For previous exercise – Surface Area and Volumes – Exercise 13.2 – Class IX

Assume π = ^{22}/_{7} , unless stated otherwise.

**Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.**

Solution:

r = ^{10.5}/_{2 }cm = 5.25 cm , l = 10 cm

Curved surface area of the cone = πrl = ^{22}/_{7} x 5.25 x 10 = 165 cm^{2}

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

Solution:

l = 21 m and d = 24 m ; r = ^{d}/_{2 }= ^{24}/_{2} = 12 m

Total surface area of the cone = πr(l + r) = ^{22}/_{7} x 12 x (21 + 12) = 1244. 57 m^{2}

**Curved surface area of a cone is 308 cm**^{2}and its slant height is 14 cm. Find

**(i) radius of the base **

**(ii) total surface area of the cone.**

Solution:

Curved surface area of a cone = 308 cm^{2}

l = 14 cm

(i) Curved surface area of a cone = πrl

308 = ^{22}/_{7} x r x 14

r = ^{308 x 7}/_{22×14} = 7

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

= ^{22}/_{7} x 7 x(14 + 7)

= 462 cm

**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 m ^{2} canvas is Rs 70.**

Solution:

l = 10 cm ; r = 24 m

(i) l^{2} = h^{2} + r^{2}

l^{2}= 10^{2} + 24^{2}

l^{2} = 100 + 576 = 676

l = √676 = 26 m

(ii) curved surface area of the tent = πrl = ^{22}/_{7} x 24 x 26 = 1961.143 m^{2}

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

Cost of ^{ }1961.143m^{2} of canvas = 70 x 1961.143 = Rs. 137280

**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 approximately 20 cm (Use π = 3.14).**

Solution:

r = 6m ; h = 8m

We have, l^{2} = r^{2} + h^{2}

l^{2} = 6^{2} + 8^{2}

l^{2} = 36 + 64

l^{2} = 100

l = 10 m

Curved surface area of the tent = πrl = ^{3.14 x 6 x 10}/_{3} m + 20 m = 62.8 m + 0.2 m = 63 m

**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 Rs 210 per 100 m**^{2}.

Solution:

l = 25 m

d =14 m

r = ^{d}/_{2} = ^{14}/_{2} = 7 m

Curved surface area = πrl = ^{22}/_{7} x 7 x 25 = 550 m^{2}

The cost of white-washing its curved surface at the rate for 100 m^{2} = Rs 210

The cost of white-washing its curved surface at the rate for 550 m^{2} = ^{550 x 210}/_{100} = Rs. ^{115500}/_{100} = Rs. 1155

**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:

r = 7 cm

h = 24

l^{2} = h^{2} + r^{2}

l^{2} = 24^{2} + 7^{2} = 625

l = 25 cm

Total cured surface of 1 cap = πrl =^{22}/_{7} x 7x 25 = 550 cm^{2}

Area of sheet required to make 10 such caps = 10 x 550 = 5500 cm^{2}

**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 Rs 12 per m**^{2 }, what will be the cost of painting all these cones? [Use π = 3.14 and take √(1.04) = 1.02]

Solution:

r = ^{40}/_{2} = 20 cm = 0.20 m

h = 1 m

l^{2} = h^{2} + r^{2} = 1^{2} + (0.2)^{2} = 1.04

l = 1.02 m

Curved surface area of 1 cone = πrl

Curved surfacearea of 50 cones = 50 x 3.14 x 0.2 x 1.02 m^{2} = 32.028 m^{2}

Cost of painting an area of 1m^{2} =Rs. 12

Cost of painting an area of 32.028 m^{2} = Rs. 12 x 32.028 = Rs. 384.34

For next exercise – Surface Area and Volumes – Exercise 13.4 – Class IX

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