Class 10 Mathematics

Area of Circle

Exercise 12.3 (NCERT Book) Part 1

Question: 1. Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

10 circle area exercise solution

Solution: Here, OR is the hypotenuse of ∆PQR, because lines from two ends of diameter always make a right angle when they meet at circumference.

So, OR2 = PQ2 + PR2

= 242 + 72

= 576 + 49 = 625

Or, OR = 25 cm

Or radius = 12.5 cm



Area of triangle = ½ x base x height

= ½ x 24 x 7 = 84 sq cm

Area of semicircle = ½ x πr2

= ½ x π x 12.52 = 245.3125 sq cm

So, area of shaded region = 245.3125 – 84 = 161.3125 sq cm

Question: 2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and angle AOC = 40°.

10 circle area exercise solution

Solution: Area of shaded region

= Area of Bigger Sector – Area of Smaller Sector

If R and r are the two radii, then area of shaded region

10 circle area exercise solution

Question: 3. Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

10 circle area exercise solution

Solution: Area of Square = Side2 = 142 = 196 sq cm

Area of two semicircles = πr2

= π x 72 = 154 sq cm

Area of shaded region = 196 – 154 = 42 sq cm

Question: 4. Find the area of the shaded region in the given 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.

10 circle area exercise solution

Solution: Area of sector outside triangle

10 circle area exercise solution

Area of triangle

10 circle area exercise solution

Area of shaded region = 94.28 + 62.352 = 156.632 sq cm

Question: 5. 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 the figure. Find the area of the remaining portion of the square.

10 circle area exercise solution

Solution: Area of square = Side2 = 42 = 16 sq cm

Area of cut portion = Area of two circles = 1 circle + 4quadrants

= 2 x π x 12 = 6.28 sq cm

So, area of remaining portion of square = 16 – 6.28 = 22.28 sq cm

Question: 6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

10 circle area exercise solution

Solution: In ∆OCB; OB = 32 cm, ∠OBC = 30o

10 circle area exercise solution


Hence, area of equilateral triangle

10 circle area exercise solution

Area of circle = πr2 = π x 322 = 3215.36 sq cm

Area of shaded region = 3215.36 – 1330.176 = 1885.184 sq cm

Question: 7. In the given figure, ABCD is a square of side 14 cm. With centres 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.

10 circle area exercise solution

Solution: Area of square = Side2 = 142 = 196 sq cm

Area of four quadrants = Area of 1 circle

= πr2 = π x 72 = 154 sq cm

Area of shaded portion = 196 – 154 = 42 sq cm

Question: 8. The following 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.

Solution: Distance = Length of straight track + circumference of inner loop

Circumference = 2πr = 3.14 x 60 = 188.4

Hence, distance = 188.4 + 106 + 106 = 400.4 m

Area of straight portion = 2 x length x width

= 2 x 106 x 10 = 2120 sq m

Area of circular portion = π(R2 – r2)

Where, R = radius of outer circle and r = radius of inner circle

Area = π(402 – 302) = 2200 sq m

So, Total area of track = 2200 + 2120 = 4320 sq m

Question: 9. In the given 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.

10 circle area exercise solution

Solution: Area of smaller circle = πr2 = π x (3.5)2 = 38.5 sq cm

Area of bigger circle = 4 x 38.5 (because radius is double that of smaller circle)

Hence, area of bigger semicircle = 2 x 38.5 = 77 sq cm

Area of ∆ABC = ½ x AB x OC

= ½ x 14 x 7 = 49 sq cm

Area of shaded portion in semicircle = 77 – 49 = 28 sq cm

Total area of shaded portion = 28 + 38.5 = 66.5 sq cm


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Exercise 1

Exercise 2 (Part 1)

Exercise 2 (Part 2)

Exercise 3 (Part 2)