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. The apothem $a = 26$. 5. To find the perimeter $P$, we need the side length $s$. The apothem relates to the side length by $$a = \frac{s}{2 \tan(\pi/n)}$$ 6. Rearranging to find $s$: $$s = 2a \tan\left(\frac{\pi}{n}\right) = 2 \times 26 \times \tan\left(\frac{\pi}{8}\right)$$ 7. Calculate $\tan(\pi/8)$: $$\tan\left(\frac{\pi}{8}\right) = \tan(22.5^\circ) \approx 0.4142$$ 8. So, $$s = 2 \times 26 \times 0.4142 = 52 \times 0.4142 = 21.53$$ 9. The perimeter is $$P = n \times s = 8 \times 21.53 = 172.24$$ 10. Now calculate the area: $$A = \frac{1}{2} \times 172.24 \times 26 = 86.12 \times 26 = 2239.12$$ --- 11. **Problem 11:** Find the area of the shaded region inside a circle of radius 11 with an inscribed regular polygon. 12. Since the polygon is inscribed, the radius $r = 11$ is the circumradius. 13. The problem does not specify the polygon type, so assume a regular polygon with $n$ sides (not given). Without $n$, we cannot find exact area. 14. If the polygon is regular with $n$ sides, the area is $$A = \frac{1}{2} n r^2 \sin\left(\frac{2\pi}{n}\right)$$ 15. Without $n$, the problem is incomplete. Assuming the polygon is a regular hexagon (common), $n=6$: 16. Calculate $$A = \frac{1}{2} \times 6 \times 11^2 \times \sin\left(\frac{2\pi}{6}\right) = 3 \times 121 \times \sin(60^\circ)$$ 17. Since $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$: 18. So, $$A = 3 \times 121 \times 0.866 = 363 \times 0.866 = 314.36$$ --- 19. **Problem 12:** Find the area of the shaded region inside a circle of radius 20 with an inscribed regular pentagon. 20. Given $r = 20$ and $n = 5$. 21. Use the formula for area of a regular polygon inscribed in a circle: $$A = \frac{1}{2} n r^2 \sin\left(\frac{2\pi}{n}\right)$$ 22. Calculate $$A = \frac{1}{2} \times 5 \times 20^2 \times \sin\left(\frac{2\pi}{5}\right) = \frac{5}{2} \times 400 \times \sin(72^\circ)$$ 23. $\sin(72^\circ) \approx 0.9511$ 24. So, $$A = 2.5 \times 400 \times 0.9511 = 1000 \times 0.9511 = 951.06$$ **Final answers:** - Problem 10 area: $2239.12$ - Problem 11 area (assuming hexagon): $314.36$ - Problem 12 area: $951.06$