NCERT Solutions: Surface Areas & Volumes (Exercise 11.1)

# Surface Areas & Volumes (Exercise 11.1) NCERT Solutions - Mathematics (Maths) Class 9

Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π = 22 / 7)
Ans:

Diameter of the base of the cone is 10.5 cm. To find the radius, we need to divide the diameter by 2.
Radius of the base of the cone, r = diameter / 2 = 10.5 / 2 = 5.25 cm
The slant height of the cone is given as 10 cm. Let's denote it as l.
Slant height of the cone, l = 10 cm
Now, we can find the curved surface area (CSA) of the cone using the formula:
CSA = π  r  l
where π (pi) is approximately equal to 22 / 7.
CSA = (22 / 7) × 5.25 × 10
CSA = 165 cm²
Hence, the curved surface area of the cone is 165 cm².

Q2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22 / 7)
Solution:

The radius of the cone, r = 24 / 2 m = 12m
Slant height, l = 21 m
To find the total surface area of a cone, we use the formula: Total Surface area of the cone = πr(l + r)
Total Surface area of the cone = (22 / 7) × 12 × (21 + 12) m2
= (22 / 7) × 12 × 33 m2
= 1244.57 m2
Thus, the total surface area of the cone is 1244.57 m2.

Q3. 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. (Assume π = 22 / 7)

Ans: Slant height of cone, l = 14 cm
Let the radius of the cone be r.
(i) We know, CSA of cone = πrl
Given: Curved surface area of a cone is 308 cm2
(308) = (22 / 7) × r × 14
308 = 44r
r = 308 / 44 = 7cm
The radius of the cone base is 7 cm.

(ii) To find the total surface area of the cone, we need to add the curved surface area (CSA) and the area of the base (πr2).
Total surface area of cone = CSA of cone + Area of base (πr2)
Total surface area of cone = 308 + (22 / 7) × 72 = 308 + 154
Therefore, the total surface area of the cone is 462 cm2.

Q4. 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 Rs 70. (Assume π = 22 / 7)
Ans:
Let ABC be a conical tent with vertex A, base center O, and base circumference point B.
Height of conical tent (AO), h = 10 m
Radius of conical tent (BO), r = 24m
Let the slant height of the tent (AB) be l.
(i) In right triangle ABO, we have
AB2 = AO2 + BO2 (using Pythagoras theorem)
l2 = h2 + r2
= (10)2 + (24)2
= 100 + 576
= 676
l = √676 = 26m
Therefore, the slant height of the tent is 26 m.

(ii) The curved surface area (CSA) of the conical tent can be calculated using the formula:
CSA = πrl
= (22/7) × 24 × 26 m2
=13728/7 m2
Now, let's calculate the cost of the canvas required to make the tent.
Cost of 1 m2 canvas = Rs 70
Cost of (13728 / 7) m2 canvas is equal to Rs (13728 / 7) × 70 = Rs 137280
Therefore, the cost of the canvas required to make such a tent is Rs 137280.

Q5. 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]
Ans: Height of conical tent, h = 8m
Radius of base of tent, r = 6m
Slant height of tent, l2 = (r2 + h2)
l2 = (62 + 82) = (36 + 64) = (100)
or l = 10
Again, CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4m2
Let the length of tarpaulin sheet required be L
As 20 cm will be wasted, therefore,
Effective length will be (L - 0.2m).
Breadth of tarpaulin = 3m (given)
Area of sheet = CSA of tent
[(L – 0.2) × 3] = 188.4
L - 0.2 = 62.8
L = 63
Therefore, the length of the required tarpaulin sheet will be 63 m.

Q6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. (Assume π = 22 / 7).
Ans: Slant height of conical tomb, l = 25m
Base radius, r = diameter / 2 = 14 / 2 m = 7m CSA of conical tomb = πrl= (22 / 7) × 7 × 25 = 550
CSA of conical tomb = 550m2
Cost of white-washing 550 m2 area, which is Rs (210 × 550) / 100
= Rs.1155
Therefore, cost will be Rs. 1155 while white-washing tomb.

Q.7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π = 22 / 7)
Ans:
Radius of conical cap, r = 7 cm
Height of conical cap, h = 24cm Slant height, l2 = (r2 + h2)
= (72 + 242)
= (49 + 576)
= (625)
Or l = 25 cm
CSA of 1 conical cap = πrl
= (22 / 7) × 7 × 25
= 550
CSA of 10 caps = (10 × 550) cm2 = 5500 cm2
Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

Q8. 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 m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) = 1.02).
Ans:
Given:
Radius of cone, r = diameter / 2 = 40 / 2 cm = 20cm = 0.2 m
Height of cone, h = 1m
Slant height of cone is l, and l2 = (r2 + h2)
Using given values, l2 = (0.22 + 12)
= (1.04)
Or l = 1.02
Slant height of the cone is 1.02 m
Now,
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02)
= 0.64056m2
CSA of 50 such cones = (50 × 0.64056)
CSA of 50 such cones = 32.028 m2
Again,
Cost of painting 1 m2 area = Rs 12 (given)
Cost of painting 32.028 m2 area
= Rs (32.028 × 12)
= Rs.384.336
= Rs.384.34 (approximately)
Therefore, the cost of painting all these cones is Rs. 384.34.

The document Surface Areas & Volumes (Exercise 11.1) NCERT Solutions | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
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## Mathematics (Maths) Class 9

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## Mathematics (Maths) Class 9

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