Let a be the length of a side of a square OABC with O being the o… - Vidyadip

MCQ

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

✓
48
B
32
C
16
D
24

✓

Answer

Correct option: A.

48

(A) 48

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

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