# Lengths of sides and diagonals of quadrilaterals and area

This question follows from another question I've asked on triangles Length of sides of a triangle and area

Let $P_1$ and $P_2$ be two convex quadrilaterals such that $P_1\neq P_2$ and $Area(P_1)\ge Area(P_2)$. Is it true that it is not possible that all sides and diagonals of $P_1$ are shorter than the corresponding sides and diagonals of $P_2$?

This sentence seems true to me, I'll explain my reasoning. Suppose that all four sides and one diagonal of $P_1$ are shorter than the corresponding sides and diagonal of $P_2$. Then, as it is explained in the answer to my older question, the two triangles which triangulate $P_2$ should be "flatter" (otherwise it's not possible $Area(P_1)\ge Area(P_2)$). From this it should follow that the other diagonal of $P_1$ is longer than the corresponding diagonal of $P_2$.

Do you think it's correct? Can you help me formalizing it?

Edit: following user Raffaele's hint I checked out the formulas in this wiki page https://en.wikipedia.org/wiki/Quadrilateral#cite_note-10, but none of them seems useful to solve this problem

## 1 Answer

It is possible that $Area(P_1)\ge Area(P_2)$ while each side and each diagonal of $P_1$ is strictly smaller than the corresponding one in $P_2$.

Take for example $P_1$ to be a square of side $1\,$. Take $P_2$ to be an isosceles trapezoid with the small base and the non-parallel sides all equal to $2$. Flatten the trapezoid while keeping those three sides at the same length $2$. In the limit the large base tends to $3\cdot 2=6$, the diagonals tend to $2 \cdot 2 = 4\,$, and the area tends to $0\,$, yet all the sides and diagonals of $P_2$ are strictly larger than those of $P_1$. Even stronger, any side or diagonal of $P_2$ is strictly larger than any other side or diagonal of $P_1$.