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

## List 8
If each diagonal divides the quadrilateral into two regions of the same area, we have:

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

But,

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

![](https://cdn.mathpix.com/cropped/2024_05_01_eead07372ad548e635bag-30.jpg?height=512&width=593&top_left_y=286&top_left_x=1010)

Thus,

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

and therefore, \( Z=X \). Consequently, we also have \( Y=W \).

Since the opposite areas are equal, it follows from the similarity of triangles that:

$$
A E \times E D=B E \times E C \text { and } A E \times E B=C E \times E D
$$

Dividing these two equations, we get:

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

Similarly, we can show that \( E A=E C \). Therefore, the diagonals intersect at the midpoint, and consequently, the quadrilateral is a parallelogram, where the perimeter is \( 2 \times 10 + 2 \times 15 = 50 \text{ cm} \).

## List 8