1. **State the problem:** We have two circles: a smaller circle centered at A with radius $r$, and a larger circle centered at C with radius $s$. Points B and D lie on both circles, and chord BD has length $s$. We need to find an expression for the area of the shaded region between the two circles in the form $(a + b\pi)r^2$. 2. **Given:** - Smaller circle radius: $r$ - Larger circle radius: $s$ - $s = \sqrt{3}r$ (from part (a)) - Chord BD length: $s$ 3. **Find:** Area of the shaded region between the two circles. 4. **Approach:** The shaded region is the lens-shaped area formed by the intersection of the two circles. The area of the lens is the sum of the areas of two circular segments minus the overlapping triangle. 5. **Calculate the area of the smaller circle segment:** - The chord BD subtends an angle $\theta$ at center A. - Using the chord length formula: $BD = 2r\sin(\theta/2)$ - Given $BD = s = \sqrt{3}r$, so: $$\sqrt{3}r = 2r\sin(\theta/2) \implies \sin(\theta/2) = \frac{\sqrt{3}}{2}$$ - Therefore, $\theta/2 = \frac{\pi}{3}$ and $\theta = \frac{2\pi}{3}$. - Area of the segment in smaller circle: $$A_1 = \frac{r^2}{2}(\theta - \sin\theta) = \frac{r^2}{2}\left(\frac{2\pi}{3} - \sin\frac{2\pi}{3}\right)$$ - Calculate $\sin\frac{2\pi}{3} = \sin 120^\circ = \frac{\sqrt{3}}{2}$: $$A_1 = \frac{r^2}{2}\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = r^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$$ 6. **Calculate the area of the larger circle segment:** - The chord BD subtends an angle $\phi$ at center C. - Since BD is radius $s$ of the larger circle, the chord length equals the radius, so the angle subtended is $120^\circ$ or $\frac{2\pi}{3}$ (isosceles triangle with sides $s, s, s$). - Area of the segment in larger circle: $$A_2 = \frac{s^2}{2}(\phi - \sin\phi) = \frac{(\sqrt{3}r)^2}{2}\left(\frac{2\pi}{3} - \sin\frac{2\pi}{3}\right)$$ - Simplify $s^2 = 3r^2$: $$A_2 = \frac{3r^2}{2}\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = 3r^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$$ 7. **Calculate the area of triangle BCD:** - Triangle BCD is equilateral with side length $s = \sqrt{3}r$. - Area: $$A_\triangle = \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(\sqrt{3}r)^2 = \frac{\sqrt{3}}{4} \times 3r^2 = \frac{3\sqrt{3}}{4}r^2$$ 8. **Calculate the shaded area:** - The shaded region is the sum of the two segments minus the triangle: $$A = A_1 + A_2 - A_\triangle$$ - Substitute values: $$A = r^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) + 3r^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) - \frac{3\sqrt{3}}{4}r^2$$ - Factor $r^2$ and terms: $$A = r^2\left[\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) + 3\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) - \frac{3\sqrt{3}}{4}\right]$$ $$= r^2\left[4\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) - \frac{3\sqrt{3}}{4}\right]$$ - Distribute 4: $$= r^2\left(\frac{4\pi}{3} - \sqrt{3} - \frac{3\sqrt{3}}{4}\right)$$ - Combine $-\sqrt{3} - \frac{3\sqrt{3}}{4} = -\frac{4\sqrt{3}}{4} - \frac{3\sqrt{3}}{4} = -\frac{7\sqrt{3}}{4}$: $$A = r^2\left(\frac{4\pi}{3} - \frac{7\sqrt{3}}{4}\right)$$ 9. **Final answer:** $$\boxed{A = \left(\frac{-7\sqrt{3}}{4} + \frac{4\pi}{3}\right) r^2}$$ This matches the form $(a + b\pi)r^2$ with $a = -\frac{7\sqrt{3}}{4}$ and $b = \frac{4}{3}$.