7. Given is a parallelogram $A B C D,|A B|=24 \mathrm{dm},|B C|=16 \mathrm{dm},|\angle B A D|=60^{\circ}$. The bisector of angle $\angle C D A$ intersects the bisector of angle $\angle B C D$ at point $F$, and side $\overline{A B}$ at point $K$. The bisector of angle $\angle A B C$ intersects the bisector of angle $\angle B C D$ at point $G$. Calculate the area of quadrilateral $B G F K$.
7. Since quadrilateral $ABCD$ is a parallelogram and $|\angle BAD|=60^{\circ}$, then $|\angle DCB|=60^{\circ}$, $|\angle CBA|=|\angle ADC|=120^{\circ}$.

The angle bisectors divide the angles, so the image is as follows

![](https://cdn.mathpix.com/cropped/2024_05_30_6e21dae8fd7ad64a6f32g-11.jpg?height=560&width=928&top_left_y=417&top_left_x=427)

Thus, $|\angle BGC|=90^{\circ}$ and $|\angle CFD|=90^{\circ}$, so $|\angle FGB|=90^{\circ}$ and $|\angle KFG|=90^{\circ}$, which means that quadrilateral $BGFK$ is a right trapezoid.

Triangle $\triangle BCG$ is half of an equilateral triangle, so

$|BG|=\frac{|BC|}{2}=8 \text{ dm}$ and $|CG|=\frac{|BC| \sqrt{3}}{2}=8 \sqrt{3} \text{ dm}$.

Triangle $\triangle CDF$ is half of an equilateral triangle, so

$|DF|=\frac{|DC|}{2}=12 \text{ dm}$ and $|CF|=\frac{|DC| \sqrt{3}}{2}=12 \sqrt{3} \text{ dm}$.

Since $|\angle KAD|=|\angle ADK|=60^{\circ}$, then $\triangle AKD$ is equilateral

so $|KD|=|AD|=16 \text{ dm}$.

$1 \text{ POINT}$

Further, $|KF|=|KD|-|DF|=4 \text{ dm}$ and $|FG|=|CF|-|CG|=4 \sqrt{3} \text{ dm}$.

Finally, for the area $P$ of the right trapezoid $BGFK$,

$P=\frac{|BG|+|KF|}{2} \cdot|FG|=24 \sqrt{3} \text{ dm}^2$.

$1 \text{ POINT}$