A rhombus has an area of $266 \mathrm{dm}^{2}$, and the sum of its diagonals is $47 \mathrm{dm}$. Determine the sides and angles of the rhombus.
Let the side of the rhombus be denoted by $a$, the diagonals by $d_{1}$ and $d_{2}$, and the angle opposite to the diagonal $d_{1}$ by $\alpha$. According to the problem,

$$
d_{1}+d_{2}=47
$$

and

$$
d_{1} d_{2}=2 \cdot 266
$$

from which $d_{1}=19 \mathrm{dm}$ and $d_{2}=28 \mathrm{dm}$. But

$$
a=\sqrt{\frac{d_{1}^{2}}{4}+\frac{d_{2}^{2}}{4}}=16.918 \mathrm{dm}
$$

and

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{19}{28}
$$

from which

$$
\alpha=68^{\circ} 19^{\prime} 6.8^{\prime \prime}
$$

(József Steiger, Eger.)

Number of solutions: 22.