4. A polygon has 5 fewer sides than another polygon and 50 fewer diagonals than that other polygon. What is the polygon (the one with fewer sides)?
4. Let $n$ be the number of sides (vertices) of the sought polygon.

From one vertex of this polygon, $n-3$ diagonals can be drawn,

and the total number of diagonals of this polygon is $\frac{n(n-3)}{2}$.

In a polygon with $n+5$ sides,

from one vertex, $n+5-3=n+2$ diagonals can be drawn,

and the total number of diagonals of this polygon is $\frac{(n+5)(n+2)}{2}$.

According to the condition of the problem, we have: $\frac{n(n-3)}{2}+50=\frac{(n+5)(n+2)}{2}$

By solving this equation, we get in turn

$n(n-3)+100=(n+5)(n+2)$

$n^{2}-3 n+100=n^{2}+7 n+10$

$10 n=90$

and finally $n=9$.

$1 \text{ POINT}$

TOTAL 6 POINTS

Note: A solution obtained by the method of successive approximation (e.g., filling in a table) can be scored with 10 points. If there are no calculations, but only the correct final result, then it is scored with 2 points.