## NCERT solutions for class 9 Maths

**Surface Areas and Volumes**

**Chapter 13 **

**Exercise 13.7**

**EX 13.7 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**

**Solution:**

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

Height (h) = 7 cm

**Ex 13.7 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**

**Solution:**

(i)Radius of conical vessel (r) = 7 cm and slant height (l) =25 cm

Thus, the capacity of the conical vessel = 1.232 litres.

(ii) Height of conical vessel (h) = 12 cm and slant height (l) = 13 cm

∴ Capacity of the conical vessel

Thus, the capacity of the conical vessel = litres.

**Ex 13.7 Question 3.**

**The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)**

**Solution:**

Volume of the cone = 1570 cm^{3}

Height of the cone (h) = 15 cm

Volume of cone = πr^{2}h

⇒1570= x 3.14 x r² x 15

⇒1570 = 3.14 x 5 x r²

r² = 1570 / 3.14 x 5 = 100

r = 10 cm

Thus, the radius of the base of cone = 10 cm.

**Ex 13.7 Question 4.**

**If the volume of a right circular cone of height 9 cm is 48 cm³, find the diameter of its base.**

**Solution:**

Volume of the cone = 48 cm^{3}

Height of the cone (h) = 9 cm

Volume of cone = πr^{2}h

Diameter = 2 x radius .

∴ Diameter of the base of the cone = (2 x 4)cm = 8 cm

**Ex 13.7 Question 5.**

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

**Solution:**

Radius of pit (r) = m

Depth (h) = 12m

⇒ Volume (capacity) = πr^{2}h

Thus, the capacity of the pit = 38.5 kilolitres.

**Ex 13.7 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**

**Solution:**

Volume of the cone V = 9856 cm

^{3}

Radius of the base (r) = = 14 cm

(i) Let the height of the cone be h cm.

Thus, the height of cone = 48 cm.

(ii) Height of cone h = 48 cm and radius r = 14 cm. Let the slant height = l cm.

⇒ l^{2} = r^{2} +h^{2}

⇒ l^{2} = 14^{2} + 48^{2} = 196 + 2304 = 2500

∴ l = 50

Thus, the slant height of cone = 50 cm.

(iii) Slant height of cone l = 50cm and radius r =. 14 cm Curved surface area of a cone = πrl

∴ Curved surface area = x 14 x 50 cm^{2}

= 2200 cm^{2}

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

**Ex 13.7 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.**

**Solution:**

If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the radius of of cone r = 5 cm Height(h) = 12cm and slant height l = 13 cm

∴ Volume of the cone so obtained = πr^{2}h

= x π x (5)^{2} x 12cm^{3}

= 100 π cm^{3}

Thus, the volume of the cone = 100πcm^{3}.

**Ex 13.7 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.**

**Solution:**

Since the right triangle is revolved about the side 5 cm.

∴ Height of the cone so obtained (h) = 5 cm

Radius of the cone (r) = 12 cm

Volume of solid = πr^{2}h = x π x 12 x 12 x 5 = 240πcm³

Thus, the volume of solid = 240πcm³

**Ex 13.7 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.**

**Solution:**

Thus, the volume of heap of wheat = 86.625m^{3}

Let, the slant height of heap of. Wheat = l cm

Thus the required area of the canvas to protect wheat = 99.825 m^{2}