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Question Number 184352 by cherokeesay last updated on 05/Jan/23

Answered by som(math1967) last updated on 05/Jan/23

$$△ABC\sim △ACD$$$$\Rightarrow x^2=6\times 4\Rightarrow x=\sqrt{24}=2\sqrt{6}$$$$BC=\sqrt{4\times 2}=2\sqrt{2}$$$$△AFE\sim △ABC$$$$\frac{x+y}{4}=\frac{x}{2\sqrt{2}}$$$$2\sqrt{6}+y=4\times \frac{2\sqrt{6}}{2\sqrt{2}}$$$$y=\frac{4\sqrt{6}-2\sqrt{12}}{\sqrt{2}}=\frac{4\sqrt{6}-4\sqrt{3}}{\sqrt{2}}$$$$\frac{x}{y}=\frac{2\sqrt{6}\times \sqrt{2}}{4(\sqrt{6}-\sqrt{3})}=\frac{\sqrt{3}}{\sqrt{3}(\sqrt{2}-1)}=\sqrt{2}+1$$

Commented by som(math1967) last updated on 05/Jan/23

Commented by cherokeesay last updated on 05/Jan/23

$$nice!thankyousir.$$

Answered by mr W last updated on 05/Jan/23

Commented by mr W last updated on 05/Jan/23

$$h^2=4\times 2\Rightarrow h=2\sqrt{2}$$$$k=4-h=4-2\sqrt{2}$$$$\frac{x}{y}=\frac{h}{k}=\frac{2\sqrt{2}}{4-2\sqrt{2}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$$

Commented by cherokeesay last updated on 05/Jan/23

$$thankyousir.$$

Answered by HeferH last updated on 05/Jan/23

$$h^2=4\cdot 2\Rightarrow h=2\sqrt{2}$$$$\frac{4}{4-2\sqrt{2}}=\frac{x+y}{y}\left(similartriangles\right)$$$$\frac{2}{2-\sqrt{2}}-1=\frac{x}{y}\Rightarrow$$$$\frac{x}{y}=\frac{\sqrt{2}}{2-\sqrt{2}}=\frac{(2+\sqrt{2})\sqrt{2}}{2}=\frac{2\sqrt{2}+2}{2}=1+\sqrt{2}$$

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