side of base = x cm then volume of pyramid = A h= xh 1 i.e. 75 = (5) x and 75 x = = 9 5 from which, length of each side of base = 9 = 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 , John Bird

10 π = π Volume of cone = rh ( 40) = 681 mm 7. Determine (a) the volume and (b) the surface area of a sphere of radius 40 mm. (a) Volume of sphere = ( ) (b) Surface are of sphere = ( ) 4 4 πr = π 40 = mm or cm 4πr = 4π 40 = mm or cm 8. The volume of a sphere is 5 cm. Determine its diameter. Volume of sphere = d π π 4 4 r = Hence, 5 = 4 d π from which, d 5 = 4π d and = 5 = 4.65 cm and diameter, d = 4.65 = 8.5 cm 4π 9. Given the radius of the Earth is 680 km, calculate, in engineering notation (a) its surface area in km and (b) its volume in km. 4πr = 4π 680 = km (a) Surface area of Earth = ( ) 4 4 πr = π 680 = km (b) Volume of earth = ( ) 10. An ingot whose volume is 1.5 m 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 = ( ) Volume of x bearings = cm 4 4 πr = π 8 Let number of ball bearings = x , John Bird

11 Hence, [ x ] 10 = π ( ) from which, number of bearings, x = π 8 = 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 cm, determine the tank capacity in litres. Volume of storage tank = π π r = = = 9 m, correct to the nearest cubic metre Volume of tank = 9 106cm If 1 litre = 1000 cm, then capacity of storage tank = litres = litres , John Bird

12 EXERCISE 16 Page Find the total surface area of a hemisphere of diameter 50 mm. Total surface area = [ ] 1 π r + 4 π r = π r + π r = π r = 50 π = 5890 mm or cm. Find (a) the volume and (b) the total surface area of a hemisphere of diameter 6 cm. Volume of hemisphere = 1 (volume of sphere) = πr = π 6.0 = cm Total surface area = curved surface area + area of circle = 1 (surface area of sphere) + πr = 1 (4πr ) + πr = πr + πr = πr 6.0 = π = 84.8 cm. Determine the mass of a hemispherical copper container whose external and internal radii are 1 cm and 10 cm, assuming that 1 cm of copper weighs 8.9 g. Volume of hemisphere = π r = π[ ] 1 10 cm Mass of copper = volume density = [ ] π 1 10 cm 8.9 g/cm = 1570 g = 1.57 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 , John Bird

13 1 1 Volume of bob = π rh+ π r = π ( ) ( 5 ) + π ( ) = 16 4π + π = 9. cm 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.5 m, with a diameter of 15 m. Calculate the surface area of material needed to make the marquee, assuming 1% of the material is wasted in the process The marquee is shown sketched below. Surface area of material for marquee = πrl πrh Hence, surface area = π(7.5)(7.9057) + π(7.5)(.5) = = m + where l = ( 7.5.5) If 1% of material is wasted then amount required = = 9.4 m + = m 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.0 cm , John Bird

14 (iv) a.5 cm by.5 cm square pyramid of perpendicular height 5.0 cm (v) a 4.0 cm by 6.0 cm rectangular pyramid of perpendicular height 1.0 cm (vi) a 4. cm by 4. 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 0 cm (i) A sketch of the cone is shown below. 1 1 = = 670 cm (a) Volume of cone = π r h π ( 8.0) ( 10) (b) Total surface area = πr + πrl where l = ( ) = cm = π ( 8.0) + π ( 8.0)( ) = = 5 cm (ii) (a) Volume of sphere = π = 180 cm (b) Surface area = 7.0 4πr = 4π = 154 cm (iii) (a) Volume of hemisphere = ( ) πr = π.0 = 56.5 cm 1 (4 πr ) + πr = πr = π.0 = 84.8 cm (b) Surface area = ( ) (iv) A sketch of the square pyramid is shown below, where AB = 5.0 cm , John Bird

15 1 (a) Volume of pyramid = (.5 ) ( 5.0 ) = 10.4 cm (b) In the diagram, AC = ( AB BC ) ( ) Surface area = ( ) 1 + = + = = =.0 cm (v) A sketch of the rectangular pyramid is shown below = 96.0 cm (a) Volume of rectangular pyramid = ( )( ) (b) In the diagram, AC = ( 1.0.0) and AD = ( 1.0.0) + = 1.69 cm + = cm Hence, surface area = ( ) = = 146 cm (vi) The square pyramid is shown sketched below , John Bird

16 Diagonal on base = ( ) = cm hence, BC = = cm Hence, perpendicular height, h = ( ) = cm 1 (a) Volume of pyramid = ( 4. ) ( ) (b) AD = ( ) + = = 86.5 cm Hence, surface area = ( ) = = 14 cm (vii) A pyramid having an octagonal base is shown sketched below. One sector is shown in diagram (p) below, where from which, x =.5 tan.5 = cm.5 tan.5 = x 1 Hence, area of whole base = = cm 1 (a) Volume of pyramid = ( )( 0 ) = 805 cm (p) (q) (b) From diagram (q) above, y = ( ) + = cm , John Bird

17 Total surface area = = = 59 cm 7. A metal sphere weighing 4 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/m 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. mass 4 kg Volume of sphere = = = 0.00m = cm = 000 cm density 8000 kg/m 4 (a) Volume of sphere = π r i.e. 000 = 4 π r and radius, r = 000 4π = cm Hence, the diameter of the sphere, d = r = = 17.9 cm (b) Volume of cone = = 550 from which, perpendicular height of cone, h = 1 1 cm = πrh = π ( 8.0) ( 8.0) h 550 = 8.0 cm π 8. Find the volume of a regular hexagonal pyramid if the perpendicular height is 16.0 cm and the side of base is.0 cm. The hexagonal base is shown sketched below. From the diagram, tan 0 = 1.5 h from which, h = 1.5 tan 0 =.598 cm , John Bird

18 Hence, area of hexagonal base = =.87 cm 1 and volume of hexagonal pyramid = (.87 )( 16.0 ) = 15 cm 9. A buoy consists of a hemisphere surmounted by a cone. The diameter of the cone and hemisphere is.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 = ( ) =.80 m 1 1 Volume of buoy = π r + π rh= π ( 1.5) + π ( 1.5) (.80) = = 10. m πrl πr = π π 1.5 Surface area = ( ) ( )( ) ( ) = 5π +.15π = 8.15π = 5.5 m 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. m determine the capacity of the tank in litres (1 litre = 1000 cm ). The petrol container is shown sketched below , John Bird

19 4 + = + Volume of container = π r π rh π ( 0.6) π ( 0.6) ( 5.0) = 0.88π + 1.8π = m = cm and tank capacity = cm cm /litre = 6560 litres 11. The diagram below shows a metal rod section. Determine its volume and total surface area ( ) = ( ) Volume of rod = πrh l b w π ( ) ( ) 1 1 = 50π +500 = cm Surface area = ( π rh) + π r + (.50.0) + (.5 100) + (.0 100) = π(1.0)(100) + π( 1.0) = 107 cm , John Bird

20 1. Find the volume (in cm ) of the die-casting shown below. The dimensions are in millimetres. 1 π Volume = ( 0 50) = π = mm = cm = 0.7 cm 1. 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 cm ), (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 per square metre, assuming that 5% extra metal is required due to wastage , John Bird

21 (a) In cm, volume of air = π ( 00) + π + π ( 150) + π ( 150) = π π π π = π cm = π cm 1000 cm /litre = 1458 litres, correct to the nearest litre (b) In m, cross-sectional area of the sheet metal 1 4 = π ( 0.5)( ) + 4π ( 0.5 ) + π ( 0.5)( 1.5) + π ( 0.4)( 1.5) + π ( ) = π π π + 1.π π =.11π = m = 9.77 m correct to significant figures (c) Sheet metal required = m Cost of sheet metal = = , John Bird

22 EXERCISE 17 Page The radii of the faces of a frustum of a cone are.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. 1 h R Rr r Volume of frustum = π ( + + ) = = 147 cm = π ( 5.0)( 4.0 (4.0)(.0).0 ) π ( 5.0)( 8.0) From the diagram below, slant length, l = ( ) = 9 Total surface area = ( ) πl R+ r + πr + πr = π ( 9 )( ) + π (.0) + π ( 4.0) =.1π + 4π + 16π = 164 cm. 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 BG BH AH = from which, height, CG = ( ) (.5) BH 8.0 BG = = 10.0 cm AH , John Bird

23 Height of complete pyramid = = 18.0 cm 1 Volume of large pyramid = ( 9.0 ) ( 18.0 ) = 486 cm 1 Volume of small triangle cut off = ( 5.0 ) ( 10.0 ) Hence, volume of frustum = = 40 cm A cross-section of the frustum is shown below. = 8. cm BC = ( 8 ) + = 8.46 cm = cm Area of four trapeziums = ( )( ) Total surface area of frustum = = 7 cm. A cooling tower is in the form of a frustum of a cone. The base has a diameter of.0 m, the top has a diameter of 14.0 m and the vertical height is 4.0 m. Calculate the volume of the tower and the curved surface area. A sketch of the cooling tower is shown below , John Bird

24 1 h R Rr r Volume of frustum = π ( + + ) 1 π + + = π = m = ( 4.0)( 16.0 (16.0)(7.0) 7.0 ) 8 ( 417) ( ) Slant length, l = ( AB BC ) ( ) + = = 5.6 m Curved surface area = l R r ( )( ) π ( + ) = π = π = 185 m 4. A loudspeaker diaphragm is in the form of a frustum of a cone. If the end diameters are 8.0 cm and 6.00 cm and the vertical distance between the ends is 0.0 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 = ( AC AB ) ( ) + = = 1.95cm Curved surface area = πl (R + r) = π(1.95)( ) = 1707 cm , John Bird

25 5. A rectangular prism of metal having dimensions 4. cm by 7. cm by 1.4 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 cm and 8 cm respectively, find the thickness of the frustum. Volume of frustum of pyramid = 90% of volume of rectangular prism = 0.9( ) = cm. A cross-section of the frustum of the square pyramid is shown below (not to scale). By similar triangles: CG BG BH AH BH h BG = AH.5 = 0.6 h = from which, height, CG = ( ) ( 1.5) = 4.1h cm Volume of large pyramid = ( ) ( h h) h = 1.8 h cm Volume of small triangle cut off = ( ) ( ) Hence, = 4.1h 1.8h =.h Thus, thickness of frustum, h = = cm 6. Determine the volume and total surface area of a bucket consisting of an inverted frustum of a cone, of slant height 6.0 cm and end diameters 55.0 cm and 5.0 cm. A sketch of the bucket is shown below , John Bird

26 Thickness of frustum, h = ( 6.0 ( ) ) 1 h R Rr r Volume of frustum = π ( + + ) Total surface area = ( ) = 4.58 cm 1 π + + = cm correct to 4 = ( 4.58)( 7.5 (7.5)(17.5) 17.5 ) πl R+ r + π r π π 17.5 = ( )( ) ( ) = 160π π = 196.5π = 6051 cm significant figures 7. A cylindrical tank of diameter.0 m and perpendicular height.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.0 m, respectively, determine the vertical height required. Volume of cylinder = rh ( ) ( ) π = π = π m A sketch of the frustum of a cone is shown below , John Bird

27 1 h R Rr r Volume of frustum = π = π ( + + ) 1 1 = π ( h)( (1.0)(0.5) ) = π h( 1.75) from which, thickness of frustum = vertical height, h = π 9 = = 5.14 m 1 π 1.75 ( 1.75) , John Bird

28 EXERCISE 18 Page Determine the volume and surface area of a frustum of a sphere of diameter cm, if the radii of the ends of the frustum are 14.0 cm and.0 cm and the height of the frustum is 10.0 cm. π h h r r 6 Volume of frustum of sphere = ( ) = π ( 10.0) ( ( 14.0 ) + (.0 ) ) 6 = 1105 cm = cm correct to 4 significant figures Surface area of frustum = πrh = π ( ) = 150 cm. Determine the volume (in cm ) and the surface area (in cm ) of a frustum of a sphere if the diameter of the ends are 80.0 mm and 10.0 mm and the thickness is 0.0 mm. The frustum is shown shaded in the cross-section sketch below (in cm units). π h h r r 6 Volume of frustum of sphere = ( ) = ( ) π in cm units π = ( ) + + = 59. cm Surface area of frustum = πrh From the above diagram, r = 6 + OP (1) , John Bird

29 r = 4 + OQ Now OQ = + OP Hence, ( ) Equating equations (1) and () gives: r 6 OP = OP () + = 4 + (.0 + OP) i.e. 6 + OP = ( OP) + OP Thus, 6 = 5 + 6(OP) from which, OP = = 6 6 From equation (1), r 11 = and radius, r = = 6.74 cm Surface area of frustum = πrh = π(6.74)(.0) = 118. cm. 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 1 5 of the surface area of the sphere, find the height h and the volume of the frustum. 4 4 πr = π 6.50 = 1150 cm Volume of sphere = ( ) 4πr = 4π 6.50 = 51 cm Surface area = ( ) The frustum is shown shaded in the sketch below , John Bird

30 1 Curved surface area = πrh = ( 51 ) cm 5 in this case 1 i.e. π(6.50)h = ( 51 ) 5 from which, height, h = 1 ( 51 ) 5 π 6.50 ( ) =.60 cm r = = cm From the diagram, 1 ( ) π h h r r 6 Volume of frustum of sphere = ( ) = π (.60 ) (.60 + ( 6.50 ) + ( ) ) 6 = 6.7 cm 4. A sphere has a diameter of.0 mm. Calculate the volume (in cm ) of the frustum of the sphere contained between two parallel planes distances 1.0 mm and 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, 1 ( ) r = = mm and ( ) r = = mm π h h r r 6 Volume of frustum of sphere = ( ) , John Bird

31 = π (.0) (.0 + ( ) + ( ) ) 6 = 1487 mm = cm 5. A spherical storage tank is filled with liquid to a depth of 0.0 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 cm ). A cross-section of the storage tank is shown sketched below. Volume of water = π r + volume of frustum From the diagram, 1 ( ) r = = 1.1 cm π h h r r 6 Volume of frustum of sphere = ( ) = π ( 7.5 ) ( (.50 ) + ( 1.1 ) ) 6 = cm Hence, total volume of water = (.50) π = 5 41 cm Number of litres of water = 5 41cm 1000cm /litre = 5.4 litres , John Bird

32 EXERCISE 19 Page 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 = ( ) + = 8.06 cm x A A A 6 Using the prismoidal rule, volume of frustum = [ ] π π π = ( 7.0) + 4 ( 8.06) + ( 7.0) = 1500 cm. Determine the volume of a cone of perpendicular height 16.0 cm and base diameter 10.0 cm by using the prismoidal rule. x A A A 6 Using the prismoidal rule: Volume, V = [ ] Area, 10.0 A 1 = πr1 = π = 5π Area, A 5.0 = πr = π = 6.5π Area, ( ) A = πr = π 0 = 0 and x = 16.0 cm , John Bird

33 x 16.0 A A A π π 6 6 Hence, volume of cylinder, V = [ ] = [ 5 + 4(6.5 ) + 0] = π = cm. A bucket is in the form of a frustum of a cone. The diameter of the base is 8.0 cm and the diameter of the top is 4.0 cm. If the length is.0 cm, determine the capacity of the bucket (in litres) using the prismoidal rule (1 litre = 1000 cm ). The bucket is shown in the sketch below. The radius of the midpoint is = 17.5 cm x A A A 6 Using the prismoidal rule, volume of frustum = [ ].0 6 = π ( 1.0) + 4π ( 17.5) + π ( 14.0) = 1 00 cm Hence, capacity of bucket = 1 00cm 1000cm /litre = 1.0 litres 4. Determine the capacity of a water reservoir, in litres, the top being a 0.0 m by 1.0 m rectangle, the bottom being a 0.0 m by 8.0 m rectangle and the depth being 5.0 m (1 litre = 1000 cm ). The water reservoir is shown sketched below , John Bird

34 A mid-section will have dimensions of = 5 m by = 10 m x A A A 6 Using the prismoidal rule, volume of frustum = [ ] = ( 0 1) + 4( 5 10) + ( 0 8) = m = cm Hence, capacity of water reservoir = cm cm /litre = litre , John Bird

35 EXERCISE 10 Page The diameter of two spherical bearings are in the ratio :5. What is the ratio of their volumes? Diameters are in the ratio :5 Hence, ratio of their volumes = :5 i.e. 8:15. An engineering component has a mass of 400 g. If each of its dimensions are reduced by 0%, determine its new mass. New mass = ( 0.7) 400 = = 17. g , John Bird

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