CAT 2022 Slot 3 Paper

$BC$ is the diameter of circle $C2$ so we can say that $\angle BAC = 90^{\circ}$ as angle in the semi circle is $90^{\circ}$ Therefore overlapping area $=\frac{1}{2}(\text{Area of circle } C2) + \text{Area of the minor sector made be } BC$ in $C1$ $AB = AC = 8\,\text{cm}$ and as $\angle BAC = 90^{\circ}$, so we can conclude that $BC = 8\sqrt{2\,\text{cm}}$ Radius of $C2 =$ Half of length of $BC = 4\sqrt{2}\,\text{cm}$ Area of $C2 = \pi (4\sqrt{2})^2 = 32\pi \,\text{cm}^2$ $A$ is the Centre of $C1$ and $C1$ passes through $B$, so $AB$ is the radius of $C1$ and is equal to $8\,\text{cm}$ Area of the minor sector made be $BC$ in $C1 = \frac{1}{4}(\text{Area of circle } C1) - \text{Area of triangle}$ $ABC = \frac{1}{4} \pi (8)^2 - \left(\frac{1}{2} \times 8 \times 8\right) = 16\pi - 32\,\text{cm}^2$ Therefore, Overlapping area between the two circles $= \frac{1}{2} (\text{Area of circle } C2) + \text{Area of the minor sector made be } BC$ in $C1$ $= \frac{1}{2}(32\pi) + (16\pi - 32) = 32(\pi - 1) \,\text{cm}^2$