Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3
Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3
GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3
Question 1.
The diameter of the base of the cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
∴ Diameter of the base = 10.5 cm
∴ Radius of the base (r) = 10.5/7 cm = 5.25 cm
Slant height (l) = 10 cm
∴ Curved surface area of the cone = πrl
= 22/7 x 5.25 x 10 = 165 cm2
Question 2.
Find the total surface area of a cone, if its slant height is 21 m and the diameter of its base is 24 m.
Solution:
Slant height (l) = 21 m
Diameter of base = 24 m
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 approximately 20 cm. (Use it = 3.14)
Solution:
For conical tent
h = 8m
r = 6 m
Question 6.
The slant height and a base diameter of a conical tomb is 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2.
Solution:
Slant height (l) = 25 m
Base diameter (d) = 14 m
∴ Base radius (r) = 14/2 m = 7m
∴ Curved surface area of the tomb
= πrl = 22/7 x 7 x 25 = 550 m2
∴ Cost of white-washing the curved surface area of the tomb at the rate of 210 per 100 m2
= ₹ 210/100 x 550 = ₹ 1155
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:
Base radius (r) = 7 cm
Height (h) = 24 cm
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 a 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?
∴ Curved surface area = πrl
= 3.14 x 0.2 x 1.02 = 0.64056 m2
∴ Curved surface area of 50 cones
= 0.64056 x 50 m2 = 32.028 m2
∴ Cost of painting all these cones at ₹ 12 per m2
= ₹ 32.028 x 12
= ₹ 384.336 = ₹ 384.34 (approximately)
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