In the diagram, the circle has radius $\sqrt{5}$. Rectangle $A B C D$ has $C$ and $D$ on the circle, $A$ and $B$ outside the circle, and $A B$ tangent to the circle. What is the area of $A B C D$ if $A B=4 A D$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_b6707dc71c8b7ccd30afg-3.jpg?height=444&width=463&top_left_y=1499&top_left_x=885)
Label the centre of the circle by $O$, the point of tangency of $A B$ with the circle by $E$, and the point of intersection of $O E$ with $C D$ by $F$ :

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-13.jpg?height=439&width=455&top_left_y=2187&top_left_x=889)

Let $E F=x$. Then $A D=x$, so $A B=4 x$, which means $C D=4 x$ because $A B C D$ is a rectangle.

Using circle properties, $O E$ is perpendicular to $A B$, and since $A B$ is parallel to $C D$, we also have that $O E$, and hence, $O F$, is perpendicular to $C D$.

We now have that $\triangle O F D$ is right-angled at $F$, and $\triangle O F C$ is also right-angled at $F$.

Since $O C$ and $O D$ are radii of the same circle, they must be equal.

Triangles $O F D$ and $O F C$ also share side $O F$, so they are congruent by hypotenuse-side similarity.

Therefore, $F$ is the midpoint of $C D$, so $D F=2 x$.

It is given that the radius of the circle is $\sqrt{5}$, which implies that $O F=\sqrt{5}-x$ and $O D=\sqrt{5}$. By the Pythagorean theorem, $O F^{2}+D F^{2}=O D^{2}$, and after substituting the lengths computed earlier, this leads to the following equivalent equations:

$$
\begin{aligned}
(\sqrt{5}-x)^{2}+(2 x)^{2} & =\sqrt{5}^{2} \\
5-2 \sqrt{5} x+x^{2}+4 x^{2} & =5 \\
5 x^{2}-2 \sqrt{5} x & =0 \\
x(5 x-2 \sqrt{5}) & =0
\end{aligned}
$$

this implies that either $x=0$ or $5 x=2 \sqrt{5}$.

We cannot have $x=0$ since $x$ is the side-length of a rectangle. Therefore, $5 x=2 \sqrt{5}$ or $x=\frac{2}{\sqrt{5}}$.

The area of $A B C D$ is $x(4 x)=\frac{2}{\sqrt{5}} \times \frac{8}{\sqrt{5}}=\frac{16}{5}$.

ANSWER: $\frac{16}{5}$

#