Task B-1.5. Side $B C$ of the square $A B C D$ is also the leg of the right triangle $B C E$, the length of whose hypotenuse $B E$ is $12 \mathrm{~cm}$. The diagonal of the square $B D$ and the hypotenuse enclose an angle of $75^{\circ}$. Calculate the area of triangle $D B E$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6e5f8ba1616e2c4b87f4g-03.jpg?height=566&width=808&top_left_y=736&top_left_x=629)

Since the diagonal of the square forms a $45^{\circ}$ angle with the side of the square, the angles of the right triangle are $30^{\circ}$ and $60^{\circ}$, and it is half of an equilateral triangle with a side length of $12 \mathrm{~cm}$. (sketch)

Then the length of the leg opposite the $30^{\circ}$ angle is $x=6 \mathrm{~cm}$.

(1 point)

The other leg is also a side of the square and measures $a=\frac{12 \sqrt{3}}{2}=6 \sqrt{3} \mathrm{~cm}$.

$y=a=6 \sqrt{3} \mathrm{~cm}$

$P=\frac{(x+y) \cdot a}{2}$

$P=\frac{(6+6 \sqrt{3}) \cdot 6 \sqrt{3}}{2}$

(1 point)

$P=18 \sqrt{3}+54$

$P=18(3+\sqrt{3}) \mathrm{cm}^{2}$.