Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  RD Sharma Solutions: Surface Area and Volume of a Sphere- 1

Surface Area and Volume of a Sphere- 1 RD Sharma Solutions | Mathematics (Maths) Class 9 PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


Q u e s t i o n : 1
Find the surface area of a sphere of radius:
i
10.5 cm
ii
5.6 cm
iii
14 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different radii
We solve it using the formula,
Curved surface area of the sphere
i Radius = 10.5cm
Therefore, the surface area of the sphere of radius 10.5cm is 
ii Radius = 5.6cm
Therefore, the surface area of the sphere of radius 5.6cm is 
iii Radius = 14cm
Therefore, the surface area of the sphere of radius 14cm is 
Q u e s t i o n : 2
Find the surface area of a sphere of diameter.
i
14 cm
ii
21 cm
iii
3.5 cm
S o l u t i o n :
Page 2


Q u e s t i o n : 1
Find the surface area of a sphere of radius:
i
10.5 cm
ii
5.6 cm
iii
14 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different radii
We solve it using the formula,
Curved surface area of the sphere
i Radius = 10.5cm
Therefore, the surface area of the sphere of radius 10.5cm is 
ii Radius = 5.6cm
Therefore, the surface area of the sphere of radius 5.6cm is 
iii Radius = 14cm
Therefore, the surface area of the sphere of radius 14cm is 
Q u e s t i o n : 2
Find the surface area of a sphere of diameter.
i
14 cm
ii
21 cm
iii
3.5 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different diameter.
We solve it using the formula,
Curved surface area of the sphere = 
i Diameter = 14 cm
Therefore, the surface area of the sphere of diameter 14 cm is .
ii Diameter = 21 cm
Therefore, the surface area of the sphere of diameter 21 cm is .
iii Diameter = 3.5 cm
Therefore, the surface area of the sphere of diameter 3.5 cm is .
Q u e s t i o n : 3
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm (Use p = 3. 14
)
S o l u t i o n :
In the given problem,
Radius of the hemisphere = 10 cm
Now, the total surface area of a hemisphere = 
Therefore, the total surface area of the hemisphere is 
Also,
The total surface area of the solid hemisphere = 
Therefore, the total surface area of the hemisphere is 
Page 3


Q u e s t i o n : 1
Find the surface area of a sphere of radius:
i
10.5 cm
ii
5.6 cm
iii
14 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different radii
We solve it using the formula,
Curved surface area of the sphere
i Radius = 10.5cm
Therefore, the surface area of the sphere of radius 10.5cm is 
ii Radius = 5.6cm
Therefore, the surface area of the sphere of radius 5.6cm is 
iii Radius = 14cm
Therefore, the surface area of the sphere of radius 14cm is 
Q u e s t i o n : 2
Find the surface area of a sphere of diameter.
i
14 cm
ii
21 cm
iii
3.5 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different diameter.
We solve it using the formula,
Curved surface area of the sphere = 
i Diameter = 14 cm
Therefore, the surface area of the sphere of diameter 14 cm is .
ii Diameter = 21 cm
Therefore, the surface area of the sphere of diameter 21 cm is .
iii Diameter = 3.5 cm
Therefore, the surface area of the sphere of diameter 3.5 cm is .
Q u e s t i o n : 3
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm (Use p = 3. 14
)
S o l u t i o n :
In the given problem,
Radius of the hemisphere = 10 cm
Now, the total surface area of a hemisphere = 
Therefore, the total surface area of the hemisphere is 
Also,
The total surface area of the solid hemisphere = 
Therefore, the total surface area of the hemisphere is 
Q u e s t i o n : 4
The surface area of a sphere is 5544 cm
2
, find its diameter
S o l u t i o n :
In the given problem,
Surface area of a sphere = 5544 cm
2
Also, surface area of a sphere = 
So according to the problem,
Now,
Therefore, diameter of the sphere = cm
= 42 cm
Therefore, the diameter of the sphere is 
Q u e s t i o n : 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 cm
2
.
S o l u t i o n :
In the given problem, we have a hemispherical bowl.
Here,
Diameter of the bowl = 10.5 cm
So, Radius of the bowl = cm
So, the curved surface area of the bowl 
Now, the rate of tin plating per 100 cm
2 
= Rs 4
The rate of tin plating per 1 cm
2 
= Rs 
So, the cost of tin plating the bowl 
= Rs 6.93
Therefore, the cost of tin plating the hemispherical bowl from inside is 
Q u e s t i o n : 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
Page 4


Q u e s t i o n : 1
Find the surface area of a sphere of radius:
i
10.5 cm
ii
5.6 cm
iii
14 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different radii
We solve it using the formula,
Curved surface area of the sphere
i Radius = 10.5cm
Therefore, the surface area of the sphere of radius 10.5cm is 
ii Radius = 5.6cm
Therefore, the surface area of the sphere of radius 5.6cm is 
iii Radius = 14cm
Therefore, the surface area of the sphere of radius 14cm is 
Q u e s t i o n : 2
Find the surface area of a sphere of diameter.
i
14 cm
ii
21 cm
iii
3.5 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different diameter.
We solve it using the formula,
Curved surface area of the sphere = 
i Diameter = 14 cm
Therefore, the surface area of the sphere of diameter 14 cm is .
ii Diameter = 21 cm
Therefore, the surface area of the sphere of diameter 21 cm is .
iii Diameter = 3.5 cm
Therefore, the surface area of the sphere of diameter 3.5 cm is .
Q u e s t i o n : 3
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm (Use p = 3. 14
)
S o l u t i o n :
In the given problem,
Radius of the hemisphere = 10 cm
Now, the total surface area of a hemisphere = 
Therefore, the total surface area of the hemisphere is 
Also,
The total surface area of the solid hemisphere = 
Therefore, the total surface area of the hemisphere is 
Q u e s t i o n : 4
The surface area of a sphere is 5544 cm
2
, find its diameter
S o l u t i o n :
In the given problem,
Surface area of a sphere = 5544 cm
2
Also, surface area of a sphere = 
So according to the problem,
Now,
Therefore, diameter of the sphere = cm
= 42 cm
Therefore, the diameter of the sphere is 
Q u e s t i o n : 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 cm
2
.
S o l u t i o n :
In the given problem, we have a hemispherical bowl.
Here,
Diameter of the bowl = 10.5 cm
So, Radius of the bowl = cm
So, the curved surface area of the bowl 
Now, the rate of tin plating per 100 cm
2 
= Rs 4
The rate of tin plating per 1 cm
2 
= Rs 
So, the cost of tin plating the bowl 
= Rs 6.93
Therefore, the cost of tin plating the hemispherical bowl from inside is 
Q u e s t i o n : 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.
S o l u t i o n :
In the given problem, the dome of the building is in a form of a hemisphere.
Here,
Radius of the hemisphere = 63 dm
= 6.3 m
So, to find the cost of painting, we first find the curved surface area of the hemisphere.
Curved surface area of the sphere 
Now, the rate of painting per m
2 
= Rs 2
So, the cost of painting the dome 
Therefore, the cost of painting the dome is 
Q u e s t i o n : 7
Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-
fourth of the earth's surface is covered by water?
S o l u t i o n :
In the given problem, let us assume that the earth is a sphere and the radius of the earth is given equal to 6370 km.
The total surface area of earth can be calculated using the formula,
So, the total surface area of earth is equal to 510109600 km
2
.
Now, according to the question three-fourth of the earth’s surface is covered with water. This means that one-fourth
of the surface is covered with land.
Therefore, we get,
The area of land covered by water 
Therefore the area of earth’s surface covered by land is equal to 
Q u e s t i o n : 8
A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the
Page 5


Q u e s t i o n : 1
Find the surface area of a sphere of radius:
i
10.5 cm
ii
5.6 cm
iii
14 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different radii
We solve it using the formula,
Curved surface area of the sphere
i Radius = 10.5cm
Therefore, the surface area of the sphere of radius 10.5cm is 
ii Radius = 5.6cm
Therefore, the surface area of the sphere of radius 5.6cm is 
iii Radius = 14cm
Therefore, the surface area of the sphere of radius 14cm is 
Q u e s t i o n : 2
Find the surface area of a sphere of diameter.
i
14 cm
ii
21 cm
iii
3.5 cm
S o l u t i o n :
Here we need to find the surface area of spheres of different diameter.
We solve it using the formula,
Curved surface area of the sphere = 
i Diameter = 14 cm
Therefore, the surface area of the sphere of diameter 14 cm is .
ii Diameter = 21 cm
Therefore, the surface area of the sphere of diameter 21 cm is .
iii Diameter = 3.5 cm
Therefore, the surface area of the sphere of diameter 3.5 cm is .
Q u e s t i o n : 3
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm (Use p = 3. 14
)
S o l u t i o n :
In the given problem,
Radius of the hemisphere = 10 cm
Now, the total surface area of a hemisphere = 
Therefore, the total surface area of the hemisphere is 
Also,
The total surface area of the solid hemisphere = 
Therefore, the total surface area of the hemisphere is 
Q u e s t i o n : 4
The surface area of a sphere is 5544 cm
2
, find its diameter
S o l u t i o n :
In the given problem,
Surface area of a sphere = 5544 cm
2
Also, surface area of a sphere = 
So according to the problem,
Now,
Therefore, diameter of the sphere = cm
= 42 cm
Therefore, the diameter of the sphere is 
Q u e s t i o n : 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 cm
2
.
S o l u t i o n :
In the given problem, we have a hemispherical bowl.
Here,
Diameter of the bowl = 10.5 cm
So, Radius of the bowl = cm
So, the curved surface area of the bowl 
Now, the rate of tin plating per 100 cm
2 
= Rs 4
The rate of tin plating per 1 cm
2 
= Rs 
So, the cost of tin plating the bowl 
= Rs 6.93
Therefore, the cost of tin plating the hemispherical bowl from inside is 
Q u e s t i o n : 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.
S o l u t i o n :
In the given problem, the dome of the building is in a form of a hemisphere.
Here,
Radius of the hemisphere = 63 dm
= 6.3 m
So, to find the cost of painting, we first find the curved surface area of the hemisphere.
Curved surface area of the sphere 
Now, the rate of painting per m
2 
= Rs 2
So, the cost of painting the dome 
Therefore, the cost of painting the dome is 
Q u e s t i o n : 7
Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-
fourth of the earth's surface is covered by water?
S o l u t i o n :
In the given problem, let us assume that the earth is a sphere and the radius of the earth is given equal to 6370 km.
The total surface area of earth can be calculated using the formula,
So, the total surface area of earth is equal to 510109600 km
2
.
Now, according to the question three-fourth of the earth’s surface is covered with water. This means that one-fourth
of the surface is covered with land.
Therefore, we get,
The area of land covered by water 
Therefore the area of earth’s surface covered by land is equal to 
Q u e s t i o n : 8
A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the
shape if the length of the shape be 7 cm.
S o l u t i o n :
In the given problem, we are given a shape in which a cylinder is placed over a hemisphere. The radius of the
hemisphere and the cylinder is equal and also the height of the cylinder is equal to the radius.
So, let draw a figure representing this shape and take the radius as r cm.
So, let us find the value of r first. From the figure, we can see that,
Also, height of the sphere = 3.5 cm
asradiusandheightofthecylinderareequal
Now, the surface area of the shape is equal to the surface area of the hemisphere and surface area of cylinder
together.
Surface area = surface area of hemisphere + surface area of cylinder
Therefore, the surface area of the given shape is 
Q u e s t i o n : 9
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface
areas.
S o l u t i o n :
In the given problem, we have two objects; moon and earth.
Let us take the diameter of the earth = d km
Therefore, the radius of the earth = km
So, the surface area of the earth (S
e
) = 
Now, according to the problem, the diameter of moon is one-fourth of the diameter of earth.
So, the diameter of the moon =  km
Therefore, the radius of the moon = km
Read More
44 videos|412 docs|54 tests

Top Courses for Class 9

44 videos|412 docs|54 tests
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

video lectures

,

Objective type Questions

,

Surface Area and Volume of a Sphere- 1 RD Sharma Solutions | Mathematics (Maths) Class 9

,

study material

,

Exam

,

Surface Area and Volume of a Sphere- 1 RD Sharma Solutions | Mathematics (Maths) Class 9

,

Summary

,

Important questions

,

pdf

,

Extra Questions

,

Viva Questions

,

mock tests for examination

,

Free

,

Semester Notes

,

practice quizzes

,

past year papers

,

Surface Area and Volume of a Sphere- 1 RD Sharma Solutions | Mathematics (Maths) Class 9

,

shortcuts and tricks

,

ppt

,

Sample Paper

,

Previous Year Questions with Solutions

,

MCQs

;