1. **Problem 19:** Find the area of the shaded region inside a circle with radius 6. 2. The formula for the area of a circle is $$A = \pi r^2$$ where $r$ is the radius. 3. Substitute $r = 6$ into the formula: $$A = \pi \times 6^2$$ 4. Calculate the square of 6: $$6^2 = 36$$ 5. So the area is: $$A = \pi \times 36 = 36\pi$$ 6. The shaded region fills the entire circle, so the area of the shaded region is $36\pi$. 1. **Problem 21:** Circles X and Y have radii 6 and 2 respectively and are tangent to each other. $\overline{AB}$ is a common external tangent. Find the area of the shaded triangular region bounded by $\overline{AB}$ and the points of tangency. 2. The distance between the centers of the two circles is the sum of their radii: $$d = 6 + 2 = 8$$ 3. The length of the common external tangent segment $\overline{AB}$ can be found using the formula for the length of the external tangent between two circles: $$AB = \sqrt{d^2 - (r_1 - r_2)^2}$$ where $r_1 = 6$ and $r_2 = 2$. 4. Calculate the difference of the radii: $$r_1 - r_2 = 6 - 2 = 4$$ 5. Substitute values into the formula: $$AB = \sqrt{8^2 - 4^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3}$$ 6. The shaded region is a triangle formed by the tangent segment $AB$ and the two radii drawn to the points of tangency. The height of this triangle is the difference in radii, $4$. 7. The area of the triangle is: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\sqrt{3} \times 4 = 8\sqrt{3}$$ **Final answers:** - Problem 19: The area of the shaded region is $36\pi$. - Problem 21: The area of the shaded triangular region is $8\sqrt{3}$.