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Question Number 167152 by cortano1 last updated on 08/Mar/22

Commented by som(math1967) last updated on 08/Mar/22

$$x^2=28?$$

Commented by cortano1 last updated on 08/Mar/22

$$\mathrm{yes}$$

Answered by som(math1967) last updated on 08/Mar/22

Commented by som(math1967) last updated on 08/Mar/22

$$letSB=lsideofsquare=a$$$$A=\frac{al}{2}$$$$2A=\frac{a\times AQ}{2}$$$$\therefore a\times AQ=2al\Rightarrow AQ=2l$$$$\therefore BR=a-lAR=a-2l$$$$3A=\frac{(a-l)(a-2l)}{2}$$$$\frac{3al}{2}=\frac{a^2-3al+2l^2}{2}$$$$\Rightarrow a^2-6al+2l^2=0$$$$\Rightarrow \frac{a^2}{l^2}-\frac{6a}{l}+2=0$$$$\therefore \frac{a}{l}=\frac{6+\sqrt{28}}{2}=3+\sqrt{7}[\because \frac{a}{l}>1\therefore -verejected]$$$$A=\frac{al}{2}=\frac{a^2}{2(3+\sqrt{7})}$$$$ar.ofyellow△=Y\left(let\right)$$$$Y=a^2-6A$$$$=a^2-\frac{3a^2}{(3+\sqrt{7})}=\frac{\sqrt{7}{a}^2}{(3+\sqrt{7})}$$$$\frac{Y}{A}=2\sqrt{7}\Rightarrow Y=2\sqrt{7}A$$$$\therefore x=2\sqrt{7}\Rightarrow x^2=28$$

Commented by Tawa11 last updated on 10/Mar/22

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

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