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

NCERT Solutions for Class 10 Maths chapter 12 exercise 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. NCERT solutions for class 10 maths chapter 12 exercise 12.3 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

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In NCERT solutions for Class 10 Maths exercise 12.3. We have interesting designs like a flower bed, drain cover, window designs are such examples. Along with NCERT syllabus Class 10 Maths chapter 12 exercise 12.3 the following exercises are also present.

Area Related To Circles Exercise 12.1

Area Related To Circles Exercise 12.2

**Areas Related to Circles**** Class 10 Chapter 12**** Exercise: 12.3 ** ** **

** Q1 ** Find the area of the shaded region in Fig, if PQ = 24 cm, PR = 7 cm, and O is the center of the circle.

**Answer: **

We know that RPQ is 90 ^{ o } as ROQ is the diameter.

RQ can be found using the Pythagoras theorem.

or

or

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

The area of a semicircle is:-

or

or

And, the area of triangle PQR is :

Hence the area of the shaded region is : _{}

** Q2 ** Find the area of the shaded region in Fig, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and .

** Answer: **

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

** Q3 ** Find the area of the shaded region in Fig, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

** Answer: **

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

Area of square is :

And the area of the semicircle is:-

Hence the area of the shaded region is given by : _{}

** Q4 ** Find the area of the shaded region in Fig, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.

** Answer: **

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

Area of the sector is : -

or

And, the area of the triangle is :

And, the area of the circle is :

or

or

Hence the area of the shaded region is:-

or

** Q5 ** From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. Find the area of the remaining portion of the square.

** Answer: **

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:-

or

And the area of the square is :

And, the area of the circle is:-

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

or

or

** Q6 ** In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design.

** Answer: **

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

Then we can write:-

or

Thus

Consider ABD,

or

or

Thus the area of an equilateral triangle is:-

or

And the area of the circle is :

or

Hence the area of the design is:-

** Q7. ** In Fig, ABCD is a square of side 14 cm. With centers A, B, C, and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

** Answer: **

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:-

or

And, the area of the square is :

Hence the area of the shaded region is :

** Q8 ** Fig. depicts a racing track whose left and right ends are semicircular.

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

** Answer: **

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

Length of the arc is -

Thus the length of the inner track is :

** Q8 ** Fig. depicts a racing track whose left and right ends are semicircular.

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.

** Answer: **

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

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

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

Thus the area of the track is :

Hence the area of the track is 4320 m ^{ 2 } .

** Q9 ** In Fig, AB, and CD are two diameters of a circle (with center O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

** Answer: **

Firstly, the area of the smaller circle is :

Now, the area of :-

or

And, the area of the bigger semicircle is :

Hence the area of the shaded region is:-

Therefore the area of the shaded region is 66.5 cm ^{ 2 } .

** Q10 ** The area of an equilateral triangle ABC is 17320.5 cm ^{ 2 } . With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see Fig.). Find the area of the shaded region. (Use and )

** Answer: **

Area of an equilateral triangle is:-

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

Thus the area of the sector:-

Thus the area of the shaded region is :

** Q11 ** On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief.

** Answer: **

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

Area of the square :

And, area of a circle :

Hence the area of the remaining portion is :

** Q12 ** In Fig, OACB is a quadrant of a circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB

** Answer: **

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

Thus the area of the quadrant is:-

Hence the area of the quadrant is _{} .

** Q12 ** In Fig, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (ii) shaded region.

** Answer: **

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

Area of triangle is:-

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

** Q13 ** In Fig, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use )

** Answer: **

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

Consider OAB,

Thus area of quadrant:-

Also, the area of the square is :

Area of the shaded region is :

** Q14 ** AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If , find the area of the shaded region.

** Answer: **

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

Hence the area of the shaded region is

** Q15 ** In Fig, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

** Answer: **

Consider ABC,

Area of triangle is :

Now, area of sector is :

And area of semicircle is : -

Hence the area of the shaded region is : _{}

** Q16 ** Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

** Answer: **

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:-

And, area of the triangle:-

Hence the area of the designed region is :

**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 Maths Chapter 12
NCERT Exemplar Solutions Class 10 Maths Chapter 12

**NCERT Solutions for Class 10 Subject Wise**

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Science

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