5. Square $A B C D$ has an area of $80 \mathrm{~cm}^{2}$. Points $E, F, G$ and $H$ are located on the sides of the square and it is given that $|A E|=|B F|=|C G|=|D H|$. If $|B E|=3|E A|$, what is the area of the shaded part?

![](https://cdn.mathpix.com/cropped/2024_06_03_84e4bc15d0ba66a9883eg-1.jpg?height=414&width=420&top_left_y=2009&top_left_x=858)

Each problem is scored out of 10 points.

The use of a pocket calculator or any reference materials is not allowed.

## COUNTY COMPETITION IN MATHEMATICS
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_32e0003636127c6bb5b9g-07.jpg?height=400&width=406&top_left_y=311&top_left_x=534)

Let $a$ be the length of the side of the square.

Then, $|A E|=|B F|=|C G|=|D H|=\frac{1}{4} a, |B E|=|C F|=|D G|=|A H|=\frac{3}{4} a$.

1 POINT

Triangles $\triangle H A E, \triangle E B F, \triangle F C G$, and $\triangle G D H$ are congruent (right triangles with congruent sides of lengths $\frac{1}{4} a$ and $\frac{3}{4} a$, according to the S-S-S congruence theorem).

From the congruence of these triangles, it follows that $|F E|=|G F|=|H G|=|E H|=x$.

1 POINT

Consider triangle $\triangle H A E$.

Let $|\measuredangle A E H|=\alpha$ and $|\measuredangle E H A|=\beta$. It holds that $\alpha+\beta=90^{\circ}$.

Due to the congruence of triangles $\triangle H A E$ and $\triangle E B F$, we have $|\measuredangle B E F|=\beta$.

Therefore, $|\measuredangle H E F|=180^{\circ}-(\alpha+\beta)=180^{\circ}-90^{\circ}=90^{\circ}$.

Similarly, it can be shown that $|\measuredangle E F G|=|\measuredangle F G H|=|\measuredangle G H E|=90^{\circ}$.

2 POINTS

Quadrilateral $E F G H$ is a square because it has all sides of equal length and all angles are right angles.

1 POINT

The shaded area is half of the square $E F G H$.

Applying the Pythagorean theorem to triangle $\triangle H A E$ gives $\left(\frac{1}{4} a\right)^{2}+\left(\frac{3}{4} a\right)^{2}=x^{2}$. \quad 1 POINT

After squaring and adding, we get $x^{2}=\frac{5}{8} a^{2}$.

1 POINT

Since $a^{2}=80$ (the area of the given square), it follows that $x^{2}=\frac{5}{8} \cdot 80=50$.

1 POINT

The area of square $E F G H$ is $50 \mathrm{~cm}^{2}$, and the shaded area is $25 \mathrm{~cm}^{2}$.

1 POINT

Note: A solution without proving that quadrilateral $E F G H$ is a square is scored with a maximum of 6 points.