Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle with the positive x-axis and the equations of its diagonals are and . Then is equal to: [2025]

Question

Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle $α$ 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: [2025]

Smart Achievers · Accepted Answer

(1)

Slope of OB = $\frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \tan 105 °$

Now, $α + 45 ° = 105 ° \Rightarrow α = 60 °$

$∴$ Coordinates of A are (a cos 60°, a sin 60°)

i.e., $(\frac{a}{2}, \frac{\sqrt{3} a}{2})$

Now, A lies on diagonal AC.

$\Rightarrow (\sqrt{3} - 1) \frac{a}{2} - (\sqrt{3} + 1) \frac{\sqrt{3} a}{2} + 8 \sqrt{3} = 0$

$\Rightarrow \frac{a}{2} [\sqrt{3} - 1 - 3 - \sqrt{3}] + 8 \sqrt{3} = 0$

$\Rightarrow \frac{a}{2} [- 4] + 8 \sqrt{3} = 0$

$\Rightarrow a = 4 \sqrt{3}$

$\Rightarrow a^{2} = 48$ .

Solution

1 48

2 32

3 24

4 16

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