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

NCERT solutions for class 10 maths chapter 12 Areas Related to Circles - This chapter is in continuation with the chapter 'Circles' of class 9 which you have already studied. There will be a lot of questions in which class 9 concepts will be used. Solutions of NCERT class 10 maths chapter 12 Areas Related to Circles cover the elaborated solutions to each and every question which is available in the practice exercises including the optional exercises. This topic is a part of the mensuration unit, and it holds 10 marks in class 10 board examinations. From the trend of the last years' exams, you can expect at least 1 question from this chapter in the board examinations. With the help of CBSE NCERT solutions for class 10 maths chapter 12 Areas Related to Circles, you won't miss any mark if a question directly comes from the practice exercises. In this chapter, you will get to know about the terms like- Sector, Chord, Segment, Radius, Diameter, etc. There is a total of 3 exercises with 35 questions in them. NCERT solutions for class 10 maths chapter 12 Areas Related to Circles are constructed in a manner so that a student can get 100% out of it. Apart from the solutions to this particular chapter, NCERT solutions are also available for different classes and subjects which can be downloaded by clicking on the link.

## Some important formulae used in NCERT solutions for class 10 maths chapter 12 Areas Related to Circles

•  Circumference of a circle = $2\pi r$.

•  Area of a circle = $\pi r^2$.

•  The area of a sector of a circle with radius $\dpi{100} 'r'$ and angle with degree measure $\theta$ is  $\dpi{100} \frac{\theta }{360}\times \pi r^2$.

• Length of an arc of a sector of a circle with radius $\dpi{100} 'r'$ and angle with degree measure $\theta$ is  $\dpi{100} \frac{\theta }{360}\times 2\pi r$

• Area of the segment of a circle= Area of the corresponding sector – Area of the corresponding triangle.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas related to circles Excercise: 12.1

Circumference of 1st circle is given :

$\\=\ 2\pi r_1\\\\=\ 2\pi \times 19\ =\ 38 \pi\ cm$

And the circumference of the 2nd circle is :       $=\ 2\pi \times 9\ =\ 18 \pi\ cm$

Thus the circumference of the new circle is   $=\ 18\pi\ +\ 38\pi\ =\ 56\pi\ cm$

or                      $2\pi r =\ 56\pi$

or                      $r\ =\ 28\ cm$

Hence the radius of the new circle is 28 cm.

We know that the area of the circle is   :    $=\ \pi r^2$

Thus area of a 1st circle  :    $=\ \pi r^2_1$

or                                       $=\ \pi \times 8^2\ =\ 64 \pi\ cm^2$

And the area of a 2nd circle is   :  $=\ \pi r^2_2$

or                                         $=\ \pi \times 6^2\ =\ 36\pi\ cm^2$

According to the question area of the new circle is   $=\ 64\pi\ +\ 36\pi\ =\ 100\pi\ cm^2$

or                                           $\pi r^2\ =\ 100 \pi$

or                                              $r\ =\ 10\ cm$

Hence the area of the new circle is 10 cm.

The radius of each scoring region can be found by adding 10.5 in respective colors.

The area of the golden region is :      $=\ \pi r^2\ =\ \pi \times 10.5^2\ =\ 346.5\ cm^2$

Similarly area of the red region is   =   Area of a red score  -  Area of the golden region

$\\=\ \pi \times 21^2\ -\ \pi \times 10.5^2\\\\=\ 1039.5\ cm^2$

Then the area of the blue region is :

$\\=\ \pi \times (31.5)^2\ -\ \pi \times 21^2\\\\=\ 1732.5\ cm^2$

And the area of the black region is :

$\\=\ \pi \times 42^2\ -\ \pi \times (31.5)^2\\\\=\ 2425.5\ cm^2$

Area of the white region is :

$\\=\ \pi \times 52.5^2\ -\ \pi\times 42^2\\\\=\ 3118.5\ cm^2$

## Q4 The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?

Let the number of revolution of the wheel be n.

The circumference of the wheel is given by:-

$\\=\ 2\pi r\\\\=\ 2\pi \times 40\ =\ 80\pi\ cm$

Now, the speed of the car is given by:-

$\\=\ 66\ Km/hr\ \\\\=\ 66\times \frac{100000}{60}\ cm/min\ \\\\=\ 110000\ cm/min.$

Thus distance traveled in 10 minutes is:  $=\ 1100000\ cm$

According to the question, we get;

$n\times 80\pi\ =\ 1100000$

or                                               $n\ =\ 4375$

Hence the number of revolutions made by the wheel is 4372.

(A)    2 units
(B)    $\pi$ units
(C)    4 units
(D)    7 units

Let the radius of the circle be r.

Then according to question, we can write:-

Perimeter      =     Area

$\\2\pi r\ =\ \pi r^2\\\\r\ =\ 2\ units$

Hence option (A) is correct.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas related to circles Excercise: 12.2

### Q1 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

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 360o 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 90o.

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

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  360o  -  90o  =  270o.

Thus the area is : -

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

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.

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 cm2.

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 60o, or we can say that all angles are 60o (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 600.

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., 60o.)

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:$=\ 150.72\ -\ 62.28\ =\ 88.44\ cm^2$.

## Q8 A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig.). Find (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

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

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

$=\ 19.62\ m^2$

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  :

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

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 36o.

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 45o.

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 80o.

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 60o 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.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas related to circles Excercise: 12.3

We know that $\angle$RPQ  is 90o 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

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 90o 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 90o 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 :

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 m2.

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 cm2

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 60o 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 90o 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 :

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

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

## NCERT solutions for class 10 maths chapter wise

 Chapter No. Chapter Name Chapter 1 CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers Chapter 2 NCERT solutions for class 10 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables Chapter 4 CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations Chapter 5 NCERT solutions for class 10 chapter 5 Arithmetic Progressions Chapter 6 Solutions of NCERT class 10 maths chapter 6 Triangles Chapter 7 CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry Chapter 8 NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Chapter 9 Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry Chapter 10 CBSE NCERT solutions class 10 maths chapter 10 Circles Chapter 11 Solutions of NCERT class 10 maths chapter 11 Constructions Chapter 12 NCERT Solutions for class 10 Maths chapter 12 Areas Related to Circles Chapter 13 CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes Chapter 14 NCERT solutions for class 10 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 10 maths chapter 15 Probability

## How to use NCERT solutions for class 10 maths chapter 12 Areas Related To Circles?

• Before coming to this chapter please make sure that you are comfortable with the basic concepts of circles, squares, rectangles, triangles, etc.

• Learn some formulae related to sectors and segments from the NCERT textbook.

• Go through some examples given in the textbook to understand the approach of solving the problems.

• Once you have done the above-said points then its time to put your efforts into practice by practicing the practice exercises.

• During the practice, you can take the help of NCERT solutions for class 10 maths chapter 12 Areas Related to Circles.

Keep working hard & happy learning!