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Mathematics, (Chapter – 13) (Surface Areas and Volumes), (Class – IX), , EXERCISE 13.1, Q.1. A plastic box 1.5 m long, 1.25 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 m2 costs Rs 20., Sol. Here, l = 1.5 m, b = 1.25 m, h = 65 cm = 0.65 m., Since the box is open at the top, it has only five faces., (i) So, surface area of the box = lb + 2(bh + hl), = 1.5 × 1.25 m2 + 2 (1.25 × 0.65 + 0.65 × 1.5) m2, = 1.875 + 2 (1.7875) m2, = (1.875 + 3.575) m2 = 5.45 m2, Hence, 5.45 m2 of sheet is required Ans., (ii) Cost of 1 m2 of the sheet = Rs 20, ∴ cost of 5.45 m2 of the sheet = Rs 20 × 5.45 m2 = Rs 109 Ans., Q.2. The length, breadth and height of a room are 5 m, 4 m and, 3 m respectively. Find the cost of white washing the walls of the room and, the ceiling at the rate of Rs 7.50 per m2., Sol. Here, l = 5 m, b = 4 m, h = 3 m, Surface area of the walls of the room and the ceiling, = 2h (l + b) + lb, = [2 × 3 (5 + 4) + 5 × 4] m2, = (6 × 9 + 20) m2 = 74 m2, Cost of white washing = Rs 7.50 per m2, ∴ total cost of white washing the walls and the ceiling of the room, = Rs 74 × 7.50 = Rs 555 Ans., Q.3. The floor of a rectangular hall has a perimeter 250 m. If the cost of, painting the four walls at the rate of Rs 10 per m2 is Rs 15000, find the, height of the hall., Sol. Let length, breadth and height of the hall be l, b and h respectively., Perimeter of the floor of the hall = 2 (l + b) = 250 m., Area of the four walls of the hall = 2h (l + b), ... (i), 15000, Also, area of the four walls of the hall =, m2, 10, = 1500 m2, ... (ii), From (i) and (ii), we have, 2h (l + b) = 1500, ⇒ h × 250 = 1500, [ 2(l + b) = 250], 1500, = 6, 250, Hence, height of the hall is 6 m Ans., , ⇒, , h =

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Q.4. The paint in a certain container is sufficient to paint an area equal to, 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can, be painted out of this container?, Sol. Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm., Total surface area of 1 brick = 2 (lb + bh + hl), = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm2, = 2(225 + 75 + 168.75) cm2 = 937.5 cm2, 937.5, m2 = 0.09375 m2., =, 100 × 100, 9.375, ∴, required number of bricks =, = 100 Ans., 0.09375, Q.5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 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 total surface area and by how much?, Sol., Here, a = 10 cm, l = 12.5 cm, b = 10 cm, h = 8 cm, (i) Lateral surface area of the cubical box = 4a2, = 4 × (10)2 cm2 = 400 cm2, Lateral surface area of the cuboidal box = 2h (l + b), = 2 × 8 (12.5 + 10) cm2, = 16 × 22.5 cm2 = 360 cm2, Difference in the lateral surface areas of the two boxes, = (400 – 360) cm2 = 40 cm2., Hence, the cubical box has greater lateral surface area by 40 cm2. Ans., (ii) Total surface area of the cubical box = 6a2, = 6 × (10)2 cm2 = 600 cm2, Total surface area of the cuboidal box = 2(lb + bh + hl), = 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm2, = 2(125 + 80 + 100) cm2, = 2 × 305 cm 2 = 610 cm2, Difference in the total surface areas of the two boxes = (610 – 600) cm2, = 10 cm2, Hence, the cubical box has smaller total surface area by 10 cm2 Ans., Q.6. A small indoor greenhouse (herbarium) is made entirely of glass panes, (including base) held together with tape. It is 30 cm long, 25 cm wide and, 25 cm high., (i) What is the area of the glass?, (ii) How much of tape is needed for all the 12 edges?, Sol., , Here, l = 30 cm, b = 25 cm, h = 25 cm., (i) Total surface area of the herbarium = 2(lb + bh + hl), = 2(30 × 25 + 25 × 25 + 25 × 30) cm2, = 2(750 + 625 + 750) cm2, = 2 × 2125 cm 2 = 4250 cm2, Hence, area of the glass = 4250 cm2 Ans., (ii) A cuboid has 12 edges. These consist of 4 lengths, 4 breadths and 4, heights.

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∴ length of the tape required = 4l + 4b + 4h, = (4 × 30 + 4 × 25 + 4 × 25) cm, = (120 + 100 + 100) cm = 320 cm Ans., Q.7. Shanti Sweets Stall was placing an order for making cardboard boxes for, packing their sweets. Two sizes of boxes were required. The bigger of, dimensions 25 cm × 20 cm × 5 cm and the smaller of dimesnsions 15 cm, × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is, required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, find the, cost of cardboard required for supplying 250 boxes of each kind., Sol. For bigger boxes :, l = 25 cm, b = 20 cm, h = 5 cm, Total surface area of 1 bigger box = 2(lb + bh + hl), = 2(25 × 20 + 20 × 5 + 5 × 25) cm2, = 2 (500 + 100 + 125) cm2 = 1450 cm2, Area of cardboard required for overlaps, = 5% of 1450 cm2 =, , 1450 × 5, cm2 = 72.5 cm2., 100, , Total area of cardboard needed for 1 bigger box, = (1450 + 72.5) cm2 = 1522.5 cm2, Total area of cardboard needed for 250 bigger boxes = 1522.5 × 250 cm2, = 380625 cm2., For smaller boxes :, l = 15 cm, b = 12 cm, h = 5 cm, Total surface area of 1 smaller box = 2 (lb + bh + hl), = 2(15 × 12 + 12 × 5 + 5 × 15) cm2, = 2 (180 + 60 + 75) cm2 = 630 cm2, Area of cardboard required for overlaps, 630 × 5, cm2 = 31.5 cm2, 100, Total area of cardboard needed for 1 smaller box = (630 + 31.5) cm 2, = 661.5 cm2, Total area of cardboard needed for 250 smaller boxes, = 661.5 × 250 cm2 = 165375 cm2, Now, total area of cardboard needed for 500 boxes (250 bigger and 250, smaller boxes) = (380625 + 165375) cm2 = 546000 cm2, Cost of 1000 cm2 of cardboard = Rs 4, , = 5% of 630 cm2 =, , 4, × 546000 = Rs 2184 Ans., 1000, Q.8. Parveen wanted to make a temporary shelter for her car, by making a boxlike structuer 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 2.5 m, with, base dimesions 4 m × 3 m?, ∴ Cost of 546000 cm2 of cardboard = Rs, , Sol. Here, l = 4 m, b = 3 m, h = 2.5 m, The tarpaulin is needed to cover 5 faces only (excluding the floor)

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EXERCISE 13.2, Q.1. The curved surface area of a right circular cylinder of height 14 cm is, 88 cm2. Find the diameter of the base of the cylinder., Sol. Here, h = 14 cm, curved, surface area = 88 cm2, r = ?, Curved surface area of the cylinder = 2πrh, 22, ⇒ 88 = 2 ×, × r × 14, 7, ⇒ 88 = 44 × 2 × r, ⇒, , r =, , 88, = 1, 44 × 2, , Hence, base diameter of the cylinder = 1 × 2cm = 2 cm Ans., Q.2. It is required to make a closed cylindrical tank of height 1 m and base, diameter 140 cm from a metal sheet. How many square metres of the sheet, are required for the same?., 140, Sol. Here, h = 1 m, r =, cm = 70 cm = 0.7 m, 2, Total surface area of the cylinder = 2πr (h + r), 22, = 2 ×, × 0.7 (1 + 0.7) m2, 7, = 44 × 0.1 × 1.7 m2 = 7.48 m2, Hence, 7.48 m2 of sheet is required Ans., Q.3. A metal pipe is 77 cm long. The inner diameter of a, cross section is 4 cm, the outer diameter being 4.4 cm, (see figure). Find its., (i) inner curved surface area,, (ii) outer curved surface area,, (iii) total surface area., Sol., Here, h = 77 cm,, 4 .4, cm = 2.2 cm,, Outer radius (R) =, 2, 4, cm = 2 cm, 2, (i) Inner curved surface area of the pipe, = 2πrh, 22, = 2 ×, × 2 × 77 cm2, 7, = 2 × 22 × 22 cm2 = 968 cm2 Ans., , Inner radius (r) =

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(ii) Outer curved surface area of the pipe = 2πRh, 22, = 2 ×, × 2.2 × 77 cm2 = 44 × 24.2 cm2, 7, = 1064.80 cm2 Ans., (iii) Total surface area of the pipe = inner curved surface area + outer, curved surface area + areas of the two base rings., = 2πrh + 2πRh + 2π (R2 – r2), = 968 cm2 + 1064.80 cm2 + 2 ×, , 22, [(2.2)2 – 22] cm2, 7, , = 2032.80 cm2 + 5.28 cm2 = 2038.08 cm2 Ans., Q.4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500, complete revolutions to move once over to level a playground. Find the area, of the playground in m2., , 84, cm = 42 cm, 2, Length of the roller (h) = 120 cm, Curved surface area of the roller = 2πrh, 22, = 2 ×, × 42 × 120 cm2 = 44 × 720 cm2 = 31680 cm2, 7, ∴ area covered by the roller in 1 revolution = 31680 cm2, ∴ area covered by the roller in 500 revolutions = 31680 × 500 cm2, = 15840000 cm2, 15840000, Hence, area of the playground =, m2 = 1584 m2 Ans., 100 × 100, Q.5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost, of painting the curved surface of the pillar at the rate of Rs 12.50 per m2., Sol. Radius of the roller (r) =, , 50, cm = 25 cm = 0.25 m, h = 3.5 m, 2, Curved surface area of the pillar = 2πrh, 22, = 2 ×, × 0.25 × 3.5 m2 = 5.5 m2, 7, Cost of painting 1 m2 = Rs 12.50, ∴ Total cost of painting the curved surface of the pillar, = Rs 12.50 × 5.5 = Rs 68.75 Ans., Q.6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of, the base of the cylinder is 0.7 m, find its height., Sol. Curved surface area of the cylinder = 4.4 m2, r = 0.7 m, h = ?, Curved surface area of the cylinder = 2πrh., 22, ⇒ 4.4 = 2 ×, × 0.7 × h, 7, 4 .4, ⇒ h =, = 1, 4 .4, Hence, height of the cylinder is 1 m Ans., Sol. Here, r =

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Q.7. The inner diameter of a circular well is 3.5 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 m2., 3 .5 m, Sol., Here, r =, , h = 10 m, 2, (i) Inner curved surface area of the well, 3 .5, 22, = 2πrh = 2 ×, ×, × 10m2, 2, 7, = 22 × 5 m2 = 110 m2 Ans., (ii) Cost of plastering 1 m2 = Rs 40, ∴ Cost of plastering the curved surface area of the well, = Rs 110 × 40 = Rs 4400 Ans., Q.8. In a hot water heating system. there is a cylindrical pipe of length 28 m, and diameter 5 cm. Find the total radiating surface in the system., 5, Sol. Here, r =, cm = 2.5 cm = 0.025 m, h = 28 m., 2, Total radiating surface in the system = total surface area of the cylinder, = 2π r(h + r), 22, = 2 ×, × 0.025 (28 + 0.025) m2, 7, 44 × 0.025 × 28.025, =, m2 = 4.4 m2 (approx) Ans., 7, Q.9. Find, (i) the lateral or curved surface area of a closed cylindrical petrol storage, tank that is 4.2 m in diameter and 4.5 m high., 1, (ii) how much steel was actually used, if, of the steel actually used was, 12, wasted in making the tank., 4 .2, Sol. Here, r =, m = 2.1 m, h = 4.5 m, 2, (i) Curved surface area of the storage tank = 2πrh, 22, = 2 ×, × 2.1 × 4.5 m2 = 59.4 m2 Ans., 7, (ii) Total surface area of the tank = 2πr (h + r), 22, × 2.1 (4.5 + 2.1) m2, 7, = 44 × 0.3 × 6.6 m2 = 87.12 m2, Let the actual area of steel used be x m2., 1, x, Area of steel wasted =, of x m2 =, m2., ... (i), 12, 12, 11, x ⎞ 2, ⎛, 2, ∴ area of the steel used in the tank = ⎜ x −, ⎟ m = 12 x m, 12, ⎝, ⎠, 11, ⇒ 87.12 =, x, 12, 87.12 × 12, ⇒, x =, = 95.04 m2, 11, Hence, 95.04 m2 of steel was actually used Ans., , = 2 ×

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Q.10. In the figure, you see the frame of a lampshade. It is to, be covered with a decorative cloth. The frame has a, base diameter of 20 cm and height of 30 cm. A margin, of 2.5 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., 20, cm = 10 cm, 2, Height = 30 cm, , Sol. Here, r =, , Circumference of the base of the, frame = 2πr, = 2π × 10 cm = 20π cm, Height of the frame = 30 cm, Height of the cloth needed for covering the frame, (including the margin) = (30 + 2.5 + 2.5) cm = 35 cm, Also, breadth of the cloth = circumference of the base of the frame., ∴ Area of the cloth required for covering the lampshade = length × breadth, 22, = 35 × 20π cm2 = 35 × 20 ×, cm2 = 2200 cm2 Ans., 7, Q.11. 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 3 cm and height 10.5, cm. The Vidyalaya was to supply the competitors with cardboard. If there, were 35 competitors, how much cardboard was required to be bought for, the competition?, Sol. Here, r = 3 cm, h = 10.5 cm, The penholders have only one base i.e., these are open at one end., Total surface area of 1 penholder, = 2πrh + πr2 = πr (2h + r), 22, =, × 3 (2 × 10.5 + 3) cm2, 7, 22, =, × 3 × 24 cm2, 7, 22, Total surface area of 35 penholders =, × 3 × 24 × 35 cm2 = 7920 cm2, 7, Hence, 7920 cm2 of cardboard is needed Ans.

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EXERCISE 13.3, Q.1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. find, its curved surface area., 10.5 cm = 5.25 cm, l = 10 cm., Sol. Here, r =, 2, Curved surface area of the cone = πrl, =, , 22, 7, , × 5.25 × 10 cm2 = 165 cm2 Ans., , Q.2. Find the total surface area of a cone, if its slant height is 21 m and, diameter of its base is 24 m., 24, m = 12 m, 2, Total surface area of the cone = πr(l + r), 22, =, × 12 (21 + 12) m2, 7, , Sol. Here, l = 21 m, r =, , 22, × 12 × 33 m2 = 1244.57 m2 Ans., 7, Q.3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm., Find (i) radius of the base and (ii) total surface area of the cone., Sol. Here, l = 14 cm, curved surface area = 308 cm2, r = ?, , =, , (i) Curved surface area of the cone = πrl, 22, × r × 14, 7, 308, ⇒ r = 22 × 2 = 7, Hence, base radius of the cone = 7 cm., (ii) Total surface area of the cone = πr (l + r), 22, =, × 7 (14 + 7) cm2 = 22 × 21 cm2 = 462 cm2 Ans., 7, Q.4. A conical tent is 10 m high and the radius of its base is 24 m. Find, (i) slant height of the tent., (ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas, is Rs 70., Sol., Here, h = 10 m, r = 24 m, (i) We have, l2 = h2 + r2, = (10)2 + (24)2, = 100 + 576 = 676, , ⇒ 308 =, , ⇒, , l =, , 676 = 26 m Ans., (ii) Curved surface area of the tent = πrl, 22, × 24 × 26 m2, 7, Cost of 1 m2 canvas = Rs 70, 22, 22, ∴ Cost of, × 24 × 26 m2 of canvas = Rs 70 ×, × 24 × 26, 7, 7, = Rs 137280 Ans., =

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Q.5. What length of tarpaulin 3 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 29 cm (use π, = 3.14), Sol. Here h = 6 m, r = 8 m, We have, l2 =, , =, , r 2 + h2, , 36 + 64 =, , 100 = 10 m, , Curved surface area of the tent = πrl, = 3.14 × 6 × 10 m2, ∴ required length of tarpaulin =, , 3.14 × 6 × 10, m + 20 cm, 3, , = 62.8 m + 0.2 m = 63 m Ans., Q.6. The slant height and base diameter of a conical tomb are 25 m and 14 m, respectively. Find the cost of white washing its curved surface at the rate, of Rs 210 per 100 m2., 14, m = 7 m, 2, Curved surface area of the tomb = πrl, 22, =, × 7 × 25 m2 = 550 m2, 7, Cost of white washing 100 m2 = Rs 210, 210, ∴ Cost of white washing 550 m2 = Rs, × 550 = Rs 1155 Ans., 100, Q.7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and, height 24 cm. Find the area of the sheet required to make 10 such caps., Sol. Here, r = 7 cm, h = 24 cm, , Sol. Here, l = 25 m, r =, , We have, l =, =, , h2 + r 2 =, , (24)2 + 72, , 576 + 49 =, , 625 = 25 cm, , Total curved surface area of 1 cap = πrl, 22, =, × 7 × 25 cm2 = 550 cm2, 7, Area of sheet required to make 10 such caps = 10 × 550 cm2 = 5500 cm2 Ans., Q.8. A bus stop is 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 12 per m2, what will be the cost of, painting all these cones? (Use π = 3.14 and take, Sol. Here, r =, l =, , 40, cm = 20 cm = 0.20 m, h = 1 m, 2, , h2 + r 2 =, , 12 + (0.2)2 =, , 1.04 = 1.02 m, , 1.04 = 1.02)

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Curved surface area of 1 cone = πrl, Curved surface area of 50 cones = 50 × 3.14 × 0.2 × 1.02 m 2, = 32.028 m2, 2, Cost of painting an area of 1 m = Rs 12, ∴ Cost of painting an area of 32.028 m2 = Rs 12 × 32.028, = Rs 384.34 (approx) Ans.

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EXERCISE 13.4, Q.1. Find the surface area of a sphere of radius :, (i) 10.5 cm, (ii) 5.6 cm, (iii) 14 cm, Sol. (i) r = 10.5 cm, Surface area of the sphere = 4πr2, 22, = 4 ×, × (10.5)2 cm2, 7, = 4 ×, , 22, × 10.5, 7, , × 10.5 cm2 = 1386 cm2 Ans., , (ii) r = 5.6 cm, Surface area of the sphere = 4πr2, 22, = 4 ×, × (5.6)2 cm2, 7, 22, × 5.6 × 5.6 cm2 = 394.24 cm2 Ans., 7, (iii) r = 14 cm, Surface area of the sphere = 4πr2, 22, = 4 ×, × 14 × 14 cm2, 7, = 88 × 28 cm2 = 2464 cm2 Ans., Q.2. Find the surface area of sphere of a diameter :, (i) 14 cm, (ii) 21 cm, (iii) 3.5 m, 14, Sol., (i) r =, cm = 7 cm, 2, Surface area of the sphere = 4πr2, 22, = 4 ×, × 72 cm2, 7, , = 4 ×, , = 4 ×, , 22, × 7 × 7 cm2, 7, , 88 × 7 cm2 = 616 cm2 Ans., 21, cm = 10.5 cm, 2, Surface area of the sphere =, 22, = 4 ×, × (10.5)2 cm2, 7, 22, = 4 ×, × 10.5 × 10.5 cm2, 7, 3 .5, (iii) r =, m = 1.75 m, 2, Surface area of the sphere =, 22, = 4 ×, × (1.75)2 m2, 7, 22, = 4 ×, × 1.75 × 1.75 m2, 7, , (ii) r =, , 4πr2, , = 1386 cm2 Ans., , 4πr2, , = 38.5 m2 Ans.

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Q.3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14), Sol. r = 10 cm, Total surface area of the hemisphere = 3πr2, = 3 × 3.14 × (10)2 cm2, = 3 × 3.14 × 100 cm2 = 942 cm2 Ans., Q.4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is, being pumped into it. Find the ratio of surface areas of the balloon in the, two cases., Sol. When r = 7 cm, Surface area of the balloon = 4πr2, = 4 × π × 7 × 7 cm2, When R = 14 cm :, Surface area of the balloon = 4πr2, = 4 × π × 14 × 14 cm2, Required ratio of the surface areas of the balloon, =, , 4×π×7×7, 1, = 1 : 4 Ans., =, 4 × π × 14 × 14 4, , Q.5. A hemispherical 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 cm2., 10.5, Sol. Here r =, cm = 5.25 cm, 2, Inner surface area of the bowl = 2πr2, 22, = 2 ×, × (5.25)2 cm2, 7, = 44 × 0.75 × 5.25 cm2 = 173.25 cm2, Cost of tin plating 100 cm2 = Rs 16, Cost of tin plating 173.25 cm2 = Rs, , 16, × 173.25 = Rs 27.72 Ans., 100, , Q.6. Find the radius of a sphere whose surface area is 154 cm2., Sol. Surface area of the sphere = 4πr2, 22, ⇒ 154 = 4 ×, × r2, 7, 154 × 7 7 × 7, ⇒ r2 = 4 × 22 = 4, 7, = 3.5, 2, Hence, radius of the sphere = 3.5 cm Ans., , ⇒ r=, , Q.7. The diameter of the moon is approximately one fourth of the diameter of, the earth. Find the ratio of their surface areas., Sol. Let diameter of the earth = 2r, Then radius of the earth = r, 2r, r, =, ∴ Diameter of the moon =, 4, 2

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∴ Radius of the moon =, , r, 4, , ⎛r⎞, Now, surface area of the moon = 4π ⎜ ⎟, ⎝4⎠, , πr 2, 4, Surface area of the earth = 4πr2, =, , 2, , ... (i), ... (ii), , πr 2, πr 2, 1, 4 =, ∴ Required ratio =, = 1 : 16 Ans., =, 2, 2, 16, 4 πr, 4 × 4 πr, Q.8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of, the bowl is 5 cm. Find the outer curved surface area of the bowl., Sol. Inner radius of the bowl (r) = 5 cm, Thickness of the steel = 0.25 cm, ∴ Outer radius of the bowl (R) = (5 + 0.25) cm = 5.25 cm, Outer curved surface area of the bowl, , 22, = 2πR2 = 2 ×, × (5.25)2 cm2 = 173.25 cm2 Ans., 7, Q.9. A right circular cylinder just encloses a sphere of, radius r (see figure). Find, (i) surface area of the sphere,, (ii) curved surface area of the cylinder,, (iii) ratio of the areas obtained in (i) and (ii)., Sol. Here, radius of the sphere = r, Radius of the cylinder = r, And, height of the cylinder = 2r, (i) Surface area of the sphere = 4πr2, (ii) Curved surface area of the cylinder = 2πrh, 2π × r × 2r =4πr2 Ans., (iii) Required ratio, , = 4π r2 = 1, 4πr2, 1

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(Class – IX)

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EXERCISE 13.6, Q.1. The circumference of the base of a cylindrical vessel is 132 cm and its, height is 25 cm. How many litres of water can it hold? (1000 cm3 = 1l), Sol. Here, h = 25 cm, 2πr = 132 cm., 2πr = 132, 22, ⇒, 2 ×, × r = 132, 7, 132 × 7, ⇒, r =, cm = 21 cm, 2 × 22, Volume of the cylinder = πr2h =, , Q.2., , Sol., , Q.3., , Sol., , 22, × 21 × 21 × 25 cm3, 7, , = 34650 cm3, 34650, =, litres = 34.65 litres Ans., 1000, The inner diameter of a cylindrical wooden pipe is 24 cm and its outer, diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the, pipe, if 1 cm3 of wood has a mass of 0.6 g., 24, Here, inner radius (r) =, cm = 12 cm, 2, 28, Outer radius (R) =, cm = 14 cm, h = 35 cm, 2, Volume of the wood used in the pipe = π(R2 – r2) h, 22, =, [(14) 4 – (12)2] × 35 cm3, 7, 22, =, × 26 × 2 × 35 cm3 = 5720 cm3, 7, Mass of 1 cm3 of wood = 0.6 g, ∴ Mass of 5720 cm3 of wood = 0.6 × 5720 g = 3432 g = 3.432 kg Ans., 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 7 cm and height 10 cm., Which container has greater capacity and by how much?, For tin can with rectangular 6 base., l = 5 cm, b = 4 cm, h = 15 cm, Volume of the tin can = lbh = 5 × 4 × 15 cm3 = 300 cm3, For plastic cylinder with circular base., 7, r =, cm = 3.5 cm, h = 10 cm, 2, Volume of the plastic cylinder = πr2h, 22, =, × 3.5 × 3.5 × 10 cm3 = 385 cm3, 7, Difference in the capacities of the two containers, = (385 – 300) cm3 = 85 cm3, Hence, the plastic cylinder with circular base has greater capacity by, 85 cm3 Ans.

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Q.4. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then, find (i) radius of its base (ii) its volume (Use π = 3.14), Sol. Here, h = 5 cm, 2πrh = 94.2 cm2., (i) 2πrh = 94.2, ⇒ 2 × 3.14 × r × 5 = 94.2, ⇒ r =, , 94.2, = 3, ×, 2 3.14 × 5, , Hence, base radius of the cylinder = 3 cm Ans., (ii) Volume of the cylinder = πr2h, = 3.14 × 3 × 3 × 5 cm3 = 141.3 cm3 Ans., Q.5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10, m deep. If the cost of painting is at the rate of Rs 20 per m2, find, (i) Inner curved surface area of the vessel,, (ii) radius of the base,, (iii) capacity of the vessel., Sol. Here, h = 10 m, Total cost, (i) Inner curved surface area =, Cost of painting per m 2, =, (ii) We have, 2πrh = 110, 22, ⇒ 2 ×, × r × 10 = 110, 7, ⇒ r = 2, , 2200 2, m = 110 m2 Ans., 20, , 110 7, 22 10 = 1.75 m Ans., , (iii) Capacity of the vessel = πr2h, , 22, × 1.75 × 1.75 × 10 m3 = 96.25 m3, 7, = 96.25 kl Ans. [1 m3 = 1 kl], =, , Q.6. 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?, Sol. Here, h = 1 m, volume = 15.4 litres, =, , 15.4, m3 = 0.0154 m3, 1000, , Also, volume of the cylinderical vessel = πr2h, ⇒ 0.0154 =, ⇒ r2 =, , 22, × r2 × 1, 7, , 0.0154 × 7, = 0.0049, 22

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⇒ r = 0.07 m, ∴ Total surface area of the cylinder = 2πr (h + r), 22, = 2 ×, × 0.07 (1 + 0.07) m2, 7, = 44 × 0.01 × 1.07 m2 = 0.4708 m2, Hence, 0.4708 m2 of metal sheet would be needed Ans., Q.7. 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 7 mm and the, dimeter of the graphite is 1 mm. If the length of the pencil is 14 cm, find, the volume of the wood and that of the graphite., Sol. Here, h = 14 cm., 7, Radius of the pencil (R) =, mm = 0.35 cm., 2, 1, Radius of the graphite (r) =, mm = 0.05 cm., 2, Volume of the the graphite = πr2h, 22, =, × 0.05 × 0.05 × 14 cm3 = 0.11 cm3, 7, Volume of the the wood = π (R2 – r2)h, 22, =, × [(0.35)2 – (0.05) 2] × 14 cm3, 7, 22, =, × 0.4 × 0.3 × 14 cm3 = 5.28 cm3, 7, Hence, volume of the wood = 5.28 cm3 and volume of the graphite, = 0.11 cm3 Ans., Q.8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter, 7 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 250 patients?, , 7, cm = 3.5 cm, h = 4 cm, 2, Capacity of 1 cylindrical bowl = πr2h, , Sol. Here, r =, , 22, × 3.5 × 3.5 × 4 cm3 = 154 cm3, 7, Hence, soup consumed by 250 patients per day, , =, , = 250 × 154 cm3 = 38500 cm3 Ans.

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= 4 ×, , 22, × 10.5, 7, , × 10.5 cm2, , = 1386 cm2, , Area of the base of the cylinder (support) = πR2, = π × (1.5)2 =, , 22, × 1.5, 7, , × 1.5 cm2, , = 7.07 cm2, Area of a sphere to painted silver, = (1386 – 7.07) cm2, = 1378.93 cm2, Area of spheres to be painted silver = 8 × 1378.93 cm2, 8 × 1378.93 × 25, 100, = Rs 2757.86, Curved surface area of a cylinder (support), 22, = 2 ×, × 1.5 × 7 cm2, 7, 22, Curved surface area of 8 supports = 8 × 2 ×, × 1.5 × 7 cm2, 7, , ∴ cost of painting the spheres, , = Rs, , Cost of painting the supports = Rs 8 × 2 ×, , 5, 22, × 1.5 × 7 ×, 100, 7, , = Rs 26.40, Total cost requiredof paint = Rs (2757.86 + 26.40) = Rs 2784.26 Ans., Q.3. The diameter of a sphere is decreased by 25%. By what per cent does its, curved surface area decrease?, Sol. Let originally the diameter of the sphere be 2r., Then, radius of the sphere = r, Surface area of the sphere = 4πr2, New diameter of the sphere, ∴, , ... (i), , = 2r – 2r ×, , New radius of the sphere =, , 3r, 4, , ⎛ 3r ⎞, Surface area of the new sphere = 4π ⎜, ⎟, ⎝ 4 ⎠, , Decrease in surface area, , Per cent decrease, , =, , 25, 3r, =, 100, 2, , = 4πr2 –, , 2, , 9πr 2, =, 4, , =, , 9πr 2, 4, 7πr 2, 4, , 7πr 2, × 100, 175, 7, 4, × 100 =, =, = 43.75, 2, 4, 16, 4 πr, , Hence, the surface area decreases by 43.75% Ans.