What is the maximum area of quadrilateral such that max|sides, diagonals|< x ?
I know that the maximum area would be the square area of diagonals of length $x$ which has area $x^2/2$, but how to prove that?
2026-03-25 18:56:25.1774464985
Maximum area of quadrilateral with sides and diagonals not exceeding a given value
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Let the two diagonals be $l, m\le x$ with angle $\alpha$. Then $$ Area = \frac{1}{2}lm\sin \alpha \le \frac{1}{2}x^2 $$