2. The sum of the diagonals of a rhombus is equal to $8 \mathrm{~cm}$, and its area is equal to $7 \mathrm{~cm}^{2}$. Determine the perimeter of the rhombus?
Solution. Let $a$ be the length of the side, and $d_{1}$ and $d_{2}$ be the lengths of the diagonals of the rhombus. From $d_{1}+d_{2}=8$ and $\frac{d_{1} \cdot d_{2}}{2}=7$, it follows that

$$
64=\left(d_{1}+d_{2}\right)^{2}=d_{1}^{2}+d_{2}^{2}+2 d_{1} d_{2}=d_{1}^{2}+d_{2}^{2}+28
$$

hence $\quad d_{1}^{2}+d_{2}^{2}=36$. Therefore, $a^{2}=\frac{d_{1}^{2}+d_{2}^{2}}{4}=\frac{36}{4}=9$, thus $\quad a=3$. Therefore, the perimeter of the rhombus is $L=4 a=12 \mathrm{~cm}$.