Hard Geometry Problem

The perimeter of the quarter-circle is 12+3π, and to get the perimeter of the shaded area, you subtract half of the rectangle's perimeter and add 6. 12+3π-8+6=10+3π.

Radius of the circle is $r=6$ . Let the length and the width of the rectangle be $l$ and $w$ respectively. Then, since the triangle ARC is right angled, therefore $l^2+w^2=r^2=36$ and $l+w=8$ . Solving we get $l=4+√2$ and $w=4-√2$ . Therefore the perimeter required is $(6-4-√2)+6+(6-4+√2)+\dfrac{6π}{2}=10+(3*π)$

Let the length of the rectangle be $\ell$ and the width be $w$ . The formula for the perimeter ( $P$ )is basically just $\begin{aligned}P&=\text{arc}ABC+\overline{SA}+\overline{CT}+\overline{AC} \\&= \frac 14(12\pi)+(6-w)+\sqrt{\ell^2+w^2}+(6-\ell) \\ &= 3\pi+12-w-\ell+\sqrt{\ell^2+w^2}. \end{aligned}$ This may be looking ugly at first, but now we have to notice a few things:

• $AC\cong RB=\text{radius}=6$

• $-w-\ell = -(w+\ell)=-8$ .

Thus, we have our answer as $3\pi+12+6-8=\boxed{10+3\pi}.$