[tex]\begin{gathered} \text{Given C=16}\pi \\ \text{But C= 2}\pi r \\ \text{where C= circumference and r =radius} \end{gathered}[/tex][tex]\begin{gathered} r=\frac{C}{2\pi} \\ r=\frac{16\pi}{2\pi} \\ r=8 \end{gathered}[/tex]
Then we can proceed to find Area A;
[tex]\begin{gathered} A=\pi\times r^2 \\ A=\pi\times8^2 \\ A=64\pi \end{gathered}[/tex]
The area of the circle is 64pie