# Let P_1 and P_2 be two convex quadrilaterals such that P_1≠P_2 and Area(P_1)≥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?

Let ${P}_{1}$ and ${P}_{2}$ be two convex quadrilaterals such that ${P}_{1}\ne {P}_{2}$ and $Area\left({P}_{1}\right)\ge Area\left({P}_{2}\right)$. 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 ${P}_{2}$?
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 $Area\left({P}_{1}\right)\ge Area\left({P}_{2}\right)$). 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?
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Domenigmh
It is possible that $Area\left({P}_{1}\right)\ge Area\left({P}_{2}\right)$ 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$2$. Flatten the trapezoid while keeping those three sides at the same length $1$. In the limit the large base tends to $3\cdot 2=6$, the diagonals tend to $2\cdot 2=4\phantom{\rule{thinmathspace}{0ex}}$, and the area tends to $0\phantom{\rule{thinmathspace}{0ex}}$, 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 ${P}_{1}$.