In the symmetric trapezoid $ABCD$, the diagonals $AC$ and $BD$ intersect at point $O$. The area of $\triangle AOB$ is $52 \, \text{m}^2$, and the area of $\triangle COD$ is $117 \, \text{m}^2$. Calculate the area of the trapezoid!
I. Solution. Let $t_{1}$ be the area of $\triangle AOB$, $t_{2}$ the area of $\triangle COD$, and $x$ the area of $\triangle AOD$ and $\triangle BOC$; the latter areas are equal due to their congruence.

The bases $DO$ and $OB$ of $\triangle AOD$ and $\triangle AOB$ lie on the same line, and their vertex $A$ is common; therefore, their heights are equal. The ratio of their areas is thus the same as the ratio of the bases $DO$ and $OB$, that is,

$$
x: t_{1} = DO: OB \ldots
$$

Similarly, for $\triangle BOC$ and $\triangle BOD$,

$$
x: t_{2} = OB: DO \ldots
$$

Multiplying the corresponding terms of (1) and (2):

$$
x^{2}: t_{1} t_{2} = DO \cdot OB: OB \cdot DO, \quad \text{thus} \quad x^{2} = t_{1} t_{2}
$$

(In other words: the area of $\triangle AOD$ and $\triangle BOC$ is the geometric mean of the areas of $\triangle AOB$ and $\triangle COD$). Based on this, the area of the trapezoid is

$$
\begin{gathered}
t = t_{1} + t_{2} + 2x = t_{1} + t_{2} + 2\sqrt{t_{1} t_{2}} = \left(\sqrt{t_{1}} + \sqrt{t_{2}}\right)^{2} \\
t = 52 + 117 + 2\sqrt{4 \cdot 13 \cdot 9 \cdot 13} = 52 + 117 + 2 \cdot 2 \cdot 3 \cdot 13 = 325 \text{ m}^{2}
\end{gathered}
$$

György Holló (Dobó István g. VI. o. Eger)

II. Solution. Let $AB = a$, $CD = b$ be the segments of the height of the trapezoid $m_{1}, m_{2}$. Since $\triangle AOB \sim \triangle COD$,

$$
m_{1}: m_{2} = a: b \quad \text{and} \quad t_{1}: t_{2} = a^{2}: b^{2}
$$

Therefore,

$$
\left(\frac{a}{b}\right)^{2} = \frac{t_{1}}{t_{2}} = \frac{52}{117} = \frac{4}{9}
$$

Thus,

$$
\frac{a}{b} = \frac{m_{1}}{m_{2}} = \frac{2}{3} \quad \text{or} \quad b = \frac{3a}{2} \quad \text{and} \quad m_{2} = \frac{3m_{1}}{2}
$$

The area of the trapezoid is

$$
\begin{aligned}
& t = \frac{1}{2}(a + b)(m_{1} + m_{2}) - \frac{1}{2}\left(a + \frac{3a}{2}\right)\left(m_{1} + \frac{3m_{1}}{2}\right) = \frac{25am_{1}}{8} = \frac{25 \cdot 2t_{1}}{8} \\
& t = \frac{25 \cdot 104}{8} = 326 \text{ m}^{2} \\
& \text{Gyula Kalcsó (Balassi Bálint g. VI. o. Balassagyarmat)}
\end{aligned}
$$