# Areas Related To Circles Chapter 12 Class 10 Maths NCERT PDF

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### Chapter 12: Areas Related to Circles

#### 12.1 Introduction

You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles, and circles from your earlier classes.

Many objects that we come across in our daily life are related to the circular shape in some form or other.

Cycle wheels, wheelbarrow (the law), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects So, the problem of finding perimeters and areas related to circular figures is of great practical importance.

In this chapter, we shall begin our discussion with a review of the concepts of the perimeter (circumference) and area of a circle and apply this knowledge to find

the areas of two special ‘parts’ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts.

#### 12.2 Perimeter and Area of a Circle — A Review

Recall that the distance covered by traveling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that The circumference of a circle bears a constant ratio to its diameter.

This constant ratio is denoted by the Greek letter π (read as ‘pi’). In other words, the circumference
diameter = π
or, circumference = π × diameter
= π × 2r (where r is the radius of the circle)
= 2πr

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

1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Solution:

The radius of the 1st circle = 19 cm (given)

∴ Circumference of the 1st circle = 2π×19 = 38π cm

The radius of the 2nd circle = 9 cm (given)

∴ Circumference of the 2nd circle = 2π×9 = 18π cm

So,

The sum of the circumference of two circles = 38π+18π = 56π cm

Now, let the radius of the 3rd circle = R

∴ The circumference of the 3rd circle = 2πR

It is given that some of the circumferences of two circles = circumference of the 3rd circle

Hence, 56π = 2πR

Or, R = 28 cm.

2. The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.

Solution:

Radius of 1st circle = 8 cm (given)

∴ Area of 1st circle = π(8)2 = 64π

Radius of 2nd circle = 6 cm (given)

∴ Area of 2nd circle = π(6)2 = 36π

So,

The sum of the 1st and 2nd circles will be = 64π+36π = 100π

Now, assume that the radius of the 3rd circle = R

∴ Area of the circle 3rd circle = πR2

It is given that the area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle

Or, πR2 = 100πcm2

R2 = 100cm2

So, R = 10cm

3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the center outwards as Gold, Red, Blue, Black, and White. The diameter of the region representing the Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:

The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm)

So, area of gold region = π r1= π(10.5)= 346.5 cm2

Now, it is given that each of the other bands is 10.5 cm wide,

So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2

= (π(21)2 − 346.5) cm2

= 1386 − 346.5

= 1039.5 cm2

Similarly,

The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm

The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm

The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm

For the area of nth region,

A = Area of circle n – Area of circle (n-1)

∴ Area of blue region (n=3) = Area of third circle – Area of second circle

= π(31.5)2 – 1386 cm2

= 3118.5 – 1386 cm2

= 1732.5 cm2

∴ Area of black region (n=4) = Area of fourth circle – Area of third circle

= π(42)2 – 1386 cm2

= 5544 – 3118.5 cm2

= 2425.5 cm2

∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle

= π(52.5)2 – 5544 cm2

= 8662.5 – 5544 cm2

= 3118.5 cm2