Two consecutive sides of a quadrilateral measure 10 and $15 \mathrm{~cm}$. If each diagonal divides the quadrilateral into two regions of the same area, calculate its perimeter.
As each diagonal divides the quadrilateral into two regions of the same area, we have

$$
\text { Area }(\triangle A B D)=\text { Area }(\triangle B C D) \quad \text { and Area }(\triangle A B C)=\text { Area }(\triangle A C D)
$$

Let's denote the areas of the four regions determined by the diagonals by $X, Y, Z$ and $W$, as shown in the figure, so that

Area $(\triangle A B D)=X+W$,

Area $(\triangle B C D)=Y+Z$,

Area $(\triangle A B C)=Z+W$ and

Area $(\triangle A D C)=X+Y$.

Thus,

$$
\begin{aligned}
Z-X & =\text { Area }(\triangle A B C)-\text { Area }(\triangle A B D) \\
& =\text { Area }(\triangle A C D)-\text { Area }(\triangle B C D)=X-Z
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-105.jpg?height=517&width=603&top_left_y=2157&top_left_x=1252)
and, therefore, $Z=X$. Consequently, we also have $Y=W$. Let $E$ be the point of intersection of the diagonals. Since the areas of the regions opposite to $E$ are equal, it follows from the similarity of triangles that $E A \times E D=E B \times E C$ and $E A \times E B=E C \times E D$.

Dividing these two equations, we get

$$
\frac{E D}{E B}=\frac{E B}{E D}
$$

thus, $E D=E B$. Similarly, we can show that $E A=E C$. Therefore, the diagonals intersect at the midpoint and, consequently, the quadrilateral is a parallelogram. Since the sides measure 10 and $15 \text{ cm}$, the perimeter of the quadrilateral is $2 \times 10+2 \times 15=50 \text{ cm}$.