CBSE Class 9

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.
Find the surface area of a sphere of radius
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
(i) We have, r = 105 cm

Question 2.
Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Solution:

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
We have, r = 10 cm
Total surface area of a hemisphere = 3πr²
= 3 x 3.14 x (10)²
= 9.42 x 100
= 942 cm²

Question 4.
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
Let initial radius, r₁ = 7 cm
After increases, r₂ = 14 cm
Surface area for initial balloon = 4πr₁² = 4 x \(\frac { 22 }{ 7 }\) x 7 x 7 = 88 x 7
A₁ = 616 cm²
Surface area for increasing balloon = 4πr₂² = 4x \(\frac { 22 }{ 7 }\) x 14 x 14 = 88 x 28
A₂ = 2464 cm²
∴ Required ratio = A₁ : A₂ = 616 : 2464 = 1 : 4

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².
Solution:
We have, inner diameter = 10.5 cm
Inner radius = \(\frac { 10.5 }{ 2 }\) cm = 5.25 cm
Curved surface area of hemispherical bowl of inner side = 2πr²

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
Surface area of a sphere = 154 cm²

Hence, the radius of the sphere is 3.5 cm.

Question 7.
The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas.
Solution:
Let diameter of the Earth = d₁

Question 8.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Outer radius of the bowl = (Inner radius + Thickness)
= ( 5 + 0.25) cm = 5.25 cm

Question 9.
A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Solution:

The radius of the sphere = r
Radius of the cylinder = Radius of the sphere = r
Height of the cylinder = Diameter = 2r
(i) Surface area of the sphere A₁ = 4πr²
(ii) Curved surface area of the cylinder = 2πrh
A₂ = 2π x r x 2r
A₂ = 4πr²
(iii) Required ratio = A₁ :A₂ = 4πr² : 4πr² = 1 : 1

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are part of NCERT Solutions for Class 9 Maths. Here we have given 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

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:
We have, diameter = 10.5 cm
Radius (r) = \(\frac { 10.5 }{ 2 }\) = 5.25 cm
and slant height l= 10 cm
Curved surface area = πrl= \(\frac { 22 }{ 7 }\) x 5.25 x 10 = 165 cm²

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:
We have, slant height l = 21 m
and diameter = 24 cm

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:
We have, slant height, l= 14cm
Curved surface area of a cone = 308 cm²

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:
We have, h = 10 m

Hence, the slant height of the canvas tent is 26 m.
(ii) Canvas required to make the tent = Curved surface area of tent
= πrl
= π x 24 x 26 = 624π m²
∵ Cost of 1 m2 canvas = ₹ 70
Cost of 624n m2 canvas = ₹ 70 x 624π
= ₹ 70x 624 x \(\frac { 22 }{ 7 }\)
= ₹ 10 x 624 x 22
= ₹ 137280
Hence, the cost of the canvas is ₹ 137280.

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:
Let r, h and l be the radius, height and slant height of the tent, respectively.

The extra material required for stitching margins and cutting = 20 cm = Q² m Hence, the total length of tarpaulin required = 62.8 + Q² = 63 m

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:
We have, slant height, l = 25m
and diameter = 14m
∴ Radius, r= 7m

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:

∵ The sheet required to make 1 cap = 550 cm²
∴ The sheet required to make 10 caps = 550 x 10= 5500 cm²

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 \( \sqrt{104} \) = 102)
Solution:

Curved surface area of a cone = πrl = 3.14 x 0.2 x 1.02 = 0.64056 m²
Cost of painting per m2 = ₹12
Cost of painting 0.64056 m2 = ₹12 x 0.64056= ₹ 7.68672
Cost of painting for 1 cone = ₹ 7.68672
Cost of painting 50 cones = ₹ 7.68672x 50= ₹ 384.336= ₹ 384.34

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes 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:
We have, height = 14 cm
Curved surface area Of a right circular cylinder = 88 cm²

r = 1 cm
Diameter = 2 x Radius = 2 x 1 = 2 cm

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:
Let r be the radius and h be the height of the cylinder.
Given, r = \(\frac { 140 }{ 2 }\) = 70 cm = 0.70 m
and h = 1 m
Metal sheet required to make a closed cylindrical tank
= Total surface area
= 2πr (h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 0.7(1 + 0.70)
= 2 x 22 x 0.1 x 170 = 7.48m²
Hence, the sheet required to make a closed cylindrical tank = 7.48m²

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:
We have, h = 77cm
Outer diameter (d₁) = 4.4 cm
and inner diameter (d₂) = 4 cm
Outer radius (r₁) = 2.2 cm
Inner radius (r₂) = 2cm
(i) Inner curved surface area = 2πr²h = 2x \(\frac { 22 }{ 7 }\) x 2 x 77 = 88 x 11 = 968 cm²
(ii) Outer curved surface area = 2πr₁h
= 2 x \(\frac { 22 }{ 7 }\) x 2.2 x 77
= 44 x 2.2 x 11= 1064.8cm²
(iii) Total surface area = Inner curved surface area + Outer curved surface area + Areas of two bases
= 968 + 1064.8 + 2\(\frac { 22 }{ 7 }\) (r₁² – r²)
= 968 + 1064.8 + 2 x \(\frac { 22 }{ 7 }\) [(2.2) – r²]
= [2032.8 + 2 x \(\frac { 22 }{ 7 }\) (4.84 – 4)]
= 2032.8 +\(\frac { 44 }{ 7 }\) x 0.84 = 2032.8+ 44x 0.12
= 2032.8 + 5.28 cm² = 2038.08 cm²

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:
We have, diameter of a roller = 84 cm
r = radius of a roller = 42 cm
h = 120 cm
To cover 1 revolution = Curved surface area of roller
= 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 42 x 120
= 44 x 720 cm²
= 31680 cm²

Area of the playground = Takes 500 complete revolutions = 500 x 3.168 m²
= 1584 m²

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:
Given, Diameter = 50 cm

Curved surface area of the pillar = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 0.25 x 35
= 2 x 22 x 0.25 x 0.5 = 55 m²
Cost of painting per m2 = ₹ 12.50
Cost of painting 5.5 m2 = ₹ 12.50×5.5 = ₹ 68.75

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:
We have, curved surface area of a right circular cylinder = 4.4m²
∴ 2πrh = 4.4

Hence, the height of the right circular cylinder is 1 m

Question 7.
he 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:
We have, inner diameter = 3.5 m
∴ inner radius = \(\frac { 3.5 }{ 2 }\) m
and h= 10m
(i) Inner curved surface area = 2πrh = 2x \(\frac { 22 }{ 7 }\) x \(\frac { 3.5 }{ 2 }\) x10 = 22 x 5 = 110m²
(ii) Cost of plastering perm2 = ₹40
Cost of plastering 110 m2 = ₹40x 110= ₹4400

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:
We have, h = 28 m
Diameter = 5 cm

Total radiating surface in the system = Curved surface area of the cylindrical pipe
= 2πrh =2 x \(\frac { 22 }{ 7 }\) x 0.025 x 28 = 4.4 m²

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:
(i) We have, diameter = 4.2 m

Since, \(\frac { 1 }{ 2 }\) of the actual steel used was wasted, therefore the area of the steel which was actually used for making the tank

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:
Given r = \(\frac { 20 }{ 2 }\) cm = 10cm
h = 30 cm
Since, a margin of 2.5 cm is used for folding it over the top and bottom so the total height of frame,
h1 = 30 + 25 + 25
h1 = 35 cm
∴ Cloth required for covering the lampshade = Its curved surface area

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:
Cardboard required by each competitor
= Base area + Curved surface area of one penholder
= πr2 + 2πrh [Given, h = 10.5 cm, r = 13cm]
= \(\frac { 22 }{ 7 }\) x (3)2 + 2 x \(\frac { 22 }{ 7 }\) x 3 x 105
= (28.28 + 198) cm²
= 22828 cm²
For 35 competitors cardboard required = 35 x 22628 = 7920 cm²
Hence, 7920 cm² of cardboard was required to be bought for the competition.

We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, drop a comment below and we will get back to you at the earliest.