1. **Problem 9:** Given points C and D on a circle with diameter AB, and BC = AC = 4 cm. (a) Find length AD that maximizes the area of quadrilateral ACBD. (b) When the area is maximal, find the perimeter of ACBD. 2. **Problem 10:** Triangle ABC is inscribed in a semicircle with diameter AC = 12 cm. From B, a height is dropped to AC, meeting at point with distance $x$ from A. (a) Express the area of triangle ABC in terms of $x$. (b) Find $x$ that maximizes the area. (c) When the area is maximal, find the area between the triangle and the semicircle. --- ### Problem 9 1. Since AB is diameter, circle center is midpoint of AB. 2. Given BC = AC = 4 cm, triangle ABC is isosceles with AC = BC. 3. Quadrilateral ACBD is cyclic; area maximized when points are positioned to maximize area. 4. Using properties of cyclic quadrilaterals and given lengths, express AD in terms of known lengths. 5. By Ptolemy's theorem for cyclic quadrilateral ACBD: $$AC \cdot BD = BC \cdot AD + AB \cdot CD$$ 6. Since BC = AC = 4, and AB is diameter, let AB = d. 7. Using the circle and chord properties, find AD that maximizes area. 8. The maximal area occurs when D is such that AD = 4 cm (equal to AC and BC). 9. Thus, $\boxed{AD = 4}$ cm. 10. To find perimeter when area is maximal: - Sides: AC=4, BC=4, AD=4, AB=d. - Since AB is diameter, and AC=4, by Pythagoras in triangle ABC: $$AB = 2 \times AC = 8$$ cm. - Perimeter: $$P = AC + CB + BD + DA = 4 + 4 + BD + 4$$ - BD equals AB (diameter) = 8 cm. - So perimeter: $$P = 4 + 4 + 8 + 4 = 20$$ cm. ### Problem 10 1. Diameter AC = 12 cm, so radius $r=6$ cm. 2. Let $x$ be distance from A to foot of height from B on AC. 3. Height from B to AC is $h$. 4. Since B lies on semicircle, coordinates: - A at 0, - C at 12, - B at $(x,h)$ with $h = \sqrt{r^2 - (x-6)^2} = \sqrt{36 - (x-6)^2}$. 5. Area of triangle ABC: $$S = \frac{1}{2} \times AC \times h = \frac{1}{2} \times 12 \times h = 6h = 6 \sqrt{36 - (x-6)^2}$$ 6. To maximize area, maximize $h$. 7. $h$ is maximal when $(x-6)^2$ is minimal, i.e., $x=6$. 8. Max area: $$S_{max} = 6 \times \sqrt{36 - 0} = 6 \times 6 = 36$$ 9. Area between triangle and semicircle: - Area of semicircle: $$A_{semi} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi 36 = 18\pi$$ - Area difference: $$A_{diff} = A_{semi} - S_{max} = 18\pi - 36$$ --- **Final answers:** - 9a: $AD = 4$ cm - 9b: Perimeter $= 20$ cm - 10a: $S = 6 \sqrt{36 - (x-6)^2}$ - 10b: $x = 6$ cm - 10c: Area difference $= 18\pi - 36$