1. **Problem statement:** We have two circles: one 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. The length $BD$ is also $s$. We need to show that $s = \sqrt{3} r$ and 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 A is the center of the smaller circle, $AB = AC = AD = r$. - Points B, C, and D lie on the smaller circle, so triangle $BCD$ is inscribed in the circle with radius $r$. - The larger circle has center C and radius $s$, passing through B and D, so $CB = CD = s$. - Given $BD = s$, triangle $BCD$ is isosceles with sides $CB = CD = BD = s$. - Therefore, triangle $BCD$ is equilateral with side length $s$. 3. **Find $s$ in terms of $r$:** - Since $A$ is the center of the smaller circle and $B, C, D$ lie on it, triangle $BCD$ is inscribed in a circle of radius $r$. - The circumradius $R$ of an equilateral triangle with side length $s$ is given by $$R = \frac{s}{\sqrt{3}}.$$ - Here, $R = r$, so $$r = \frac{s}{\sqrt{3}} \implies s = \sqrt{3} r.$$ 4. **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 (center C, radius $s$) but outside the smaller circle (center A, radius $r$). - The area of the larger circle is $$\pi s^2 = \pi (\sqrt{3} r)^2 = 3 \pi r^2.$$ - The area of the smaller circle is $$\pi r^2.$$ - The shaded region is the part of the larger circle outside the smaller circle but inside the overlapping region. 5. **Calculate the overlapping area:** - The distance between centers A and C is $AC = r$ (since $C$ lies on the smaller circle centered at $A$). - The two circles have radii $r$ and $s = \sqrt{3} r$ and center distance $d = r$. - The area of intersection of two circles with radii $r$ and $s$ and center distance $d$ is given by: $$ A = r^2 \cos^{-1}\left(\frac{d^2 + r^2 - s^2}{2 d r}\right) + s^2 \cos^{-1}\left(\frac{d^2 + s^2 - r^2}{2 d s}\right) - \frac{1}{2} \sqrt{(-d + r + s)(d + r - s)(d - r + s)(d + r + s)} $$ - Substitute $r = r$, $s = \sqrt{3} r$, and $d = r$: Calculate each term: - $$\frac{d^2 + r^2 - s^2}{2 d r} = \frac{r^2 + r^2 - 3 r^2}{2 r \cdot r} = \frac{-r^2}{2 r^2} = -\frac{1}{2}.$$ - $$\cos^{-1}(-\frac{1}{2}) = \frac{2 \pi}{3}.$$ - $$\frac{d^2 + s^2 - r^2}{2 d s} = \frac{r^2 + 3 r^2 - r^2}{2 r \cdot \sqrt{3} r} = \frac{3 r^2}{2 r^2 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}.$$ - $$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}.$$ Calculate the square root term: - $$(-d + r + s) = (-r + r + \sqrt{3} r) = \sqrt{3} r,$$ - $$(d + r - s) = (r + r - \sqrt{3} r) = (2 - \sqrt{3}) r,$$ - $$(d - r + s) = (r - r + \sqrt{3} r) = \sqrt{3} r,$$ - $$(d + r + s) = (r + r + \sqrt{3} r) = (2 + \sqrt{3}) r.$$ Multiply: $$ \sqrt{(\sqrt{3} r)(2 - \sqrt{3}) r (\sqrt{3} r)(2 + \sqrt{3}) r} = r^2 \sqrt{3 (2 - \sqrt{3}) \cdot \sqrt{3} (2 + \sqrt{3})}. $$ Simplify inside the root: - $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$ - So the product inside the root is $3 \times \sqrt{3} \times 1 = 3 \sqrt{3}$ Therefore: $$ \sqrt{3 \sqrt{3}} = \sqrt{3} \cdot 3^{1/4} = 3^{3/4}. $$ But to keep it simple, note that the product is $3 \sqrt{3}$, so the square root is $\sqrt{3 \sqrt{3}}$. Alternatively, calculate numerically: - $3 \sqrt{3} \approx 3 \times 1.732 = 5.196$ - $\sqrt{5.196} \approx 2.28$ But since we want exact form, keep as $r^2 \sqrt{3 \sqrt{3}}$. 6. **Calculate the intersection area:** $$ A = r^2 \cdot \frac{2 \pi}{3} + 3 r^2 \cdot \frac{\pi}{6} - \frac{1}{2} r^2 \sqrt{3 \sqrt{3}} = r^2 \left(\frac{2 \pi}{3} + \frac{3 \pi}{6} - \frac{1}{2} \sqrt{3 \sqrt{3}}\right). $$ Simplify the $ \pi$ terms: $$ \frac{2 \pi}{3} + \frac{3 \pi}{6} = \frac{2 \pi}{3} + \frac{\pi}{2} = \frac{4 \pi}{6} + \frac{3 \pi}{6} = \frac{7 \pi}{6}. $$ So: $$ A = r^2 \left(\frac{7 \pi}{6} - \frac{1}{2} \sqrt{3 \sqrt{3}}\right). $$ 7. **Find the shaded area:** - The shaded region is the larger circle area minus the intersection area: $$ \text{Shaded area} = 3 \pi r^2 - A = 3 \pi r^2 - r^2 \left(\frac{7 \pi}{6} - \frac{1}{2} \sqrt{3 \sqrt{3}}\right) = r^2 \left(3 \pi - \frac{7 \pi}{6} + \frac{1}{2} \sqrt{3 \sqrt{3}}\right). $$ Simplify the $ \pi$ terms: $$ 3 \pi - \frac{7 \pi}{6} = \frac{18 \pi}{6} - \frac{7 \pi}{6} = \frac{11 \pi}{6}. $$ So the shaded area is: $$ r^2 \left(\frac{11 \pi}{6} + \frac{1}{2} \sqrt{3 \sqrt{3}}\right). $$ 8. **Final answer:** - The shaded area is in the form $(a + b \pi) r^2$ where: - $a = \frac{1}{2} \sqrt{3 \sqrt{3}}$ - $b = \frac{11}{6}$ --- **Summary:** - (a) $s = \sqrt{3} r$ - (b) Shaded area $= \left(\frac{1}{2} \sqrt{3 \sqrt{3}} + \frac{11 \pi}{6}\right) r^2$