# 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, 2 r = 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

= 152p20.4375

= 3.14 × 225 – 20.4375

= 686.0625 cm2
Solution 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  ## Why to choose our CBSE Class 10 Maths Study Materials?

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