A wire of $32\,\mathrm{cm}$ in length is bent into the figure given below. $APD$ is a semicircle and $AB=BC=CD$. Find the radius and area of the circle.
Answer: Radius${}=3.5\mathrm{cm}$ and Area${}=68.25\mathrm{cm}^2$
I tried using $\pi r+ 2r$ to find the perimeter of the semicircle and $4r^2$ for the three sides of the square (dotted line is not counted in perimeter) and got $\pi r+ 2r + 4r^2= 32$, but I'm not getting the answer. Please help! Let me know if my logic is wrong.
Hint: if the radius is $\require{color}r$, then the perimeter is $\pi r+\textcolor{red}{6}r=32$.