4. The base of a right quadrilateral prism is a rhombus whose diagonals differ by $14 \mathrm{~cm}$. If the smaller diagonal is increased by 2, and the larger diagonal is decreased by 4, the area of the rhombus remains the same. Calculate the surface area of the prism if its height is twice the side of the rhombus.
Solution. Let the diagonals of the rhombus be denoted by $d_{1}$ and $d_{2}, d_{1}>d_{2}$. From the conditions of the problem, we have:

$$
d_{1}-d_{2}=14, \frac{d_{1} d_{2}}{2}=\frac{\left(d_{1}-4\right)\left(d_{2}+2\right)}{2}
$$

from which we obtain $d_{1}=24, d_{2}=10$.

Since the diagonals of any rhombus are mutually perpendicular (draw a diagram), it follows that

$$
a^{2}=\left(\frac{d_{1}}{2}\right)^{2}+\left(\frac{d_{2}}{2}\right)^{2}=12^{2}+5^{2}=169, \text { i.e. } a=13
$$

Finally, for the surface area $P$ of the prism, we will have

$$
P=2 \frac{d_{1} d_{2}}{2}+4 a H=1592
$$

## 2nd year