SURFACE AREAS AND VOLUMES

CHAPTER 1 Points to Remember : 1. Cuboid (i) Volume = lbh (ii) Curved surface area = h (l + b) (iii) Total surface area = (lb + bh + lh) (iv) Diagonal l b h. Cube (i) Volume = a (ii) Curved surface area = 4a (iii) Total surface area = 6a (iv) Diagonal. Cylinder (i) Volume = r h. a (ii) Curved surface area = rh (iii) Total surface area = r (r + h) 4. Cone 1 (i) Volume r h SURFACE AREAS AND VOLUMES (ii) slant height, l h r (iii) curved surface area r l (iv) Total surface area r ( l r) MATHEMATICS IX SURFACE AREAS AND VOLUMES 1

5. Sphere 4 (i) Volume r (ii) Total surface area = 4 r 6. Hemi-sphere (i) Volume r (ii) Curved surface area = r (iii) Total surface area = r ILLUSTRATIVE EXAMPLES Example 1. A cubical box has each edge 10 cm and another cuboidal box is 1.5 cm long, 10 cm wide and 8 cm high. (i) Which box has the greater lateral surface area and by how much? (ii) Which box has the smaller surface area and by how much? NCERT Solution. (i) Lateral surface area (L 1 ) of cubical box = 4 (edge) = 4 (10) cm = 400 cm and, lateral surface area (L ) of cuboidal box = (length + breadth) height = (1.5 + 10) 8 cm = 60 cm Clearly, L 1 > L. Now, L 1 L = 400 cm 60 cm = 40 cm The cubical box has larger lateral surface area and is greater by 40 cm. (ii) Total surface area of the cubical box (S 1 ) = 6 (edge) = 6 (10) cm = 600 cm Total surface area of the cuboidal box (S ) = (lb + lh + bh) = (1.5 10 + 10 8 + 8 1.5) cm = (15 + 80 + 100) cm = 610 cm Clearly, S > S 1. S S 1 = 610 cm 600 cm = 10 cm Thus, the cuboidal box has greater surface area and is greater by 10 cm. Example. Two cubes each of 15 cm edge are joined end to end. Find the surface area of the resulting cuboid. Solution. here, l = length of resulting cuboid = 15 cm + 15 cm = 0 cm b = breadth of resulting cuboid = 15 cm h = height of resulting cuboid = 15 cm SURFACE AREAS AND VOLUMES MATHEMATICS IX

Example. Solution. Surface area of resulting cuboid = (lb + bh + lh) = (0 15 + 15 15 + 0 15) cm = (450 + 5 + 450) cm = (115) cm = 50 cm Ans. A plastic box 1.5 m long, 1.5 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box. (ii) The cost of sheet for it, if a sheet measuring 1 m costs Rs. 0. NCERT We have, Length; l = 1.5 m, Breadth, b = 1.5 m and Depth = Height, h = 0.65 m (i) Since the plastic box is open at the top, Plastic sheet required for making such box = (l + b) h + lb = (1.5 + 1.5) 0.65 m + 1.5 1.5 m =.5 0.65 m + 1.85 m =.55 m + 1.85 m = 5.45 m (ii) Cost of 1 m of sheet = Rs. 0 Total cost of 5.45 m of sheet = Rs. 5.45 0 = Rs. 109 Ans. Example 4. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height.5 m, with base dimensions 4 m m? NCERT Solution. here, l = 4m, b = m, h =.5 m. Since there is no tarpaulin for the floor. Tarpaulin required = [ (l + b) h + lb] = [ (4 + ).5 + 4 ] m Example 5. Solution. = (.5 + 1) m = (5 + 1) m = 4 m Ans. The sum of length, breadth and height of a cuboid is 1 cm and the length of its diagonal is 1 cm. Find the surface area of the cuboid. Let the length, breadth and height of the cuboid be l cm, b cm and h cm respectively. Then, l + b + h = 1...(1) Now, diagonal = 1 cm l b h 1 l + b + h = 144...() Now, l + b + h = 1 Squaring both sides, we get (l + b + h) = (1) l + b + h + lb + bh + lh = 441 144 + (lb + bh + lh) = 441 ( using ()) (lb + bh + lh) = 441 144 = 9 Surface area of the cuboid is 9 cm Ans. MATHEMATICS IX SURFACE AREAS AND VOLUMES

Example 6. Aggarwal sweets stall was placing an order for making card board boxes for packing their sweets. Two size of boxes were required. The bigger of dimensions 5 cm 0 cm 5 cm and the smaller of dimensions 15 cm 1 cm 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs. 5 for 1000 cm, find the cost of cardboard required for supplying 00 boxes of each kind. Solution. Surface area of Ist box = (5 0 + 0 5 + 5 5) cm = (500 + 100 + 15) cm = 1450 cm Surface area of IInd box = (15 1 + 1 5 + 15 5) cm = (180 + 60 + 5) cm = 60 cm Total combined surface area = 1450 cm + 60 cm = 080 cm Example. Solution. Example 8. Area of overlaps = 5% of 080 cm 5 080 cm 104 cm 100 Total surface area of boxes (one of each kind) = (080 + 104) cm = 184 cm Surface area of 00 boxes of each kind = 00 184 cm = 65500 cm Now, cost of cardboard for 1000 cm = Rs. 5 5 Cost of cardboard for 1 cm Rs. 1000 5 Cost of cardboard for 65500 cm Rs. 65500 1000 = Rs. 6 Ans. A cylindrical pillar is 50 cm in diameter and.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 1.50 per m. NCERT Diameter of cylindrical pillar = 50 cm 50 radius (r) cm 5 cm 0.5 m also, height (h) =.5 m Now, curved surface = rh 0.5.5 m = 5.5 m Cost of painting 1 m = Rs. 1.50 Cost of painting 5.5 m = Rs. 1.50 5.5 = Rs. 68.5 Ans. In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 0 cm and height of 0 cm. A margin of.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. NCERT 4 SURFACE AREAS AND VOLUMES MATHEMATICS IX

0 Solution. Here, radius ( r) cm 10 cm and, height (h) = 0 cm +.5 cm = 5 cm (for margin) Cloth required for covering the lampshade = Its curved surface area = rh Example 9. Solution. 10 5 cm = 00 cm Ans. It is required to make a closed cylindrical tank of height 10 cm and base diameter 140 cm from a metal sheet. How many square metres of the sheet is required for the same? We have, diameter of base = 140 cm. 140 radius of base cm 0cm and, height of cylinder = 10 cm. Total surface area of required tank = r (r + h) 0 (0 10) cm 8600 440 190 cm 8600 cm m 8.6 m Ans. 10000 Example 10. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 5 competitors, how much cardboard was required to be bought for the competition? NCERT Solution. Cardboard required by each competitor = curved surface area of one penholder + base area = rh + r, where r = cm, h = 10.5 cm 10.5 () MATHEMATICS IX SURFACE AREAS AND VOLUMES 5 cm = (198 + 8.8) cm = 6.8 cm (approx) Cardboard required for 5 competitors = 5 6.8 cm = 90 cm (approx) Ans. Example 11. The diameter of a roller is 84 cm and its length is 100 cm. It takes 00 complete revolution to move once over to level a playground. Find the area of the playground in m. 84 Solution. Radius of roller ( r) cm 4 cm Length of the cylindrical roller (h) = 100 cm. Area moved by the roller in one revolution rh 4100 cm

Area moved in 00 revolutions 4100 00 cm = 44 180000 cm = 44 18 m = 9 m Ans. Example 1. Find (i) the curved surface area of a cylindrical petrol storage tank that is 4. m in diameter and 1 4.5 m high. (ii) how much steel was actually used if of the steel actually used was wasted in 1 making the closed tank? NCERT 4. Solution. Here, r m. 1 m, h 4. 5 m (i) Curved surface area = r h. 1 4. 5 m = 59.4 m (ii) Total surface area of closed tank r h r r ( r h). 1(. 1 4. 5) m = 8.1 m Let the total sheet used for making the cylindrical tank be x m. Given, wastage m. 1 x according to given question, x 8. 1 1 11 8. 11 x 8. 1 x 95. 04 m 1 11 Steel used for making closed tank including wastage = 95.04 m Ans. Example 1. Curved surface area of a cone is 08 cm and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. NCERT Solution. (i) Here, r l = 08 cm, l = 14 cm. 14 r 08 08 r cm. and, (ii) Total surface area = r (r + l) ( 14) cm = 1 cm = 46 cm Example 14. How many meters of cloth, 5 m wide, will be required to make a conical tent, the radius of whose base is m and height is 4 m? Solution. Radius of the tent, r = m height of the tent, h = 4 m slant height, l r h 4 m 65 m 5 m 6 SURFACE AREAS AND VOLUMES MATHEMATICS IX x

curved surface area = r l i.e., area of the cloth = 550 m 5 m 550 m area 550 Now, length of cloth required m 110 m width 5 length of cloth required = 110 m. Example 15. A conical tent is 10 m high and the radius of its base is 4 m. Find : (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m canvas is Rs. 0. NCERT Solution. (i) here, r = 4 m, h = 10 m Let l be the slant height of the cone. then, l = h + r l h r 4 10 56 100 66 6 m. (ii) Canvas required to make the conical tent = curved surface of the cone rl 4 6 cm Now, Rate of canvas for 1 m = Rs. 0 Total cost of canvas Rs. 4 6 0 = Rs. 180 Ans. Example 16. A joker s cap is in the form of a right circular cone of base radius cm and height 4 cm. Find the area of the sheet required to make 10 such caps. NCERT Solution. here, radius of cap (r) = cm height of cap (h) = 4 cm Let l be the slant height. then, l h r 4 MATHEMATICS IX SURFACE AREAS AND VOLUMES 56 49 65 5 cm Sheet required for one cap = curved surface of the cone = rl 5 cm 550 cm Sheet required for 10 such caps = 10 550 cm = 5500 cm. Ans. Example 1. A bus stop in barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 1 per m, what will be the cost of painting all these cones? (Use =.14 and 1. 04 1. 0 ). NCERT 40 Solution. here, radius ( r ) cm 0 cm 0. m and height (h) = 1 m slant height ( l) r h 0. 04 1 1. 04 1. 0 m

Now, curved surface of 1 cone = rl 0. 1. 0 m Curved surface of 50 such cones 50 0. 1. 0 m Now, cost of painting 1 m = Rs. 1 Total cost of painting Rs. 150 0.1.0 = Rs. 84.68 (approx) Ans. Example 18. A corn cob (see figure), shaped somewhat like a cone, has the radius of its broadest end as.1 cm and length as 0 cm. If each 1 cm of the surface of the cob carries an average of four grains, find how many grains you would find on entire cob? (NCERT) Solution. We have, r =.1 cm, h = 0 cm let, slant height be l cm. then, l r h (.1) (0) cm 404.41 cm 0.11 cm curved surface area of corn cob = r l 4.41 400 cm.1 0.11 cm = 1.6 cm Now, number of grains on 1 cm = 4 number of grains on 1.6 cm = 4 1.6 = 50.904 51 Hence, total number of grains on the corn cob = 51 Ans. Example 19. The surface area of a sphere is 154 cm. Find its radius. NCERT Solution. Let the radius of the sphere be r cm. then, 4 r = 154 (given) r r 154 154 49 4 4 4 49 cm cm 4 radius of the sphere cm. 8 SURFACE AREAS AND VOLUMES MATHEMATICS IX

Example 0. The radius of a spherical baloon increases from cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. NCERT Solution. Let S 1 and S be the total surface area in two cases of r = cm and R = 14 cm. S 1 = 4 r = 4 () cm and S = 4 R = 4 (14) cm S Required ratio 1 4π 1 i.e.1: 4 Ans. S 4π1414 4 Example 1. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. NCERT Solution. Let the diameter of earth be R and that of the moon will be 4 R R R The radii of moon and earth are and respectively. 8 R 4 1 Ratio of their surface area 8 64 1 4 1 1 i.e. 1 : 16 Ans. R 64 1 16 4 4 Example. A hemispherical bowl is made of steel, 0.5 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of bowl. NCERT Solution. Inner radius, r = 5 cm Thickness of Steel = 0.5 cm Outer radius, R = (r + 0.5) cm = (5 + 0.5) cm = 5.5 cm Outer curved surface r 5. 55. 5cm = 1.5 cm Example. The internal and external diameters of a hollow hemispherical vessel are 0 cm and 8 cm respectively. Find the cost of painting the vessel all over at 15 paisa per cm. Solution. Outer radius of vessel, R = 14 cm Inner radius of vessel, r = 10 cm Area of the outer surface = R = (14) cm = 9 cm Area of the inner surface = r = (10) cm = 00 cm Area of the ring at the top = (R r ) = (14 10 ) cm = (14 10) (14 + 10) cm = 96 cm Total area to be painted = 9 cm + 00 cm + 96 cm = 688 cm Now, cost of painting 1 cm = 15 paisa 15 100 MATHEMATICS IX SURFACE AREAS AND VOLUMES 9 Rs.

15 Cost of painting 688 cm 688 π Rs 100 15 688 Rs 100 = 4.4 Rs. Ans. Example 4. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m = 1000 l). NCERT Solution. Here, l = 6m, b = 5m and h = 4.5 m Volume of the tank = lbh = (6 5 4.5) m = 15 m The tank can hold = 15 1000 litres = 15000 litres of water. ( 1 m = 1000 litres) Example 5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively.5 m and 10 m. NCERT Solution. Given capacity of a cuboidal tank 50000 50000 l m 50 m 1000 Let the breadth of cuboidal tank be b m. according to given question, we have.5 b 10 = 50 50 5 b 50 b b 5 breadth of the tank is m. Ans. Example 6. A river m deep and 40 m wide is flowing at the rate of km per hour. How much water will fall into the sea in a minute? NCERT Solution. Volume of water that flows in 1 hour (60 minutes) = volume of water of a cuboid whose dimensions are m, 40 m and 000 m. ( km = 000 m) = 40 000 m Volume of water that flows in 1 minute 40 000 m 4000 m Ans. 60 Example. Three cubes whose edges are cm, 4 cm and 5 cm respectively are melted and recast into a single cube. find the surface area of the new cube. Solution. Let x cm be the edge of new cube. Then, volume of the new cube = sum of the volumes of three cubes. x = + 4 + 5 = + 64 + 15 = 16 = (6) x = 6 cm. Edge of the new cube is 6 cm. and, surface area of the new cube = 6 (6) cm = 16 cm Ans. Example 8. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 0 m 15 m 6 m. For how many days will the water of this tank last? NCERT Solution. Here l = 0 m, b = 15 m, and h = 6 m Capacity of the tank = lbh = (0 15 6) m = 1800 m Water requirement per person per day = 150 litres 10 SURFACE AREAS AND VOLUMES MATHEMATICS IX

Water required for 4000 person per day = (4000 150) l 4000150 m 1000 Number of days the water will last 600 m Capacity of tank Total water required per day 1800 0 600 Thus, the water will last for 0 days Ans. Example 9. A godown measures 60 m 5 m 10 m. Find the maximum number of wooden crates each measuring 1.5 m 1.5 m 0.5 m that can be stored in the godown. NCERT Solution. Volume of the godown = (60 5 10) m = 15000 m Volume of 1 crate = (1.5 1.5 0.5) m = 0.95 m Number of crate that can be stored in the godown Volume of the godown Volume of 1 crate 15000 16000 Ans. 0. 95 Example 0. If the lateral surface of cylinder is 94. cm and its height is 5 cm, then find (i) radius of its base (ii) its volume (use =.14). NCERT Solution. (i) Let r be the radius of the base and h be the height of the cylinder. Then, Lateral surface = 94. cm rh = 94..14 r 5 = 94. 94. r. 145 Thus, the radius of its base = cm. (ii) Volume of the cylinder = r h = (.14 5) cm = 141. cm Ans. Example 1. It costs Rs. 00 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 0 per m, Find : (i) inner curved surface area of the vessel (ii) radius of the base (iii) capacity of the vessel NCERT Solution. Total cost of painting (i) inner curved surface area of the vessel Rate of painting 00 m 0 110 m MATHEMATICS IX SURFACE AREAS AND VOLUMES 11

(ii) Let r be the radius of the base and h be the height of the cylindrical vessel. rh = 110 r 10 110 110 r 1. 5 10 4 Thus, the radius of the base = 1.5 m (iii) Capacity of the vessel = r h 10 m 4 4 = 96.5 m Example. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it? NCERT Solution. Capacity of a closed cylindrical vessel = 15.4 litres 1 15. 4 m 0.0154 m 1000 Let r be the radius of the base and h be the height of the vessel. Then, Volume = r h = r 1 = r ( h = 1m) r = 0.0154 r 0. 0154 r 0. 0154 0. 0049 r 0. 0049 0. 0 Thus, the radius of the base of vessel = 0.0 m. Metal sheet needed to make the vessel = Total surface area of the vessel = rh + r = r (h + r) 0.0 (1 0.0) m = 44 0.01 1.0 m = 0.408 m Example. A patient in a hospital in given soup daily in a cylindrical bowl of diameter cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 50 patients? NCERT Solution. Diameter of the cylindrical bowl = cm Radius cm Height of serving bowl = 4 cm. 1 SURFACE AREAS AND VOLUMES MATHEMATICS IX

Soup served to 1 patient = Volume of the bowl = r h 4 cm = 154 cm Soup served to 50 patients = (50 154) cm = 8500 cm = 8.5 l. Ans. Example 4. The inner diameter of a cylindrical wooden pipe is 4 cm and its outer diameter is 8 cm. The length of the pipe is 5 cm. Find the mass of the pipe, if 1 cm of wood has a mass of 0.6 gm. NCERT 4 Solution. here, inner radius ( r) cm 1 cm 8 and outer radius (R) = cm 14 cm h = length of the pipe = 5 cm. volume of wood used in making the pipe R h r h h ( R r 5[14 ) 1 ]cm 5 (14 1) (14 1) cm 5 6 cm 50 cm Now, 1 cm of wood = 0.6 g 50 cm of wood = 0.6 50 g = 4.0 g =.4 kg. Ans. Example 5. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is mm and the diameter of the graphite is 1 mm. If the length of the pencil is 10 cm, find the weight of the whole pencil if the specific gravity of the wood is 0. gm/cm and that of the graphite is.1 gm/cm. NCERT Solution. For graphite cylindrical rod : radius (r) of graphite cylinder 1 1 1 cm cm 10 0 and, length of graphite rod (h) = 10 cm. volume of graphite cylindrical rod = r h 1 0 10 cm Weight of graphite used for pencil = volume specific gravity 1 1 10. 1 gm ( 0 0 = 0.165 gm..1gm) MATHEMATICS IX SURFACE AREAS AND VOLUMES 1 1 cm

Again, for pencil including graphite rod, we have, radius of pencil (R) mm 0 and, length of pencil (h) = 10 cm volume of pencil = R h 0 10 cm volume of wood used for pencil R h r h cm 10 cm 0 0 11 1 48 cm 0 1 1 1 cm 0 0 10 11 1 weight of wood 48 0. gm ( 1 cm 0. gm) 0 =.64 gm. Total weight =.64 gm + 0.165 gm =.805 gm Ans. Example 6. A well of diameter m is dug 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment. Solution. Radius of the well ( r) m, height (h) = 14 m Volume of the earth taken out of the well r h 14 m Outer radius of the embankment 99 m 11 R m 4 m m Area of embankment = outer area inner area R r 11 4 88 m Height of the embankment 11 11 m 9 8 m 1.15 m Ans. 14 SURFACE AREAS AND VOLUMES MATHEMATICS IX Volume Area 99 88

Example. The height of a cone is 15 cm. If its volume is 150 cm. Find the radius of the base (Use =.14). NCERT Solution. Here h = 15 cm and volume = 150 cm Let the radius of the base of the cone be r cm. Now, Volume = 150 cm 1 r h 150 1. 14 r 15 150 150 r 100. 14 5 r 100 10 Thus, the radius of the base of cone is 10 cm Ans. Example 8. The radius and height of a right circular cone are in the ratio of 5 : 1. If its volume is 51 cm, find the slant height and radius of the base of the cone. (use =.14) Solution. Let the radius of cone be 5x and height be 1 x. Volume of the cone 1 according to question, we get 1. 14 (5x) r h (1 x) 51 1 14 5x 1x 51 100 51 x 8 x 14 radius of base = 5x = 5 cm = 10 cm and, height = 1 x = 1 cm = 4 cm slant height r h 10 4 100 56 66 6 radius of cone = 10 cm and slant height = 6 cm Ans. Example 9. The volume of a right circular cone is 9856 cm. If the diameter of the base is 8 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone NCERT Solution. Given, volume of cone = 9856 cm 8 and, radius of base (r) cm 14 cm (i) we know, volume of cone 1 1 1414 h 9856 r h MATHEMATICS IX SURFACE AREAS AND VOLUMES 15

9856 h cm 48 cm. 14 (ii) Slant height of a cone (l) h r (48) (iii) Curved surface area of a cone = r l 16 SURFACE AREAS AND VOLUMES MATHEMATICS IX (14) 04 196 cm cm 500 cm 50 cm 1450 cm 00 cm. Example 40. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. NCERT Solution. Diameter of the base of the cone = 10.5 m radius 10. 5 r m 5.5 m Height of the cone = m 1 Volume of the cone To find the slant height l r h 1 5. 5 5. 5 m = 86.65 m We have, l h r (5.5) = 9 +.565 = 6.565 l 6. 565 6. 046 m (approx.) Canvas required to protect wheat from rain = Curved surface area rl 5. 5 6. 046 m = 99. m (approx) Example 41. A right triangle ABC with sides 5 cm, 1 cm and 1 cm is revolved about the side 1 cm. Find the volume of the solid so obtained? Solution. The solid obtained is a cone with r = 5 cm and h = 1 cm. Volume 1 r h 1.14 551 cm 100.14 cm = 14 cm Ans.

Example 4. If the surface area of a sphere is 616 cm, find its volume. Solution. Let r be the radius of sphere. then, 4 r = 616 616 616 r 49 4 4 r 49 cm cm volume of sphere 4 4 r cm = 14. cm Ans. Example 4. The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? NCERT Solution. Let the radius of earth be R. R then, radius of moon. 4 4 volume of earth R 4 and, volume of moon. 4 volume of earth volume of moon R 4 R 4 R 4 64 1 i.e., volume of earth is 64 times the volume of the moon. 1 i.e., volume of moon is times that of earth. 64 Example 44. The diameter of a metallic ball is 4. cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm? NCERT Solution. Diameter of the ball = 4. cm Radius 4. cm.1 cm 4 Volume of the ball r 4.1.1.1 cm 8.808 cm Now, Density of metal = 8.9 gm per cm Mass of the ball = 8.808 8.9 g = 45.91 g = 45.4 g (approx) Ans. MATHEMATICS IX SURFACE AREAS AND VOLUMES 1

Example 45. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the volume of the iron used to make the tank. NCERT Solution. Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel, then R = 1.01 m and r = 1 m ( as thickness = 1cm = 0.01 m) Volume of iron used = External volume Internal volume R r [(1.01) ( R r (1) ] m 44 (1.00011) m 1 ) 44 0.0001 m 1 = 0.0648 m (approx) Example 46. A dome of a building is in the form of hemi-sphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is Rs. per square meter, find: (i) the inside surface area of the dome (ii) the volume of the air inside the dome. NCERT Solution. Let r be the inner radius of the hemispherical dome. Then, inside surface area of the hemisphere = r. Since, at the rate of Rs. per square metre, the total cost of white-wash is Rs. 498.96, 498.96 surface area of hemisphere m 49.48 m according to question, r 49. 48 49. 48 r 9. 69 r 9. 69 m 6. m (i) Inside surface are of the dome = r = 49.48 m (calculated above) (ii) Inside volume of the dome r (6.) m = 5.908 m Ans. Example 4. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface are S. Find the: (i) radius r of the new sphere. (ii) ratio of S and S. NCERT 4 Solution. (i) Volume of solid spheres of radius r r...(1) 4 Volume of the new sphere of radius r r...() 18 SURFACE AREAS AND VOLUMES MATHEMATICS IX

According to the problem, we have, r 4 4 r r r (r) r r S 4r r r 1 (ii) Required ratio 1 : 9 S 4r (r) 9r 9 Example 48. A wooden bookshelf has external dimensions as follows : Height = 110 cm, Depth = 5 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 0 paisa per cm and the rate of painting is 10 paisa per cm, find the total expenses required for polishing and painting the surface of the bookshelf. NCERT Solution. Area to be polished = (110 85 + 85 5 + 5 110 + 4 5 5 + 110 5) cm = (950 + 450 + 5500 + 1500 + 1100) cm = 100 cm 0 cost of polishing @ 0 paisa per cm = 100 Rs. 440 100 Also, Area to be painted = (6 5 0 + 90 0 + 5 90) cm Cost of painting @ 10 paisa per cm = (9000 + 600 + 650) cm = 1950 cm 10 1950 Rs. Rs. 195 100 Total expense = Rs. 440 + Rs. 195 = Rs. 65 Ans. Example 49. The front compound wall of a house is decorated by wooden spheres of diameter 1 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height cm and is to be painted black. Find the cost of paint required if silver paint costs 5 paisa per cm and black paint costs 5 paisa per cm. NCERT MATHEMATICS IX SURFACE AREAS AND VOLUMES 19

Solution. Clearly, we have to subtract the area of the circle on which sphere is resting while calculating the cost of silver paint. Surface area to be painted silver = 8 (curved surface area of the sphere area of circle on which sphere is resting) 1 = 8 (4 R r ) where R cm, r 1. 5 cm 441 8π 4.5 cm 4 Cost of silver paint @ 5 paisa per cm 8π (441.5) cm 5 Rs. 8 48.5 Rs. 5.86 (approx) 100 Again, surface area to be painted black = 8 curved surface area of cylinder = 8 rh = 8 1.5 cm 58 cm Cost of black paint @ 5 paisa per cm 5 Rs. 58 Rs. 6.40 100 Total cost of painting = Rs. 5.86 + Rs. 6.40 0 SURFACE AREAS AND VOLUMES MATHEMATICS IX 8π (48.5) cm = Rs. 84.6 (approx) Ans. Example 50. The diameter of a sphere is decreased by 5%. By what percent does its curved surface decrease? NCERT Solution. Let d be the diameter of the sphere. Then, its surface area d 4 d On decreasing its diameter by 5%, New diameter d 5 5 d 5% of d d d d 100 100 1 d 4

d1 1 d New surface area 4 4 4 Decrease in surface area d 9d 9 4.. d 64 16 9 d 16 MATHEMATICS IX SURFACE AREAS AND VOLUMES 1 9 1 d 16 Percentage decrease in surface area decrease in surface area 100% original surfacearea d 16 d 16 100% 100% = 4.5% Ans. d 16 Questions based on Surface area of cuboid and cube PRACTICE EXERCISE 1. Find the surface area of a cuboid 16 m long, 14 m broad and m high.. Find the length of the longest pole that can be placed in a room 1 m long, 8 m broad and 9 m high.. The length, breadth and height of a room are 5 m, 4 m and m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 8.50 per m. 4. Find the percentage increase in the surface area of a cube when each side is doubled. 5. The paint in a certain container is sufficient to paint an area equal to 9.5 m. How many bricks of dimensions.5 cm 10 cm.5 cm can be painted out of this container? NCERT 6. A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 0 cm long, 5 cm wide and 5 cm high. (i) What is the area of the glass? (ii) How much of tape is needed for all the 1 edges? NCERT. Three cubes each of side 6 cm are joined end to end. Find the surface of the resulting cuboid. 8. A plastic box 1.5 m long, 1.5 m wide and 65 cm deep is to be made. It is to be open at top. Ignoring the thickness of the plastic sheet, find: (i) area of the sheet required to make the box. (ii) the cost of the sheet for it, if a sheet measuring 1 m cost Rs.. 9. If the surface area of the cube is 96 cm, find its edge and length of its diagonal. 10. The dimensions of a rectangular box are in the ratio of : : 4 and the difference between the cost of covering it with sheet of paper at the rate of Rs. 4 and Rs. 4.50 per m is Rs. 650. Find the dimensions of the box. 11. Mary wants to decorate her christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santaclaus on it (see figure). She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 0 cm respectively, how many square sheets of paper of side 40 cm would she require? NCERT

1. The length and breadth of a hall are in the ratio 4 : and its height is 5.5 meters. The cost of decorating its walls (including doors and windows) at Rs. 6.60 per m is Rs. 508. Find the length and breadth of the room. Questions based on Surface area of a cylinder 1. The curved surface area of a right circular cylinder of height 14 cm is 16 cm. Find the diameter of the base of the cylinder. 14. A metal pipe is cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. (see figure). Find its : NCERT (i) inner curved surface area (ii) outer curved surface area (iii) Total surface area 15. The inner diameter of a circular well is.5 m. It is 10 m deep. Find: (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m. 16. An iron pipe 0 cm long has exterior diameter 5 cm. If the thickness of the pipe is 1 cm, find the total surface area of the pipe. 1. A rectangular sheet of paper 88 cm 50 cm is rolled along its length and a cylinder is formed. Find curved surface area of the cylinder formed. 18. A solid cylinder has total surface area of 46 cm. Its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder. 19. The total surface area of a hollow metal cylinder, open at both the ends of external radius 8 cm and height 10 cm is 8 cm. Find thickness of the metal in the cylinder. SURFACE AREAS AND VOLUMES MATHEMATICS IX

0. In the given figure, you see the frame of a lamp shade. It is to be covered with a decorative cloth. The frame has a base diameter of 0 cm and height of 0 cm. A margin of.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. NCERT Question based on surface area of a Cone 1. Find the curved surface area of a cone, if its slant height is 50 cm and the diameter of its base is 8 cm.. Find the total surface area of a cone, if its slant height is 1 cm and diameter of its base is 4 cm. NCERT. The curved surface area of a cone is 400 cm and its radius is 5 cm. What is its slant height? use π 4. The radius and slant height of a cone are in the ratio 4 :. If its curved surface area is 9 cm, find its radius. use π 5. The circumference of the base of a 10 m high conical tent is 44 m. Calculate the length of canvas used in making the tent if width of canvas is m. use π 6. What length of tarpaulin m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 0 cm. (use =.14) NCERT. The slant height and base diameter of a conical tomb are 5 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 10 per 100 m? NCERT 8. The curved surface area of a cone of radius 6 cm is 188.4 cm. Find its height. Questions based on surface area of sphere and hemi-sphere 9. The surface area of a sphere is 5544 cm. Find its radius. 0. A hemi-spherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm. NCERT MATHEMATICS IX SURFACE AREAS AND VOLUMES

1. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. NCERT. In the given figure, a right cylinder just encloses a sphere of radius r. Find (i) Surface area of the sphere (ii) curved surface area of the cylinder (iii) ratio of the areas obtained in (i) and (ii).. The internal and external diameters of a hollow hemi-spherical vessel are 0 cm and 8 cm respectively. Find the cost of painting the vessel all over at 15 paisa per cm. 4. A toy is in the form of a cone mounted on a hemi-sphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Find the surface area of the toy. (use =.14) Question based on Volume of Cuboid and Cube 5. The total surface area of a cube is 150 cm. Find its volume. 6. 500 persons took dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is 4 m?. Three cubes of a metal with edges 6 cm, 8 cm and 10 cm respectively are melted and formed into a single cube. Find the edge of the new cube formed. Also, find its volume. 8. The volume of a cuboid is 156 m. Its length is 16 m, and its breadth and height are in the ratio :. Find surface area of the cuboid. 9. A field is 0 m long and 40 m broad. In one corner of the field, a pit which is 10 m long, 8 m broad and 5 m deep, has been dug out. The earth taken out of it is evenly spread over the remaining part of the field. Find the rise in the level of the field. 40. The areas of three adjacent faces of a cuboid are 15 cm, 40 cm and 4 cm. Find the volume of the cuboid. 41. How many bricks, each measuring 5 cm 15 cm 8 cm will be required to build a wall 10 m 4 dm 5 m when one-tenth of its volume is occupied by mortar? 4. A rectangular reservoir is 10 m long and 5 m wide. At what speed per hour must water flow into it through a square pipe of 0 cm wide so that the water rises by.4 m in 18 hours. Question based on Volume of a Cylinder 4. The radius of a cylinder is 14 cm and its height is 40 cm. Find (i) curved surface area (ii) the total surface area (iii) volume of the cylinder. 44. The total surface area of a cylinder is 46 cm. Its curved surface is one-third of its total surface area. Find the volume of the cylinder. 45. The curved surface area and the volume of a pillar are 64 m and 96 m respectively. Find the diameter and the height of the pillar. 4 SURFACE AREAS AND VOLUMES MATHEMATICS IX

46. The sum of the height and radius of the base of a solid cylinder is m. If the total surface area of the cylinder is 168 m, find its volume. 4. A cylindrical tube, open at both ends, is made up of metal. The internal radius of the tube is 5. cm and its length is 5 cm. The thickness of the metal is 8 mm. Calculate the volume of the metal. 48. A soft drink is available in two packs : (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter cm and height 10 cm. Which container has greater capacity and by how much? NCERT 49. If the diameter of the cross-section of a wire is decreased by 5%, how much percent will the length be increased so that the volume remains the same? NCERT 50. Water flows out through a circular pipe, whose internal diameter is cm, at the rate of 0 cm per second into a cylindrical tank, the radius of whose base is 40 cm. By how much time will the level of water rise in half an hour? 51. A rectangular piece of paper is cm long and 1 cm wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder. 5. A well, with inner radius 4m, is dug 14 m deep. The earth taken out of it has been spread evenly all round it to a width of m to form an embankement. Find the height of this embankement. use π Question based on Volume of a Cone 5. The base radii of two cones of the same height are in the ratio : 4. Find the ratio of their volumes. 54. A cone of height 4 cm has curved surface area 550 cm. Find its volume. use π 55. The radius and height of a right circular cone are in the ratio of 5 : 1. If its volume is 14 cm, find the slant height and radius of the base of the cone. (use = 1.4) 56. Find the slant height and curved surface area of a cone whose volume is 195 cm and the radius of the base is 1 cm. 5. A semi-circular thin sheet of metal of diameter 8 cm is bent and an open conical cup is made. Find the capacity of the cup. 58. Monica has a piece of canvas whose area is 551 m. She uses it to have a conical tent made with a base radius of m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m, find the volume of the tent that can be made with it. NCERT 59. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm. 60. A right angled triangle of which the sides containing the right angle are 6. cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area. Question based on Volume of a Sphere and Hemi-Sphere 61. Find the surface area of a sphere whose volume is 606.5 m. 6. A solid sphere of radius cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm. Find the number of small balls thus obtained. 6. The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes. 64. The diameter of a metallic sphere is 6 cm. It is melted and drawn into a wire having diameter of the crosssection as mm. Find the length of the wire. MATHEMATICS IX SURFACE AREAS AND VOLUMES 5

65. The radii of the internal and external surfaces of a metallic spherical shell are cm and 5 cm respectively. It is melted and recast into a solid cylinder of height 10 cm. Find the diameter of the base of the cylinder. 66. The largest sphere is carved out of a cube of side cm. Find the volume of the sphere. (use =.14) 6. Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube. 68. A hemi-spherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of the steel used in making it. 69. A hemi-spherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter cm and height 4 cm. How many bottles are required to empty the bowl? 0. A hemi-sphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Find the height of the cone. Miscellaneous Questions 1. The height of a cone is 0 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume 1 be of the volume of the given cone, at what height above the base is the section made?. A circus tent consists of a cylindrical base surmounted by a conical roof. The radius of the cylinder is 0 m. The height of the tent is 6 m and that of the cylindrical base is 4 m. Find the volume of air contained in the tent and the area of canvas used for making it.. How many litres of water flows out of pipe having an area of cross-section of 5 cm in one minute, if the speed of water in the pipe is 0 cm/sec? 4. A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 6 cm. If the sphere is completely submerged in water, find by how much will the surface level of water be raised. 5. An ice-cream cone has a hemispherical top. If the height of the conical portion is 9 cm and base radius.5 cm, find the volume of ice-cream in the ice-cream cone. use π 6. In the given figure, a solid is made of a cylinder with hemispherical ends. If the entire length of the solid is 108 cm and the diameter of the hemispherical ends is 6 cm, find the cost of polishing the surface of the solid at the rate of paisa per cm.. A spherical copper ball of diameter 9 cm is melted and drawn into a wire, the diameter of whose thickness is mm. Find the length of the wire in meters. 8. The difference between the inside and outside surfaces of a cylindrical water pipe 14 m long is 88 m. If the volume of pipe be 16 m. Find the inner and outer radii of the water pipe. 6 SURFACE AREAS AND VOLUMES MATHEMATICS IX

9. Water is flowing at the rate of.5 km/hr through a circular pipe 0 cm internal diameter, into a circular cistern of diameter 0 m and depth.5 m. In how much time will the cistern be filled? 80. A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered into the water (see figure), and its size is such that when it touches the sides of the conical vessel, it is just immersed. How much water will remain in the cone after the overflow? M.M. : 0 General Instructions : PRACTICE TEST Q. 1-4 carry marks, Q. 5-8 carry marks and Q. 9-10 carry 5 marks each. Time : 1 hour 1. The floor of a rectangular hall has a perimeter 50 m. If the cost of painting the four walls at the rate of Rs. 10 per m is Rs. 15000, find the height of the hall.. A cylindrical pillar is 50 cm in diameter and.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 150 per m.. A right triangle PQR with sides 5 cm, 1 cm and 1 cm is revolved about the side 1 cm. Find the volume of the solid so obtained. 4. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? 5. A joker s cap is in the form of a right circular cone of base cm and height 4 cm. Find the area of the sheet required to make 10 such caps. 6. A hemispherical bowl is made of steel, 0.5 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.. A village, having a population of 4000, requires 150 l of water per head per day. It has a tank measuring 0 m 15 m 6 m. For how many days will the water of this tank last? 8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 50 patients? 9. The radius and height of a cone are in the ratio 4 :. The area of the base is 154 cm. Find the area of the curved surface. 10. The diameter of a sphere is decreased by 5%. By what percent its curved surface area decrease? ANSWERS OF PRACTICE EXERCISE 1. 868 cm. 1 m. Rs. 69 4. 00% 5. 100 6. (i) 450 cm (ii) 0 cm. 60 cm 8. (i) 5.45 m (ii) 119.90 Rs. 9. 4 cm, 4 cm 10. 10 m, 15 m, 0 m 11. 1. 40 m, 0 m 1. 4 cm 14. (i) 968 cm (ii) 1064.80 cm (iii) 08.08 cm 15. (i) 110 m (ii) Rs. 4400 MATHEMATICS IX SURFACE AREAS AND VOLUMES

16. 168 cm 1. 440 cm 18. r = cm, h = 14 cm 19. cm 0. 00 cm 1. 00 cm. 144.5 cm. cm 4. 1 cm 5. 14. m 6. 6 m. Rs. 1155 8. 8 cm 9. 1 cm 0. Rs.. 1. 1 : 16. (i) 4 r (ii) 4r (iii) 1 : 1. 4.4 Rs 4. 10.6 cm 5. 5 cm 6. 50 cm. 1 cm, 18 cm 8. 8 cm 9. 14. cm 40. 10 cm 41. 6000 4. 0 km/hr 4. (i) 50 cm (ii) 45 cm (iii) 4640 cm 44. 59 cm 45. 6 m, 14 m 46. 460 m 4. 04 cm 48. cylindrical tin, 85 cm 49. 10.8% 50. 8.5 cm 51. 46 m 5. 6.8 m (approx) 5. 9 : 16 54. 1 cm 55. 1 cm, 5 cm 56. 5 cm, 10 cm 5. 6. cm 58. 1 m 59. 190.9 cm 60. 415.8 cm,.9 cm 61. 46.5 m 6. 1000 6. 1 : 8 64. 6 m 65. cm 66. 19.5 cm 6. 6 : 68. 56.8 cm 69. 54 0. 8.44 cm 1. 0 cm. 61600 m, 10.85 m. 9 l 4. 1 cm 5. 91.66 cm 6. Rs. 855.6. 11.5 m 8. 1.5 m,.5 m 9. 10 hours 80. 188.5 cm ANSWERS OF PRACTICE TEST 1. 6 m. Rs. 68.5. 100 cm 4. 0.0 l (approx.) 5. 5500 cm 6. 1.5 cm. days 8. 8.5 l 9. 19.5 cm 10. 4.5% 8 SURFACE AREAS AND VOLUMES MATHEMATICS IX