Exercise 3. In the figure below, \( AB = AF = CD = DE = 18 \), \( BC = EF = a \), and \( BF = 24 \). Additionally, the quadrilateral \( BCEF \) is a rectangle.

![](https://cdn.mathpix.com/cropped/2024_06_04_d69c62b301e4ba44b0f0g-2.jpg?height=238&width=512&top_left_y=1949&top_left_x=796)

We assume that the area of the hexagon \( ABCDEF \) is equal to the area of a rectangle whose two consecutive sides have lengths \( a \) and 36. Find \( a^2 \). Justify your answer.

Note. You may use the Pythagorean theorem, which states that if \( XYZ \) is a right triangle at \( X \), then \( XY^2 + XZ^2 = YZ^2 \).
Solution to Exercise 3 Let $H$ be the midpoint of $[B F]$. We have $B H=12$, so $A H^{2}=A B^{2}-B H^{2}=$ $18^{2}-12^{2}=6^{2}\left(3^{2}-2^{2}\right)=5 \times 36$, which gives $A H=6 \sqrt{5}$.

We deduce that the area of $A B F$ is $1 / 2 \times \sqrt{5} \times 6 \times 24$, so the area of the hexagon is equal to $\sqrt{5} \times 6 \times 24+24 \times a=24(6 \sqrt{5}+a)$.

Therefore, $24(6 \sqrt{5}+a)=36 a$ which simplifies to $2(6 \sqrt{5}+a)=3 a$, or again $a=12 \sqrt{5}$. Finally, $a^{2}=720$.

## COMMON EXERCISES