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.

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, OR^2 = PQ^2 + PR^2

= 24^2 + 7^2

= 576 + 49 = 625

Or, OR = 25  cm

Area of triangle = ½ xx text(base) xx text(height)

= ½ xx 24 xx 7 = 84  sq  cm

Area of semicircle = ½ xx πr^2

= ½ xx π xx 12.5^2 = 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°.

= Area of Bigger Sector – Area of Smaller Sector

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

=(40°)/(360°)π(R^2-r^2)

=(1)/(9)xx(22)/(7)xx(14^2-7^2)

=51.33 sq cm

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.

Solution: Area of Square = text(Side)^2 = 14^2 = 196  sq  cm

Area of two semicircles = πr^2

= π xx 7^2 = 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.

Solution: Area of sector outside triangle

=(300°)/(360°)πxx6^2

=94.28 sq cm

Area of triangle

=(sqrt3)/(4)xxtext(side)^2

=(sqrt3)/(4)xx12^2

=62.352 sq cm

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.

Solution: Area of square = text(Side)^2 = 4^2 = 16  sq  cm

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

= 2 xx π xx 1^2 = 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).

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

text(cos)30°=(BC)/(OB)

Or, (sqrt3)/(2)=(BC)/(32)

Or, BC=16sqrt3

Hence, area of equilateral triangle

=(sqrt3)/(4)xxtext(side)^2

=(sqrt3)/(3)xx(32sqrt3)^2

=1330.176 sq cm

Area of circle = πr^2 = π xx 32^2 = 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.

Solution: Area of square = text(Side)^2 = 14^2 = 196  sq  cm

Area of four quadrants = Area of 1 circle

= πr^2 = π xx 7^2 = 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 xx 60 = 188.4

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

Area of straight portion = 2 xx text(length) xx text(width)

= 2 xx 106 xx 10 = 2120  sq  m

Area of circular portion = π(R^2 – r^2)

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

Area = π(40^2 – 30^2) = 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.

Solution: Area of smaller circle = πr^2 = π xx (3.5)^2 = 38.5  sq  cm

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

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

Area of ∆ABC = ½ xx AB xx OC

= ½ xx 14 xx 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)