aera of rectangle

Let the radius of the semicircle be $r$ . Then the height of rectangle $ABCD$ , $AB=CD=r$ and $PC=2r$ . Let the base of rectangle $ABCD$ , $BC=b$ . As $PC$ the diameter of the semicircle, $\angle PQC = 90^\circ$ . Then $\triangle PQC$ and $\triangle QBC$ are similar.

Therefore $\dfrac {BC}{QC} = \dfrac {QC}{PC} \implies \dfrac b {10} = \dfrac {10}{2r} \implies b = \dfrac {50}r$ . The area of rectangle $ABCD$ , $A=br = \dfrac {50}r \times r = \boxed{50}$ .

Let $\angle {QCB}=α$ . Then, since $\angle {PQC}$ is a semicircular angle, therefore $\cos α=\dfrac{10}{2b}$ , where $b=|\overline {CD}|=|\overline {SD}|=$ the radius of the semicircle.

Also, $\cos α=\dfrac{l}{10}$ where $l=|\overline {BC}|$ .

Hence, the required area is $lb=\dfrac{5}{\cos α}\times 10\cos α=\boxed {50}$ .