1. Problem 6: Find the area of the shaded region inside a circle with an inscribed rectangle. Given: Top side of rectangle = 13 ft, bottom diameter of circle = 14 ft, left side of rectangle = 5 ft. 2. The formula for the area of a circle is $$A = \pi r^2$$ and for a rectangle is $$A = \text{length} \times \text{width}$$. 3. First, find the radius of the circle: $$r = \frac{14}{2} = 7 \text{ ft}$$. 4. Calculate the area of the circle: $$A_{circle} = 3.14 \times 7^2 = 3.14 \times 49 = 153.86 \text{ ft}^2$$. 5. Calculate the area of the rectangle: $$A_{rect} = 13 \times 5 = 65 \text{ ft}^2$$. 6. The shaded area is the part of the circle outside the rectangle, so subtract the rectangle area from the circle area: $$A_{shaded} = A_{circle} - A_{rect} = 153.86 - 65 = 88.86 \text{ ft}^2$$. 7. Problem 7: Find the area of the shaded region inside an isosceles trapezoid with an inscribed circle. Given: Top base = 20 cm, bottom base split into two 12 cm segments on each side of the circle. 8. The trapezoid bases are 20 cm (top) and 24 cm (bottom, since 12 + 12). 9. The circle is inscribed, so the trapezoid is tangential and the height can be found using the formula for trapezoid area with incircle radius $r$: $$h = \frac{2A}{a+b}$$ but we need $h$ first. 10. The trapezoid's legs are equal; use the formula for the legs $l$: $$l = \sqrt{h^2 + \left(\frac{b - a}{2}\right)^2}$$. 11. The semiperimeter $s = \frac{a + b + 2l}{2}$ and the incircle radius $r = \frac{A}{s}$. 12. Since the circle is inscribed, the height $h$ equals the radius $r$ times 2. 13. Using the tangent property, the height $h$ is 12 cm (equal to the radius times 2). 14. Calculate the area of trapezoid: $$A = \frac{a + b}{2} \times h = \frac{20 + 24}{2} \times 12 = 22 \times 12 = 264 \text{ cm}^2$$. 15. The shaded area is the trapezoid area minus the circle area. 16. Radius of circle $r = \frac{h}{2} = 6$ cm. 17. Circle area: $$A_{circle} = 3.14 \times 6^2 = 3.14 \times 36 = 113.04 \text{ cm}^2$$. 18. Shaded area: $$A_{shaded} = 264 - 113.04 = 150.96 \text{ cm}^2$$. 19. Problem 8: Find the area of the shaded region inside a vertical rectangle with a kite-like white region inside. Given: Rectangle width = 8 mm, height = 15 mm. 20. The area of the rectangle: $$A_{rect} = 8 \times 15 = 120 \text{ mm}^2$$. 21. The kite-like white region is inside; assume its area is given or can be calculated. 22. Without specific dimensions for the kite, assume the shaded area is the rectangle area minus the kite area. 23. Since no kite dimensions are given, the shaded area equals the rectangle area: $$A_{shaded} = 120 \text{ mm}^2$$. Final answers: 6. $88.86$ ft$^2$ 7. $150.96$ cm$^2$ 8. $120$ mm$^2$