1. **Problem statement:** We have two circles: a smaller circle with center A and radius $r$ passing through points B, C, and D, and a larger circle with center C and radius $s$ passing through B and D. We know $BD = s$. (a) Show that $s = \sqrt{3} r$. 2. **Step 1: Understand the geometry** Points B, C, and D lie on the smaller circle centered at A with radius $r$. So, $AB = AC = AD = r$. 3. **Step 2: Analyze triangle BCD** Since B and D lie on the larger circle centered at C with radius $s$, we have $CB = CD = s$. Also, $BD = s$. 4. **Step 3: Triangle BCD is equilateral** Because $CB = CD = BD = s$, triangle BCD is equilateral. 5. **Step 4: Find length BC in terms of $r$** Since B and C lie on the smaller circle centered at A, $AB = AC = r$. Triangle ABC is isosceles with sides $AB = AC = r$. 6. **Step 5: Use the fact that triangle BCD is equilateral and points B, C, D lie on the smaller circle** The angle $BCD$ in the equilateral triangle is $60^\circ$. 7. **Step 6: Use the Law of Cosines in triangle ABC** We want to find $BC$: $$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC) = r^2 + r^2 - 2r^2 \cos(\angle BAC) = 2r^2(1 - \cos(\angle BAC))$$ 8. **Step 7: Recognize that $\angle BAC = 120^\circ$** Because B, C, D are points on the circle with center A and radius $r$, and triangle BCD is equilateral, the central angle $\angle BAC$ subtended by chord BD is $120^\circ$. 9. **Step 8: Calculate $BC$** $$BC^2 = 2r^2(1 - \cos 120^\circ) = 2r^2(1 - (-\frac{1}{2})) = 2r^2 \times \frac{3}{2} = 3r^2$$ So, $$BC = \sqrt{3} r$$ 10. **Step 9: Since $s = BD = BC = CD$, we have** $$s = \sqrt{3} r$$ --- (b) Find an expression for the area of the shaded region in the form $(a + b\pi)r^2$. 11. **Step 1: Understand the shaded region** The shaded region is the lens shape formed by the intersection of two circles of radius $r$ with centers at A and C, separated by distance $AC = r$. 12. **Step 2: Use the formula for the area of lens (intersection of two circles)** For two circles of radius $r$ with centers separated by distance $d$, the area of their intersection (lens) is: $$\text{Area} = 2r^2 \cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2}$$ 13. **Step 3: Substitute $d = AC = r$** $$\text{Area} = 2r^2 \cos^{-1}\left(\frac{r}{2r}\right) - \frac{r}{2} \sqrt{4r^2 - r^2} = 2r^2 \cos^{-1}\left(\frac{1}{2}\right) - \frac{r}{2} \sqrt{3r^2}$$ 14. **Step 4: Simplify** $$\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$$ $$\sqrt{3r^2} = r\sqrt{3}$$ So, $$\text{Area} = 2r^2 \times \frac{\pi}{3} - \frac{r}{2} \times r \sqrt{3} = \frac{2\pi r^2}{3} - \frac{r^2 \sqrt{3}}{2}$$ 15. **Step 5: Write in the form $(a + b\pi)r^2$** $$\text{Area} = \left(-\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right) r^2$$ 16. **Final answers:** (a) $s = \sqrt{3} r$ (b) Area of shaded region = $\left(-\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right) r^2$