NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3

Chapter 13 Surface Areas and Volumes Exercise 13.3

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:

Given
Diameter of the base of the cone = 10.5 cm
Slant height of the cone = 10 cm
Curved Surface Area of Cone = ?
Radius = Diameter/2 = 10.5/2 cm

CSA of cone = πrl
= 22/7 × 10.5/2 × 10
= 22 × 1.5 × 5
= 165 cm²
Therefore, the CSA of cone is 165 cm².

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:

Given
Slant height of the cone = 21 m
Diameter of the base of the cone = 24 m
TSA of the cone = ?
Radius = Diameter/2 = 24/2 = 12 m

TSA of cone = πr(r + l)
= 22/7 × 12(12 + 21)
= 22/7 × 12 × 33
= 8712/7
= 1244.57 m²
Therefore, the TSA of cone is 1244.57 m².

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:

Given
CSA of cone = 308 cm²
Slant height of cone = 14 cm
Radius of the cone = ?
TSA of the cone = ?

CSA of cone = πrl
308 = 22/7 × r × 14
308 = 44r
r = 308/44
r = 7 cm

TSA of cone = πr(r + l)
= 22/7 × 7 × (7 + 14)
= 22 × 21
= 462 cm²
Therefore, the radius of cone is 7 cm and the TSA of the cone is 462 cm².

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.
ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70.
Answer:

Given
Height of the conical tent = 10 m
Radius of the base = 24 m
Slant height of the tent = ?
Cost of canvas required to make the tent = ?
l = √r² + h²
= √(24)² + (10)²
= √576 + 100
= √676
= 26 m

Area of canvas required = ?
CSA of cone = πrl
= 22/7 × 24 × 26
= 13728/7 m²
Cost of 13728/7 m² canvas = ?
Cost of 1 m² canvas = ₹70
= 13728/7 × 70
= ₹137280
Therefore, the cost of canvas required for tent is ₹137280.

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 π = 3.14).
Answer:

Given
Width of tarpaulin cloth = 3 m
Height of conical tent = 8 m
Radius of the base of conical tent = 6 m
Length of tarpaulin cloth required = ?

l = √r² + h²
= √(6)² + (8)²
= √36 + 64
= √100
= 10 m
CSA of cone = πrl
= 3.14 × 6 × 10
= 188.4 m²

Let the length of tarpaulin cloth be x.
20 cm of cloth will be wasted for stitching margins. So, the length of tarpaulin will be (x - 0.2) m
188.4 = (x - 0.2) × 3
62.8 = x - 0.2
x = 62.8 + 0.2 = 63 m
Therefore, the length of tarpaulin cloth required is 63 m.

Question 6: 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.
Answer:

Given
Slant height of conical tomb = 25 m
Diameter of the base = 14 m
Cost of white-washing its curved surface = ?
Radius = Diameter/2 = 14/2 = 7 m

CSA of cone = πrl
= 22/7 × 7 × 25
= 22 × 25
= 550 m²
Cost of white washing per 100 m² = ₹210
Cost of white washing 550 m² = ?
= 550 × 210/100
= 115500/100
= ₹1155
Therefore, the cost of white washing its curved surface area is ₹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.
Answer:

Given
Radius of the base of joker’s cap = 7 cm
Height of joker’s cap = 24 cm
Area of sheet required to make 10 caps = ?

l = √r² + h²
= √(7)² + (24)²
= √49 + 576
= √625
= 25 cm
CSA of cone = πrl
= 22/7 × 7 × 25
= 22 × 25
= 550 cm² of sheet to make 1 cap
Area of sheet required to make 10 caps = 550 × 10 = 5500 cm².

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 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? (Use π = 3.14 and take √1.04 = 1.02)
Answer:

Given
Bus stop is barricaded from road by using 50 hollow cones.
Diameter of the base of the cone = 40 cm (0.4 m)
Height of the cone = 1 m
Cost of painting outer surface of 50 hollow cones = ?
Radius = Diameter/2 = 0.4/2 = 0.2 m
l = √r² + h²
= √(0.2)² + (1)²
= √0.04 + 1
= √1.04
= 1.02 m (approx)

CSA of cone = πrl
= 3.14 × 0.2 × 1.02
= 0.64056 m² of cardboard is required for 1 cone
Area of cardboard required for 50 cone = 0.64056 × 50 = 32.028 m²

Cost of painting per m² = ₹12
Cost of painting 32.028 m² = ?
= 32.028 × 12
= ₹384.336
= ₹384.34 (approx)
Therefore, the cost of painting 50 hollow cones is ₹384.34.

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