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 cm^{2} 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 m^{2} 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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