In the trapezoid $K L M N$, the base $K L$ has a length of $40 \mathrm{~cm}$ and the base $M N$ has a length of $16 \mathrm{~cm}$. Point $P$ lies on the segment $K L$ such that the segment $N P$ divides the trapezoid into two parts with equal areas.

Determine the length of the segment $K P$.

(L. Hozová)

Hint. Modify the shapes so that their areas can be more easily compared.
Segment $N P$ divides trapezoid $K L M N$ into triangle $K P N$ and trapezoid $P L M N$. Trapezoid $K L M N$ has the same area as triangle $K O N$, where point $O$ on line $K L$ is the image of point $N$ with respect to central symmetry about the midpoint of segment $L M$. Similarly, trapezoid $P L M N$ has the same area as triangle $P O N$.

![](https://cdn.mathpix.com/cropped/2024_04_17_3ab73cf5e47d7567d598g-4.jpg?height=431&width=714&top_left_y=2000&top_left_x=677)

Segment $N P$ divides trapezoid $K L M N$ into two parts of equal area if and only if $P$ is the midpoint of segment $K O$. Given that $|K O|=|K L|+|L O|$ and $|L O|=|M N|$, we have

$$
|K P|=\frac{40+16}{2}=28(\mathrm{~cm})
$$

Note. In the initial part of the above solution, a derivation of the well-known formula $S=\frac{1}{2}(a+c) v$ for the area of a trapezoid with bases $a$ and $c$ and height $v$ is hinted at. By substitution, the requirement from the problem can be expressed using the equation

$$
\frac{x \cdot v}{2}=\frac{(40+16-x) v}{2}
$$

where $x=|K P|$.