Let a be the length of a side of a square OABC with $O$ being the origin. Its side OA makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3}+1) x+(\sqrt{3}-1) y=0$ and $(\sqrt{3}-1) x-(\sqrt{3}+1) y+8 \sqrt{3}=0$. Then $\mathrm{a}^2$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Slope of diagonal OB }=\frac{\sqrt{3}+1}{1-\sqrt{3}} \ & 	herefore=	an 105^{\circ} \ & 	herefore \alpha=60^{\circ} \ & 	herefore \mathrm{A}\left(\operatorname{acos} 60^{\circ}, \operatorname{asin} 60^{\circ}ight) \ & 	herefore \mathrm{A}\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}ight)\end{aligned}$ A Lies on other diagonal $\begin{aligned} & 	herefore\left(\frac{\sqrt{3}-1}{2}ight) a-\left(\frac{\sqrt{3}+1}{2}ight) \cdot \sqrt{3} a+8 \sqrt{3}=0 \ & a\left[\frac{\sqrt{3}-1-3-\sqrt{3}}{2}ight]=-8 \sqrt{3} \ & a=4 \sqrt{3} \ & 	herefore a^2=48 \end{aligned}$ *

Let a be the length of a side of a square OABC with $O$ being the origin. Its side OA makes an acute angle…

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