NCERT Solutions for Class 10 Maths Chapter 12 - Areas Related to Circles
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Chapter 12 - Areas Related to Circles Exercise Ex. 12.1
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Let the radius of the circle be r
Circumference of circle = 2
r
Area of circle = r2
Given that circumference and area of the circle are equal.
So, 2r = r2
2 = r
Hence, the radius of the circle will be 2 units
Chapter 12 - Areas Related to Circles Exercise Ex. 12.2
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Radius (r) of the circle = 15
Area of sector OPRQ =
=
=
= 117.75 cm2
In ∆OPQ
.... (Since OP = OQ)
\
∆OPQ is an equilateral triangle.
Area of ∆OPQ =
Area of segment PRQ = Area of sector OPRQ –Area of ∆OPQ
= 117.75 – 97.3125
= 20.4375 cm2
Area of major segment PSQ
= Area of circle – Area of segment PRQ
= 152p – 20.4375
= 3.14 × 225 – 20.4375
= 686.0625 cm2Solution 7
Draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ∆OVS
OV = 6
ST = 2SV = 
Area of ∆OST = 
= 
= 
= 36 × 1.73
= 62.28
Area of sector OSUT = 
= 150.72
Area of segment SUT = Area of sector OSUT
= 150.72 – 62.28
= 88.44 cm2
Solution 8
The horse can graze a sector of 90° in a circle of 5 m radius.
i. So area that can be grazed by horse = area of sector OACB
= 
= 
= 19.63 m2
ii. Area that can be grazed by the horse when the length of rope is 10 m long = 
= 
= 78.5
Change in grazing area = 78.5 – 19.63 = 58.87 cm2
Solution 9
Solution 10
Solution 11
The figure shows that each blade of the wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.
Solution 12
Solution 13
Solution 14
Chapter 12 - Areas Related to Circles Exercise Ex. 12.3
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
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