JEE MAIN - Mathematics (2025 - 8th April Evening Shift - No. 4)
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 $a$2 is equal to :
48
16
24
32
Explanation
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$$\begin{aligned} &\begin{aligned} & \text { Slope of diagonal } \mathrm{OB}=\frac{\sqrt{3}+1}{1-\sqrt{3}} \\ & =\tan 105^{\circ} \\ & \therefore \alpha=60^{\circ} \\ & \therefore \mathrm{A}\left(\operatorname{acos} 60^{\circ}, \mathrm{asin} 60^{\circ}\right) \\ & \therefore \mathrm{A}\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}\right) \end{aligned}\\ &\text { A Lies on other diagonal }\\ &\begin{aligned} & \therefore\left(\frac{\sqrt{3}-1}{2}\right) a-\left(\frac{\sqrt{3}+1}{2}\right) \cdot \sqrt{3} a+8 \sqrt{3}=0 \\ & a\left[\frac{\sqrt{3}-1-3-\sqrt{3}}{2}\right]=-8 \sqrt{3} \end{aligned} \end{aligned}$$
$$\begin{aligned} & \mathrm{a}=4 \sqrt{3} \\ & \therefore \mathrm{a}^2=48 \end{aligned}$$
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