A plane is going through a cube so the cross section is a pentagon. Prove that the area of the pentagon is smaller than the product of the pentagon's two largest sides.
The first thing I tried to do was to express some of the sides through pythagorean theorem, however i failed to do so as not all sides are a part of a right triangle. I don't have any idea after this.
Sharing an approach that you can turn into a more formal proof -
The point to observe is that if the cross section of the cube cut by the plane is a pentagon, it must go through $5$ sides of the cube and two pairs of sides must be parallel (that are on parallel sides of the cube). Here is a visualization -
In the diagram, we have $AB \parallel ED, AE \parallel BC$ and $AB \gt DE$ and $AE \gt BC$.
If we extend line $ED$ and line $BC$ and say they meet at point $M$ then we have a parallelogram $ABME$.
WLOG, $AB$ is the longer side and $AE$ is the smaller side of the parallelogram $ABME$.
Area of pentagon $ \lt$ area of parallelogram ABME $ \leq \ AB \cdot AE \leq P, $ where $P$ is the product of the two largest sides of the pentagon.
So the inequality clearly hold even if $DE \gt AE \ $ or $CD \gt AE$ or $AB$.