ch - 13 Surface Areas and Volumes 9TH...∴ Total surface area of a brick = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) = 2(225 + 75 + 168.75) = 2(468.75) = 937.5 cm2

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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 1m2 costs Rs.20.

• Solution: (i) l = 1.5 m

b = 1.25 m h = 65 cm = 0.65 m

∴ The area of the sheet required for making the box

= lb + 2(bh + hl) = (1.5)(1.25) + 2[(1.25)(0.65) + (0.65)(1.5)] = 1.875+ 2(0.8125+0.975) = 1.875 + 2(1.7875) = 1.875 + 3.575 = 5.45 m2.

(ii) The cost of sheet for it = Rs. 5.45 × 20 = Rs. 109.

2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find

the cost of white washing the walls of the room and the ceiling at the rate of Rs.7.50 Per m2.

• Solution: l = 5 m

b = 4 m h = 3m Area of the walls of the room = 2(l + b) h

= 2( 5 + 4) 3 = 54 m2

Area of the ceiling = lb = (5) (4) = 20 m2

∴ Total area of the walls of the room and the ceiling = 54 m2 + 20 m2 = 74 m2

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∴ Cost of white washing the walls of the room and the ceiling = 74 × 7.50 = Rs. 555.

3. The floor of a rectangular hall has a perimeter 250m. If the cost of painting the

four walls at the rate of Rs.10 perm2 is Rs.15000, Find the height of the hall. [Hint: Area of the four walls = Lateral Surface area.]

• Solution: Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.

Perimeter = 250 m

⇒ 2(1 +b) = 250 ⇒ l + b= 125 ...(1)

Area of the four walls = = 1500 m2

⇒ 2(l + b) h = 1500 ⇒ (l + b) h = 750

⇒ 125 h = 750 Using (1)

⇒ h = ⇒ h = 6m

Hence the height of the hall is 6 m.

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?

• Solution: For a brick

l = 22.5 cm b= 10 cm h = 7.5 cm

∴ Total surface area of a brick = 2(lb + bh + hl)

= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) = 2(225 + 75 + 168.75)

= 2(468.75) = 937.5 cm2 = 0.09375 m2







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∴ Number of bricks that can be painted out = = 100.

5. A cubical box has each edge 10cm and another cuboidal box is 12.5 cm long,

10cm wide and 8cm 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?

• Solution: (i) Each edge of the cubical box (a) = 10 cm ∴ Lateral surface area of the cubical box = 4a2 = 4(10)2 = 400cm2.

For cuboidal box

l = 12.5 cm

b = 10 cm h = 8cm

∴ Lateral surface area of the cuboidal box = 2(l + b) h = 2(12.5 + 10)(8) = 360 cm2. ∴ Cubical box has the greater lateral surface area than the cuboidal box by (400 - 360) cm2,

i.e., 40 cm2. (ii) Total surface area of the cubical box = 6a = 6(10)2 = 600cm2

Total surface area of the cuboidal box = 2(lb + bh + hl)

= 2[(12.5)(10) +)(10)(8) +(8)(12.5)] = 2[125 + 80 + 100] = 610 cm2.

∴ Cubical box has the smaller total surface area than the cuboidal box by (610 - 600)cm2, i.e., 10 cm2.

6. A small indoor greenhouse (herbarium) is made entirely of glass panes

(Including base) held together with tape. It is 30cm long, 25cm 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?







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• Solution: (i) For herbarium

l = 30 cm

b = 25 cm

h = 25 cm ∴ Area of the glass = 2(lb + bh + hi) = 2[(30)(25) + (25)(25) + (25)(30)] = 2[750 + 625 + 750] = 4250 cm2. (ii) The tape needed for all the 12 edges

= 4(l +b + h)

= 4(30 + 25 + 25) = 320 cm.

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 25cm 20cm 5cm and the smaller of dimensions 15 cm 12cm 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.

• Solution: For bigger box

l = 25cm

b = 20 cm

h = 5 cm ∴ Total surface area of the bigger box = 2(lb + bh + hl)

= 2[(25)(20) + (20)(5) + (5)(25)] = 2[500 + 100 + 125] = 1450 cm2

Cardboard required for all the overlaps = 1450 × = 72.5 cm2

∴ Net surface area of the bigger box = 1450 cm2 + 72.5 cm2 = 1522.5 cm2 ∴ Net surface area of 250 bigger boxes = 1522.5 × 250 = 380625 cm2







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∴ Cost of cardboard = = Rs. 1522.50.

For smaller box

l =15 cm b = 12 cm h = 5 cm

∴ Total surface area of the smaller box = 2(lb + bh + hl)

= 2[180 + 60 + 75] = 630 cm2

Cardboard required for all the overlaps = 630 × =31.5 cm2

∴ Net surface area of the smaller box = 630 cm2 + 31.5 cm2

= 661.5 cm2

∴ Net surface area of 250 smaller boxes = 661.5 × 250 = 165375 cm2

∴ Cost of cardboard = × 165375 = Rs. 661.50.

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 2.5m, with base dimensions 4m 3m?

• Solution: For shelter

l = 4 m b = 3 m h = 2.5m

∴ Total surface area of the shelter = lb + 2(bh + hl)

= (4)(3) + 2[(3)(2.5) + (2.5(4)] = 12 + 2[7.5 + 10] = 47 m2







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Hence 47 m2 of tarpaulin will be required.

∴ Cost of cardboard required for supplying 250 boxes of each kind

= Rs. 1522.50 + Rs. 661.50 = Rs. 2184.

9. The curved surface area of a right circular cylinder of height 14cm is 88cm2.

Find the diameter of the base of the cylinder.

• Solution: Let the radius of the base of the cylinder be r cm

h = 14 cm | Given

Curved surface area = 88 cm2

⇒ 2π rh = 88 ⇒ = 88

⇒ r = ⇒ r = 1

⇒€2r = 2

Hence the diameter of the base of the cylinder is 2 cm.

10. It is required to make a closed cylindrical tank of height 1 m and base

diameter 140cm from a metal sheet. How many Square metres of the sheet are required for the same?

• Solution: h = 1m = 100 cm

2r = 140 cm

⇒€r = cm = 70 cm

∴ Total surface area of the close cylindrical tank = 2π r(h + r)

=

= 74800 cm2 = m2







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= 7.48 m2

Hence 7.48 square metres of the sheet are required.

11. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its

(i) inner curved surface area.

(ii) Outer curved surface area.

(iii) Total surface area.

• Solution: h = 77 cm

2r = 4 cm

⇒ r = 2 cm

2R = 4.4 cm

⇒ R = 2.2 cm

(i) Inner curved surface area = 2π rh

= = 968 cm2

(ii) Outer curved surface area = 2π Rh

= = 1064.8 cm2

(iii) Total surface area = 2π Rh + 2π rh + 2π (R2 - r2)

= 1064.8+







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= 1064.8 + 968 +

= 1064.8 + 968 + 2 x

= 1064.8 + 968 + 5.28 = 2038.08 cm2.

12. The diameter of a roller is 84cm and its length is 120cm. It takes 500

complete revolutions to move once over to level A playground. Find the area of the playground in m2.

• Solution: 2r = 84 cm ⇒ r = 42 cm

h = 120 cm

∴ Area of the playground levelled in taking 1 complete revolution

= 2π rh

= = 31680 cm2

∴ Area of the playground = 31680 × 500

= 15840000 cm2 = = 1584 m2

Hence the area of the playground is 1584 m2.

13. A cylindrical pillar is 50cm in diameter and 3.5 m in height. Find the cost of

painting the curved surface of the pillar at the rate 12.50 per m2.

• Solution: 2r = 50 cm

r = 25 cm = 0.25 m

h = 3.5 m

∴ Curved surface area of the pillar = 2π rh







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= = 5.5 m2

∴ Cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2

= Rs. 5.5 × 12.50 = Rs. 68.75.

14. 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.

• Solution: Let the height of the right circular cylinder be h m.

r = 0.7m

Curved surface area = 4.4 m2

⇒ 2π rh = 4.4 ⇒ = 4.4

⇒ 4.4h = 4.4 ⇒ h = 1m

Hence the height of the right circular cylinder is 1m.

15. The inner diameter of a circular well is 3.5m. It is 10m deep. Find

(i) its inner curved surface area. (ii) the cost of plastering this curved surface at the rate of Rs.40 per m2.

• Solution: (i)2r = 3.5 m

⇒

⇒ r = 1.75 m

h = 10m

∴ Inner curved surface area of the circular well = 2π rh

= = 110 m2.

(ii) Cost of plastering the curved surface at the rate of Rs. 40 per m2

= Rs.110 × 40 = Rs. 4400.







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16. 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.

• Solution: h = 28

2r = 5 cm

∴ r = = = =

∴ Total radiating surface in the system = 2π rh

= = 4.4 m2.

17. 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.

(ii) How much steel was actually used, if of the steel actually used was wasted in making the tank.

• Solution: (i) 2r = 4.2 m

∴ r = = 2.1 m

h = 4.5 m

∴ Lateral or curved surface area = 2π rh

= = 59.4 m2

(ii) Total surface area = 2π r (h + r)

= = = 87.12 m2

Let the actual area of steel used be xm2.







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Since of the actual steel used was wasted, the area of the steel which has gone into the

tank = of x

∴ = 87.12 ∴ x = = 95.04 m2

∴ Steel actually used = 95.04 m2.

18. 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 20cm and height of 30cm. A margin of 2.5cm 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.

• Solution: 2r = 20 cm

⇒ r = 10 cm h = 30 cm

∴ Cloth required = 2π r(h + 2.5 + 2.5)

= 2π r(h + 5) = = 2200 cm2

19. 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 3cm and height 10.5cm. 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?

• Solution: r = 3 cm







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h = 10.5 cm

∴ Cardboard required for 1 competitor = 2π rh + π r2

=

= =

=

∴ Cardboard required for 35 competitors to be bought for the competition.

Hence 7920 cm2 of cardboard was required to be bought for the competition.

20. Diameter of the base of a cone is 10.5 cm and its slant height is 10cm. Find its

curved surface area.

• Solution: Diameter of the base = 10.5 cm

∴ Radius of the base (r) = cm = 5.25 cm

Slant height (l) = 10 cm

∴ Curved surface area of the cone = π rl

= 165 cm2.

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

• Solution: Slant height (l) = 21 m Diameter of base = 24 m

∴ Radius of base (r) = m = 12 m

∴ Total curved surface aera of the cone = π r(l + r)







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=

= =

= 1244 m2.

22. Curved surface area of a cone is 308 cm2 and its slant height is 14cm. Find

(i) radius of the base and (ii) Total surface area of the cone.

• Solution: (i) Slant height (l) = 14 cm Curved surface area = 308 cm2

⇒ π rl = 308 ⇒ = 308

⇒ r = ⇒ r = 7 cm

Hence the radius of the base is 7 cm.

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

=

= = 462 cm2

Hence the total surface area of the cone is 462 cm2.

23. A conical tent is 10 m high and the radius of its base is 24m. 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.

• Solution: (i) h = 10 m







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r = 24m

l = =

= = = 26 m

Hence the slant height of the tent is 26 m.

(ii) Curved surface area of the tent = π rl = m2

∴ Cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs.70.

= Rs = Rs. 137280. Hence the cost of the canvas is Rs. 137280.

24. What length of tarpaulin 3m wide will be required to make conical tent of

height 8m and base radius 6m? Assume that the extra length of material that will be

required for stitching margins and wastage in cutting is approximately 20cm (use = 3.14).

• Solution: For conical tent

h = 8 m r = 6 m

∴€l =

= = = = 10 m

∴ Width surface area = π rl = 3.14 × 6 × 10 = 188.4 m2

Width of tarpaulin = 3 m

∴ Length of tarpaulin = = 62.8 m

Extra length of the material required = 20 cm = 0.2 m

∴ Actual length of tarpaulin required = 62.8 m + 0.2 m = 63 m.







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25. 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.

• Solution: Slant height (l) = 25 m Base diameter = 14 m

∴ Base radius (r) = = 7 m

∴ Curved surface area of the tomb = π rl = =550 m2

∴ Cost of white-washing the curved surface of the tomb at the rate of Rs. 210 per

100 m2

= Rs. = Rs. 1155.

26. A Joker’s cap is in the form of a right circular cone of base radius 7cm and

height 24cm. Find the area of the sheet required to make 10 such caps.

• Solution: Base radius (r) = 7 cm Height (h) = 24 cm

∴€Slant height (l) = = = = = 25 cm

∴ Curved surface area of a cap = π rl = = 550 cm2

∴ Curved surface area of 10 caps = 550 × 10 = 5500 cm2

Hence the area of the sheet required to make 10 such caps is 5500 cm2.

27. 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 1m. 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 =

3.14 and take = 1.02).







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• Solution: Base diameter = 40 cm

∴ Base radius (r) = cm = 20 cm = m = 0.2m

Height (h) = 1 m

∴ l = = = = = 1.02 m (approximately).

∴∴∴∴ Curved surface area = π rl = 3.14 × 0.2 × 1.02 = 0.64056 m2

∴ Curved surface area of 50 cones = 0.64056 × 50 m2 = 32.028 m2

∴ Cost of painting all these cones = 32.028 × 12 = 384.336 = Rs. 384.34 (approximately).

28. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14cm

• Solution: (i) r = 10.5 cm

∴ Surface area = 4π r2 = 4 x = 1386 cm2.

(ii) r = 5.6 cm

∴ Surface area = 4π r2 = 4 x = 394.24 cm2.

(iii) r = 14 cm


29. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21cm (iii) 3.5m

• Solution: (i) Diameter - 14 cm

∴ Radius (r) = cm = 7 cm








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(ii) Diameter = 21 cm

∴ Radius (r) = cm

∴ Surface area = 4π r2 = = 1386 cm2.

(iii) Diameter = 3.5 cm

∴ Radius (r) = cm = 1.75 cm

∴ Surface area = π r2 = = 38.5 cm2.

30. Find the total surface area of a hemisphere of radius 10cm. (use = 3.14).

• Solution: r = 10 cm.

∴ Total surface area of the hemisphere = 3π r2 = 3 × 3.14 × (10)2=942 cm2.

31. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

• Solution: Case I. r = 7 cm

Surface area = 4π r2 = = 616 cm2.

Case II. r = 14 cm

∴ Surface area = 4π r2 = = 2464 cm2

∴ Ratio of surface areas of the balloon = 616 : 2464

= = = 1 : 4

32. 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 100cm2.







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• Solution: Inner diameter = 10.5 cm

∴ Inner radius (r) = cm = 5.25 cm

∴ Inner surface area = 2π r2 = = 173.25 cm2

∴ Cost of tin-plating at the rate of Rs. 16 per 100 cm2

= Rs. = Rs. 27.72.

33. Find the radius of a sphere whose surface area is 154cm2.

• Solution: Let the radius of the sphere be r cm.

Surface area = 154 cm2

⇒ 4π r2 = 154 ⇒ = 154

⇒ r2 = ⇒ r2 =

⇒ r = ⇒ r = = 3.5 cm

Hence the radius of the sphere is 3.5 cm.

34. The diameter of the moon is approximately one fourth of the diameter of the

earth. Find the ratio of their surface areas.

• Solution: Let the diameter of the earth be 2r.

Then diameter of the moon = (2r) =







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∴ Radius of the earth = = r

and, Radius of the moon = =

∴ Surface area of the earth = 4π r2

and, Surface area of the moon = 4π r2

Ratio of their surface areas =

= = = 1 : 16

35. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the

bowl is 5cm. Find the outer curved surface area of the bowl.

• Solution: Inner radius of the bowl = 5 cm

Thickness of steel = 0.25 cm

∴ Outer radius of the bowl = 5 + 0.25 = 5.25 cm

∴ Outer curved surface of the bowl = 4π r2

= = 346.5 cm2.

36. A right circular 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).

• Solution: (i) Surface area of the sphere = 4π r2 (ii) For cylinder







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• Radius of the base = r

• Height = 2r

• ∴ Curved surface area of the cylinder = 2π (r)(2r) = 4π r2 (iii) Ratio of the areas obtained in (i) and (ii)

• =

• = = = 1 : 1.

37. A match box measures 4 cm ×2.5 cm × 1.5 cm. What will be the volume of a

packer containing 12 such boxes?

• Solution: Volume of a matchbox = 4 × 2.5 × 1.5 cm2 = 15 cm2

Volume of a packet containing 12 such boxes = 15 × 12 cm2 = 180 cm2

38. A cuboidal water tank is 6m long, 5m wide and 4.5 m deep. How many litres of

water can it hold?(1 m3 = 1000 l)

• Solution: Capacity of the tank = 6 × 5 × 4.5 m3 = 135 m3

∴ Volume of water it can hold = 135 m3 = 135 × 1000 l =135000 l.

39. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to

hold 380 cubic metres of a liquid?

• Solution: Let the height of the cuboidal vessel be h m.

l = 10m b = 8m







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Capacity of the cuboidal vessel = 380 m3

⇒ lbh = 380 ⇒ (10)(8)h = 380

⇒ h = ⇒ h =

⇒ h = 4.75 m

Hence the cuboidal vessel must be made 4.75 m high.

40. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the

rate of Rs. 30 per m3.

• Solution: l = 8 m b = 6 m h = 3 m ∴ Volume of the cuboidal pit = lbh = 8 × 6 × 3 m3 = 144 m3

∴ Cost of digging the cuboidal pit @ Rs. 30 per m3 = Rs. 144 × 30 = Rs. 4320.

41. 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 10m.

• Solution: Let the breadth of the cuboidal tank be b m.

l = 2.5m

h = 10 m

Capacity of the cuboidal tank = 50000 litres = m3 = 50 m3

⇒ lbh = 50 ⇒ 2.5 × b × 10 = 50

⇒ 25b = 50 ⇒ b = = 20 m

Hence the breadth of the cuboidal tank is 20 m.







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

• Solution: Requirement of water per head per day = 150 litres ∴ Requirement of water for the total population of the village per day

= 150 × 4000 litres

= 600000 litres = m3 = 600 m3

For tank

l = 20m

b=15m h = 6m

∴ Capacity of the tank = 20 × 15 × 6 m3 = 1800 m3

∴ Number of days for which the water of this tank last

=

= = 3

Hence the water of this tank will last for 3 days.

43. A godown measures 40m × 25m × 10m. Find the maximum number of wooden

crates each measuring 1.5 m × 1.25m × 0.5m that can be stored in the godown.

• Solution: For godown l = 40m 6 = 25 m h= 10 m

∴ Capacity of the godown = lbh = 40 × 25 × 10m3 =10000m3







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For a wooden crate l =1.5m b= 1.25m h = 0.5 m Capacity of a wooden crate = lbh = 1.5 × 1.25 × 0.5 m3 = 0.9375 m3

We have,

= 10666.66

Hence the maximum number of wooden crates that can be stored in the godown is 10666.

44. A solid cube of side 12cm 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.

• Solution: Side of the solid cube (a) = 12 cm Volume of the solid cube = a3

= (12)3 =12 × 12 × 12 cm3 = 1728 cm3

∴ It is cut into eight cubes of equal volume.

∴ Volume of a new cube = cm3 = 216 cm3

Let the side of the new cube be x cm.

Then, volume of the new cube = x3 cm3.

According to the question,

x3 = 216 ⇒ x = (216)1/3 ⇒ x = (6 × 6 × 6)1/3

⇒ x - 6 cm

Hence the side of the new cube will be 6 cm. Surface area of the original cube = 6x2 = 6(12)2 cm2

Surface area of the new cube = 6x2 = 6(6)2 cm2







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∴ Ratio between their surface areas =

= = = 4 : 1 Hence the ratio between their surface areas is 4 : 1.

45. A river 3 m deep and 40m wide is flowing at the rate of 2 Km per hour. How

much water will fall into the sea in a minute?

• Solution: In one hour

l = 2 km = 2 × 1000 m = 2000 m b = 40m h = 3m

∴ Water fell into the sea in one hour = lbh = 2000 × 40 × 3 m3

∴ Water fell into the sea in a minute = 4000 m3

Hence 4000 m3 of water will fall into the sea in a minute.

46. The circumference of the base of a cylindrical vessel is 132cm and its height is

25cm. How many litres of water can it hold? (1000 cm3=1L)

• Solution: Let the base radius of the cylindrical vessel be r cm

Then, circumference of the base of the cylindrical vessel = 2π r cm


2π r = 132

⇒ = 132 ⇒ r =

⇒ r = 21 cm; h = 25 cm

∴ Capacity of the cylindrical vessel = π r2h







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= cm3

= 34650 cm3 = = 34.65 l

⇒ Hence the cylindrical vessel can hold 34.65 l of water.

47. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer

diameter is 28 cm. The length of the pipe is 35cm. Find the mass of the pipe, If 1 cm3 of wood has a mass of 0.6g.

• Solution: Inner diameter = 24 cm

∴ Inner radius (r) = cm = 12 cm

∴ Outer diameter = 28 cm

∴ Outer radius (R) = cm = 14 cm

Length of the pipe (h) = 35 cm

Outer volume = π R2h

= = 15840 cm3

Inner volume = π r2h

= = 15840 cm3

∴ Volume of the wood used = Outer volume - Inner volume

= 21560 cm3 - 15840 cm3

∴ Mass of the pipe = 5720 × 0.6 g = 3432 g = 3.432 kg.

48. A soft drink is available in two packs – (i) a tin can with a rectangular base of

length 5 cm and width 4cm, having a height of 15cm and (ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much?







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• Solution: (i) For tin can

l= 5 cm b = 4 cm h=15 cm

∴ Capacity = l × b × h = 5 × 4 × 15 cm3 = 300 cm3.

(ii) For plastic cylinder Diameter = 7 cm

∴ Radius (r) = cm

Height (h) = 10 cm

∴ Capacity = π r2h

= = 385 cm3

Clearly the second container i.e., a plastic cylinder has greater capacity than the first container i.e., a tin can by 385 – 380 = 85 cm3.

49. 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 ππππ = 3.14)

• Solution: (i) Let the radius of the base of the cylinder be r cm.

h = 5 cm

Later surface = 94.2 cm2

⇒ 2π rh = 94.2 ⇒ 2 × 3.14 × r × 5 = 94.2

⇒ r = ⇒ r =

⇒ r = 3 cm

Hence the radius of the base is 3 cm.







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(ii) r = 3 cm

h = 4 cm

∴ Volume of the cylinder = π r2h = 3.14 × (3)2 × 5 = 141.3 cm3.

50. It costs RS.2200 to paint the inner curved surface of a cylindrical vessel 10m

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.

• Solution:

(i) Inner curved surface area of the vessel = = 110 m2

(ii) Let the radius of the base b r m

h = 10 m

Inner curved surface area = 110 m2

⇒ 2π rh = 110 ⇒ = 110

⇒ r = ⇒ r =

⇒€r = 1.75 m

Hence the radius of the base is 1.75 m.

(iii) r = 1.75 m

h = 10 m

∴ Capacity of the vessel = π r2h

=

= 96.25 m3

Hence the capacity of the vessel is 96.25 m3(or 96.25 kl).







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51. The capacity of a closed cylindrical vessel of height 1m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

• Solution: h = l m

Capacity = 15.4 litres = = 0.0154 m3

Let the radius of the base be r m.

Capacity = 0.0154 m3

⇒ π r2h = 0.0154 ⇒ = 0.0154

⇒ r2 = ⇒ r2 = 0.0049

⇒ r = ⇒ r = 0.07 m

∴ Curved surface area = 2π rh + 2π r2 = 2 ×

= 0.44 + 0.0308 = 0.4708 m2

Hence 0.47 m2 of metal sheet should be needed.

52. 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 7mm and the diameter of the graphite is 1mm. If the length of the pencil is 14cm, find the volume of the wood and that of the graphite.

• Solution: For solid cylinder of graphite

∴ Radius (r) =

Length of the pencil (h) = 14 cm = 140 mm

∴ Volume of the graphite = π r2h = = 110 mm3







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= cm3 = 0.11 cm3

For cylinder of wood

Diameter = 7 mm

∴ Radius(R) =

Length of the pencil(h) = 14 cm = 140 mm

∴ Volume of the wood = π (R2 - r2)h = 140

= 5280 mm3 = cm3 = 5.28 cm3.

53. A patient in a hospital is given soup daily in a cylindrical bowl of diameter

7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients?

• Solution: Diameter = 7 cm

∴ Radius (r) =

Height (h) = 4 cm

∴ Volume of soup in the cylindrical bowl = π r2h = = 154 cm3

∴ Volume of soup to be prepared daily to serve 250 patients

= 154 × 250 cm3 = 38500 cm3 (or 38.5l)

Hence the hospital has to prepare 38500 cm3(or 38.5l) of soup daily to serve 250 patients.

54. Find the volume of the right circular cone with

(i) radius 6 cm, height 7cm (ii) radius 3.5 cm, height 12 cm







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• Solution: i) r = 6 cm

h =7 cm

∴ Volume of the right circular cone = = = 264 cm3

(ii) r = 3.5 cm

h = 12 cm

∴ Volume of the right circular cone = = = 154 cm3

55. Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25cm (ii) height 12cm,slant height 13cm

• Solution: (i) r = 7 cm

l = 25 cm

r2 + h2 = l2

⇒ (7)2 + h2 = (25)2 ⇒ h2 = (25)2 – (7)2

⇒ h2 = 625 – 49 ⇒ h2 = 576

⇒ h = ⇒ h = 24 cm

∴ Capacity = = = 1232 cm3 = 1.232 l. (ii) h = 12 cm

l = 13 cm

r2+h2 = l2

⇒ r2+(12)2 = (13)2 ⇒ r2 + 144 = 169

⇒ r2 = 169 – 144 ⇒ r2 = 25

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⇒ r = ⇒ r = 5 cm

∴ Capacity =

= = =

56. The height of a cone is 15cm. If its volume is 1570cm3, find the radius of the

base. (use ππππ= 3.14)

• Solution: Let the radius of the base of the cone be r cm.

h = 15 cm

volume = 1570 cm3

⇒ = 1570 ⇒ = 1570

⇒ r2 = ⇒ r2 = 100

⇒ r = ⇒ r = 10 cm

57. If the volume of a right circular cone of height 9cm is 48 cm3, find the

diameter of its base.

• Solution: Let the radius of the base of the right circular cone be r cm.

h = 9 cm

volume = 48π cm3

⇒ = 48π ⇒ = 48

⇒ = 48 ⇒ r2 =







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⇒ r2 = 16 ⇒ r = = 4 cm

⇒ 2r = 2(4) = 8 cm

Hence the diameter of the base of the right circular cone is 8 cm.

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

kilolitres?

• Solution: Diameter = 3.5 cm

∴ Radius (r) = = 1.75 m

Depth (h) = 12 m

∴ Capacity of the conical pit =

= = 38.5 m3 = 38.5 × 1000 l = 38.5 kl.

59. 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 (ii) slant height of the cone (iii) curved surface area of the cone.

• Solution: (i) Diameter of the base = 28 cm

∴ Radius of the base (r) = = 14 cm

Let the height of the cone be h cm.

Volume = 9856 cm3

⇒ = 9856 ⇒ = 9856







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⇒ h = ⇒ h = 48 cm

Hence the height of the cone is 48 cm.

(ii) r = 14 cm

h = 48 cm

∴ l = = = = = 50 cm

Hence the slant height of the cone is 50 cm.

(iii) r = 14 cm

l = 50 cm

∴ Curved surface area = π rl = = 2200 cm2

Hence the curved surface area of the cone is 2200 cm2.

60. A right triangle ABC with sides 5cm, 12cm and 13cm is resolved about the side

12cm. Find the volume of the solid so obtained.

• Solution: The solid obtained will be a right circular cone whose radius of the base is 5 cm and height is 12 cm.

∴ r = 5 cm

h = 12 cm

∴ Volume = = = 100π cm3







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Hence the volume of the solid so obtained is 100π cm3

61. If the triangle ABC in the Question 7 above is revolved about the side 5cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in question 7 and 8.

• Solution: The solid obtained will be a right circular cone whose radius of the base is 12 cm and height is 5 cm.

∴ r =12 cm

h = 5 cm

∴ Volume = = = 240π cm3

Ratio of the volumes of the two solids obtained.

62. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height

is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

• Solution: For heap of wheat

Diameter = 10.5 m

∴ Radius (r) = cm = 5.25 m

Height (h) = 3m

∴ Volume = = = 86.625 m3

Slant height (l) = = = 6.05 m (approx.)

∴ Curved surface area = π rl = = 99.825 m2

Hence the area of the canvas required is 99.825 m2

63. Find the volume of a sphere whose radius is (i) 7cm (ii) 0.63m.







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• Solution: (i) r = 7 cm

∴ Volume = = = = 1437

(ii) r = 0.63 m

∴ volume = = = 1.05 m3(approx.)

64. Find the amount of water displaced by a solid spherical ball of diameter (i)

28cm (ii) 0.21m

• Solution: (i) Diameter = 28 cm

∴ Radius (r) = cm = 14 cm

∴€Amount of water displaced = = = cm3 = 11498

(ii) Diameter = 0.21 m

∴ Radius (r) = = 0.105 m

∴ Amount of water displaced = = = 0.004581 m3

65. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the

density of the metal is 8.9 g percm3?

• Solution:

Diameter (r) = = 2.1 cm

∴ Volume = = = 38.808 cm3

Density = 8.9 g per cm3







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∴ Mass of the bass = Volume × Density = 38.808 × 8.9 = 345.39 (approx.)

66. The diameter of the moon is approximately one – fourth of the diameter of the

earth. What fraction of the volume of the earth is the volume of the moon?

• Solution: Let the radius of the earth = 2r

Then, diameter of the earth = 2r

∴ Diameter of the moon = =

∴ Radius of the moon = =

Volume of the earth (v1) =

Volume of the moon(v2) = =

= (volume of the earth)

67. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

• Solution: Diameter = 10.5 cm

∴ Radius (r) = = 5.25 cm

∴ Amount of milk = = = 303 cm3 (approx.) = 0.303 l (approx).

68. 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.







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• Solution: Inner radius (r) = 1m

Thickness of iron sheet = 1 cm = 0.01 m

∴ Outer radius (R) = Inner radius (r) + Thickness of iron sheet

∴€Volume of the iron used to make the tank = = = 0.06348 m3 (approx.)

69. Find the volume of a sphere whose surface area is 154 cm2.

• Solution: Let the radius of the sphere be r cm.

Surface area = 154 cm2

⇒ 4π r2 = 154 ⇒ = 154

⇒ = ⇒ =

⇒ h = ⇒ h = cm

∴ Volume of the sphere = = = = 179 cm3.

70. A dome of a building is in the form of a hemisphere. From inside, it was white-

Washed at the cost of Rs. 498.96. If the cost of white- washing is Rs. 2.00 per square metre, Find the (i) inside surface area of the dome (ii) Volume of the air inside the dome.

• Solution:

(i) Inside surface area of the dome = = 249.48 m2







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(ii) Let the radius of the hemisphere be r m. Inside surface area = 249.48 m3

⇒ 2π r2 = 249.48 ⇒ = 249.48

⇒ = ⇒ = 39.69

⇒ h = ⇒ h =6.3 cm

∴ Volume of the air inside the done = = = 523.9 m3 (approx.)

71. Twenty seven solid iron spheres, each of radius r and surface area s are melted to form a sphere with surface area S’. Find the

(i) radius r’ of the new sphere (ii) ratio of S and S’.

• Solution:

(i) Volume of a solid iron sphere =

∴ Volume of 27 solid iron spheres = 27

∴ Volume of the new sphere = 36π r3

Let the radius of the new sphere be r’.

Then,

Volume of the new sphere =


= 36π r’3







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⇒ r’3 = ⇒ r’3 = 27r3

⇒ r’ = (27r3)1/3 ⇒ r’ = (3 × 3 × 3 × r3)1/3

⇒ r’ = 3r

Hence the radius r’ of the new-sphere is 3r.

(ii) S = 4π r2

S’ = 4π (3r)2

∴ = = = 1 : 9

Hence the ratio of S and S’ is 1: 9

72. 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?

• Solution: ∴ Diameter of the capsule = 3.5 mm

∴ Radius of the capsule (r) = = 1.75 mm

∴ Capacity of the capsule = = = 22.46 mm3 (approx.)

Hence 22.46 mm3 (approx.) of medicine is needed to fill this capsule.

73. The outer diameter of a spherical shell is 10 cm and the inner diameter is 9

cm. Find the volume of the metal contained in the shell

• Solution: .∴ Outer diameter = 10 cm

∴ Outer radius (R) = cm = 5 cm

∴ Inner diameter = 9 cm







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∴ Inner radius (r) = cm

Volume of the metal contained in the shell =

= =

=

= = .

74. If the number of square centimetres on the surface of a sphere is equal to the

number of cubic centimetres in its volume, what is the diameter of the sphere?

• Solution: Let the radius of the sphere be r cm. Then,

Surface area = 4π r2cm2

and, volume

According to the question, 4π r2 =

⇒ r = 3

⇒ 2r = 6

Hence the diameter of the sphere is 6 cm.

75. A cone and a hemisphere have equal bases and equal volumes. Find the ratio

of their heights.

• Solution:

Volume of the cone =







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Volume of the hemisphere =

According to the question, =

⇒ h = 2r

⇒ Height of the cone = 2r cm.

Height of the hemisphere = r cm.

∴ Ratio of their heights = 2r : r = 2 : 1







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Documents

ch - 13 Surface Areas and Volumes 9TH...∴ Total surface area of a brick = 2(lb + bh + hl) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) = 2(225 + 75 + 168.75) = 2(468.75) = 937.5 cm2