Question

(geometry question!) use the (figure in comments) circle to answer the question.

Answer 1

$1. \\A_{\triangle{QDE}}= \frac{15\times15\times\sin{72^o}}{2} \approx 107 \\ \\A_{sector}= \frac{15^2\pi\times 72^o}{360^o } \approx 141.37 \\ \\A=141.37-107=34.37 \\ \\2. \\C=2\times 15 \times \pi=30\pi \\ \\3. \\A=15^2\pi=225\pi \\ \\4. \\l= \frac{2\times 15 \pi \times 72^o}{360^o} =6\pi \\ \\5. \\A_{DQ}= \frac{15^2\pi}{2} = \frac{225\pi}{2}$