NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

# NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 - Areas related to circles

Edited By Ramraj Saini | Updated on Nov 27, 2023 05:06 PM IST

## NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2

NCERT Solutions for Exercise 12.2 Class 10 Maths Chapter 12 Areas related to circles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 Maths ex 12.2 lists a few practice numerical problems on, area of a quadrant of the circle, the area swept by the hands of the clock, etc. The Class 10 Maths chapter 12 exercise 12.2 covers the topics like the concept of finding major and minor segment/sectors subtended by the chord of the circle. Also, some properties and technical terms of circular curves are enclosed with some special conditions

Exercise 12.2 Class 10 Maths which is another exercise followed by exercise 12.1 that includes the rest of the part of areas related to circles, and is important as per the exams. Questions based on the area of sector out of some given conditions like radius and angle of the sector, and is like a base for the exam. 10th class Maths exercise 12.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

## Assess NCERT Solutions for Class 10 Maths chapter 12 exercise 12.2

Areas Related to Circles Class 10 Chapter 12 Exercise: 12.2

We know that the area of a sector having radius r and angle $\Theta$ is given by:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\times \pi r^2$

Thus the area of the given sector is:-

$\\=\ \frac{60^{\circ}}{360^{\circ}}\times \pi \times 6^2\\\\=\ \frac{132}{7}\ cm^2$

We are given the circumference of the circle.

Thus,

$\\2\pi r\ =\ 22\\\\r\ =\ \frac{11}{\pi}\ cm$

Also, we know that the area of a sector is given by :

$Area\ =\ \frac{\Theta}{360^{\circ}}\times \pi r^2$

It is given that we need to find the area of a quadrant thus $\Theta\ =\ 90^{\circ}$

Hence the area becomes:-

$\\=\ \frac{90}{360^{\circ}}\times \pi \left ( \frac{11}{\pi} \right )^2\\\\=\ \frac{77}{8}\ cm^2$

The minute hand rotates 360 o in one hour.

We need to find rotation in 5 min. :-

$=\ \frac{360^{\circ}}{60}\times 5\ =\ 30^{\circ}$

The area of the sector is given by :

$\\Area\ =\ \frac{\Theta}{360^{\circ}}\times \pi r^2\\\\Area=\ \frac{30}{360^{\circ}}\times \pi \times 14^2\\\\Area=\ \frac{154}{3}\ cm^2$

Hence the area swept by minute hand in 5 minutes is $\frac{154}{3}\ cm^2$ .

The angle in the minor sector is 90 o .

Thus the area of the sector is given by:-

$\\Area\ =\ \frac{\Theta}{360^{\circ}}\times \pi r^2\\\\Area=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 10^2\\\\Area=\ \frac{1100}{14}\ cm^2\ =\ 78.5\ cm^2$

Now the area of a triangle is:-

$Area\ =\ \frac{1}{2}bh\ =\ \frac{1}{2}\times 10\times 10\ =\ 50\ cm^2$

Thus the area of minor segment = Area of the sector - Area of a triangle

or $=\ 78.5\ -\ 50\ =\ 28.5\ cm^2$

(ii) major sector. (Use π = 3.14)

The area of the major sector can be found directly by using the formula :

$Area\ =\ \frac{\Theta}{360^{\circ}}\times \pi r^2$

In the case of this, the angle is 360 o - 90 o = 270 o .

Thus the area is : -

$\\=\ \frac{270^{\circ}}{360^{\circ}}\times \pi \times 10\times 10\\\\=\ \frac{3300}{14}\ =\ 235.7\ cm^2$

(i) the length of the arc

The length of the arc is given by:-

$Length\ of\ arc\ =\ \frac{\Theta}{360^{\circ}} \times 2\pi r$

$\\=\ \frac{60^{\circ}}{360^{\circ}} \times 2\times \pi \times 21\\\\=\ 22\ cm$

Hence the length of the arc is 22 cm.

(ii) area of the sector formed by the arc

We know that the area of the sector is given by:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

$\\=\ \frac{60}{360^{\circ}}\ \times \pi \times 21^2\\\\=\ 231\ cm^2$

Thus the area of the sector is 231 cm 2 .

(iii) area of the segment formed by the corresponding chord

For the area of the segment, we need to subtract the area of the triangle attached with the area of arc.

Thus consider the triangle:-

It is given that the angle of arc is 60 o , or we can say that all angles are 60 o (since two sides are equal). Hence it is an equilateral triangle.

Area of triangle is:-

$\\=\ \frac{\sqrt{3}}{4}\times a^2\\\\\\=\ \frac{\sqrt{3}}{4}\times 21^2\\\\=\ \frac{441\sqrt{3}}{4}\ cm^2$

Hence the area of segment is:-

$=\ \left (231\ -\ \frac{441\sqrt{3}}{4} \right )\ cm^2$

The area of the sector is :

$\\=\ \frac{\Theta }{360^{\circ}}\times \pi r^2\\\\=\ \frac{60^{\circ} }{360^{\circ}}\times \pi \times 15^2\\\\=\ 117.85\ cm^2$

Now consider the triangle, the angle of the sector is 60 0 .

This implies it is an equilateral triangle. (As two sides are equal so will have the same angle. This possible only when all angles are equal i.e., 60 o .)

Thus, the area of the triangle is:-

$=\ \frac{\sqrt{3}}{4}\times 15^2$

or $=\ 56.25 \sqrt{3}\ =\ 97.31\ cm^2$

Hence area of the minor segment : $=\117.85\ -\ 97.31\ =\ 20.53\ cm^2$

And the area of the major segment is :

$=\ \pi r^2\ -\ 20.53$

or $=\ \pi\times 15^2\ -\ 20.53$

or $=\ 707.14\ -\ 20.53$

or $=\ 686.6\ cm^2$

For the area of the segment, we need the area of sector and area of the associated triangle.

So, the area of the sector is :

$=\ \frac{120^{\circ}}{360^{\circ}}\times \pi \times 12^2$

or $=\ 150.72\ cm^2$

Now, consider the triangle:-

Draw a perpendicular from the center of the circle on the base of the triangle (let it be h).

Using geometry we can write,

$\frac{h}{r}\ =\ \cos 60^{\circ}$

or $h\ =\ 6\ cm$

Similarly, $\frac{\frac{b}{2}}{r}\ =\ \sin 60^{\circ}$

or $b\ =\ 12\sqrt{3}\ cm$

Thus the area of the triangle is :

$=\ \frac{1}{2}\times 12\sqrt{3}\times 6$

or $=\ 62.28\ cm^2$

Hence the area of segment is: $=\ 150.72\ -\ 62.28\ =\ 88.44\ cm^2$ .

(i) the area of that part of the field in which the horse can graze.

The part grazed by the horse is given by = Area of sector

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

$=\ \frac{90^{\circ}}{360^{\circ}}\ \times \pi \times 5^2$

$=\ 19.62\ m^2$

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\pi = 3.14$ )

When the length of the rope is 10 m, the area grazed will be:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

$=\ \frac{90^{\circ}}{360^{\circ}}\ \times \pi \times 10^2$

$=\ 25 \pi\ m^2$

Hence the change in the grazing area is given by :

$=\ 25 \pi\ -\ \frac{25 \pi}{4}\ =\ 58.85\ m^2$

(i) the total length of the silver wire required.

The total wire required will be for 5 diameters and the circumference of the brooch.

The circumference of the brooch:-

$\\=\ 2\pi r\\\\=\ 2\times \pi \times \frac{35}{2}\\\\=\ 110\ mm$

Hence the total wire required will be:- $=\ 110\ mm\ +\ 5 \times 35\ mm\ =\ 285\ mm$ .

The total number of lines present in the brooch is 10 (line starting from the centre).

Thus the angle of each sector is 36 o .

The area of the sector is given by:-

$\\Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2\\\\Area=\ \frac{36^{\circ}}{360^{\circ}}\ \times \pi \times \left ( \frac{35}{2} \right )^2\\\\Area=\ \frac{385}{4}\ mm^2$

It is given that the umbrella has 8 ribs so the angle of each sector is 45 o .

Thus the area of the sector is given by:-

$\\Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2\\\\Area=\ \frac{45^{\circ}}{360^{\circ}}\ \times \pi \times 45^2\\\\Area=\ \frac{22275}{28}\ cm^2$

Hence the area between two consecutive ribs is $\frac{22275}{28}\ cm^2$ .

The area cleaned by one wiper is:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

or $=\ \frac{115^{\circ}}{360^{\circ}}\ \times \pi \times 25^2$

or $=\ \frac{158125}{252}\ cm^2$

Hence the required area (area cleaned by both blades) is given by:-

$=\ 2\times \frac{158125}{252}\ =\ \frac{158125}{126}\ cm^2$

The area of the sector is given by:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

In this case, the angle is 80 o .

Thus the area is:-

$=\ \frac{80^{\circ}}{360^{\circ}}\ \times \pi \times 16.5^2$

or $=\ 189.97\ Km^2$

The angle of each of the six sectors is 60 o at the center. $\left ( \because \frac{360^{\circ}}{6}\ =\ 60^{\circ} \right )$

Area of the sector is given by:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

or $=\ \frac{60^{\circ}}{360^{\circ}}\ \times \pi \times 78^2$

or $=\ 410.66\ cm^2$

And the area of the equilateral triangle associated with segment:-

$=\ \frac{\sqrt{3}}{4}\times a^2$

or $=\ \frac{\sqrt{3}}{4}\times 28^2\ =\ 333.2\ cm^2$

Hence the area of segment is : $=\ 410.66\ -\ 333.2\ =\ 77.46\ cm^2$

Thus the total area of design is : $=\6\times 77.46\ =\ 464.76\ cm^2$

So, the total cost for the design is:- $=\ 0.35\times 464.76\ =\ Rs. 162.66$

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) $\frac{p}{180}\times 2\pi R$

(B) $\frac{p}{180}\times \pi R^2$

(C) $\frac{p}{360}\times 2\pi R$

(D) $\frac{p}{720}\times 2\pi R^2$

We know that the area of the sector is given by:-

$Area\ =\ \frac{\Theta}{360^{\circ}}\ \times \pi r^2$

$=\ \frac{p}{360^{\circ}}\ \times \pi r^2$

Hence option (d) is correct.

## More About NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

Moving towards more information about given exercise 12.2 Class 10 Maths. The numerical present in NCERT syllabus for Class 10 Maths chapter 12 exercise 12.2 are the problems of finding area within the sector and perimeter enclosed in a circular arc, length, and area of the sector around it as discussed. And later on questions of Class 10 Maths chapter 12 exercise, 12.2 is to find values in real scenario algebraic problems of horse and goat grazing a particular area if attached with a rope of definitive length. In Class 10 Maths chapter 12 exercise 12.2 there is also coverage of problems of the angle subtended by tangents and chords to the circles.

Also Read| Areas Related to Circles Class 10 Notes

## Benefits of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

• Initial benefits of exercise 12.2 Class 10 Maths and the NCERT solutions for Class 10 Maths chapter 12 exercise 12.2 is that as it covers a variety of questions from the basic to a higher level which gives a better understanding of the play
• Taking about higher level in exercise 12.2 Class 10 Maths understanding the concepts makes students work on related topics which is helpful to solve better examples under given conditions as mentioned in Class 10 Maths chapter 12 exercise 12.2.
• For Class 10 final exams students might find some MCQs, short numerical or long numerical questions from all types are covered in the Class 10 Maths chapter 12 exercise 12.2

Also, see-

## NCERT Solutions for Class 10 Subject Wise

1. How exercise 12.2 is different from 12.1 ?

Exercise 12.2 consist of many numericals based on real life examples and their concepts are little bit different from the previous exercise.

2. What are the important points to be covered in exercise 12.2?

The exercise deals with a question related to the area of circles. These are important topics which will be important and covered in higher classes

3. What is the formula of the area of a segment of the circle?

Area of a segment of circle = area of the corresponding sector – the area of the corresponding triangle

4. From How many points we can draw a circle, and can we trace these points in straight line and form triangle with them?

From three non-collinear points we can draw a circle, talking about triangle then yes we can make depending upon coordinates of point we  draw a triangle out of it.

5. Total how many exercises are there in chapter 12 areas related to circle?

There is a total of 3 exercises there in chapter 12 namely exercise 12.2 exercise 12.2 and exercise 12.3 which have a variety of questions to give a brief analysis of the topic as well as subject.

6. What is the area of the right angled triangle ABC, the right angle at point B ?

Ans:  The area of the right-angle triangle is equal to  0.5*base*(height)

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