What is the area of a quadrilateral whose sides are: $a=52 \mathrm{~m}, b=56 \mathrm{~m}, c=33 \mathrm{~m}, d=39 \mathrm{~m}$ and one of its angles $\alpha=112^{\circ} 37^{\prime} 12^{\prime \prime} ?$
Let $ABCD$ be the quadrilateral, $AB=a, BC=b, CD=c, DA=d, \angle BAD=\alpha$ and $BD=e$. Then the area of the quadrilateral is:

$$
t=\frac{ad \sin \alpha}{2}+\sqrt{s(s-b)(s-c)(s-e)}
$$

where $e$ can be determined from the equation

$$
e^{2}=a^{2}+d^{2}-2ad \cos \alpha
$$

Substituting the given values,

$$
t=1774 \text{ m}^{2}
$$

(Brichta Lajos, Nyitra.)

Note. If $\angle ADC=\alpha$, then $t=1938 \text{ cm}^{2}$; if $\angle BCD=\alpha$, then $t=1848 \text{ cm}^{2}$. Finally, if $\angle ABC=\alpha$, then the problem is impossible.