3. Given is a rectangle $A B C D$. For the square $D E F G$, it is true that point $D$ is a vertex of the rectangle, point $E$ lies on the segment $\overline{A B}$, point $F$ lies on the segment $\overline{B C}$, and $|\angle B E F|=30^{\circ}$. If the area of the square $D E F G$ is $36 \mathrm{~cm}^{2}$, calculate the perimeter and the area of the intersection of the rectangle $A B C D$ and the square $D E F G$.
3. Given is a rectangle $A B C D$. For the square $D E F G$, it is given that point $D$ is a vertex of the rectangle, point $E$ lies on the segment $\overline{A B}$, point $F$ lies on the segment $\overline{B C}$, and $|\angle B E F|=30^{\circ}$. If the area of the square $D E F G$ is $36 \mathrm{~cm}^{2}$, calculate the perimeter and the area of the intersection of the rectangle $A B C D$ and the square $D E F G$.

## Solution.

Point $H$ is the intersection of segments $\overline{D C}$ and $\overline{F G}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5715e7b36c946fa665d7g-34.jpg?height=457&width=462&top_left_y=1185&top_left_x=1208)

The area of the square is $36 \mathrm{~cm}^{2}$, so the length of the side of the square is $6 \mathrm{~cm}$.

1 POINT

$|\angle B E F|=30^{\circ}$ and $|\angle F E D|=90^{\circ}$, so $|\angle D E A|=60^{\circ}$.

$|\angle B E F|=30^{\circ}$ and $|\angle F B E|=90^{\circ}$, so $|\angle E F B|=60^{\circ}$.

$|\angle D E A|=60^{\circ}$ and $|\angle E A D|=90^{\circ}$, so $|\angle A D E|=30^{\circ}$.

Triangles $\triangle A E D$ and $\triangle E B F$ are halves of an equilateral triangle where the hypotenuse is twice the length of the shorter leg.

1 POINT

Since $|D E|=|E F|=6$, it follows that $|A E|=|B F|=3$.

1 POINT

Let $|A D|=|E B|=x$.

By applying the Pythagorean theorem, we get:

$x^{2}+3^{2}=6^{2}$

$x=3 \sqrt{3}$

Thus, $|A D|=|E B|=3 \sqrt{3}$.

1 POINT

$|\angle E F B|=60^{\circ}$ and $|\angle G F E|=90^{\circ}$, so $|\angle C F H|=30^{\circ}$.

$|\angle C F H|=30^{\circ}$ and $|\angle H C F|=90^{\circ}$, so $|\angle F H C|=60^{\circ}$.

Triangle $\triangle H F C$ is also half of an isosceles triangle.

1 POINT

$|F C|=|B C|-|B F|=3 \sqrt{3}-3$.

Let $|H F|=2|C H|=2 y$.

From the Pythagorean theorem, we get $y^{2}+(3 \sqrt{3}-3)^{2}=4 y^{2}$, or $(3 \sqrt{3}-3)^{2}=3 y^{2}$, so

$y^{2}=\frac{(3 \sqrt{3}-3)^{2}}{3}$

and further

$y=\frac{3 \sqrt{3}-3}{\sqrt{3}}=3-\sqrt{3}$.

1 POINT

Therefore,

$|H F|=2 y=6-2 \sqrt{3}$ and

1 POINT

$|D H|=|D C|-|H C|=3+3 \sqrt{3}-(3-\sqrt{3})=4 \sqrt{3}$.

1 POINT

The intersection of the rectangle and the square is the trapezoid $D E F H$.

The perimeter is:

$o=|D E|+|E F|+|H F|+|D H|=6+6+6-2 \sqrt{3}+4 \sqrt{3}=(18+2 \sqrt{3}) \mathrm{cm}$.

1 POINT

The area can be calculated by subtracting the areas of two congruent right triangles $\triangle A E D$ and $\triangle E B F$ and the area of the right triangle $\triangle F C H$ from the area of the rectangle $A B C D$:

$$
\begin{aligned}
P=(3 & +3 \sqrt{3}) \cdot 3 \sqrt{3}-2 \cdot \frac{3 \cdot 3 \sqrt{3}}{2}-\frac{(3-\sqrt{3})(3 \sqrt{3}-3)}{2} \\
& =9 \sqrt{3}+27-9 \sqrt{3}-\frac{9 \sqrt{3}-9-9+3 \sqrt{3}}{2} \\
& =27-\frac{12 \sqrt{3}-18}{2} \\
& =27-6 \sqrt{3}+9 \\
& =(36-6 \sqrt{3}) \mathrm{cm}^{2}
\end{aligned}
$$

1 POINT

(Alternatively, the area can be calculated using the formula for the area of a trapezoid:

$P=\frac{|D E|+|H F|}{2} \cdot|E F|=\frac{6+6-2 \sqrt{3}}{2} \cdot 6=(36-6 \sqrt{3}) \mathrm{cm}^{2}$.

TOTAL 10 POINTS