3. The base of a right quadrilateral prism is a rhombus whose diagonals differ by $10 \mathrm{~cm}$. If the larger diagonal is increased by $2 \mathrm{~cm}$, and the smaller one is decreased by $1 \mathrm{~cm}$, the area of the rhombus remains the same.

Calculate the area of the prism if its height is twice the side of the rhombus.
Solution. If $d_{1}$ and $d_{2}$ are the diagonals of the rhombus, then according to the condition of the problem, we have:

$$
d_{1}-d_{2}=10, \quad \frac{d_{1} d_{2}}{2}=\frac{\left(d_{1}+2\right)\left(d_{2}-1\right)}{2}
$$

from which we obtain $d_{1}=22 \mathrm{~cm}, d_{2}=12 \mathrm{~cm}$. The diagonals of any rhombus are mutually perpendicular, so for the side $a$ of the rhombus, we have:

$$
a^{2}=\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}=157, \quad a=\sqrt{157} \mathrm{~cm}
$$

The height of the prism is $H=2 \sqrt{157} \mathrm{~cm}$, so for the surface area $P$ of the prism, we have:

$$
P=2 B+M=2 \frac{d_{1} d_{2}}{2}+4 a H=1520 \mathrm{~cm}^{2}
$$