1. **Problem statement:** We have two circles with centers A and B each of radius 10 cm. B lies on circle A, so AB = 10 cm. Points C and D are intersections of the circles. E is on circle B so that ABE is a diameter. 2. **Explain why triangle ABC is equilateral:** - AB is the radius of circle A, so $AB=10$ cm. - Since B lies on circle A, $AB=10$ cm. - AC is also a radius of circle A so $AC=10$ cm. - BC is the radius of circle B, so $BC=10$ cm. - All sides of triangle ABC are equal: $AB=BC=CA=10$ cm. - Therefore, triangle ABC is equilateral. 3. **Write down angle CBE in terms of $\pi$:** - Since ABE is a diameter of circle B, angle $CBE$ is an angle in a semicircle. - Angle in a semicircle is a right angle, hence $\angle CBE = \frac{\pi}{2}$. 4. **Find the perimeter of the shaded region:** - The shaded region perimeter consists of arcs $CD$ from both circles and the segment $AC$. - Triangle ABC is equilateral with side 10. - Since $AB=BC=10$, chord $CD$ corresponds to the intersection of the circles. - The angle subtended by chord $CD$ at center A (and B) is $120^\circ$ or $\frac{2\pi}{3}$ radians (because triangle ABC is equilateral, the central angle subtending chord $CD$ is $360^\circ - 120^\circ = 240^\circ$ for major arc, so minor arc is $120^\circ$). - Length of arc $CD$ in each circle is $\text{arc length} = r\times \theta = 10 \times \frac{2\pi}{3} = \frac{20\pi}{3}$ cm. - The shaded perimeter = arc $CD$ from circle A + arc $CD$ from circle B + segment $AC$. - Total perimeter = $\frac{20\pi}{3} + \frac{20\pi}{3} + 10 = \frac{40\pi}{3} + 10$ cm. 5. **Find the area of the shaded region:** - The shaded region is the union of the two circle segments minus triangle ABC. - Area of each circle segment with chord $CD$ and radius 10 and angle $\theta = \frac{2\pi}{3}$ is given by: $$\text{segment area} = \frac{r^{2}}{2}(\theta - \sin \theta) = \frac{10^{2}}{2}\left(\frac{2\pi}{3} - \sin \frac{2\pi}{3}\right) = 50\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)$$ - Total segment areas (both circles) = $$2 \times 50 \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) = 100 \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)$$ - Area of equilateral triangle ABC with side 10: $$\text{area} = \frac{\sqrt{3}}{4} \times 10^{2} = 25\sqrt{3}$$ - The shaded area = sum of segments + triangle area (the shaded looks like lens shape plus triangle? Actually, since triangle is inside the overlapping region, the shaded region is the union of two segments minus the overlapping triangle area, so shaded area = sum of two lens segments + area of triangle ABC (since the shaded includes the triangle seen in the figure.) - However, from the problem wording, the shaded appears as the lens formed by intersecting circles plus triangle ABC. - So shaded area = sum of two segments + area of triangle ABC $$\text{shaded area} = 100 \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) + 25\sqrt{3} = \frac{200\pi}{3} - 50\sqrt{3} + 25\sqrt{3} = \frac{200\pi}{3} - 25\sqrt{3}$$ **Final Answers:** - Triangle ABC is equilateral because $AB=BC=CA=10$ cm. - $\angle CBE = \frac{\pi}{2}$. - Perimeter of shaded region = $\frac{40\pi}{3} + 10$ cm. - Area of shaded region = $\frac{200\pi}{3} - 25\sqrt{3}$ cm$^{2}$.