The points $E$ and $F$ are on the sides $A D$ and $B C$, respectively, of the square $A B C D$. Knowing that $B E=E F=F D=30$, find the area of the square.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-09.jpg?height=422&width=434&top_left_y=2099&top_left_x=914)

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Solution

Let $G$ and $H$ be the feet of the perpendiculars drawn from $E$ and $F$ to the sides $B C$ and $A D$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_01_10107d2fba0ec26bc845g-10.jpg?height=419&width=425&top_left_y=453&top_left_x=724)

Since $A B=C D$ and $B E=F D$, applying the Pythagorean Theorem, we have:

$$
\begin{aligned}
A E & =\sqrt{B E^{2}-A B^{2}} \\
& =\sqrt{F D^{2}-C D^{2}} \\
& =F C
\end{aligned}
$$

Furthermore, since $E F=F D$, again by the Pythagorean Theorem, we can conclude that

$$
\begin{aligned}
E H & =\sqrt{E F^{2}-H F^{2}} \\
& =\sqrt{F D^{2}-H F^{2}} \\
& =H D \\
& =F C
\end{aligned}
$$

Consequently, $A E=E H=H D=x$ and $A B=A D=3 x$. Finally, applying the Pythagorean Theorem to triangle $A E B$, we have

$$
900=B E^{2}=A E^{2}+A B^{2}=x^{2}+9 x^{2}
$$

Thus, the area of the square is $A B^{2}=9 x^{2}=810$.