## EXERCISE 13.7

#### Page No 233:

#### Question 1:

Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

#### Answer:

(i) Radius (*r*) of cone = 6 cm

Height (*h*) of cone = 7 cm

Volume of cone

Therefore, the volume of the cone is 264 cm^{3}.

(ii) Radius (*r*) of cone = 3.5 cm

Height (*h*) of cone = 12 cm

Volume of cone

Therefore, the volume of the cone is 154 cm^{3}.

#### Question 2:

Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

#### Answer:

(i) Radius (*r*) of cone = 7 cm

Slant height (*l*) of cone = 25 cm

Height (*h*) of cone

Volume of cone

Therefore, capacity of the conical vessel

=

= 1.232 litres

(ii) Height (*h*) of cone = 12 cm

Slant height (*l*) of cone = 13 cm

Radius (*r*) of cone

Volume of cone

Therefore, capacity of the conical vessel

=

= litres

#### Question 3:

The height of a cone is 15 cm. If its volume is 1570 cm^{3}, find the diameter of its base. [Use π = 3.14]

#### Answer:

Height (*h*) of cone = 15 cm

Let the radius of the cone be *r*.

Volume of cone = 1570 cm^{3}

⇒ *r* = 10 cm

Therefore, the diameter of the base of cone is

10×2=20 cm.

#### Question 4:

If the volume of a right circular cone of height 9 cm is 48π cm^{3}, find the diameter of its base.

#### Answer:

Height (*h*) of cone = 9 cm

Let the radius of the cone be *r*.

Volume of cone = 48π cm^{3}

Diameter of base = 2*r* = 8 cm

#### Question 5:

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

#### Answer:

Radius (*r*) of pit

Height (*h*) of pit = Depth of pit = 12 m

Volume of pit

= 38.5 m^{3}

Thus, capacity of the pit = (38.5 × 1) kilolitres = 38.5 kilolitres

#### Question 6:

The volume of a right circular cone is 9856 cm^{3}. 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

#### Answer:

(i) Radius of cone =

Let the height of the cone be *h*.

Volume of cone = 9856 cm^{3}

*h* = 48 cm

Therefore, the height of the cone is 48 cm.

(ii) Slant height (*l*) of cone

Therefore, the slant height of the cone is 50 cm.

(iii) CSA of cone = π*rl*

= 2200 cm^{2}

Therefore, the curved surface area of the cone is 2200 cm^{2}.

#### Question 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.

#### Answer:

When right-angled ΔABC is revolved about its side 12 cm, a cone with height (*h*) as 12 cm, radius (*r*) as 5 cm, and slant height (*l*) 13 cm will be formed.

Volume of cone

= 100π cm^{3}

Therefore, the volume of the cone so formed is 100π cm^{3}.

#### Question 8:

If the triangle ABC in the 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.

#### Answer:

When right-angled ΔABC is revolved about its side 5 cm, a cone will be formed having radius (*r*) as 12 cm, height (*h*) as 5 cm, and slant height (*l*) as 13 cm.

Volume of cone

Therefore, the volume of the cone so formed is 240π cm^{3}.

Required ratio

#### Question 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.

#### Answer:

Radius (*r*) of heap

Height (*h*) of heap = 3 m

Volume of heap

Therefore, the volume of the heap of wheat is 86.625 m^{3}.

Area of canvas required = CSA of cone

Therefore, 99.825 m^{2} canvas will be required to protect the heap from rain.