NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.2)

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NCERT solutions for class 9 Maths


Surface Areas and Volumes


Chapter 13


Exercise 13.2


EX 13.2 QUESTION 1.


The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Solution:
Let the radius of base of  cylinder. = r cm
height (h) = 14 cm and curved surface area of cylinder = 88 cm2
Curved surface area of a cylinder = 2πrh
⇒ 2πrh = 88
⇒ 2 x \frac { 22 }{ 7 } x r x 14 = 88
⇒ r = \frac { 88\times 7 }{ 2\times 22\times 14 } = 1 cm
∴ Diameter = 2 x r = (2 x 1) cm = 2 cm


Ex 13.2 Question 2.


It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?
Solution:
Height (h) = 1 m
Diameter of the base = 140 cm = 1.40 m
Radius (r) = \frac { 1.40 }{ 2 }m = 0.70 m
Total surface area of the cylinder = 2πr (h + r)
= 2 x \frac { 22 }{ 7 } x 0.70(1 + 0.70)m2
= 2 x 22 x 0.10 x 1.70 m2
= 2 x 22 x \frac { 10 }{ 100 } x \frac { 170 }{ 100 }m2
\frac { 748 }{ 100 }m2 = 7.48 m2
Hence, the required sheet = 7.48 m2


Ex 13.2 Question 3.


A metal pipe is 77 cm long. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.

Solution:
Height (h)= 77 cm
Diameter = 4 cm I

Inner radius (r) = \frac { 4 }{ 2 } cm = 2 cm

Outer diameter = 4.4 cm
⇒ Outer radius (R) = \frac { 4.4 }{ 2 } cm = 2.2 cm

(i) Inner curved surface area = 2πrh
= 2 x \frac { 22 }{ 7 } x 2 x 77 cm2
= 2 x 22 x 2 x 11 cm2 = 968 cm2

(ii) Outer curved surface area = 2πRh
= 2 x \frac { 22 }{ 7 } x 2.2 x 77 cm2= 2 x 22 x 2.2 x 11 cm2 = 1064.80cm2

(iii)Inner radius of pipe r = 2cm , Outer radius R = 2.2 cm and Height h = 77 m
Area of upper ring of pipe = π(R2 – r2)
22/7 * (2.22 -22) = 22/7 × (4.84 -4) = 22 × 0.12 = 2.64cm2
Area of lower ring of pipe = 2.64 cm2

Total surface area = [Inner curved surface area] + [Outer curved surface area] + [Area of two circular ends]=

968 + 1064.80 + 2.64 + 2.64 = 2038.08 cm²


Ex 13.2 Question 4.


The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
Solution:
The roller is in the form of a cylinder of diameter = 84 cm
⇒ Radius of the roller(r) = \frac { 84 }{ 2 } cm = 42 cm = 0.42 m
Length of the roller (h) = 120 cm =1.2 m
Curved surface area of the roller = 2πrh
= 2 x \frac { 22 }{ 7 } x 0.42 x 1.2m2
= 2 x 22 x 0.06 x 1.2 m2 = 3.168m2

Area of the playground levelled in one revolution = 3.168m2
∴ Area of the playground levelled in 500 revolutions = 500 x 3.168m2 = 1584m2


Ex 13.2 Question 5.


A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 .
Solution:
Diameter of the pillar = 50 cm
∴ Radius (r) = \frac { 50 }{ 2 }cm = 25 cm = 0.25
and height (h) = 3.5m
Curved surface area of a pillar = 2πrh

=2× \frac { 22 }{ 7 } × 0.25 ×3.5 = 2× 22× 0.25 × 0.5 = 5.5 m ²
∴ Cost of painting of 1 m2 area = Rs. 12.50
∴ Cost of painting of 5.5 m2 area= ₹ ( 5.5x 12.50 )
= ₹68.75.


Ex 13.2 Question 6.


Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
Radius (r) = 0.7 m
Curved surface area of a cylinder = 2πrh
4.4 = 2 x \frac { 22 }{ 7 } x \frac { 7 }{ 10 } x h

4.4= 4.4h

h = 1m


Ex 13.2 Question 7.


The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of ₹40 per m2.
Solution:
Radius of the well (r) = \frac { 3.5 }{ 2 } m = 1.75 m
Height of the well (h) = 10 m
(i) Inner curved surface area = 2πrh
=2 * \frac { 22 }{ 7 } * 1.75 * 10 = 110 m2

(ii) The cost of plastering this curved surface at the rate of per m2.= ₹ 40
∴ Total cost of plastering the area 110 m2 = ₹(110 x 40) = ₹4400


Ex 13.2 Question 8.


In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
Length (h) = 28 m
∴ Radius (r) = \frac { 5 }{ 2 } cm = 2.5m
Total surface area of a cylindrical pipe = 2πr(r + h)
= 2* \frac { 22 }{ 7 } * 0.025 * (0.025 + 28) = 4.4 m2  (approx)
Thus, the total radiating surface in the system is 4.4 m2 .


Ex 13.2 Question 9.


Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if \frac { 1 }{ 12 } of the steel actually used was wasted in making the tank.
Solution:
Radius (r) = \frac { 4.2 }{ 2 } = 2.1 m
Height (h) = 4.5 m
(i) Curved surface area of a cylindrical petrol tank = 2πrh
= 2 x \frac { 22 }{ 7 } x 2.1 x 4.5 m2
= 2 x 22 x 0.3 x 4.5 m2 59.4 m2

(ii) Total surface area of the tank = 2πr(r + h)
= 2 x \frac { 22 }{ 7 } x 2.1(2.1 + 4.5)m2
= 44 x 0.3 x 6.6 m2 = 87.13 m2
Let  the area of the steel used to make this cylindrical petrol storage tank = x m2
∴ Area of steel that was wasted = \frac { 1 }{ 12 } x x m = \frac { x }{ 12 }m2
Area of steel used = x – \frac { x }{ 12 } m2

Thus, the required area of the steel that was actually used is 95.04 m2.


Ex 13.2 Question 10.


In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:
Radius of lampshade (r) = \frac { 20 }{ 2 } cm = 10 cm
and height = 30 cm.
Heigh of margin for folding to cover the top and bottom of the frame. H = 30 + 2.5 + 2.5 = 35 cm
∴ Total height of the cylinder, (h)
= 30 cm + 2.5 cm + 2.5 cm = 35 cm
Now, curved surface area = 2πrH
= 2 x \frac { 22 }{ 7 } x 10 x 35 cm2
= 2200 cm2


Ex 13.2 Question 11.


The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution:
Radius of  each penholder (r) = 3 cm
Height of a penholder (h) = 10.5 cm
The penholder open from the top,therefore the area of cardboard for 1 penholder = 2πrh + πr2
= 2 * π * 3 * 10.5 + π * 3  = 63π +9π = 72π cm2

∴ Surface area of 35 penholders= 35 * 72π cm2

=35 * 72 *  \frac { 22 }{ 7 } = 5 * 72 * 22 = 7920 cm2