NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.3)

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NCERT solutions for class 9 Maths


Surface Areas and Volumes


Chapter 13


Exercise 13.3


EX 13.2 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:
Radius (r) = \frac { 10.5 }{ 2 } cm = 5.25 cm
and slant height (l) =10 cm
Curved surface area of the cone = πrl
\frac { 22 }{ 7 } x \frac { 10.5 }{ 2 } x 10cm2
= 11 x 15 x 1 cm2 = 165cm2

Ex 13.3 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:
Radius (r) = \frac { 24 }{ 2 } m = 12 m
and slant height (l) = 21 m
∴ Total surface area of a cone = πr(r +l)
\frac { 22 }{ 7 }  * 12 * (12 + 21 ) = = \frac { 22 }{ 7 }  * 12 * 33 = 1244.57 m2


Ex 13.3 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:
Curved surface area = πrl = 308 cm2
Slant height (l) = 14 cm

(i) Let the radius of the base of cone = r cm
= 308 ⇒ \frac { 22 }{ 7 } x r x 14 = 308
r = \frac { 308\times 7 }{ 22\times 14 } = 7cm
Hence, the radius of the cone is 7 cm

(ii) Total surface area of the cone = πr(r + l)
\frac { 22 }{ 7 }  * 7 * (7 + 14) = 22 * 21 = 462 cm2


Ex 13.3  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:
Height tent (h) = 10 m
Radius of the cone(r) = 24 m

(i) The slant height, l m
l2 = r2 + h2l2 = 242 + 102 = 576 + 100=  676
l =\sqrt { 676 } m = 26m

(ii) Curved surface area of the cone = πrl

= \frac { 22 }{ 7 }  * 24 * 26 m2
Cost of 1m2 canvas =₹70

∴ The cost of \frac { 22 }{ 7 }  * 24 * 26 m2 canvas = 70 *  \frac { 22 }{ 7 }  * 24 * 26 = ₹137280


Ex 13.3  Question 5.


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)
Solution:
Radius of cone (r) = 6 m
Height(h) = 8m
∴ Slant height (l) = l2 = r2 + h2
l2 = 62 + 82  = 36 + 64 = 100
= \sqrt { 100 }m = 10 m
Area of tarpaulin to make the tent = πrl
= 3.14 x 6 x 10 = 1884 m2
Let the length of 3 m wide tarpaulin = L m
Therefore, the area of tarpaulin required = 3 * L
According to question,
3 * L  = 188.4 ⇒ L = \frac { 188.4 }{ 3 } = 62.80 m
Extra  tarpaulin for stitching margins and wastage = 20cm = 0.2m
Thus, total length of tarpaulin = 62.8 m + 0.2 m = 63 m


Ex 13.3 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:
Radius of conical tomb (r) = \frac { 14 }{ 2 } m = 7 m
Slant height (l) = 25 m
∴ Curved surface area of conical tomb = πrl
\frac { 22 }{ 7 } x 7 x 25 m2 = 550 m2
Cost of white-washing for 100 m2 area = Rs. 210
∴ Cost of white-washing for 550 m2 area
= Rs. \frac { 210 }{ 100 } x 550 = Rs. 1155

Ex 13.3 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 right circular cone (r) = 7 cm and height (h) = 24 cm
Slant height (l)l m
l2 = r2 + h2= l2 = 72 + 242 = 49 + 576  = 625.

l = \sqrt { 625 } cm = 25 cm
∴Area of sheet required to make 1 cap = πrl = \frac { 22}{ 7 } x 7 x 25 cm2 = 550 cm2
∴ Area of  sheet required to make 10 caps = 10 x 550 cm2
= 5500 cm2


Ex 13.3 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  \sqrt{104}  = 1.02)
Solution:
Radius of cone (r) = 40/2 = 20 cm and height (h) = 1 cm
Slant height (l)l m
l2 = r2 + h2= l2 = (0.2)2 + 12 = 0.04+ 1 = 1.04

l =  \sqrt{104} = 1.02m
Curved surface area of cone = πrl = 3.14 * 7 * 25 cm2 = 6.4056 m2
∴  Curved surface area of  50 cone s = 50 x 6.4056 = 32.028 m2
Cost of painting at rate of ₹12 per m2 = ₹12 * 32.028  = Rs. 384.34 (approx.)