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## Chapter 13: Surface Areas and Volumes

#### TEXTBOOK’S EXERCISE – 13.7

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

Sol. Here, r = 3 5 2. m = 1.75 m, h = 12 m

The capacity of the pit = 13 πr2h

= 13 × 22/7 × 1.75 × 1.75 × 12 m3

= 38.5 m3 = 38.5 kl

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

(ii) the slant height of the cone

(iii) the curved surface area of the cone.

Sol. Here, r = 28/2 cm = 14 cm,

volume = 9856 cm3

(i) Volume of the cone = 1/3 πr2h

⇒ 9856 = 1/3 × 22/7 × 14 × 14 × h

= 48

Hence, height of the cone = 48 cm

(ii) Slant height

Hence, slant height of the cone = 50 cm

(iii) Curved surface area of the cone

= πrl = 22/7 × 14 × 50 cm2 = 2200 cm2

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.

Sol.

The solid formed is a cone, whose height h = 12 cm, base radius

r = 5 cm.

∴ Volume of the cone = 13 πr2h

= 1/3 × π × 5 × 5 × 12 cm3 = 100 π cm3

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

Sol. Here, radius r of the cone = 12 cm and height h of the cone = 5 cm

∴ The volume of the cone = 1/3 πr2h

= 1/3 π × 12 × 12 × 5 = 240 πcm3

Hence, the required ratio

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.

Sol. Here, radius r = 10.5/ 2 m = 5.25 m, h = 3 m

Volume of the heap = 1/3 πr2h

= 1/3 × 22/7 × 5.25 × 5.25 × 3 m3 = 86.625 m3

Curved surface area of the cone = πrl

= 22/7 × 5.25 × 6.05 m2 = 99.825 m2

Hence, 99.825 m2 of canvas is needed.

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