c=2πr a=πr² common thing is r hmm, solve for r in circumfernce c=2π \(\frac{c}{2 \pi}\)=r sub that for r in other a=πr² a=π \(( \frac{c}{2 \pi})^2\) a=\(\frac{c^2}{4\pi}\)
so c=8π a=\(\frac{(8\pi)^2}{4\pi}\) a=\(\frac{64\pi^2}{4\pi}\) a=16π square inches
