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Question Number 218493 by Spillover last updated on 10/Apr/25

Answered by mr W last updated on 11/Apr/25

Commented by mr W last updated on 11/Apr/25

$$s=sidelengthoflargestsquare$$$$\tan \alpha =\frac{1}{2}\Rightarrow \sin \alpha =\frac{1}{\sqrt{5}},\cos \alpha =\frac{2}{\sqrt{5}}$$$$s=\frac{x}{\cos \alpha }+x\sin \alpha$$$$s=x\cos \alpha +\frac{3}{\cos \alpha }+3\sin \alpha$$$$x\cos \alpha +\frac{3}{\cos \alpha }+3\sin \alpha =\frac{x}{\cos \alpha }+x\sin \alpha$$$$\Rightarrow x=\frac{3(\frac{1}{\cos \alpha }+\sin \alpha )}{\frac{1}{\cos \alpha }+\sin \alpha -\cos \alpha }$$$$=\frac{3(\frac{\sqrt{5}}{2}+\frac{1}{\sqrt{5}})}{\frac{\sqrt{5}}{2}+\frac{1}{\sqrt{5}}-\frac{2}{\sqrt{5}}}=7$$$$area?=x^2=7^2=49$$

Commented by Spillover last updated on 11/Apr/25

$$correct$$

Answered by Spillover last updated on 11/Apr/25

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