Overlapping Area of Intersecting Circles

Since the circle with $BC$ as diameter passes through $A$, the angle $\angle BAC = 90^\circ$. $AB = AC = 8\text{ cm}$, so $\triangle ABC$ is a right isosceles triangle. $BC = \sqrt{8^2 + 8^2} = 8\sqrt{2}\text{ cm}$.

1. First circle $C_1$: Diameter $BC = 8\sqrt{2}$, radius $R_1 = 4\sqrt{2}$. Center is midpoint of $BC$. Since $\angle BAC = 90^\circ$, $A$ lies on $C_1$.
2. Second circle $C_2$: Center $A$, radius $R_2 = 8$.

The overlapping region consists of two segments:

• The semicircle of $C_1$ bounded by diameter $BC$ has area $\frac{1}{2} \pi R_1^2 = \frac{1}{2} \pi (4\sqrt{2})^2 = 16\pi$.
• The segment of $C_2$ cut off by chord $BC$: Sector area of $C_2$ (angle $90^\circ$) $= \frac{1}{4} \pi R_2^2 = \frac{1}{4} \pi (8^2) = 16\pi$. Area of $\triangle ABC = \frac{1}{2} \times 8 \times 8 = 32$. Segment area of $C_2 = 16\pi - 32$.

Sum of areas of the two parts forming the lens/overlap: $\text{Overlapping Area} = \text{Area of semicircle } C_1 + \text{Area of segment } C_2 - \text{Area}(\triangle ABC)$ Wait, the overlap is equal to (Area of semicircle $C_1$) + (Area of segment of $C_2$) which equals: $\text{Area of semicircle } C_1 = 16\pi$ $\text{Segment of } C_1 \text{ over } BC = 16\pi - 32$ Combined overlapping region $= 16\pi + (16\pi - 32) = 32\pi - 32 = 32(\pi - 1)$.