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 cm2 = 165 cm2
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) m2
= 22/7 ×12 × 33 m2 = 1244.57 m2
3. Curved surface area of a cone is 308 cm2 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 cm2, 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) cm2 = 22 × 21 cm2 = 462 cm2
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, l2 = h2 + r2 = (10)2 + (24)2
= 100 + 576 = 676
(ii) Curved surface area of the tent
= πrl = 22/7 × 24 × 26 m2
Cost of 1 m2 canvas = Rs 70
∴ Cost of 22/7 × 24 × 26 m2 of canvas
= Rs 70 × 22/7 × 24 × 26 = Rs 137280.