1. Minko and Binko are preparing for an athletic competition, so they are training on a field in the shape of a parallelogram with vertices $A B C D$ and the following data: $S=28 \sqrt{3} \mathrm{~km}^{2}, \beta=120^{\circ}, a+b=15 \mathrm{~km}$. Minko has to run the distance between points $A$ and $C$, while Binko has to run the distance between points $B$ and $D$. By how many kilometers is Minko's route longer?
1. In the formula for the area of a parallelogram $S=a \cdot b \cdot \sin \alpha$, we substitute the given data:

$28 \sqrt{3}=a(15-a) \sin 60^{\circ}$, rearrange the equation to $a^{2}-15 a+56=0$ and solve it. We get the solutions $a=7 \mathrm{~km}$ and $b=8 \mathrm{~km}$. We use the cosine rule to calculate the lengths of the diagonals: $e^{2}=a^{2}+b^{2}-2 a b \cos \beta$ and $f^{2}=a^{2}+b^{2}-2 a b \cos \alpha$. We calculate the lengths of both diagonals $e=13$ $\mathrm{km}$ and $f=7.5 \mathrm{~km}$. Mink's path is thus longer by approximately $5.5 \mathrm{~km}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-09.jpg?height=54&width=1639&top_left_y=1778&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-09.jpg?height=57&width=1636&top_left_y=1828&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-09.jpg?height=55&width=1636&top_left_y=1880&top_left_x=210)
Writing or using the cosine rule $e^{2}=a^{2}+b^{2}-2 a b \cos \beta$ and $f^{2}=a^{2}+b^{2}-2 a b \cos \alpha$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_7bf3e9b12130e254269ag-09.jpg?height=54&width=1639&top_left_y=2029&top_left_x=206)
Answer: Mink's path is thus longer by approximately $5.5 \mathrm{~km}$. 1 point