CERT Tetual Eercise Solve - StudyMate€¦ · CERT Tetual Eercise Solve Surface Areas and Volumes EXERCISE – 5.1 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be

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NCERT Textual Exercise (Solved)

Surface Areas and Volumes

EXERCISE – 5.1 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 the sheet for it, if a sheet measuring 1 m² costs ` 20. 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 ` 7.50 per m².

3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ` 10 per m² is ` 15000, find the height of the hall.

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?

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? 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? 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 dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, extra 5% of the total surface area is required extra. If the cost of the cardboard is ` 4 per 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

8. 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 2.5 m, with base dimensions 4 m × 3 m?

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Test Yourself – SAV 1 1. Two cubes each of 10 cm edge are joined end to end. Find the surface area

of the resulting cuboid. 2. The length of a hall is 20 m and width 16 m. The sum of the areas of the

floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall.

3. A classroom is 7 m long, 6.5 m wide and 4 m high. It has one door 3 m × 1.4 m and three windows each measuring 2 m × 1 m . The interior walls are to be colour washed. The contractor charges ` 5.25 per m2. Find the cost of colour washing.

4. The cost of papering four walls of a room at ` 0.70 per m2 is ` 157.50. The height of the room is 5 m. Find the length and the breadth of the room if they are in the ratio 4 : 1.

5. Find the surface area of the cuboid whose length is 16 cm, breadth 15 cm and height 8.5 cm.

6. A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of ` 5 per metre, iron sheet being 2 m wide.

7. What is the area of the cardboard required to make a rectangular box 12 cm long, 8 cm wide and 6 cm high?

8. The surface area of a rectangular block is 846 cm². Find the length, breadth and height, if these dimensions are in the ratio of 5 : 4 : 3.

9. An open box is made of wood of 3 cm thickness. Its external length is 1.48 m, breadth 1.16 m and height 8.3 dm. Find the cost of painting the inner surface at ` 5 per m2.

EXERCISE – 5.2 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. 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 metal sheet are required for the same?

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3. A metal pipe is 77 cm long. The inner diameter of its 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. 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500

complete revolutions to level a playground. Find the area of the playground in sq. m.

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 ` 12.50 per m².

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.

7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find (i) its inner curved surface area and (ii) the cost of plastering this curved surface at the rate of ` 40 per m². 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. 9. Find (i) the lateral or curved surface area of a cylindrical petrol storage tank

which is 4.2 m in diameter and 4.5 m high.

(ii) how much steel was actually used, in making the tank if 1

12 of the

steel actually used was wasted in making the closed tank. 10. In 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.

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 had to supply the cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

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Test Yourself – SAV 2 1. Find the (i) curved surface area, and (ii) total surface area of a closed right

circular cylinder whose height is 7 cm and radius of the base being 3 cm. 2. The radii of two right circular cylinders are in the ratio 2 : 3 and their heights

are in the ratio 5 : 4. Find the ratio of their curved surface areas. 3. The diameter of a garden roller is 1.4 m and it is 2 m long. How much area

will it cover in 5 revolutions? 4. The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete

revolutions to level a playground, determine the cost of levelling it at the rate of ` 0.30 per m2.

5. An iron pipe 20 cm long has exterior diameter equal to 25 cm. If the thickness of the pipe is 1 cm, find the whole surface of the pipe.

6. 30 circular plates, each of the radius 14 cm and thickness 3 cm are placed one above the other to form a solid. Find the total surface area.

7. The area of curved surface of a cylinder is 2200 cm² and circumference of its base is 220 cm. Find the height of the cylinder.

8. The height of a cylinder is 3 m. Three times the sum of the areas of its two circular faces is twice the area of its curved surface. Find the radius of its base.

9. A roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m². How many revolutions did it make?

10. The total surface area of a hallow cylinder which is open from both sides is 4620 cm², area of base ring is 115.5 cm² and height 7 cm. Find the thickness of the cylinder.

EXERCISE – 5.3

Assume π = 227

, unless stated otherwise.

1. The diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

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.

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4. A conical tent is 10 m high and the radius of its base is 24 m. Find (i) the slant height of the tent, and (ii) the cost of the canvas required to make the tent, if the cost of 1 m²

canvas is ` 70. 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 20 cm. (use π = 3.14)

6. The slant height and the 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 ` 210 per 100 m2.

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.

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 ` 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and 1 04. = 1.02)

Test Yourself – SAV 3 1. The radius of the base and the height of a right circular cone are respectively

21 cm and 28 cm. Find the curved surface area and total surface area of the cone.

2. The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height?

3. The radius and the slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm², find its radius.

4. The circumference of the base of a 10 m high conical tent is 44 m. Find the length of the canvas used in making the tent if width of canvas is 2 m.

5. How many metres of cloth 5 m wide will be required to make a conical tent, with the radius of the base is 7 m and height 24 m?

6. The height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.

7. A cloth having an area of 165 cm² is shaped into a conical tent of radius

5 m. How many students can sit in the tent if a student occupies 57

m²?

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8. The radius and height of a cone are in the ratio 4 : 3. The area of the base is 154 cm². Find the area of the curved surface.

9. The curved surface area of a right circular cone of radius 6 cm is 188.4 cm². Find its height.

10. Find what length of canvas of 4 m wide is required to make a conical tent of 8 m in diameter and 5.6 m in slant height. Also, find the cost of the canvas at the rate of ` 19.20 per metre.

EXERCISE – 5.4

Assume π = 227


1. Find the surface area of a sphere of radius (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm 2. Find the surface area of a sphere of diameter (i) 14 cm (ii) 21 cm (iii) 3.5 cm 3. Find the total area of a hemisphere of radius 10 cm. (Use π = 3.14) 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 5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the

cost of tin plating on the inside at the rate of ` 16 per 100 cm². 6. Find the radius of a sphere whose surface area is 154 cm². 7. The diameter of the moon is approximately one fourth of the diameter of

the earth. Find the ratio of their surface areas. 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. 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, and (iii) ratio of the areas obtained in (i) and (ii).

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Test Yourself – SAV 4 1. Find the surface area of the sphere whose radius is (i) 4.2 cm (ii) 4.3 cm. 2. The internal and external diameters of a hollow spherical vessel are 24 cm

and 25 cm respectively. The cost to paint one sq. cm of the surface is ̀ 0.07. Find the total cost to paint the vessel all over (ignore the area of the edge).

3. The surface area of a sphere is 5544 cm², find its diameter. 4. The internal and external diameters of a hollow hemispherical vessel are

21 cm and 25.2 cm respectively. The cost of painting 1 sq. cm of the surface is 10 paise. Find the total cost to paint the vessel all over.

5. Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.

6. The dome of a building is in the form of a hemisphere. Its radius is 63 cm. Find the cost of painting at the rate of ` 2 per m2.

7. The surface area of a sphere is the same as the curved surface area of a cone having the radius of base as 120 cm and height 160 cm. Find the radius of the sphere. (use 15 = 3.873 approx)

8. The surface area of a sphere is 346.5 m². Find its radius. 9. If the diameter of a sphere is d and curved surface area s, show that

s = πd². Hence, find the surface area of a sphere whose diameter is 4.2 cm.

10. The surface area of a sphere is 45247

cm². What is its radius?

EXERCISE – 5.5 1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of

a packet containing 12 such boxes ? 2. 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 m³ = 1000 L ) 3. A cuboidal vessel is 10 m long and 8 m wide. How much high must it be

made to hold 380 m3 of a liquid? 4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at

the rate of ` 30 per m³. 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 2.5 m and 10 m.

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6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?

7. A godown measures 60 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Test Yourself – SAV 5 1. The volume of a cuboid is 440 cm³ and the area of its base is 88 cm². Find

the height. 2. A cube of edge 9 cm is immersed completely in a rectangular vessel

containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel.

3. How many planks each of which is 2 m long, 2.5 cm broad and 4 cm thick can be cut off from a wooden block 6 m long, 15 cm broad and 40 cm thick?

4. A 4 cm cube is cut into 1 cm cubes, Find the number of cubes formed. Calculate the total surface area of small cubes? What is the ratio of surface areas of small cubes to that of the large cubes?

5. A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5 m. The earth dug out is spread evenly over the remaining portion of the field. How much is the level of the field raised?

6. 100 persons can sleep in a room 25 m by 9.8 m. If each person required 12.25 m3 of air, find the height of the room?

7. A river 2.5 m deep and 44 m wide is flowing at the rate of 2.4 km per hour. How much water runs into the sea per minute?

8. The areas of three adjacent faces of a cuboid are x, y and z. If the volume is v, prove that v² = xyz.

9. The external dimensions of a closed wooden box are 48 cm, 36 cm and 30 cm. The wood of which its made is 1.5 cm thick. How many bricks of size 6 cm × 3 cm × 0.75 cm can be put in this box?

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EXERCISE – 5.6

Assume π = 227


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 = 1 L )

2. 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.

3. 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?

4. If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, find (i) radius of its base, and (ii) its volume (use π = 3.14).

5. The costs to paint the inner curved surface of a cylindrical vessel of 10 m deep is ` 2200. If the cost of painting is at the rate of ` 20 per m², find

(i) inner curved surface area of the vessel, (ii) radius of the base, and (iii) capacity of the vessel. 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? 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 diameter 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.

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?

Test Yourself – SAV 6 1. Two cylinder cans have bases of the same size. The diameter of each is

14 cm. One of the cans is 10 cm high and the other is 20 cm high. Find the ratio of their volumes.

2. A rectangular sheet of paper 44 cm × 18 cm is rolled along its length and a cylinder is formed. Find the volume of the cylinder.

3. The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm³. Find its radius.

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4. A solid cylinder has total surface area of 462 cm2. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder.

5. Into a circular drum of radius 4.2 m and height 3.5m, how many full bags of wheat can be emptied if the space required for wheat in each bag is 2.1 m2. (take π = 3.14)

6. A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment.

7. A hollow cylindrical pipe is 210 cm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of copper used in making the pipe.

8. Find the number of coins, 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

9. The ratio between the radius of the base and the height of a cylinder is 2 : 3. Find the total surface area of the cylinder, if its volume is 1617 cm³.

10. Two cylindrical vessels are filled with oil. The radius of one vessel is 15 cm and its height is 25 cm. The radius and height of the other vessel are 10 cm and 18 cm respectively. Find the radius of the cylindrical vessel 30 cm in height, which will just contain the oil of the two given vessels.

EXERCISE – 5.7

Assume π = 227


1. Find the volume of the right circular cone with (i) radius 6 cm and height 7 cm. (ii) radius 3.5 cm and height 12 cm. 2. Find the capacity in litres of a conical vessel with (i) radius 7 cm and slant height 25 cm, and (ii) height 12 cm and slant height 13 cm. 3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of

the base if (use π = 3.14) (i) radius 7 cm and slant height 25 cm, and (ii) height 12 cm and slant height 13 cm. 4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the

diameter of its base.

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5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find

(i) height of the cone, and (ii) curved surface area of the cone. 7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about

the side 12 cm. Find the volume of the solid so obtained. 8. If the triangle ABC in the Question 7 above is revolved about the side

5 cm, find the volume of the solid so obtained. Also find, the ratio of the volumes of the solids obtained in Question 7 and the new volume so obtained.

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m, Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Test Yourself – SAV 7 1. The radius and the height of a cone are in the ratio 3 : 4. If its volume is

301.44 cm³, what is its diameter and the slant height? 2. The base radii of two right circular cones of the same height are in the ratio

3 : 5. Find the ratio of their volumes. 3. A cone of height 24 cm has a curved surface area 550 cm². Find its volume. 4. A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It

is melted and recast into a right circular cone with the radius of its base as 1.2 cm. Find its height.

5. Into a conical tent of radius 8.4 m and vertical height 3.5 m, how many full bags of wheat can be emptied, if space required for the wheat in each bag is 1.96 m³. (take π = 3.14)

6. Find the volume of the largest circular cone that can be cut off from a cube whose edge is 9 cm.

7. A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (take π = 3.14)

8. The circumference of the base of a cone is 22 cm. If the height be 8 cm, find the volume.

9. The volume of a conical tent is 616 cm³ and area of its base 77 cm2. How much cloth will be required to build it?

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10. A cone of radius 5 cm is filled with water. If the water is poured in a cylinder of radius 10 cm, the height of water rises 2 cm, find the height of the cone.

EXERCISE – 5.8

Assume π = 227


1. Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m 2. Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m 3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the

density of the metal is 8.9 g per cm³? 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? 5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm

hold? 6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner

radius is 1 m, find the volume of the iron used to make the tank. 7. Find the volume of a sphere whose surface area is 154 cm². 8. A dome of a building is in the form of a hemisphere. From inside, it was

white washed at the cost of ` 498.96. If the cost of white washing is ` 2.00 per m2, find the

(i) inside surface area of the dome, and (ii) volume of the air inside the dome. 9. 27 solid iron spheres, each of radius r and surface area, s are melted to form

a sphere with surface area s1. Find the (i) radius r1 of the new sphere, (ii) ratio of s and s1. 10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How

much medicine (in mm³) is needed to fill this capsule?

Test Yourself – SAV 8 1. Find the volume of the hemisphere of radius 3.5 cm. 2. A solid sphere of radius 3 cm is melted and then cast into small spherical

balls each of diameter 0.6 cm. Find the number of balls thus obtained.

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3. The volume of two spheres are in the ratio 64 : 27. Find their radii if the sum of their radii is 21 cm.

4. The largest sphere is carved out of a cube of edge 7 cm. Find the volume of the sphere.

5. The volume of a metal sphere is 38808 cm³, find its surface area. 6. If the radius of a sphere is doubled, what is the ratio of the volume of the

first sphere to that of the second sphere. 7. The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn

into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

8. A drop of soap water of radius r is converted into a bubble of radius 2 cm. If the thickness of the bubble is 0.02 cm, find r. [given, (1.98)³ = 7.76 (approx.)]

9. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?

10. Three solid spheres of iron whose diameters are 2 cm, 12cm and 16 cm respectively, are melted into a single solid sphere. Find the radius of the new solid sphere so formed.

EXERCISE – 5.9

Optional 1. A wooden bookshelf has external dimensions

as follows: height = 110 cm, depth = 25 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 ̀ 0.20 per cm² and the rate of painting is ` 0.10 per cm², find the total expenses required for polishing and painting the surfaces of the bookshelf.

2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, each placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted in silver colour. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted in black colour. Find the cost of the paint required if the silver paint costs ` 0.25 per cm² and black paint costs ` 0.05 per cm².

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3. The diameter of a sphere is decreased by 25%. By what per cent does its

curved surface area decrease?


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NCERT Exercises and Assignments

Exercise – 5.1 1. We have, Length, l = 1.5 m, Breadth, b = 1.25 m and Depth = Height, h = 65 cm = 0.65m (i) Since the plastic box is open at the top. Therefore, Plastic sheet required for making such a box = [2 (l + b) × h + lb] = [2(1.5 + 1.25) × 0.65 + 1.875] = [2 × 2.75 × 0.65 + 1.875] = (3.575 + 1.875) = 5.45m² (ii) Cost of 1 m² of sheet = ` 20 ∴ Total cost of 5.45 m² of sheet = ` (5.45 × 20) = ` 109 2. Here, l = 5 m, b = 4 m and h = 3 m Area of four walls including ceiling = [2(l + b) × h + lb] = [2(5 + 4) × 3 + 5 × 4] = (2 × 9 × 3 + 20) = (54 + 20) = 74 m² Cost of white washing is ` 7.50 per m2. ∴ Cost of white washing = ` (74 × 7.50) = ` 555 3. Cost of painting the four walls = ` 15000

∴ Area of four walls = 15000

10

m² = 1500 m²

∴ 2 (l + b) h = 1500 ∴ Perimeter × Height = 1500 250 × Height = 1500

Height = 1500250

= 6 m

Hence, the height of the hall = 6 m. 4. Surface area of one brick = 2(lb + bh + hl)


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= 222.5100

× 10100

+ 10100

× 7.5100

+ 7.5100

× 22.5100

= 2 ×1

1001

100× (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)

= 15000

× (225 + 75 + 168.75) = 0.09375 m²

Area for which the paint is just sufficient is 9.375 m².

∴ Number of bricks that can be painted with the available paint = 9.375

0.09375 = 100

5. (i) Lateral surface area of cubical box of edge 10 cm = 4 × 10² cm² = 400 cm² Lateral surface area of cuboidal box = 2(l + b) × h = 2 × (12.5 + 10) × 8 = 2 × 22.5 × 8 = 360 cm² Thus, lateral surface area of the cubical box is greater and is more by (400 – 360) cm²,

i.e. 40 cm². (ii) Total surface area of cubical box of edge 10 cm = 6 × 10² = 600 cm² Total surface area of cuboidal box = 2 (lb + bh + hl) = 2(12.5 × 10 + 10 × 8 + 8 × 12.5) = 2(125 + 80 + 100) = (2 × 305) = 610 cm² Thus, total surface area of cuboidal box is greater by 10 cm². 6. Here, l = 30 cm, b = 25 cm and h = 25 cm. (i) Area of the glass = Total surface area = 2(lb + bh + hl) = 2(30 × 25 + 25 × 25 + 25 × 30) = 2(750 + 625 + 750) = (2 × 2125) = 4250 cm² (ii) Tape needed for all the 12 edges = The sum of all the edges = 4 (l + b + h) = 4 (30 + 25 + 25) = 4 × 80 = 320 cm. 7. In case of bigger box, l = 25 cm, b = 20 cm and h = 5 cm


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Total surface area = 2(lb + bh + hl) = 2(25 × 20 + 20 × 5 + 5 × 25) = 2(500 + 100 + 125) = (2 × 725) = 1450 cm² In case of smaller box, l = 15 cm, b = 12 cm and h = 5 cm Total surface area = 2(lb + bh + hl) = 2(15 × 12 + 12 × 5 + 5 × 15) = 2(180 + 60 + 75) = (2 × 315) = 630 cm² Total surface area of 250 boxes of each type = No. of boxes (surface area of big boxes + surface

area of small boxes) = 250(1450 + 630) = (250 × 2080) = 5,20,000 cm² Cardboard required (i.e. including 5% extra for overlaps etc.)

= 520000 × 105100

= 5,46,000 cm² Cost of 1000 cm² of cardboard = ` 4

1 cm2 cardboard = ` 4

1000

∴ Total cost of cardboard = ` 5460001000

4×

= ` 2184

8. Dimensions of the box-like structure are l = 4 m, b = 3 m and h = 2.5 m. Since there is no tarpaulin for the floor, ∴ Tarpaulin required = 2 (l + b) × h + lb = [2(4 + 3) × 2.5 + 4 × 3] = (2 × 7 × 2.5 + 12) = (35 + 12) = 47 m²

Test Yourself – SAV 1 1. 1000 cm² 2. 8.88 m 3. ` 513.45 4. 18 m, 4.5 m 5. 1007 cm² 6. ` 960


131

7. 432 cm² 8. 15 cm, 12 cm, 9 cm 9. ` 27.21

Exercise – 5.2 1. Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then, curved surface area = 2πrh

88 = 2 × 227

× r × 14

r = 88 × 72 × 22 × 14

= 1

∴ Diameter of the base = 2r = 2 × 1 = 2 cm. 2. Let r be the radius of the base and h be the height of the cylinder. Metal sheet required to make a closed cylindrical tank. = total surface area = 2πr (h + r)

= 2 × 227 × 0.7 (1 + 0.70) [ h = 1 m, r =

1402

cm = 70 cm = 0.70 m]

= 2 × 22 × 0.1 × 1.70 = 7.48 m² Hence, the sheet required = 7.48 m².

3. We have, R = external radius = 4 42.

cm = 2.2 cm

r = internal radius = 42

cm = 2 cm

h = length of the pipe = 77 cm. (i) Inner curved surface = 2πrh

= 2 × 227

× 2 × 77

= 968 cm² (ii) Outer curved surface = 2πRh

= 2 × 227

× 2.2 × 77

= 1064.8 cm² (iii) Total surface area of pipe = Inner curved surface area + Outer curved surface area + Areas

of two bases = 2πrh + 2πRh + 2π (R² – r²)


132

= 968 + 1064.8 + 2 × 227

(4.84 4)2

−

= 2032.8 + 447

× 0.84

= (2032.8 + 5.28) = 2038.08 cm² 4. The length of the roller = 120 cm, i.e. h = 1.2 m radius of the cylinder

(i.e. roller) = 842

cm = 42 cm = 0.42 m. Distance covered by roller in one revolution = Its curved surface area = 2πrh

= 2 × 227

× 0.42 × 1.2

= 3.168m² Area of the playground = Distance covered by roller in 500 revolution = (500 × 3.168) = 1584 m² 5. Let r be the radius of the base and h be the height of the pillar.

∴ r = 502

cm = 25 cm = 0.25 m and h = 3.5 m.

Curved surface = 2πrh

= 2 × 227

× 0.25 × 3.5

= 5.5 m² Cost of painting the curved surface at the rate of ` 12.50 per m² = ` (5.5 × 12.5) = ` 68.75 6. Let r be the radius of the base and h be the height of the cylinder. Then, Curved surface area = 4.4m² 2πrh = 4.4

2 × 227

× 0.7 × h = 4.4 [ r = 0.7 m]

h = 4.4 × 7

2 × 22 × 0.7

= 1 m Thus, the height of the cylinder = 1 m.


133

7. (i) Let r be the radius of the face and h be depth of the well. Then, Curved surface = 2πrh

= 2 × 227

× 3.52

× 10

= 110 m² (ii) Since the cost of plastering is ` 40 per m² ∴ cost of plastering the curved surface = ` (110 × 40) = ` 4400. 8. Total radiating surface in the system = Curved surface area of the pipe

= 2πrh, where r = 52

cm = 2.5cm = 2 5100

. m

= 0.025 m and h = 28 m

= 2 227

0 025 28× × ×

. m² = 4.4 m²

9. (i) Here, r = 4 22.

m = 2.1 m and h = 4.5 m.

Lateral surface area = 2πrh

= 2 227

2 1 4 5× × ×

. .

= 59.4 m²

(ii) Since 1

12 of the actual steel used was wasted,

the area of the steel which has gone into the tank = 1 112

−

of x =

1112

of x.

Steel used = (2πrh + 2πr²)

∴ 1112

× x = 87.12 = 59.4 + 2 × × 2.1 × 2.1227

∴ x = 87 12 12

11. ×

= (59.4 + 27.72)

x = 95.04 m² Hence, the actual area of the steel used = 95.04 m².

10. Here r = 202

cm = 10 cm and h = 30 cm + 2 × 2.5 cm (i.e. margin) = 35 cm.

Cloth required for covering the lampshade = Its curved surface area = 2πrh


134

= 2 × 227

× 10 × 35

= 2200 cm² 11. Cardboard required by each competitor = Curved surface area of one penholder + base area = 2πrh + πr² where r = 3cm, h = 10.5cm

= 2 × 227

× 3 × 10.5 + 227

× 9

= (198 + 28.28) = 226.28cm² Cardboard required for 35 competitors = (35 × 226.28) = 7920 cm² (approx.)

Test Yourself – SAV 2 1. (i) 132 cm² (ii) 188.57 cm² 2. 5 : 6 3. 44m² 4. ` 475.20 5. 3168 cm² 6. 9152 cm² 7. 10 cm 8. 2 cm 9. 200

10. 7

19 cm.

Exercise – 5.3

1. Here r = 10 5

2.

cm = 5.25 cm and l = 10 cm

Curved surface area of the cone = (πrl)

= 227

× 5.25 × 10

= 165 cm²

2. Here r = 242

m = 12 m and l = 21 m.

Total surface area of the cone = (πrl + πr²) = πr (l + r )

= 227

× 12 × (21 + 12)


135

= 227

× 12 × 33

= 1244.57 m² (approx.) 3. (i) Curved surface of a cone = 308 cm² slant height, l = 14cm. ∴ πrl = 308

∴ 227

× r × 14 = 308

∴ r = 308 × 722 × 14

= 7 cm

(ii) Total surface area of the cone = πr (l + r)

= 227

× 7 × (14 + 7)

= (22 × 21) = 462 cm². 4. (i) Here r = 24 m, h = 10 m. Let l be the slant height of the cone. Then, l² = h² + r²

∴ l = h r2 2+

∴ l = 24 10 576 100² ²+ = +

= 676 = 26m (ii) Canvas required to make the conical tent = Curved surface of the cone

= πrl = 227

× 24 × 26

m²

Rate of canvas per 1 m² is ` 70

∴ Total cost of canvas = ` 227

× 24 × 26 × 70

= ` 137280

5. Let r m be the radius, h m be the height and l m be the slant height of the tent. Then, r = 6 m, h = 8 m

∴ l = r h² ² ² ²+ = + = +6 8 36 64

= 100 = 10 m. Area of the canvas used for the tent = curved surface are of the tent = πrl = (3.14 × 6 × 10) = 188.4 m² Now, this area is to be covered with a tarpaulin of width 3m.

∴ Length of tarpaulin required = Area of trapaulin required

Width of tarpaulin


136

= 188 4

3.

= 62.8 m The extra material required for stitching margins and cutting = 20 cm = 0.2 m. So, the total length of tarpaulin required = (62.8 + 0.2) m = 63 m.

6. Here, l = 25 m and r = 142

m = 7 m.

Curved surface area = πrl

= 227

25 7× ×

= 550 m²

Rate of white-washing is ` 210 per 100 m²

∴ Cost of white washing the tomb = ` 550 210100

×

= ` 1155.

7. Let r cm be the radius, h cm be the height and l cm be the slant height of the joker’s cap. Then, r = 7 cm, h = 24 cm

∴ l = h r2 2+

l = 24 72 2+ = 576 49+ = 625 = 25 cm. Sheet required for one cap = Curved surface of the cone = πrl

= 227

7 25× ×

= 550 cm²

Sheet required for 10 such caps 10 × 550 cm2 = 5500cm². 8. Let r m be the radius, h m be the height and l m be the slant height of a cone.

Then, r = 402

cm = 20 cm = 0.2 m, h = 1 m

∴ l = r h2 2+ = 0.04 +1 = 1 04. = 1.02 m Curved surface of 1 cone = πrl = (3.14 × 0.2 × 1.02) m² Curved surface of such 50 cones = (50 × 3.14 × 0.2 × 1.02) m² Cost of painting at the rate of ` 12 per m² = ` (50 × 3.14 × 0.2 × 1.02 × 12) = ` 384.34 (approx.)

Test Yourself – SAV 3 1. 2310 cm², 3696 cm² 2. 37 cm 3. 12 cm 4. 134.2 m 5. 110 m 6. 1078 cm²


137

7. 110 students 8. 192.5 m² 9. 8 cm 10. ` 337.60

Exercise – 5.4 1. (i) We have r = radius of the sphere = 10.5 cm Now, surface area = 4πr²

= 4 × × 10.5 × 10.5227

= 1386 cm² (ii) We have r = radius of the sphere = 5.6 cm Now, surface area = 4πr²

= 4 × × 5.6 × 5.6227

= 394.24 cm² (iii) We have r = radius of the sphere = 14 cm

Now, surface area = 4 × × 14 × 14227

= 2464 cm²

2. (i) Here r = 142

cm = 7cm

Now, surface area = 4 πr²

= 4 × × 7 × 7227

= 616 cm²

(ii) Here r = 212

cm = 10.5 cm


= 4 × × 10.5 × 10.5227

= 1386 cm²

(iii) Here r = 3 52.

cm = 1.75 cm


= 4 × × 1.75 × 1.75227

= 38.5 cm²

3. Here r = 10 cm Total surface area of hemisphere = 3πr² = (3 × 3.14 × 10 × 10) = 942 cm²


138

4. Let r1 and r2 be the radii of balloons in the two cases. Here r1 = 7 cm and r2 = 14 cm

∴ Ratio of their surface areas = 44

12

22

12

22

ππ

rr

rr

=

= 7 714 14

14

××

=

Thus, the required ratio of their surface areas = 1 : 4.

5. Here r = 10 52.

cm = 5.25 cm

Curved surface area of the hemisphere = 2 πr²

= 2 5 25 5 25227

× × ×

. .

= 173.25 cm² Rate of tin plating is ` 16 per 100 cm².

∴ Cost of tin plating the hemisphere = ` 173 25 16100

. ×

= ` 27.72.

6. Let r be the radius of the sphere. Now, surface area = 154 cm² or 4πr² = 154

or 4 × 227

× r² = 154

or r² = 154 × 74 × 22

= 12.25

∴ r = 12 25. = 3.5 Thus, the radius of the sphere is 3.5 cm.

7. Let the diameter of the earth be R and that of the moon will be R4

.

The radii of the moon and the earth are R8

and R2

respectively.

Ratio of their surface area = 4

4

1

18

2

64

4

2

2

π

π

R

R

=

= 164

41

116

× =


139

Alternative Method Let the radius of the moon be = R ∴ Radius of the earth = 4R

Ratio of their surface area = 4

4 4

2

2

ππ

RR( )

= RR

2

2161

16=

8. Inner radius, r = 5 cm Thickness of steel = 0.25 cm ∴ Outer radius, R = (r + 0.25) = (5 + 0.25) = 5.25 cm ∴ Outer curved surface = 2πR²

= 2 × × 5.25 × 5.25227

= 173.25 cm² 9. (i) Let the radius of the sphere r, so its surface area = 4πr² (ii) Since the right circular cylinder just encloses a sphere of radius r. So the radius of cylinder

will be r and its height 2r. Curved surface of cylinder = 2πr (2r) [ r = r, h = r] = 4πr² (iii) Ratio of areas = 4πr² : 4πr² = 1 : 1

Test Yourself – SAV 4 1. (i) 221.76 cm² (ii) 232.45 cm² 2. ` 132.11 3. 42 cm 4. ` 184.34 6. ` 10 7. 77.46 cm 8. 5.25 cm 9. 55.44 cm² 10. 6 cm

Exercise – 5.5 1. Here l = 4 cm, b = 2.5 cm and h = 1.5 cm. ∴ Volume of one matchbox = l × b × h = (4 × 2.5 × 1.5) = 15 cm³ ∴ Volume of a packet containing 12 such boxes = (12 × 15) = 180 cm³ 2. Here l = 6 m, b = 5 m and h = 4.5 m ∴ Volume of the tank = lbh = (6 × 5 × 4.5) = 135 m³


140

∴ The tank can hold = 135 × 1000 litres = 1,35,000 litres of water [1 m³ = 1000 litres] 3. Here, length = 10 m, breadth = 8 m and volume = 380 m³

Height = Volume of cuboidLength × Breadth

= 38010 × 8

= 4.75 m.

4. Here, l = 8 m, b = 6 m and h = 3 m Volume of the pit = lbh = (8 × 6 × 3) = 144 m³. Rate of digging is ` 30 per m³ ∴ Cost of digging the pit = ` (144 × 30) = ` 4320. 5. Here, l = 2.5 m, d = 10 m and volume = 50000 litres

Volume = 50000 × 11000

[ 1 m³ = 1000 litres]

= 50 m³

Breadth = Volume of cuboidLength × Depth

= 50

2.5 × 10

= 2 m.

6. Here, l = 20 m, b = 15 m and h = 6 m. Capacity of the tank = lbh = (20 × 15 × 6) = 1800 m³ Water requirement per person per day = 150 litres Water required for 4000 person per day = (4000 × 150)L

= 4000 × 1501000

m³ = 600 m³

Number of days the water will last = Capacity of tank

Total water required per day

= 1800600

= 3

7. Volume of the godown = 60 × 25 × 10 = 15000 m³ Volume of 1 crate = 1.5 × 1.25 × 0.5 = 0.9375 m³


141

Number of crates that can be stored in the godown

= Volume of the godown

Volume of 1 crate

= 150000 9375. m

= 16000. 8. Let V1 = Volume of the cube of edge 12 cm = (12 × 12 × 12) cm³ and V2 = Volume of the cube cut out of the first one

= 18

× V1 = 18

× 12 × 12 × 12

cm³

= (6 × 6 × 6) cm³ ∴ side of the new cube = 6 cm

Ratio of their surface areas = 6 (side)6 (side)

2

2 = 6 × 12 × 126 × 6 × 6 = 4

1, i.e. 4 : 1.

9. Since the water flows at the rate of 2 km per hour, the water from 2 km of river flows into the sea in one hour.

Let l = 2 km = 2000 m, b = 40 m, h = 3 m ∴ The volume of water flowing into the sea in one hour = Volume of the cuboid = l × b × h = (2000 × 40 × 3) m³

∴ The volume of water flowing into the sea in one minute = 2000 × 40 × 360

m³ = 4000 m³.

Test Yourself – SAV 5 1. 5 cm 2. 4.05 cm 3. 180 4. 64 cubes, 384 cm², 4 : 1 5. 4.2 m 6. 5 m 7. 4400 m³ 9. 2970

Exercise – 5.6 1. Let r cm be the radius of the base and h cm be the height of the cylinder. Circumference of the base = 132 cm ∴ 2πr = 132

∴ 2 × 227

× r = 132


142

∴ r = 132 × 72 × 22

= 21 cm

Volume of the cylinder = π r²h

= 227

× 21 × 21 × 25

= 34650 cm³

∴ Vessel can hold = 346501000

litres, i.e. 34.65 litres of water.

2. We have h = Height of the cylindrical pipe = 35 cm

R = external radius = 282

cm = 14 cm

r = internal radius = 242

cm = 12 cm

Volume of the wood used in making the pipe = Volume of the external cylinder – Volume of the internal cylinder = πR²h – πr²h = π(R² – r²)h

= 227

× (14² – 12²) × 35

= 227

× 26 × 2 × 35

= 5720 cm³ Weight of 1 cm³ = 0.6 g ∴ Weight of 5720 cm³ = (5720 × 0.6) g

= 5720 × 0.6

1000

kg

= 3.432 kg. 3. (i) Capacity of tin can = l × b × h = (5 × 4 × 15) = 300 cm³ (ii) Capacity of plastic cylinder = πr²h

= 227

× 72

× 72

× 10

= 385 cm³

Thus, the plastic cylinder has greater capacity by 85 cm³. 4. (i) Let r cm be the radius of the base and h cm be the height of the cylinder. Then, Lateral surface = 94.2 cm²


143

∴ 2πrh = 94.2 ∴ 2 × 3.14 × r × 5 = 94.2

∴ r = 94.2

2 × 3.14 × 5 = 3

Thus, the radius of its base = 3cm. (ii) Volume of the cylinder = πr2h = (3.14 × 3² × 5) = 141.3 cm³.

5. (i) Inner curved surface area of the vessel = Total cost of painting

Rate of painting

= 220020

m² = 110m²

(ii) Let r be the radius of the base and h be the height of the cylindrical vessel. ∴ 2πrh = 110

∴ 2 × 227

× r × 10 = 110

∴ r = 110 × 7

2 × 22 × 10 =

74

= 1.75

Thus, the radius of the base = 1.75m (iii) Capacity of the vessel = πr²h

= 227

× 74

× 74

× 10

= 96.25 m³

6. Capacity of a closed cylindrical vessel = 15.4 litres [ 1m³ = 1000 litres]

= 15.4 × 11000

m³ = 0.0154 m³

Let r m be the radius of the base and h m be the height of the vessel. Then, Volume = πr²h = πr² × 1 = πr² [ h = 1m] ∴ πr² = 0.0154

∴ 227

× r² = 0.0154

∴ r² = 0.0154 × 722

= 0.0049

∴ r = 0 0049. = 0.07 Thus, the radius of the base of vessel = 0.07m Metal sheet needed to make the vessel = Total surface area of the vessel = 2πrh + 2πr² = 2πr(h + r)


144

= 2 × 227

× 0.07 × (1 + 0.07)

= 44 × 0.01 × 1.07 = 0.4708 m².

7. Diameter of the graphite cylinder = 1 mm = 1

10 cm

∴ Radius = 120

cm 1 110

m cm=

Length of the graphite cylinder = 14 cm

Volume of the graphite cylinder = 227

× 120

× 120

× 14

cm³

= 0.11 cm³

Diameter of the pencil = 7 mm = 710

cm

∴ Radius of the pencil = 720

cm

and, length of the pencil = 14 cm

∴ Volume of the pencil = 227

× 720

× 720

× 14

cm³

= 5.39 cm³ Volume of wood = Volume of the pencil – Volume of the graphite = (5.39 – 0.11) = 5.28 cm³ 8. Diameter of the cylindrical bowl = 7 cm

∴ Radius = 72

cm

Height of serving bowl = 4 cm ∴ Soup served in one serving = Volume of the bowl

= πr²h = 227

× 72

× 72

× 4

= 154 cm³ Soup served to 250 patients = (250 × 154) cm³ = 38500 cm³, i.e. 38.5l. (1000 cm³ = 1 L )

Test Yourself – SAV 6 1. 1 : 2 2. 2772 cm³ 3. 5 cm 4. 539 cm³ 5. 92 bags 6. 4.66 m


145

7. 10560 cm³ 8. 450 9. 770 cm² 10. 15.73 cm

Exercise – 5.7 1. (i) Here, r = 6 cm and h = 7cm

Volume of the cone = 13

πr²h

= 13

× 227

× 6 × 6 × 7

= 264 cm³

(ii) Here, r = 3.5 cm and h = 12cm

∴ Volume of the cone = 13 πr²h

= 13

× 227

× 3.5 × 3.5 × 12

= 154 cm³

2. (i) Here, r = 7 cm and l = 25 cm. Let the height of the cone be h cm. Then, h² = l² – r² = 25² – 7² = 625 – 49 = 576

∴ h = 576 = 24 cm

Volume of the conical vessel = 13

πr²h

= 13

× 227

× 7 × 7 × 24

= 1232 cm³

∴ Capacity of the vessel in litres = 12321000

L = 1.232 L. [1000 cm³ = 1 L ]

(ii) Here, h = 12 cm and l = 13 cm. Let the radius of the base of the cone be r cm. Then, r² = l² – h² = 13² – 12² = 169 – 144 = 25

∴ r = 25 = 5 cm

Volume of the conical vessel = 13

πr²h

= 13

× 227

× 5 × 5 × 12

=

22007

cm³

∴ Capacity of the vessel in litres = 2200

7× 1

1000

L =

1135 L (1000 cm³ = 1 L)


146

3. Here, h = 15 cm and volume = 1570 cm³ Let the radius of the base of cone be r cm. ∴ Volume = 1570 cm³

∴ 13

π r²h = 1570

∴ 13

× 3.14 × r² × 15 = 1570

∴ r² = 1570×3

3.14 × 51 = 100

∴ r = 100 = 10 Thus, the radius of the base of cone is 10 cm. 4. Here, h = 9 cm and volume = 48π cm³ Let the radius of the base of the cone be r cm. ∴ Volume = 48π cm³

or 13

π r²h = 48π

or 13

× r² × 9 = 48

or 3r² = 48

or r² = 483

= 16

∴ r = 16 = 4

Thus, the diameter of the base of the cone = 8 cm. 5. Diameter of the top of the conical pit = 3.5 m

∴ Radius = 3 52.

m

= 1.75 m Depth of the pit, i.e. height = 12 m

Volume = 13

π r²h

= 13

× 227

× 1.75 × 1.75 × 12

= 38.5 m³ Capacity of pit = 38.5 kilolitres [1 m³ = 1000 L = 1 kilolitres]


147

6. (i) Diameter of the base of the cone = 28cm

∴ Radius = 282 = 14 cm.

Volume of the cone = 9856 cm³ Let the height of the cone be h cm. Now, volume = 9856cm³

or 13 π r²h = 9856

or 13

227

´ × 14 × 14 × h = 9856

∴ h = 9856 × 3 × 722 × 14 × 14 = 48

Thus, the height of the cone = 48 cm. (ii) Here, r = 14 cm and h = 48 cm. Let l cm be the slant height of the cone. Then, l² = h² + r² = 48² + 14² = 2304 + 196 = 2500 ∴ l = 2500 = 50 Thus, the slant height of the cone = 50 cm. Here, r = 14 cm and l = 50 cm. Curved surface area = πrl

= 227

× 14 × 50

= 2200 cm².

13 cm

12 cm

C1

C

B 5 cm

A

7. On revolving the right triangle ABC about the side AB(= 12 cm), we get a cone as shown in the figure.

Volume of solid so obtained = 13

πr²h

= 13

25 12× × ×π

= 100π cm³. 8. On revolving the right triangle ABC about the side BC (= 5 cm), A

B

A1

C5 cm

13 cm

12 cm

we get a cone as shown in the figure of Q 7.

Volume of solid so obtained = 13

π r²h

= 13

× π × 12 × 12 × 5

= 240π cm³


148

∴ Ratio of their volumes = 100π : 240π (i.e. of Q 7 and Q 8) = 5 : 12. 9. Diameter of the base of the cone = 10.5 m

∴ Radius = r = 10 5

2.

m = 5.25 m.

Height of the cone = 3 m

∴ Volume of the cone (heap) = 13

π r²h

= 13

227

5 25 5 25 3× × × ×

. . m³

= 86.625 m³ To find the slant height l, we have, l² = h² + r² = 3 + (5.25)² = 9 + 27.5625 = 36.5625

∴ l = 36 5625. = 6.0467 m (approx.) Canvas required to protect wheat from rain = Curved surface area = πrl

= 227

× 5.25 × 6.0467

m³

= 99.77 m² (approx.)

Test Yourself – SAV 7 1. 12 cm; 10 cm 2. 9 : 25 3. 1232 cm³ 4. 6.4 cm 5. 132 bags 6. 190.93 cm³ 7. 74.18 m³; 80.54 m² 8. 102.66 cm³ 9. 381.209 cm² 10. 24 cm

Exercise – 5.8 1. (i) We have r = radius of the sphere = 7 cm

∴ Volume of the sphere = 43

πr³

= 43

227

7 7 7× × × ×

cm³

= 4312

3 cm³ = 143713

cm³


149

(ii) We have r = radius of the sphere = 0.63 m

∴ Volume of the sphere = 43 π r³

= 43

227

0 63 0 63 0 63× × × ×

. . . m³

= 1.05 m³ (approx.) 2. (i) Diameter of the spherical ball = 28 cm

∴ Radius = 282

cm = 14 cm.

Amount of water displaced by the spherical ball = volume of the spherical ball = 43 πr³

= 43

227

14 14 14× × × ×

cm3 =

344963 = 11498

23 cm³

(ii) Diameter of the spherical ball = 0.21 m

∴ Radius = 0 21

2.

m = 0.105 m.

Amount of water displaced by the spherical ball = volume of the spherical ball = 43

π r³

= 43

227

0 105 0 105 0 105× × × ×

. . . = 0.004851 m³

3. Diameter of the ball = 4.2 cm

∴ Radius = 4 22.

cm = 2.1 cm.

Volume of the ball = 43 πr³

= 43

227

2 1 2 1 2 1× × × ×

. . .

= 38.808 cm³ Density of the metal is 8.9 g per cm³ ∴ Mass of the ball = 38.808 × 8.9 = 345.3912 = 345.39 g (approx.)

4. Let the diameter of the moon be r. Then, the radius of the moon = r2

According to the question, the diameter of the earth is 4r, so its radius = 2r

V1 = Volume of the moon =43

π r2

3

= 43

πr³ × 18


150

∴ 8V1 = 43

πr³ ...(i)

and, V2 = Volume of the earth = 43

2 3π r( )

= 43

πr³ × 8

∴ V2

8 = 43

πr³ ...(ii)

From (i) and (ii), we have

8V1 = V2

8

∴ V1 = 164

V2

Hence, the volume of the moon is 164

of the volume of the earth.

Alternative Method Since diameter of the moon is approximately one fourth of diameter of the earth ∴ Radius of moon is = the fourth of radius of the earth. Let Radius of moon = R Radius of earth = 4R

Ratio of their volumes =

43

43

4

3

3

π

π

R

R( ) = 1 : 64

5. Diameter of a hemispherical bowl = 10.5 cm

∴ Radius = 10 5

2.

cm = 5.25 cm.

Volume of the bowl = 23

πr²

= 23

× 227

× 5.25 × 5.25 × 5.25

= 303.1875 cm³ (1000 cm³ = 1 L) Hence, the hemispherical bowl can hold 0.303 l (approx.) of milk. 6. Let R cm and r cm be the external and internal radii respectively of the hemispherical vessel.

Then, R = 1.01 m (thickness = 1 cm = 0.01 m) and r = 1 m.


151

Volume of iron used = External volume – Internal volume

= 23

πR³ – 23

πr³ = 23

π(R³ – r³ )

= 23

× 227

× [(1.01)³ – (1)³]

= 4421 × (1.030301 – 1)

= 44 × 0.03030121

= 0.06348 m³ (approx.) 7. Let r cm be the radius of the sphere. So, surface area = 154 cm² ∴ 4πr² = 154

or 4 × 227 × r² = 154

or r² = 154 × 74 × 22

= 12.25

∴ r = 12 25. = 3.5 cm.

Now, volume = 43

πr³

= 4 × 22 × 3.5 × 3.5 × 3.53 7

= 539

3 cm³ = 179

23

cm³.

8. (i) Inside surface area of the dome = Total cost of white washing

Rate of white washing

= 498 96

2 00.

.

m² = 249.48 m².

(ii) Let r be the radius of the dome. ∴ Surface area = 2πr²

or 2 × 227

× r² = 249.48

or r² = 249.48 × 7

2 × 22 = 39.69

∴ r = 39 69. = 6.3 m.


152

Volume of the air inside the dome = Volume of the dome

= 23

π r³

= 23

× 227

× 6.3 × 6.3 × 6.3 m³

= 523.9 m³.

9. (i) Volume of 27 solid spheres of radius r = 27 × 43

π r³

Volume of the new sphere of radius r1 = 43

π (r1)³

According to the problem, we have

43

π (r1)³ = 27 × 43

π r³

∴ (r1)³ = 27r³ = (3r)³ ⇒ r’ = 3r

(ii) Required ratio = SS1 = =

( )4

4 3

2

1 2

2

2

ππ

rr

rr( )

= rr

2

2919

= = 1 : 9

10. Diameter of the spherical capsule = 3.5 mm

Radius = 3 52. mm = 1.75 mm.

Medicine needed for its filling = Volume of spherical capsule

= 43

π r³

= 43

× 227

× 1.75 × 1.75 × 1.75

= 22.46 mm³ (approx).

Test Yourself – SAV 8 1. 89.83 cm³ 2. 1000 balls 3. 12 cm and 9 cm 4. 179.66 cm³ 5. 5544 cm² 6. 1 : 8 7. 0.6 cm 8. 0.6214 cm 9. 8 laddoos 10. 9 cm.


153

Exercise – 5.9

Optional 1. Area to be polished = (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5) = (9350 + 4250 + 5500 + 1500 + 1100) = 21700 cm²

Cost of polishing at the rate of ` 0.20 per cm² = 21700 × 20100

= ` 4340 Area to be painted = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90) = (9000 + 3600 + 6750) = 19350 cm²

Cost of polishing at the rate of ` 0.10 per cm² = ` 19350 × 10100

= ` 1935

∴ Total expense = ` (4340 + 1935) = ` 6275 2. Clearly, we have to subtract the area of part of the sphere that is resting on the cylinder while

calculating the cost of silver paint. surface area to be silver paint = 8 (Curved surface area of the sphere – area of circle on which sphere is resting)

= 8 (4πR2 – πr²) , where R = 212

cm, r = 1.5 cm

= 8π 4 × 441 2.254

−

= 8π (441 – 2.25) = 8π (438. 75) cm² ∴ Cost of silver paint at the rate of ` 0.25 per cm²

= ` 8 × 22 × 438.75 × 251007

= ` 19305

7

= ` 2757.86 (approx.)

Surface area to be black painted = 8 × curved surface area of cylinder = 8 × 2πrh

= 8 × 2 × 227

× 1.5 × 7 cm² = 528 cm²

Cost of black paint at the rate of ` 0.05 per cm²

= ` 528 × 5100


154

= ` 26.40 Total cost of painting = ` (2757.86 + 26.40) = ` 2784.26 (approx) 3. Let d be the diameter of the sphere. Then, its surface area

= 4π d2

2

= πd²

On decreasing its diameter by 25%, the new diameter,

d1 = 75

10034

×

=

d

d

∴ New surface area = 4π d1

2

2

= 4π12

34

2

× d

= 4π

9 2

64d

= πd² 9

16

Decrease in surface area = πd² 1 916

−

= πd²

716

∴ Percentage decrease in surface area = ππ

dd

22

716

1 100 × × ×

%

= 70016

%

= 43.75%.


155

Self Evaluation Test

Time: 1 Hour Maximum Marks: 30

1. Find the total surface area of cone whose radius is r2

and slant height is 2l. [1]

2. Find the number of planks of dimensions (4m × 50 cm × 20 cm) that can be stored in a pit which is 16 m long, 12m wide and 4 m deep. [1]

3. If the length of diagonal of a cube is 6 3 cm , then find the edge of the cube. [2] 4. Two solid spheres made of the same metal have weights 2920 g and 740g respectively. Determine

the radius of the larger sphere, if the diameter of the smaller one is 5 cm. [2] 5. A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume

of the solid so formed. [2] 6. If a sphere is inscribed in a cube, then find the ratio of the volume of the cube to the volume of

the sphere. [2] 7. The dimensions of a metallic cuboid are 100 cm × 80 cm × 64 cm. It is melted and recast into

a cube. Find [3] (i) The edge of the cube. (ii) The lateral surface area of cube. (iii) The total surface area of the cube. 8. A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a

rectangular solved of iron with dimensions 32 cm × 22 cm × 14 cm. Find the rise in the level of the water when the solid is completely submerged. [3]

9. The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of sphere. [3] 10. A spherical canon ball, 28 cm in diameter is melted and recast into a right circular conical

mould, the base of which is 35 cm is diameter. Find the height of the cone, correct to one place of decimal. [3]

11. The sum of the inner and outer curved surfaces of a hollow metallic cylinder is 1056 cm2, and the volume of material in it is 1056 cm3. Find the internal and external radii. Given that the height of the cylinder is 21 cm. [4]

12. A cylinder has a diameter of 20 cm. The area of curved surface is 100 cm2. Find [4] (i) The height of the cylinder correct to one decimal place. (ii) The volume of the cylinder correct to one decimal place. (use π = 22/7)

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CERT Tetual Eercise Solve - StudyMate€¦ · CERT Tetual Eercise Solve Surface Areas and Volumes EXERCISE – 5.1 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be