Surface Area

Exercise 13.3 Part 1

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.

Answer: d = 10.5 so, r = 5.25 cm, l = 10 cm
Curved surface area of cone `= π rl`
`= (22/7) xx 5.25 xx 10`
`= 22 xx 7.5 = 165  sq  cm`


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

Answer: d = 24 m so, r = 12 m, l = 21 m
CSA of cone `=π rl`
`= (22/7) xx 12 xx 21`
`= 66 xx 12 = 792  sq  m`

Area of base `= π r^2`
`= (22/7) xx 12 xx 12 = 452.16  sq  m`
Total surface area `= 792 + 452.16 = 1244.16  sq  m`


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.

Answer: CSA = 308 sq cm, l = 14 cm, r = ?
CSA `= π rl`
Or, `308 = (22/7) xx r xx 14`
Or, `r = 308/44 = 7  cm`

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.

Answer: Slant height can be calculated by using Pythagoras Theorem
`= l^2 = h^2 + r^2`
`= 10^2 + 24^2`
`= 100 + 576 = 676`
Or, `l = 26  cm`

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

Answer: Answer: CSA of cone `= π rl`
`= (22/7) xx 24 xx 26`
Hence, `co\st = Ar\ea xx Ra\te`
`= (22/7) xx 24 xx 26 xx 70 = Rs. 137280`



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