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, 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°.
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
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 = 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.
Solution: Area of sector outside triangle
Area of triangle
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 = 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).
Solution: In ∆OCB; OB = 32 cm, ∠OBC = 30o
Hence, area of equilateral triangle
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.
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.
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
Exercise 2 (Part 1)
Exercise 2 (Part 2)
Exercise 3 (Part 2)