[ The ratio of the areas of triangles with a common base or common height ]

In trapezoid $M P Q F$, the bases $M F=24, P Q=4$. The height of the trapezoid is 5. Point $N$ divides the lateral side into segments $M N$ and $N P$, such that $M N=3 N P$.

Find the area of triangle $N Q F$.

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Continue $F N$ until it intersects with line $Q P$ and consider the similar triangles formed.

## Solution

Extend segment $F N$ until it intersects with line $Q P$ at point $A$. From the similarity of triangles $A P N$ and $F M N$, we find that $A N: N F = P N: N M = 1: 3$,

$A P = \frac{1}{3} M F = 8$.

Thus, $A Q = A P + P Q = 12$, $S_{A Q F} = \frac{1}{2} A Q \cdot 5 = 30$.

Therefore, $S_{N Q F} = \frac{F N}{A F} \cdot S_{A Q F} = \frac{3}{4} \cdot 30 = \frac{45}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-21.jpg?height=392&width=800&top_left_y=2165&top_left_x=629)

## Answer

$22.5$.

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