NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 13 Surface Areas and Volumes Ex 13.3.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3
Ex
13.3 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 = πrl
= 22/7 × 10.5/2 × 10 cm2
= 11 × 15 × 1 cm2 = 165 cm2
Ex
13.3 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 13.3 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 = πrl = 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 13.3 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 = r2 + h2
= 242 + 102 = 576 + 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 = πrl
∴ Area of the canvas required
Ex
13.3 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 π = 3.14)
Solution:
Here, base radius (r) = 6 m
Height (h) = 8 m
∴ Slant height l2 = r2 + h2
= 62 + 82 = 36 + 64 = 100
Therefore,
l = √100 = 10 m
Now, curved surface area = πrl
= 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.
Length × Breadth = 188.4
⇒ 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
13.3 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
13.3 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 = h2 +
r2 = 242 + 72
= 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 13.3 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.).
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