Let's determine the sides and angles of a cyclic quadrilateral whose area is $100 \mathrm{~cm}^{2}$ and whose sides are in the ratio $5: 6: 7: 9$.
The area of a cyclic quadrilateral:

$$
t=\sqrt{(s-a)(s-b)(s-c)(s-d)}, \text { if } s=\frac{a+b+c+d}{2}
$$

Given that

$$
a: b: c: d=5: 6: 7: 9
$$

therefore

$$
b=\frac{6 a}{5}, c=\frac{7 a}{5}, d=\frac{9 a}{5}
$$

thus

$$
s=\frac{27 a}{10}
$$

Substituting this value into the area formula:

$$
a=7.609 \text{ cm}, b=9.131 \text{ cm}, c=10.653 \text{ cm}, d=13.696 \text{ cm}
$$

The angles are derived from the following formula:

$$
\operatorname{tg} \frac{\alpha}{2}=\sqrt{\frac{(s-a)(s-b)}{(s-c)(s-d)}}, \operatorname{tg} \frac{\beta}{2}=\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}, \text { etc. }
$$

from which we get:

$$
\alpha=68^{\circ} 13^{\prime} 34^{\prime \prime}, \beta=96^{\circ} 56^{\prime}, \gamma=111^{\circ} 46^{\prime} 26^{\prime \prime}, \delta=83^{\circ} 41^{\prime}
$$

(Neidenbach Emil, Arad.)

The problem was also solved by: Bartók I., Demjén E., Deutsch E., Deutsch I., Enyedi B., Haar A., Jánosy Gy., Kürti I., Liebner A., Moskovits Zs., Pivnyik I., Popoviciu M., Riesz K., Schuster Gy., Schwemmer J., Sonnenfeld J., Szűcs A.