Answer the following :
Two circles each of radius 7 cm, intersect each other. The distance between their centres is `7sqrt(2)` cm. Find the area common to both the circles.
Solution
Let O and O1 be the centres of two circles intersecting each other at A and B.
Then OA = OB = O1A = O1B = 7 cm
and OO12 = 98 ...(i)
Since OA2 + O1A2 = 72 + 72
= 98
= OO12 ...[From (i)]
∴ m∠OAO1 = 90°
∴ m∠OAO1B is a square.
m∠AOB = m∠AO1B = 90°
= `(90 xx pi/180)^"c"`
= `(pi/2)^"c"`
Now, A(sector OAB) = `1/2"r"^2theta`
= `1/2 xx 7^2 xx pi/2`
= `(49pi)/4"sq.cm"`
and A(sector O1AB) = `1/2"r"^2theta`
= `1/2 xx 7^2 xx pi/2`
= `(49pi)/4"sq.cm"`
A(∠OAO1B) = (side)2 = (7)2 = 49 sq.cm
∴ Required area = area of shaded portion
= A(sector OAB) + A(sector O1AB) – A(∠OAO1B)
= `(49pi)/4 + (49pi)/4 - 49`
= `(49pi)/2 - 49`
= `49(pi/2 - 1)"sq.cm"`