Ex 13.3 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev

Question 1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
 Solution: Here, diameter of the base = 10.5 cm
⇒ Radius (r) = (10.5/2)cm
Slant height (l) = 10 cm
∴ Curvered surface area of the cone = πrl

Question 2. Find the total surfce area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
 Solution: Here, diameter = 24 m

⇒ Radius (r) = (24/2)m = 12 m
Slant height (l) = 21 m
∴ Total surface area = πr(r + l)

Question 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.
 Solution: Here, curved surface area = 308 cm² 
Slant height (l) = 14 cm

(i) Let the radius of the base be ‘r’ cm.

Thus, the required radius of the cone is 7 cm.

and curved surface area = 308 cm²                [Given]
∴ Total surface area = [Curved surface area] + [Base area]
= 308 cm² + 154 cm² = 462 cm²

Question 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 70.
 Solution: Here, height of the tent (h) = 10 m
Radius of the base (r) = 24 m (i)
∵ The slant height, 

∴ The slant height of the tent =

Thus, the required slant height of the tent is 26 m.

 (ii) ∵ Curved surface area of the cone = πrl
∴ Area of the canvas required = 

Question 5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is apπroximately 20 cm (Use π = 3.14).
 Solution: Here, Base radius (r) = 6 m
Height (h) = 8 m

∴ Area of the canvas (tarpaulin) required to make the tent = 188.4 m² 
Let the length of the tarpaulin = ‘L’ m
∴ Length x Breadth = 188.4
⇒ L x 3 = 188.4
⇒ L= (188.4/3) = 62.8 m
Extra length of tarpaulin  for margins = 20 cm = (20/100) m = 0.2 m
Thus, total length of tarpaulin required = 62.8 m + 0.2 m = 63.0 m

Question 6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively.
 Find the cost of white washing its curved surface at the rate of  210 per 100 m².
 Solution: Here, Base radius (r) = (14/7) = 7 m
Slant height (l) = 25 m
∴ Curved surface area = πrl = (22/7) x 7 x 25 m² 
= 22 x 25 m² = 550 m² 
Cost of white washing: 
Rate of whitewashing = 210 per 100 m²
∴ Cost of whitewashing for 550 m² 

Question 7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
 Solution: Here, Radius of the base (r) = 7 cm  height (h) = 24 cm

⇒ Lateral surface area of 10 caps = 10 x 550 cm² = 5500 cm² 
Thus, the required area of the sheet = 5500 cm² 

Question 8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04= 1.02)
 Solution: Here, ∵ Diameter of the base = 40 cm

∴ Radius (r) = 40/2 cm = 20 cm = 10/100 m

= 2/10 m = 0.2 m

Height (h) = 1 m

= 1.02 m [∵ √1.04= 1.02             (Given)]
Now, curved surface area = πrl  

⇒ Curved surface area of 1 cone = 3.14 x 0.2 x 1.02 m²

⇒ Curved surface are of 50 cones 

Cost of painting
∵ Rate of painting = ₹ 12 per m²

SURFACE AREA OF A SPHERE

A solid generated by revolving a circular lamina about any of its diameters, is called a sphere. If ‘r’ be its radius, then its surface area = 4πr².

For the following πroblems, assume π = (22/7), unless stated otherwise.