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 points B and D. The length $BD$ is also $s$. We need to show that $s = \sqrt{3} r$ and then find the area of the shaded region in the form $(a + b\pi)r^2$. 2. **Part (a): Show that $s = \sqrt{3} r$** - Since B, C, and D lie on the smaller circle centered at A with radius $r$, we have $AB = AC = AD = r$. - Points B, C, and D form an equilateral triangle on the smaller circle because all are radius $r$ from A. - The larger circle is centered at C with radius $s$ passing through B and D, so $CB = CD = s$. - The length $BD$ is also $s$. 3. **Using triangle properties:** - Triangle BCD is isosceles with sides $CB = CD = s$ and base $BD = s$. - Therefore, triangle BCD is equilateral with all sides equal to $s$. 4. **Find $s$ in terms of $r$:** - Consider triangle ABC with points A, B, and C on the smaller circle. - Since $AB = AC = r$ and $BC$ is a chord, triangle ABC is isosceles. - The length $BC$ is a side of the equilateral triangle BCD with length $s$. 5. **Calculate $BC$ using the Law of Cosines in triangle ABC:** - Angle $BAC$ is $120^\circ$ because B, C, D are equally spaced on the circle (360°/3). - Using Law of Cosines: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(120^\circ)$$ $$= r^2 + r^2 - 2r^2 \times (-\frac{1}{2}) = 2r^2 + r^2 = 3r^2$$ - So, $BC = \sqrt{3} r$. 6. **Since $BC = s$, we have shown:** $$s = \sqrt{3} r$$ 7. **Part (b): Find the area of the shaded region in the form $(a + b\pi)r^2$** - The shaded region is the area inside the larger circle centered at C but outside the smaller circle centered at A. - Area of larger circle: $\pi s^2 = \pi (\sqrt{3} r)^2 = 3 \pi r^2$. - Area of smaller circle: $\pi r^2$. 8. **Find the area of overlap (lens) between the two circles:** - Distance between centers A and C is $AC = r$. - Radii are $r$ and $s = \sqrt{3} r$. 9. **Formula for area of intersection of two circles with radii $R$ and $r$ and center distance $d$:** $$A = r^2 \cos^{-1}\left(\frac{d^2 + r^2 - R^2}{2 d r}\right) + R^2 \cos^{-1}\left(\frac{d^2 + R^2 - r^2}{2 d R}\right) - \frac{1}{2} \sqrt{(-d + r + R)(d + r - R)(d - r + R)(d + r + R)}$$ 10. **Substitute $r = r$, $R = s = \sqrt{3} r$, and $d = r$:** - Compute each term inside the inverse cosines and the square root. 11. **Calculate each term:** - $$\frac{d^2 + r^2 - R^2}{2 d r} = \frac{r^2 + r^2 - 3 r^2}{2 r \times r} = \frac{-r^2}{2 r^2} = -\frac{1}{2}$$ - $$\frac{d^2 + R^2 - r^2}{2 d R} = \frac{r^2 + 3 r^2 - r^2}{2 r \times \sqrt{3} r} = \frac{3 r^2}{2 r^2 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$$ 12. **Calculate the square root term:** - $$(-d + r + R) = (-r + r + \sqrt{3} r) = \sqrt{3} r$$ - $$(d + r - R) = (r + r - \sqrt{3} r) = (2 - \sqrt{3}) r$$ - $$(d - r + R) = (r - r + \sqrt{3} r) = \sqrt{3} r$$ - $$(d + r + R) = (r + r + \sqrt{3} r) = (2 + \sqrt{3}) r$$ - Multiply all: $$\sqrt{(\sqrt{3} r)(2 - \sqrt{3}) r (\sqrt{3} r)(2 + \sqrt{3}) r} = r^2 \sqrt{3 (2 - \sqrt{3}) 3 (2 + \sqrt{3})}$$ - Simplify inside the root: $$3 \times 3 = 9$$ $$(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$$ - So the product inside the root is $9 \times 1 = 9$. - Therefore, the square root term is $r^2 \sqrt{9} = 3 r^2$. 13. **Calculate the area of intersection:** - $$A = r^2 \cos^{-1}(-\frac{1}{2}) + (\sqrt{3} r)^2 \cos^{-1}(\frac{\sqrt{3}}{2}) - \frac{1}{2} \times 3 r^2$$ - $$= r^2 \times \frac{2\pi}{3} + 3 r^2 \times \frac{\pi}{6} - \frac{3}{2} r^2$$ - $$= \frac{2\pi}{3} r^2 + \frac{3\pi}{6} r^2 - \frac{3}{2} r^2 = \frac{2\pi}{3} r^2 + \frac{\pi}{2} r^2 - \frac{3}{2} r^2$$ - Combine the $ \pi$ terms: $$\frac{2\pi}{3} + \frac{\pi}{2} = \frac{4\pi}{6} + \frac{3\pi}{6} = \frac{7\pi}{6}$$ - So, $$A = \frac{7\pi}{6} r^2 - \frac{3}{2} r^2$$ 14. **Find the shaded area:** - Shaded area = Area of larger circle - Area of intersection - $$= 3 \pi r^2 - \left(\frac{7\pi}{6} r^2 - \frac{3}{2} r^2\right) = 3 \pi r^2 - \frac{7\pi}{6} r^2 + \frac{3}{2} r^2$$ - Combine the $ \pi$ terms: $$3 \pi - \frac{7\pi}{6} = \frac{18\pi}{6} - \frac{7\pi}{6} = \frac{11\pi}{6}$$ - So, $$\text{Shaded area} = \frac{11\pi}{6} r^2 + \frac{3}{2} r^2 = \left(\frac{3}{2} + \frac{11\pi}{6}\right) r^2$$ 15. **Final answer:** - $a = \frac{3}{2}$ and $b = \frac{11}{6}$ - The shaded area is: $$\boxed{\left(\frac{3}{2} + \frac{11\pi}{6}\right) r^2}$$