7. The perimeter of a parallelogram is $30 \mathrm{~cm}$. The sum of the areas of the squares constructed on two adjacent sides is $113 \mathrm{~cm}^{2}$. What are the lengths of those sides?

The use of a pocket calculator or any reference materials is not allowed.
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_46538cfe21908c5128afg-7.jpg?height=502&width=502&top_left_y=1388&top_left_x=366)

Sketch or description (Let $a$ be the length of the longer side, and $b$ be the length of the shorter side.)

Then we have $2a + 2b = 30$ or $a + b = 15$

and $a^2 + b^2 = 113$.

1 POINT

Since $(a + b)^2 = a^2 + 2ab + b^2$, after substitution we get $15^2 = 113 + 2ab$ or

$2ab = 225 - 113 = 112$.

2 POINTS

Since $(a - b)^2 = a^2 - 2ab + b^2$, after substitution we get

$(a - b)^2 = (a^2 + b^2) - 2ab = 113 - 112$, or $(a - b)^2 = 1$.

2 POINTS

If $(a - b)^2 = 1$, then (assuming $a > b$ or $a - b > 0$) it must be $a - b = 1$.

The lengths of the sides satisfy the system of equations $\left\{\begin{array}{l}a + b = 15 \\ a - b = 1\end{array}\right.$

1 POINT

By solving the system, we find the solution $a = 8 \text{ cm}$ and $b = 7 \text{ cm}$.

1 POINT

TOTAL 10 POINTS