Class 9 Chapter 21 – Surface Area and Volume of A Sphere Exercise 21.1 solutions are provided here. This study material is an invaluable aid to students when they need help with their homework while preparing for exams and while learning. Here all solutions to questions in RD Sharma textbook class 9 are given in a detailed and step by step format to help the students understand the concepts in an easy way. Click on the link below to get your pdf now.
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Question 1: Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Solution:
Surface area of a sphere = 4Ï€r2
Where, r = radius of a sphere
(i) Radius = 10.5 cm
Surface area = 4 x 22/7 x (10.5)2
= 1386
Surface area is 1386 cm2
(ii) Radius= 5.6 cm
Surface area = 4×22/7×(5.6)2
= 394.24
Surface area is 394.24 cm2
(iii) Radius = 14 cm
Surface area = 4×22/7×(14)2
= 2464
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:
Surface area of a sphere = 4Ï€r2
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
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
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
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. (Ï€=3.14)
Solution:
Radius of a hemisphere = Radius of a solid hemisphere = 10 cm (Given)
Surface area of the hemisphere = 2Ï€r2
= 2×3.14×(10)2 cm2
= 628 cm2
And, surface area of solid hemisphere = 3Ï€r2
= 3×3.14×(10)2 cm2
= 942 cm2
Question 4: The surface area of a sphere is 5544 cm2, find its diameter.
Solution:
Surface area of a sphere is 5544 cm2
Surface area of a sphere = 4Ï€r2
So, 4Ï€r2 = 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:
Inner diameter of hemispherical bowl = 10.5 cm
So, radius = Diameter/2 = 10.5/2 cm = 5.25 cm
Now, Surface area of hemispherical bowl = 2Ï€r2
= 2 × 3.14 × (5.25)2
= 173.25
So, Surface area of 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, cost of tin plating the inner side of hemispherical bowl is Rs.6.93. Answer!!
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:
Radius of hemispherical dome = 63 dm or 6.3 m
Inner surface area of dome = 2Ï€r2
=2×3.14×(6.3)2
= 249.48
So, Inner surface area of dome is 249.48 m2
Now find the cost:
Cost of painting 1m2 = Rs.2 (given)
Therefore, cost of painting 249.48 m2= Rs. (249.48×2) = Rs.498.96.
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RD Sharma Solutions for Class 9 Maths Chapter 21 Exercise 21.1
Class 9 Maths Chapter 21 Surface Area and Volume of A Sphere Exercise 21.1 is based on 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