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 O_{1} be the centres of two circles intersecting each other at A and B.

Then OA = OB = O_{1}A = O_{1}B = 7 cm

and OO_{1}2 = 98 ...(i)

Since OA^{2} + O1A^{2} = 7^{2} + 7^{2 }

= 98

= OO1^{2} ...[From (i)]

∴ m∠OAO_{1} = 90°

∴ m∠OAO_{1}B is a square.

m∠AOB = m∠AO_{1}B = 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 O_{1}AB) = `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 O_{1}AB) – A(∠OAO_{1}B)

= `(49pi)/4 + (49pi)/4 - 49`

= `(49pi)/2 - 49`

= `49(pi/2 - 1)"sq.cm"`