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Question Number 217080 by hardmath last updated on 28/Feb/25

Commented by mr W last updated on 01/Mar/25

$$alsoArea\left(ABCD\right)⩽\sqrt{abcd}$$

Answered by mr W last updated on 01/Mar/25

Commented by mr W last updated on 01/Mar/25

$$saythediagonalsareeandfwith$$$$angle\theta betweenthem.$$$$accordingto{euler}^{\prime }squadrilateral$$$$theorem:$$$$a^2+b^2+c^2+d^2=e^2+f^2+4g^2$$$$\Rightarrow e^2+f^2⩽a^2+b^2+c^2+d^2$$$$\left[ABCD\right]=\frac{\left[A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }\right]}{2}$$$$=\frac{ef\sin \theta }{2}$$$$⩽\frac{ef}{2}$$$$⩽\frac{e^2+f^2}{4}$$$$⩽\frac{a^2+b^2+c^2+d^2}{4}✓$$

Answered by mr W last updated on 02/Mar/25

$$aquadrilateralwithsidesa,b,c,d$$$$hasthemaximumareawhenitis$$$$cyclic.thismaximumareais$$$$\sqrt{(s-a)(a-b)(s-c)(s-d)}with$$$$s=\frac{a+b+c+d}{2}.myproofseeQ30233.$$$$thatmeans$$$$Area{\left(ABCD\right)}_{arbitrary}$$$$⩽Area{\left(ABCD\right)}_{cyclic}$$$$=\sqrt{(s-a)(s-b)(s-c)(s-d)}$$$$=\sqrt[4]{{(s-a)}^2{(s-b)}^2{(s-c)}^2{(s-d)}^2}$$$$⩽\frac{{(s-a)}^2+{(s-b)}^2+{(s-c)}^2+{(s-d)}^2}{4}$$$$=\frac{4s^2-2s(a+b+c+d)+a^2+b^2+c^2+d^2}{4}$$$$=\frac{4s^2-4s^2+a^2+b^2+c^2+d^2}{4}$$$$=\frac{a^2+b^2+c^2+d^2}{4}✓$$

Answered by mr W last updated on 04/Mar/25

$$Area\left(ABCD\right)$$$$=\mathrm{\Delta }\left(ABC\right)+\mathrm{\Delta }\left(CDA\right)$$$$=\frac{ab\sin \mathrm{\angle }B}{2}+\frac{cd\sin \mathrm{\angle }D}{2}$$$$⩽\frac{ab}{2}+\frac{cd}{2}$$$$⩽\frac{a^2+b^2}{4}+\frac{c^2+d^2}{4}=\frac{a^2+b^2+c^2+d^2}{4}✓$$

Commented by hardmath last updated on 03/Mar/25

$$\mathrm{cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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