Question 18.

BC is the diameter of circle C2 so we can say that $∠BAC = 90°$ as angle in the semi circle is $90°$. Therefore overlapping area = $1/2$(Area of circle C2) + Area of the minor sector made be BC in C1.

AB = AC = 8 cm and as $∠BAC = 90°$, so we can conclude that BC = $8√2$ cm.

Radius of C2 = Half of length of BC = $4√2$ cm.

Area of C2 = $π(4√2)^2 = 32π\text{ cm}^2$.

A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm.

Area of the minor sector made be BC in C1 = $1/4$(Area of circle C1) – Area of triangle ABC = $1/4π(8)^2 – (1/2 × 8 × 8) = 16π – 32\text{ cm}^2$.

Therefore, overlapping area between the two circles = $1/2$(Area of circle C2) + Area of the minor sector made be BC in C1 = $1/2(32π) + (16π – 32) = 32(π – 1)\text{ cm}^2$.