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