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Question Number 186784 by Rupesh123 last updated on 10/Feb/23

Answered by a.lgnaoui last updated on 11/Feb/23

$$1\bullet$$$$soitABCD\left(Rectangle\right)$$$$BC=BM+MC(AB=aAC=b)$$$${AB}^2={AM}^2+{MC}^2=$$$$a^2{\cos }^{2}2\theta +{(c-b\cos 2\theta )}^2$$$$=a^2\cos {}^{2}2\theta +b^2\cos {}^{2}2\theta -2bc\cos 2\theta$$$$=(a^2+b^2)\cos {}^{2}2\theta -2bc\cos 2\theta +c^2$$$$c=\sqrt{a^2+b^2}$$$$Airetotale=ab=A_1+A_2+A\left(yellow\right)$$$$A_1=\frac{1}{2}b\cos 2\theta$$$$A_2+A_3=\frac{1}{2}b\sin 2\theta$$$$A\left(yellow\right)=A-(A_1+A_3)$$$$A_2=\frac{1}{4}AE\sin 2\theta$$$$A_3=A\left(Yellow\right)=\frac{1}{2}b\sin \theta$$$$=b\times \frac{ME}{2}=\frac{1}{2}AE\sin \theta$$$$=ab-(\frac{1}{4}AE\sin 2\theta +\frac{1}{2}b\cos 2\theta )$$$$\frac{1}{2}AE\sin \theta =ab-\frac{1}{4}AE\sin 2\theta -\frac{1}{2}b\cos 2\theta$$$$AE(\frac{1}{2}\sin \theta +\frac{1}{4}\sin 2\theta )=ab-\frac{1}{2}\cos 2\theta$$$$$$$$A\left(yellow\right)=\frac{1}{2}b\sin \theta \left(1\right)$$$$=\frac{1}{2}\sin \theta \left(\frac{2ab\cos 2\theta -\cos 2\theta }{\sin \theta +\frac{1}{2}\sin 2\theta }\right)$$$$=\frac{1}{4}\left(\frac{2ab-1}{1+\cos \theta }\right)\cos 2\theta \left(2\right)$$$$b\sin \theta =\frac{1}{2}\left(\frac{2ab-1}{1+\cos \theta }\right)\cos 2\theta$$$$endeduire\theta =f(a,b)$$$$A=f(a,b)$$$$Aftercalcul$$$$$$$$A\left(yellow\right)=\frac{b}{2\sqrt{2}}\sqrt{1-\frac{b}{c}}$$$$c=\sqrt{a^2+b^2}$$$$$$

Commented by a.lgnaoui last updated on 11/Feb/23

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