In the drawing below, $ABCD$ is a rectangle and the points $P$ and $Q$ lie on the diagonal $AC$ such that $AQ = PQ = PC = 1$ and $\angle AQD = \angle BPC = 90^{\circ}$. Find the area of the rectangle $ABCD$.

![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-11.jpg?height=371&width=517&top_left_y=2187&top_left_x=861)
Solution

By the metric relations in the right triangle, we have $D Q^{2}=A Q \cdot Q C=2$. By the Pythagorean Theorem in triangles $\triangle D A Q$ and $\triangle D Q C$, we have:

$$
\begin{aligned}
A D^{2} & =D Q^{2}+A Q^{2} \\
& =2+1 \\
& =3 \\
D C^{2} & =D Q^{2}+Q C^{2} \\
& =2+4 \\
& =6
\end{aligned}
$$

The area of rectangle $A B C D$ is $A D \cdot D C=\sqrt{3} \cdot \sqrt{6}=3 \sqrt{2}$.