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Question Number 47407 by ajfour last updated on 09/Nov/18

Answered by ajfour last updated on 10/Nov/18

$$letBD=x,\mathrm{\angle }A=\varphi ,\mathrm{\angle }C=\theta$$$$S_{ABCD}=\frac{ab\sin \theta }{2}+\frac{cd\sin \varphi }{2}$$$$x^2=a^2+b^2-2ab\cos \theta$$$$=c^2+d^2-2cd\cos \varphi$$$$\Rightarrow \left(ab\sin \theta \right)\frac{d\theta }{dx}=\left(cd\sin \varphi \right)\frac{d\varphi }{dx}=2x$$$$\frac{dS}{dx}=0\Rightarrow$$$$\left(ab\cos \theta \right)\frac{d\theta }{dx}+\left(cd\cos \varphi \right)\frac{d\varphi }{dx}=0$$$$\Rightarrow \cos \theta +\left(\cos \varphi \right)\left(\frac{\sin \theta }{\sin \varphi }\right)=0$$$$\Rightarrow \theta +\varphi =\pi$$$$\Rightarrow a^2+b^2-2ab\cos \theta$$$$=c^2+d^2+2cd\cos \theta$$$$or\cos \theta =\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}$$$${\left(S_{ABCD}\right)}_{max}=\frac{\sin \theta }{2}(ab+cd)$$$$=\frac{(ab+cd)}{2}\sqrt{1-{\left[\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\right]}^2}$$$$S_{max}=\frac{1}{4}\sqrt{4{(ab+cd)}^2-{[(a^2+b^2)-(c^2+d^2)]}^2}.$$

Commented by mr W last updated on 10/Nov/18

$$thankyousir!$$$$itshowstheareareachesmaximum$$$$whenthequadrilateralisinscribedin$$$$acircle.$$

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