If $M A$ and $N B$ are the altitudes of triangles $L M P$ and $N K P$, then the right triangles $M A P$ and $N B P$ are similar with a ratio of $1 / 4$.

## Solution

Since $P$ is the midpoint of $K L$, the altitudes of triangles $P L M$ and $P K N$, dropped from vertex $P$, are equal. Therefore, $S_{P L M}: S_{P K N}=L M: K N=1: 4$, and since $S_{P L M Q}: S_{P K N Q}=1: 4$, then $S_{M P Q}: S_{N P Q}=1: 4$. Therefore, $M Q: Q N=1: 4$. Let $M A$ and $N B$ be the altitudes of triangles $L M P$ and $N K P$. Since $L P=P K$, then $A M: B N=1: 4$. By the theorem of proportional segments, $A P: P B=M Q: Q N=1: 4$, so the right triangles $M A P$ and $N B P$ are similar. Therefore, $M P: P N=1: 4$, from which $P N=4 P M=16$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-30.jpg?height=913&width=1226&top_left_y=1853&top_left_x=425)

## Answer

16. 

## Problem
In trapezoid $K L M N$, the bases $K N$ and $L M$ are equal to 12 and 3, respectively. From point $Q$, lying on side $M N$, a perpendicular $Q P$ is dropped to side $K L$. It is known that $P$ is the midpoint of side $K L$, $P M=4$, and that the area of quadrilateral $P L M Q$ is four times smaller than the area of quadrilateral $P K N Q$.

Find the length of segment $P N$.