## Task B-2.2.

In a rectangular hall, there is a carpet in the shape of a rhombus. The area of the carpet is $54 \mathrm{~m}^{2}$, and the perimeter is $30 \mathrm{~m}$. The vertices of the rhombus are exactly at the midpoints of the sides of the rectangle. Determine the dimensions of the hall.
## Solution.

Notice that the dimensions of the hall (the sides of the rectangle $a$ and $b$) are also the diagonals of the rhombus.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-07.jpg?height=411&width=762&top_left_y=1571&top_left_x=567)

If $P=\frac{a \cdot b}{2}=54$, and $o=4 d=30$, then the lengths of the sides are the solution to the following system of equations:

$$
\begin{aligned}
& a b=108 \\
& a^{2}+b^{2}=225
\end{aligned}
$$

If we multiply the first equation by -2 and add it to the second, we get $(a-b)^{2}=9$.

We can assume that $a>b$, so $a=b+3$ or $b^{2}+3 b-108=0$.

This gives $b=9$ and $a=12$.

Therefore, the dimensions of the hall are 12 and 9 meters.