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Question Number 172266 by Mikenice last updated on 25/Jun/22

Answered by mr W last updated on 25/Jun/22

Commented by mr W last updated on 25/Jun/22

$$distancefromBtoCD$$$$=\sqrt{2}+\frac{3}{\sqrt{2}}=\frac{5}{\sqrt{2}}$$$$distancefromAtoCD$$$$=\sqrt{2}+\frac{3}{\sqrt{2}}+\frac{x}{\sqrt{2}}=\frac{x+5}{\sqrt{2}}$$$$\frac{areaBCD}{areaACD}=\frac{distancefromBtoCD}{distancefromAtoCD}$$$$\frac{2}{3}=\frac{5}{x+5}$$$$\Rightarrow x=\frac{5}{2}$$$$\tan \alpha =\frac{3}{x+2}=\frac{3}{\frac{5}{2}+2}=\frac{2}{3}$$$$\frac{h_1}{\tan \alpha }+h_1=x+3$$$$\frac{3h_1}{2}+h_1=\frac{5}{2}+3$$$$\Rightarrow h_1=\frac{11}{5}$$$$h_2=x\tan \alpha =\frac{5}{2}\times \frac{2}{3}=\frac{5}{3}$$$$areaR=\frac{(x+3)h_1}{2}-\frac{{xh}_2}{2}$$$$=\frac{1}{2}[(\frac{5}{2}+3)\frac{11}{5}-\frac{5}{2}\times \frac{5}{3}]=\frac{119}{30}$$$$arealargestsquare=3^2=9$$$$\frac{areaR}{arealargestsquare}=\frac{119}{30\times 9}=\frac{119}{270}\approx 44\mathrm{\%}$$

Commented by Tawa11 last updated on 25/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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