NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.1

# NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.1

NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.1 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 11 Surface Areas and Volumes Ex 11.1.

## NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes Ex 11.1

Ex 11.1 Class 9 Maths 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.

Solution:
Here, diameter of the base = 10.5 cm
Radius of the base (r) = 10.5/2 cm and slant height (l) = 10 cm
Curved surface area of the cone = Ï€rl
22/7 × 10.5/2 × 10 cm2
= 11 × 15 × 1 cm2 = 165 cm2

Ex 11.1 Class 9 Maths 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.

Solution:
Here, diameter of the base = 24 m
Radius of the base (r) = 24/2 m = 12 m and slant height (l) = 21 m
Total surface area of a cone = Ï€r(r + 1)

Ex 11.1 Class 9 Maths 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.

Solution:
Here, curved surface area = Ï€rl = 308 cm2
Slant height (l) = 14 cm

(i) Let the radius of the base be ‘r’ cm
Ï€rl = 308  22/7 × r × 14 = 308
r = 308 × 7/22 × 1/14 = 7 cm
Thus, the radius of the cone is 7 cm.

(ii) Base area = Ï€r2 = 22/7 × 72 cm2
= 22 × 7 cm2 = 154 cm2
and curved surface area = 308 cm2 [Given]
Total surface area of the cone
= [Curved surface area] + [Base area] = 308 cm2 + 154 cm2
= 462 cm2

Ex 11.1 Class 9 Maths 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.

Solution:
Here, height of the tent (h) = 10 m
Radius of the base (r) = 24 m

(i) The slant height, l2 = rh2

2410576 100 = 676

Therefore, l = √676 = 26 m
Thus, the required slant height of the tent is 26 m.

(ii) Curved surface area of the cone = Ï€rl
Area of the canvas required

Ex 11.1 Class 9 Maths 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 Ï€ = 3.14)

Solution:
Here, base radius (r) = 6 m
Height (h) = 8 m
Slant height l2 = rh2

6836 64 = 100

Therefore, l = √100 = 10 m
Now, curved surface area = Ï€rl
= 3.14 × 6 × 10 m2
= 188.4 m2
Thus, area of the tarpaulin required to make the tent = 188.4 m2
Let the length of the tarpaulin be L m.
L × 3 = 188.4

L = 188.4/3 = 62.8 m
Extra length of tarpaulin required for margins = 20 cm = 20/100 m = 0.2 m
Thus, total length of tarpaulin required = 62.8 m + 0.2 m = 63 m

Ex 11.1 Class 9 Maths 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 .

Solution:
Here, base radius (r) = 14/2 m = 7 m
Slant height (l) = 25 m
Curved surface area = Ï€rl
22/7 × 7 × 25 m2 = 550 m2
Cost of white-washing for 100 m2 area = ₹ 210
Cost of white-washing for 550 m2 area
= ₹ 210/100 × 550 = ₹ 1155

Ex 11.1 Class 9 Maths 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:
Radius of the base (r) = 7 cm and height (h) = 24 cm
Slant height, l2 = h+ r2472

576 49 = 625

Therefore, l = 25 cm
∴ Lateral surface area = Ï€rl = 22/7 × 7 × 25 cm2 = 550 cm2
Lateral surface area of 10 caps = 10 × 550 cm2 = 5500 cm2
Thus, the required area of the sheet = 5500 cm2

Ex 11.1 Class 9 Maths 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 Ï€ = 3.14 and take 1.04 = 1.02)

Solution:
Diameter of the base = 40 cm

= ₹ 384.336 = ₹ 384.34 (approx.)
Thus, the required cost of painting is ₹ 384.34 (approx.).