NCERT Solutions for Class 9 Maths Exercise 13.4

NCERT Solutions for Class 9 Maths Exercise 13.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Surface Areas and Volumes Download as PDF

NCERT Solutions for Class 9 Mathematics Surface Areas and Volumes

Assume unless stated otherwise.

1. Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Ans. (i) Radius of sphere = 105 cm

Surface area of sphere =

= = 1386 cm²

(ii) Radius of sphere = 5.6 m

Surface area of sphere =

= = 3.94.84 m²

(iii) Radius of sphere = 14 cm

Surface area of sphere =

= = 2464 cm²

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2. Find the surface area of a sphere of diameter:

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

Ans. (i) Diameter of sphere = 14 cm,

Therefore Radius of sphere = = 7 cm

Surface area of sphere =

= = 616 cm²

(ii) Diameter of sphere = 21 cm

Radius of sphere = cm

Surface area of sphere =

= = 1386 cm²

(iii) Diameter of sphere = 3.5 cm

Radius of sphere = = 1.75 cm

Surface area of sphere =

= = 38.5 cm²

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NCERT Solutions for Class 9 Maths Exercise 13.4

3. Find the total surface area of a hemisphere of radius 10 cm.

Ans. Radius of hemisphere 10 cm

Total surface area of hemisphere =

= 3 x 3.14 x 10 x 10 = 942 cm²

Hence total surface area of hemisphere is

942 cm².

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NCERT Solutions for Class 9 Maths Exercise 13.4

4. The radius of a spherical balloonincreases from 7 cm to 14 cm as air isbeing pumped into it. Find the ratio ofsurface areas of the balloon in the twocases.

Ans. I case: Radius of balloon = 7 cm

Surface area of balloon =

= cm²……….(i)

II case: Radius of balloon (R) = 14 cm

Surface area of balloon =

= cm2 …….(ii)

Now, Ratio [from eq. (i) and (ii)],

=

=

Hence, required ratio = 1 : 4

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NCERT Solutions for Class 9 Maths Exercise 13.4

5.A hemispherical bowl made of brass has inner diameter 105 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm².

Ans. Inner diameter of bowl

= 10.5 cm

Inner radius of bowl

= 5.25 cm

Now, Inner surface area of bowl

=

=

= = cm²

Cost of tin-plating per 100 cm²

= Rs. 16

Cost of tin-plating per 1 cm² =

Cost of tin-plating per cm²

== Rs. 27.72

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NCERT Solutions for Class 9 Maths Exercise 13.4

6. Find the radius of a sphere whose surface area is 154 cm².

Ans. Surface area of sphere = 154 cm²

= 154

= 3.5 cm

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NCERT Solutions for Class 9 Maths Exercise 13.4

7.The diameter of the moon is approximately one fourth the diameter of the earth. Find the ratio of their surface areas.

Ans. Let diameter of Earth =

Radius of Earth =

Surface area of Earth =

=

Now, Diameter of Moon = of diameter of Earth =

Radius of Moon =

Surface area of Moon =

= =

Now, Ratio =

= = =

Required ratio = 1 : 16

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NCERT Solutions for Class 9 Maths Exercise 13.4

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.

Ans. Inner radius of bowl = 5 cm

Thickness of steel = 0.25 cm

Outer radius of bowl (R) =

= 5 +0.25 = 5.25 cm

Outer curved surface area of bowl

= =

=

= = 173.25 cm²

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NCERT Solutions for Class 9 Maths Exercise 13.4

9. A right circular cylinder just encloses a sphere of radius (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).

Ans. (i) Radius of sphere =

Surface area of sphere

=

The cylinder just encloses the sphere in it.

The height of cylinder will be equal to diameter of sphere.

And The radius of cylinder will be equal to radius of sphere.

(ii) Curved surface area of cylinder

= =

=

(iii)

Required ratio = 1 :1

NCERT Solutions for Class 9 Maths Exercise 13.4

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