5. The diagonals of an arbitrary trapezoid divide the trapezoid into four triangles. The areas of the triangles, expressed in $\mathrm{cm}^{2}$, which contain the bases of the trapezoid are $m$ and $n$. Calculate the area of the trapezoid.
Solution. The triangles $A B O$ and $C D O$ are similar with a similarity coefficient of $\sqrt{m}: \sqrt{n}$. If $A B=a, C D=c$ and if $h^{\prime}$ and $h^{\prime \prime}$ are the heights of these triangles from vertex $O$, then

$$
\frac{h^{\prime}}{h^{\prime \prime}}=\frac{a}{c}=\frac{\sqrt{m}}{\sqrt{n}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_3c43981f47282d5136d0g-4.jpg?height=368&width=601&top_left_y=895&top_left_x=860)

From this,

$$
\frac{h^{\prime}+h^{\prime \prime}}{h^{\prime \prime}}=\frac{a+c}{c}=\frac{\sqrt{m}+\sqrt{n}}{\sqrt{n}} \text { and } \frac{\left(h^{\prime}+h^{\prime \prime}\right)(a+c)}{h^{\prime \prime} c}=\frac{(\sqrt{m}+\sqrt{n})^{2}}{n} .
$$

Since $h^{\prime \prime} c=2 n$, the area of the trapezoid is

$$
P=\frac{\left(h^{\prime}+h^{\prime \prime}\right)(a+c)}{2}=\frac{\left(h^{\prime}+h^{\prime \prime}\right)(a+c)}{h^{\prime \prime} c} \cdot \frac{h^{\prime \prime} c}{2}=\frac{(\sqrt{m}+\sqrt{n})^{2}}{n} \cdot n=(\sqrt{m}+\sqrt{n})^{2}
$$