NCERT Solutions for Exercise 12.3 Class 10 Maths Chapter 12 - Areas related to circles

# NCERT Solutions for Exercise 12.3 Class 10 Maths Chapter 12 - Areas related to circles

Edited By Ramraj Saini | Updated on Nov 27, 2023 05:14 PM IST

## NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3

NCERT Solutions for Exercise 12.3 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 12.3 introduces areas and perimeters of plane figures like square, rectangle, parallelogram etc. Previously we have calculated the areas of figures separately like circle, square, prism, pyramid etc. This ex 12.3 class 10 also includes the combination of figures. We find the combination of plane figures use in our daily life and in the form of various interesting designs. The concepts related to areas figures and their combinations are well explained in this NCERT book Class 10 Maths chapter 12 exercise 12.3

In this class 10 ex 12.3, we have interesting designs like a flower bed, drain cover, window designs are such examples. 10th class Maths exercise 12.3 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 12.3

Areas Related to Circles Class 10 Chapter 12 Exercise: 12.3

We know that $\angle$ RPQ is 90 o as ROQ is the diameter.

RQ can be found using the Pythagoras theorem.

$RP^2\ +\ PQ^2\ =\ QR^2$

or $7^2\ +\ 24^2\ =\ QR^2$

or $QR\ =\ \sqrt{625}\ =\ 25\ cm$

Now, the area of the shaded region is given by = Area of semicircle - Area of $\Delta$ PQR

The area of a semicircle is:-

$=\ \frac{1}{2}\pi r^2$

or $=\ \frac{1}{2}\pi \times \left ( \frac{25}{2} \right )^2$

or $=\ 245.53\ cm^2$

And, the area of triangle PQR is :

$=\ \frac{1}{2}\times 24\times 7\ =\ 84\ cm^2$

Hence the area of the shaded region is : $=\ 245.53\ -\ 84\ =\ 161.53\ cm^2$

The area of a shaded region can be easily found by using the formula of the area of the sector.

Area of the shaded region is given by : Area of sector OAFC - Area of sector OBED

$=\ \frac{40^{\circ}}{360^{\circ}}\times \pi (14)^2\ -\ \frac{40^{\circ}}{360^{\circ}}\times \pi (7)^2$

$=\ \frac{616}{9}\ -\ \frac{154}{9}\ =\ \frac{462}{9}$

$=\ 51.33\ cm^2$

Area of the shaded region is given by = Area of the square - Area of two semicircles.

Area of square is : $=\ 14^2\ =\ 196\ cm^2$

And the area of the semicircle is:-

$=\ \frac{1}{2}\pi r^2$

$=\ \frac{1}{2}\pi \times 7^2$

$=\ 77\ cm^2$

Hence the area of the shaded region is given by : $=\ 196 - 2(77) = 42\ cm^2.$

Area of the shaded region is given by = Area of triangle + Area of the circle - Area of the sector

Area of the sector is : -

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

or $=\ \frac{132}{7}\ cm^2$

And, the area of the triangle is :

$=\ \frac{\sqrt{3}}{4}a^2\ =\ \frac{\sqrt{3}}{4}\times 12^2\ =\ 36\sqrt{3}\ cm^2$

And, the area of the circle is : $=\ \pi r^2$

or $=\ \pi \times 6^2$

or $=\ \frac{792}{7}\ cm^2$

Hence the area of the shaded region is:-

$=\ 36\sqrt{3}\ +\ \frac{792}{7}\ -\ \frac{132}{7}$

or $=\ \left ( 36\sqrt{3}\ +\ \frac{660}{7} \right )\ cm^2$

Consider the quadrant in the given figure:- We have an angle of the sector as 90 o and radius 1 cm.

Thus the area of the quadrant is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi\times 1^2$

or $=\ \frac{22}{28}\ cm^2$

And the area of the square is : $=\ side^2$

$=\ 4^2$

$=\ 16\ cm^2$

And, the area of the circle is:-

$=\ \pi r^2\ =\ \pi \times 1^2\ =\ \pi\ cm^2$

Hence the area of the shaded region is: = Area of the square - Area of the circle - 4 (Area of quadrant)

or $=\ 16\ -\ \frac{22}{7}\ -\ 4\left ( \frac{22}{28} \right )$

or $=\ \frac{68}{7}\ cm^2$

Assume the center of the circle to be point C and AD as the median of the equilateral triangle.

Then we can write:-

$AO\ =\ \frac{2}{3}AD$

or $32\ =\ \frac{2}{3}AD$

Thus $AD\ =\ 48\ cm$

Consider $\Delta$ ABD,

$AB^2\ =\ AD^2\ +\ BD^2$

or $AB^2\ =\ 48^2\ +\ \left ( \frac{AB}{2} \right )^2$

or $AB\ =\ 32\sqrt{3}\ cm$

Thus the area of an equilateral triangle is:-

$=\ \frac{\sqrt{3}}{4}\times \left ( 32\sqrt{3} \right )^2$

or $=\ 768\sqrt{3}\ cm^2$

And the area of the circle is : $=\ \pi r^2\ =\ \pi\times 32^2$

or $=\ \frac{22528}{7} cm^2$

Hence the area of the design is:-

$=\ \left ( \frac{22528}{7}\ -\ 768 \sqrt{3} \right )\ cm^2$

It is clear from the figure that the area of all sectors is equal (due to symmetry).

Also, the angle of the sector is 90 o and the radius is 7 cm.

Thus the area of the sector is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi (7)^2$

or $=\ \frac{77}{2}\ cm^2$

And, the area of the square is : $=\ a^2\ =\ 14^2\ =\ 196\ cm^2$

Hence the area of the shaded region is :

$=\ 196\ -\ 4\times \frac{77}{2}$

$=\ 42\ cm^2$

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(i) the distance around the track along its inner edge

The distance around the track is Length of two straight lines + Length of two arcs.

Length of the arc is -

$=\ \frac{1}{2}\times 2\pi r$

$=\ \frac{1}{2}\times 2\pi \times 30\ =\ 30\pi\ m$

Thus the length of the inner track is :

$=\ 106\ +\ 30\pi\ +\ 106\ +\ 30\pi$

$=\ 212\ +\ 60\pi$

$=\ \frac{2804}{7}\ m$

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(ii) the area of the track.

The area of track = Area of outer structure - Area of inner structure.

Area of outer structure is: = Area of square + Area of 2 semicircles

$=\ 106\times 80\ +\ \frac{1}{2}\pi (40)^2\ +\ \frac{1}{2}\pi (40)^2\ m^2$

And, area of inner structure: = Area of inner square + Area of 2 inner semicircles

$=\ 106\times 60\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2\ m^2$

Thus the area of the track is :

$\\=\ 106\times 80\ +\ \frac{1}{2}\pi (40)^2\ +\ \frac{1}{2}\pi (40)^2\ -\ \left ( 106\times 60\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2 \right )\\\\=\ 4320\ m^2$

Hence the area of the track is 4320 m 2 .

Firstly, the area of the smaller circle is :

$=\ \pi r^2$

$=\ \pi \times \left ( \frac{7}{2} \right )^2$

$=\ \frac{77}{2}\ cm^2$

Now, the area of $\Delta ABC$ :-

$=\ \frac{1}{2}\times AB\times OC$

or $=\ \frac{1}{2}\times 14\times 7\ =\ 49\ cm^2$

And, the area of the bigger semicircle is :

$=\ \frac{1}{2}\pi r^2$

$=\ \frac{1}{2}\pi \times 7^2$

$=\ 77\ cm^2$

Hence the area of the shaded region is:-

$=\ \frac{77}{2}\ +\ 77\ -\ 49$

$=\ 66.5\ cm^2$

Therefore the area of the shaded region is 66.5 cm 2 .

Area of an equilateral triangle is:-

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

$\frac{\sqrt{3}}{4}\times a^2\ =\ 17320.5$

$a\ =\ 200\ cm$

Now, consider the sector:- Angle of the sector is 60 o and the radius is 100 cm.

Thus the area of the sector:-

$=\ \frac{60^{\circ}}{360^{\circ}}\times \pi \times 100^2$

$=\ \frac{15700}{3}\ cm^2$

Thus the area of the shaded region is :

$=\ 17320.5\ -\ 3\times \frac{15700}{3}$

$=\ 1620.5\ cm^2$

Since one side of the square has 3 circles, thus the side of the square is 42 cm.

Area of the square : $=\ a^2\ =\ 42^2\ =\ 1764\ cm^2$

And, area of a circle :

$=\ \pi r^2\ =\ \frac{22}{7}\times 7^2$

$=\ 154\ cm^2$

Hence the area of the remaining portion is : $=\ 1764\ -\ 9\times 154$

$=\ 378\ cm^2$

The quadrant OACB is a sector with angle 90 o and radius 3.5 cm.

Thus the area of the quadrant is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 3.5^2$

$=\ \frac{77}{8}\ cm^2$

Hence the area of the quadrant is $\frac{77}{8}\ cm^2$ .

For area of shaded region we need to find area of the triangle.

Area of triangle is:-

$=\ \frac{1}{2}\times 3.5\times 2$

$=\ 3.5\ cm^2$

Hence the area of the shaded region is = Area of the quadrant - Area of triangle

$=\ \frac{77}{8}\ -\ 3.5$

$=\ \frac{49}{8}\ cm^2$

In the given figure we need to find the radius of the circle:-

Consider $\Delta$ OAB,

$OB^2\ =\ OA^2\ +\ AB^2$

$=\ 20^2\ +\ 20^2$

$OB\ =\ 20\sqrt{2}\ cm$

$=\ \frac{90^{\circ}}{360^{\circ}}\times 3.14\times (20\sqrt{2})^2$

$=\ 628\ cm^2$

Also, the area of the square is : $=\ 20^2\ =\ 400\ cm^2$

Area of the shaded region is : $=\ 628\ -\ 400\ =\ 228\ cm^2$

Area of the shaded region is = Area of larger sector - Area of smaller sector

$=\ \frac{30^{\circ}}{360^{\circ}}\times \pi \times 21^2\ -\ \frac{30^{\circ}}{360^{\circ}}\times \pi \times 7^2$

$=\ \frac{30^{\circ}}{360^{\circ}}\times \pi \times (21^2\ -\ 7^2)$

$=\ \frac{308}{3}\ cm^2$

Hence the area of the shaded region is $\frac{308}{3}\ cm^2.$

Consider $\Delta$ ABC,

$BC^2\ =\ AC^2\ +\ AB^2$

$=\ 14^2\ +\ 14^2$

$BC\ =\ 14\sqrt{2}\ cm$

Area of triangle is :

$=\ \frac{1}{2}\times 14\times 14$

$=\ 98\ cm^2$

Now, area of sector is :

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 14^2$

$=\ 154\ cm^2$

And area of semicircle is : -

$=\ \frac{1}{2} \pi r^2$

$=\ \frac{1}{2} \pi \times (7\sqrt{2})^2$

$=\ 154\ cm^2$

Hence the area of the shaded region is : $=\ 154\ -\ (154\ -\ 98)\ =\ 98\ cm^2$

It is clear from the figure that the required area (designed area) is the area of the intersection of two sectors.

Area of the sector is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 8^2$

$=\ \frac{352}{7}\ cm^2$

And, area of the triangle:-

$=\ \frac{1}{2}\times 8\times 8\ =\ 32\ cm^2$

Hence the area of the designed region is :

$=\ 2\left ( \frac{352}{7}\ -\ 32 \right )$

$=\ \frac{256}{7}\ cm^2$

## More About NCERT Solutions for Class 10 Maths Exercise 12.3

Class 10 Maths chapter 12 exercise 12.3 Area Related To Circles exercise 12.3 Class 10 Maths, Includes the areas and perimeters which includes plane figures like a wheel, square, parallelogram circle, rectangle and their combinations. NCERT solutions for Class 10 Maths exercise 12.3 mainly focus on the find the shaded area between the two plane figures is given in exercise 12.3 Class 10 Maths.

Also Read| Areas Related to Circles Class 10 Notes

## Benefits of NCERT Solutions for Class 10 Maths Exercise 12.3

• NCERT solutions for Class 10 Math is considered the best material for solving Class 10 Maths chapter 12 exercise 12.3.

• NCERT Class 10 Maths chapter 12 exercise 12.3, contains all important questions from exam pov and all questions are revised from Class 10 Maths chapter 12 exercise 12.3

• Exercise 12.3 Class 10 Maths, is based on the combination of figures like circle inscribed in a square, circle inscribed in the square, square with semicircles etc.

Also see-

## NCERT Solutions for Class 10 Subject Wise

1. What is the value of π ?

We use π=3.14 and also 22/7 for easy calculations

2. What is the special case of recatngle?

When length of a rectangle is equal to breadth of a rectangle

Then area of rectangle = l*b = a²

= Area of square

3. What is the perimeter of a semicircle ?

Perimeter of semicircle= 2πr+ 2r

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