1. **Problem 10:** Find the area of a regular octagon with apothem length 26. 2. **Formula:** The area $A$ of a regular polygon is given by $$A = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}$$ 3. For a regular octagon, the number of sides $n = 8$. 4. To find the perimeter, we need the side length $s$. The apothem $a = 26$ relates to the side length by $$s = 2a \tan\left(\frac{\pi}{n}\right) = 2 \times 26 \times \tan\left(\frac{\pi}{8}\right)$$ 5. Calculate $\tan(\pi/8)$: $$\tan\left(\frac{\pi}{8}\right) \approx 0.4142$$ 6. So, $$s = 2 \times 26 \times 0.4142 = 21.53$$ 7. Perimeter $P = n \times s = 8 \times 21.53 = 172.24$ 8. Area: $$A = \frac{1}{2} \times 172.24 \times 26 = 2239.12$$ --- 9. **Problem 11:** Find the area of the shaded region inside a circle of radius 11 with an inscribed regular polygon. 10. Since the polygon is inscribed, the radius $r = 11$ is the circumradius. 11. The problem does not specify the polygon type, so we assume the polygon is a regular polygon inscribed in the circle. 12. The area of the shaded region is the area of the circle minus the area of the inscribed polygon. 13. Area of the circle: $$A_{circle} = \pi r^2 = \pi \times 11^2 = 121\pi \approx 380.13$$ 14. Without the number of sides, we cannot find the polygon area exactly, so we assume the polygon is a regular hexagon (common inscribed polygon). 15. For a regular hexagon inscribed in a circle of radius $r$, side length $s = r$. 16. Area of regular hexagon: $$A_{hex} = \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 11^2 = \frac{3\sqrt{3}}{2} \times 121 \approx 314.16$$ 17. Shaded area: $$A_{shaded} = A_{circle} - A_{hex} = 380.13 - 314.16 = 65.97$$ --- 18. **Problem 12:** Find the area of the shaded region inside a circle of radius 20 with an inscribed regular pentagon. 19. Area of the circle: $$A_{circle} = \pi \times 20^2 = 400\pi \approx 1256.64$$ 20. Area of a regular pentagon with circumradius $R$ is $$A = \frac{5}{2} R^2 \sin\left(\frac{2\pi}{5}\right)$$ 21. Calculate $\sin(2\pi/5)$: $$\sin\left(\frac{2\pi}{5}\right) \approx 0.9511$$ 22. So, $$A_{pentagon} = \frac{5}{2} \times 20^2 \times 0.9511 = \frac{5}{2} \times 400 \times 0.9511 = 950.99$$ 23. Shaded area: $$A_{shaded} = A_{circle} - A_{pentagon} = 1256.64 - 950.99 = 305.65$$