Class 10 RD SHARMA Solutions Maths Chapter 13 - Areas Related to Circles
Areas Related to Circles Exercise Ex. 13.1
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Area of a circle = πr2 = (22/7) × 28 × 28 = 2464 cm2
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Areas Related to Circles Exercise Ex. 13.2
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
*Answer does not match with textbook answer.
Solution 21
Solution 22
*Note: Answer given in the book is incorrect.
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Areas Related to Circles Exercise Ex. 13.3
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Areas Related to Circles Exercise Ex. 13.4
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
*Answer is not matching with textbook.
Solution 26
Solution 27
Solution 28
Consider the following figure:
Solution 29
(i)
According to the figure in the question, there are 6 triangles.
Area of one triangle is 9 cm2.
Area of hexagon = 6 × 9 = 54 cm2
(ii)
Area of the equilateral triangle = 9 cm2
Area of the circle in which the hexagon is inscribed
=
=
=
= 65.26 cm2
NOTE: Answer not matching with back answer.
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
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Solution 47
Solution 48
Since the data given in the question seems incomplete and inconsistent with the figure, we make the following assumptions to solve it:
1. ABCD a symmetric trapezium with AD = BC
2. AD = BC = 14 cm (the distance between AB and CD is not 14 cm)
Draw perpendiculars to CD from A and B to divide the trapezium into one rectangle and two congruent right angled triangles.
The base of the right angled triangle=(CD - AB) ÷ 2
=(32 - 18) ÷ 2=7 cm
cos∠D = base ÷ hypotenuse = 7 ÷ 14 =1/2
m∠D = 60°
Hence, m∠A = 120°
*Answer is not matching with textbook answer.
Solution 49
Solution 50
Solution 51
Solution 52
According to the question,
Side of a square is 28 cm.
Radius of a circle is 14 cm.
Required area = Area of the square + Area of the two circles - Area of two quadrants …(i)
Area of the square = 282 = 784 cm2
Area of the two circles = 2πr2
= 
= 1232 cm2
Area of two quadrants = 
= 
= 308 cm2
Required area = 784 + 1232 - 308 = 1708 cm2
NOTE: Answer not matching with back answer.
Solution 53
According to the question,
For a cylindrical tank
d = 2 m, r = 1 m, h = 5 m
Volume of the tank = πr2h
= 
= 
After recycling, this water is used irrigate a park of a hospital with length 25 m and breadth 20 m.
If the tank is filled completely, then
Volume of cuboidal park = Volume of tank
h = 0.0314 m = 3.14 cm = p cm
Solution 54
Join OB.
Here, 
is a right triangle.
By Pythagoras theorem,
Therefore, radius of the circle (r) 
Area of the square 
Area of the quadrant of a circle 
Area of the shaded region = Area of quadrant - Area of square
= 128.25 cm2
Solution 55
Join AC.
Here, 
is a right triangle.
By Pythagoras theorem,
Therefore, diameter of the circle = 4 cm
So, the radius of the circle (r) = 2 cm
Area of the square 
Area of the circle 
Area of the shaded region = Area of the circle - Area of square
= 4.56 cm2
Areas Related to Circles Exercise 13.69
Solution 1
Correct Option :- (D)
Solution 2
According to the question,
Circumference
of a circle = 
= 
= 44 cm
Solution 3
Correct option (c)
Solution 4
correct option - (c)
Solution 5
correct option - (b)
Solution 6
Correct Option: d
Solution 7
Correct Option: (c)
Solution 8
Correct option - (c)
Solution 9
Correct option (c)
Solution 10
Correct Option ( d )
Solution 11
Correct option (a)
Solution 12
Let r be the radius of the circle.
2pr = 88
Perimeter of a triangle = 30 cm
Semi-perimeter = 15 cm
Hence,
Area of a triangle = r × s …(r = incircle radius, s =semi perimeter)
= 14 × 15
= 210 cm2
Solution 13
Correct option - (c)
Areas Related to Circles Exercise 13.70
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
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Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Correct option: (d)
Diameter of circle = side of square
2r = 10
r = 5 cm
Area of circle = πr2 = 25 π cm2
Solution 25
Areas Related to Circles Exercise 13.71
Solution 26
Correct option: (b)
Inner radius = r
outer radius = r + h
area of shaded region = area of outer circle - area of inner circle
= π (r + h)2 - πr2
= π {(r + h)2 - r2 }
= π (r + h - r) (r + h + r)
= π (2r + h) h
Solution 27
Solution 28
Correct option: (b)
area = circumference
πr2 = 2πr
r = 2 units
area = πr2
= 4π sq. units
Solution 29
Solution 30
** img pending
Solution 31
Solution 32
Areas Related to Circles Exercise 13.72
Solution 33
Solution 34
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Solution 37
Areas Related to Circles Exercise 13.73
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Solution 43
Areas Related to Circles Exercise 13.74
Solution 44
Solution 45
Solution 46
Correct option: (b)
radius of Circle = 5 cm
area = π (5)2
= 25 π
rdius of circle 2 = 12 cm
area = π (12)2
= 144 π
area of larger circle = 144 π + 25π
= 169 π
πr2 = 169 π
r2 = 169
r = 13
diameter = 2r
= 26
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52
Solution 53
