Class 10 NCERT Solutions Maths Chapter 12 - Areas Related to Circles

Brush up your knowledge of circles with NCERT Solutions for CBSE Class 10 Mathematics Chapter 12 Areas Related to Circles. Learn to calculate the number of revolutions made by the wheel of a car based on the dimensions of the wheel. Practise the method to work out the area of a sector in a given Maths problem related to circles.

Also, practise the ways to calculate the area of a semicircle or the area of a specific design with the step-wise answers in our textbook solutions. To boost your Maths skills for board exam preparation, utilise TopperLearning’s CBSE Class 10 Maths videos, question papers, practice tests and more.

Ex. 12.1

Ex. 12.2

Ex. 12.3

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 $straight pi$ r

Area of circle = $straight pi$ r²

Given that circumference and area of the circle are equal.

So, 2 $straight pi$ r = $straight pi$ r²

2 = r

Hence, the radius of the circle will be 2 units

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 = $fraction numerator 60 degree over denominator 360 degree end fraction cross times πr squared$

= $1 over 6 cross times 3.14 cross times 15 squared$

= $fraction numerator 706.5 over denominator 6 end fraction$

= 117.75 cm²

In ∆OPQ

$angle O P Q equals angle O Q P$ .... (Since OP = OQ)
$angle O P Q plus angle O Q P plus angle P O Q equals 180 degree$
\ $2 angle O P Q equals 120 degree$

$therefore angle O P Q equals 60 degree$

∆OPQ is an equilateral triangle.

Area of ∆OPQ = $fraction numerator square root of 3 over denominator 4 end fraction cross times s i d e squared equals fraction numerator square root of 3 over denominator 4 end fraction cross times 15 squared equals fraction numerator 225 cross times 1.73 over denominator 4 end fraction equals 97.3125 space c m squared$

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 cm²

Solution 7

Draw a perpendicular OV on chord ST. It will bisect the chord ST.

SV = VT

In ∆OVS

$fraction numerator O V over denominator O S end fraction equals cos space 60 degree$

$fraction numerator O V over denominator O S end fraction equals 1 half$

OV = 6

$fraction numerator S V over denominator S O end fraction equals sin space 60 degree equals fraction numerator square root of 3 over denominator 2 end fraction$

$fraction numerator S V over denominator 12 end fraction equals fraction numerator square root of 3 over denominator 2 end fraction$

$S V equals 6 square root of 3$

ST = 2SV = $2 cross times 6 square root of 3 equals 12 square root of 3$

Area of ∆OST = $1 half cross times S T cross times O V$

= $1 half cross times 12 square root of 3 cross times 6$

= $36 square root of 3$

= 36 × 1.73

= 62.28

Area of sector OSUT = $fraction numerator 120 degree over denominator 360 degree end fraction cross times straight pi cross times 12 squared$

= 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

= $fraction numerator 90 degree over denominator 360 degree end fraction cross times πr squared$

= $1 fourth cross times 3.14 cross times 5 squared$

= 19.63 m²

ii. Area that can be grazed by the horse when the length of rope is 10 m long = $fraction numerator 90 degree over denominator 360 degree end fraction cross times pi cross times 10 squared$