Chapter 13 Surface Areas and Volumes Ex 13.3
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
We have, slant height l = 21 m
and diameter = 24 cm
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.
We have, slant height, l= 14cm
Curved surface area of a cone = 308 cm2
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.
We have, h = 10 m
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? 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)
Let r, h and l be the radius, height and slant height of the tent, respectively.
The extra material required for stitching margins and cutting = 20 cm = Q2 m Hence, the total length of tarpaulin required = 62.8 + Q2 = 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 ₹ 210 per 100 m2.
We have, slant height, l = 25m
and diameter = 14m
∴ Radius, r= 7m
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.
∵ The sheet required to make 1 cap = 550 cm2
∴ The sheet required to make 10 caps = 550 x 10= 5500 cm2
Curved surface area of a cone = πrl = 3.14 x 0.2 x 1.02 = 0.64056 m2
Cost of painting per m2 = ₹12
Cost of painting 0.64056 m2 = ₹12 x 0.64056= ₹ 7.68672
Cost of painting for 1 cone = ₹ 7.68672
Cost of painting 50 cones = ₹ 7.68672x 50= ₹ 384.336= ₹ 384.34