#### Class nine Maths Chapter 13 NCERT Solutions

## Chapter 13: Surface Areas and Volumes

**TEXTBOOK’S
EXERCISE – 13.3 **

**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. **

**Sol. **Here, r = 10.5/2 cm
= 5.25 cm, l = 10 cm.

The curved surface area of the cone = πrl

= 22/7 ×
5.25 × 10 cm^{2} = 165 cm^{2}

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

**Sol. **Here, l = 21 m, r = 24/2
m = 12 m

Total surface area of the cone

= πr(l +
r) = 22/7
× 12 (21 + 12) m^{2}

= 22/7 ×12
× 33 m^{2} = 1244.57 m^{2}

**3. Curved surface area of a cone is 308 cm**^{2}** and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. **

**Sol.
**Here, l = 14 cm,
curved surface area

= 308 cm^{2},
r =?

(i) Curved surface area of the cone = πrl

⇒ 308 = 22/7 × r × 14

⇒ r = 308× 22/2 = 7

Hence, base radius of the cone = 7 cm.

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

= 22/7 × 7 (14 + 7) cm^{2 }=
22 × 21 cm^{2 }= 462 cm^{2}

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

**Sol.
**Here, h = 10 m, r = 24 m

(i)
We have, l^{2} = h^{2} + r^{2} = (10)^{2} +
(24)^{2}

= 100 + 576 = 676

(ii) Curved surface area of the tent

= π*rl *= 22/7
× 24 × 26 m^{2}

Cost
of 1 m^{2} canvas = Rs 70

∴ Cost of 22/7 × 24 × 26 m^{2} of
canvas

= Rs 70 × 22/7 × 24 × 26 = Rs 137280.