7. Let $A B C D$ be a trapezoid with bases $A B$ and $C D$ inscribed in a circle $\Gamma$, such that the diagonals $A C$ and $B D$ are perpendicular. Let $P$ be the point of intersection of the diagonals $A C$ and $B D$. What is the ratio between the area of $\Gamma$ and the sum of the areas of triangles $A P B$ and $C P D$?
(A) 1
(B) $\pi / 2$
(C) $\pi$
(D) $2 \pi$
(E) The data is not sufficient to determine it
7. The answer is $\mathbf{( C )}$. Let $O$ and $r$ be the center and the radius of the circle $\Gamma$, respectively. Note that the trapezoid is necessarily isosceles (since it is inscribed in a circle), so the diagonals are equal and $A P=P B$ (thus also $P C=P D$). Therefore, triangles $A P B$ and $B P C$ are isosceles and right-angled: their areas are $\frac{B P^{2}}{2}$ and $\frac{P C^{2}}{2}$, respectively.

Moreover, since triangle $A P B$ is isosceles and right-angled, we find $45^{\circ}=\widehat{P A B}=\widehat{C A B}=\frac{\widehat{C O B}}{2}$ : that is, $\widehat{C O B}=90^{\circ}$. Using the Pythagorean theorem on triangles $B P C$ and $B O C$, we find that: $2 r^{2}=B C^{2}=B P^{2}+P C^{2}$.

The sum of the areas of triangles $A P B$ and $B P C$ is $\frac{B P^{2}+P C^{2}}{2}=r^{2}$, and the desired ratio is $\frac{\pi r^{2}}{r^{2}}=\pi$.