Area of Circle
NCERT Exercise
12.3 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.
Answer: 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`
Or radius = 12.5 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°.
Answer: 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
`=(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.
Answer: 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.
Answer: 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.
Answer: 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).
Answer: 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.
Answer: 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.
Answer: 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.
Answer: 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`