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## NCERT Class 10 Maths Textbook Chapter 12 With Answer Book PDF Free Download

### 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

Author | NCERT |

Language | English |

No. of Pages | 16 |

PDF Size | 1248 KB |

Category | Mathematics |

Source/Credits | ncert.nic.in |

### 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 1^{st} circle = 19 cm (given)

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

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

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

So,

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

Now, let the radius of the 3^{rd} circle = R

∴ The circumference of the 3^{rd} circle = 2πR

It is given that some of the circumferences of two circles = circumference of the 3^{rd} 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 1^{st} circle = 8 cm (given)

∴ Area of 1^{st} circle = π(8)^{2} = 64π

Radius of 2^{nd} circle = 6 cm (given)

∴ Area of 2^{nd} circle = π(6)^{2} = 36π

So,

The sum of th^{e }1^{st} and 2^{nd} circles will be = 64π+36π = 100π

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

∴ Area of the circle 3^{rd} circle = πR^{2}

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

Or, πR^{2} = 100πcm^{2}

R^{2} = 100cm^{2}

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 1^{st} circle, r_{1} = 21/2 cm (as diameter D is given as 21 cm)

So, area of gold region = π r_{1}^{2 }= π(10.5)^{2 }= 346.5 cm^{2}

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

So, the radius of 2^{nd} circle, r_{2} = 10.5cm+10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2^{nd} circle − Area of gold region = (πr_{2}^{2}−346.5) cm^{2}

= (π(21)^{2} − 346.5) cm^{2}

= 1386 − 346.5

= 1039.5 cm^{2}

Similarly,

The radius of 3^{rd} circle, r_{3} = 21 cm+10.5 cm = 31.5 cm

The radius of 4^{th} circle, r_{4} = 31.5 cm+10.5 cm = 42 cm

The Radius of 5^{th} circle, r_{5} = 42 cm+10.5 cm = 52.5 cm

For the area of n^{th }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 cm^{2}

= 3118.5 – 1386 cm^{2}

= 1732.5 cm^{2}

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

= π(42)^{2} – 1386 cm^{2}

= 5544 – 3118.5 cm^{2}

= 2425.5 cm^{2}

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

= π(52.5)^{2} – 5544 cm^{2}

= 8662.5 – 5544 cm^{2}

= 3118.5 cm^{2}

NCERT Class 10 Maths Textbook Chapter 12 With Answer Book PDF Free Download