RD Sharma Solutions Class 9 Maths Chapter 21 Surface Area and Volume of Sphere Exercise 21.1

RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.1 are provided here. This study material is an invaluable aid to students when they need help with their homework and while preparing for exams. Here, all the questions in RD Sharma textbook Solutions Class 9 are explained in a detailed and step-by-step format to help the students understand the concepts in an easy way. Click on the links below to get the solutions in PDFs now.

RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.1

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Access Answers to RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.1 Page Number 21.8

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

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Solution:

The surface area of a sphere = 4πr2

Where, r = radius of a sphere

(i) Radius = 10.5 cm

Surface area = 4 x 22/7 x (10.5)2

= 1386

The surface area is 1386 cm2

(ii) Radius= 5.6 cm

Surface area = 4×22/7×(5.6)2

= 394.24

The surface area is 394.24 cm2

(iii) Radius = 14 cm

Surface area = 4×22/7×(14)2

= 2464

The surface area is 2464 cm2

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

(i) 14 cm (ii) 21 cm (iii) 3.5 cm

Solution:

The surface area of a sphere = 4πr2

Where, r = radius of a sphere

(i) Diameter= 14 cm

So, Radius = Diameter/2 = 14/2 cm = 7 cm

Surface area = 4×22/7×(7)2

= 616

The surface area is 616 cm2

(ii) Diameter = 21cm

So, Radius = Diameter/2 = 21/2 cm = 10.5 cm

Surface area= 4×22/7×(10.5)2

= 1386

The surface area is 1386 cm2

(iii) Diameter= 3.5cm

So, Radius = Diameter/2 = 3.5/2 cm = 1.75 cm

Surface area = 4×22/7×(1.75)2

= 38.5

The surface area is 38.5 cm2

Question 3: Find the total surface area of a hemisphere and a solid hemisphere, each of radius 10 cm. (π=3.14)

Solution:

Radius of a hemisphere = Radius of a solid hemisphere = 10 cm (Given)

The surface area of the hemisphere = 2πr2

= 2×3.14×(10)2 cm2

= 628 cm2

And, the surface area of the solid hemisphere = 3πr2

= 3×3.14×(10)2 cm2

= 942 cm2

Question 4: The surface area of a sphere is 5544 cm2, find its diameter.

Solution:

The surface area of a sphere is 5544 cm2

The surface area of a sphere = 4πr2

So, 4πr2 = 5544

4×22/7×(r)2 = 5544

r2 = (5544 × 7)/88

r2 = 441

or r = 21cm

Now, Diameter=2(radius) = 2(21) = 42cm

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 Rs.4 per 100 cm2.

Solution:

The inner diameter of the hemispherical bowl = 10.5 cm

So, radius = Diameter/2 = 10.5/2 cm = 5.25 cm

Now, the surface area of the hemispherical bowl = 2πr2

= 2 × 3.14 × (5.25)2

= 173.25

So, the surface area of the hemispherical bowl is 173.25 cm2

Find the cost:

Cost of tin plating 100 cm2 area= Rs.4 (given)

Cost of tin plating 173.25cm2 area = Rs. 4×173.25100 = Rs. 6.93

Therefore, the cost of tin plating the inner side of the hemispherical bowl is Rs.6.93.

Question 6: The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs. 2 per sq m.

Solution:

The radius of the hemispherical dome = 63 dm or 6.3 m

The inner surface area of dome = 2πr2

=2×3.14×(6.3)2

= 249.48

So, the inner surface area of the dome is 249.48 m2

Now find the cost:

Cost of painting 1m2 = Rs.2 (given)

Therefore, the cost of painting 249.48 m2= Rs. (249.48×2) = Rs. 498.96.


RD Sharma Solutions for Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.1

RD Sharma Solutions Class 9 Maths Chapter 21 Surface Area and Volume of a Sphere Exercise 21.1 is based on the following topics:

  • Section of a sphere by a plane
  • Hemisphere
  • Spherical shell
  • Surface area of a sphere
  • Surface area of a hemisphere and
  • Surface area of a spherical shell

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