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In this page we have *NCERT Solutions for Class 10 Maths Areas Related to Circles* for
Exercise 12.3 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

Unless stated otherwise, use π =22/7Find the area of the shaded region in below figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

PQ = 24 cm and PR = 7 cm

∠P = 90° (Angle in the semi-circle)

By Pythagoras theorem,

QR

QR

QR = 25 cm

Now Diameter of the circle=QR

Therefore, Radius of the circle = 25/2 cm

Area of the semicircle = (π R

= (22/7 × 25/2 × 25/2)/2 cm

= 13750/56 cm

Area of the ΔPQR = 1/2 × PR × PQ

= 1/2 × 7 × 24 cm

= 84 cm

Find the area of the shaded region in below figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Radius inner circle = 7 cm

Radius of outer circle = 14 cm

Angle made by sector = 40°

Area of the sector OAC = (40°/360°) × π r

= 1/9 × 22/7 × 14

Area of the sector OBD = (40°/360°) × π r

= 1/9 × 22/7 × 7

= 68.44 - 17.11 = 51.33 cm

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

There are two semicircles in the figure.

Side of the square = 14 cm

Therefore, Diameter of the semicircle = 14 cm

Radius of the semicircle = 7 cm

Area of the square = 14 × 14 = 196 cm

Area of the semicircle = (π R

= (22/7 × 7 × 7)/2 cm

Area of two semicircles = 2 × 77 = 154 cm

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

OAB is an equilateral triangle with each angle equal to 60°.

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle = √3/4 × (OA)

Area of the circle = π R

Area of the sector making angle 60° = (60°/360°) × π r

= 1/6 × 22/7 × 6

= 36√3 + 792/7 - 132/7

= (36√3 + 660/7) cm

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

Side of the square = 4 cm

Radius of the circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

Area of square = (side)

Area of the quadrant = (π R

Therefore, Total area of the 4 quadrants = 4 × (11/14) cm

Area of the circle = π R

= 16 cm

= 68/7 cm

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

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

Since, AD is the median of the triangle

Therefore, AO = Radius of the circle = 2/3 AD

2/3 AD = 32 cm

AD = 48 cm

In ΔADB,

By Pythagoras theorem,

AB

AB

3/4 (AB

AB = 32√3 cm

Area of ΔABC = √3/4 × (32√3)

Area of circle = π R

= (22528/7 - 768√3) cm

In below figure, 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.

Side of square = 14 cm

Four quadrants are included in the four sides of the square.

Therefore, Radius of the circles = 14/2 cm = 7 cm

Area of the square ABCD = 14

Area of the quadrant = (π R

= 77/2 cm

Total area of the quadrant = 4 × 77/2 cm

= 196 - 154

= 42 cm

Below Figure 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

(ii) the area of the track.

Width of track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O'C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O'B

= 30 + 10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)

= 106 + 106 + (2 × πr) m = 212 + (2 × 22/7 × 30) m

= 2804/7 m

= (AB × CD) + (EF × GH) + 2 × (πr

= (106 × 10) + (106 × 10) + 2 × π/2 (r

= 4320 m

In below figure, AB and CD are two diameters of a circle (with centre 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.

Radius of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm

Area of larger circle = πR

Area of larger semicircle = 154/2 cm

Area of smaller circle = πr

= Area of larger circle - Area of triangle - Area of larger semicircle + Area of smaller circle

Area of the shaded region = (154 - 49 - 77 + 39.5)

= 66.5 cm

The area of an equilateral triangle ABC is 17320.5 cm

ABC is an equilateral triangle.

Therefore, ∠A = ∠B = ∠C = 60°

There are three sectors each making 60°.

Let x be the side of the equilateral triangle

Area of ΔABC = 17320.5 cm

√3/4 × (x)

x = 200 cm

Therefore, Radius of the circles = 200/2 cm = 100 cm

Area of the sector = (60°/360°) × π r

= 1/6 × 3.14 × (100)

= 15700/3 cm

Area of the shaded region = Area of equilateral triangle ABC – 3 × Area of sector

= 17320.5 - 15700 cm

On a square handkerchief, nine circular designs each of radius 7 cm are made . Find the area of the remaining portion of the handkerchief.

Number of circular design = 9

Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.

Therefore, Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm

Area of the square = 42 × 42 cm

Area of the circle = π r

Area of the remaining portion of the handkerchief

= Area of the square - Total area of the design

= Area of the square - 9× Area of the circle

= 1764 - 1386 = 378 cm

In below figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB

(ii) shaded region.

Radius of the quadrant = 3.5 cm

(i) Area of quadrant OACB = (πR

= (22/7 × 3.5

= 9.625 cm

(ii) Area of triangle BOD = 1/2 × 3.5 × 2 cm

= 3.5 cm

Area of shaded region = Area of quadrant - Area of triangle BOD

= (9.625 – 3.5) cm

= 6.125 cm

In below figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

Now OB is the diagonal of the square

OB = 20√2 cm

Area of the quadrant = (πR

Area of the square = 20 × 20 = 400 cm

Area of the shaded region = Area of the quadrant - Area of the square

= 628 - 400 = 228 cm

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O .If ∠AOB = 30°, find the area of the shaded region.

Radius of the larger circle, R = 21 cm

Radius of the smaller circle, r = 7 cm

Angle made by sectors of both concentric circles = 30°

Area of the larger sector = (30°/360°) × π R

= 1/12 × 22/7 × 21

= 231/2 cm

Area of the smaller circle = (30°/360°) × π r

= 1/12 × 22/7 × 7

= 77/6 cm

Area of the shaded region = 231/2 - 77/6 cm

= 616/6 cm

In below figure, 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.

Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC

BC

BC = 14√2 cm

Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC = 1/2 × 14 × 14 = 98 cm

Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm

Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm

Area of the shaded region = Area of the semicircle –(Area of quadrant - Area of ΔABC )

= 154 + 98 - 154 cm

Calculate the area of the designed region in below figure common between the two quadrants of circles of radius 8 cm each.

AB = BC = CD = AD = 8 cm

Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm

Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 8

= 352/7 cm

Area of shaded region =

(Area of quadrant AECB - Area of ΔABC) + (Area of quadrant AFCD - Area of ΔADC)

= (352/7 - 32) + (352/7 -32) cm

= 2 × (352/7 -32) cm

= 256/7 cm

Download Area Related to Circle Exercise 12.3 as pdf

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.