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 $ \alpha $ with the positive $ x $-axis and the equations of its diagonals are $\left( \sqrt{3} + 1 \right) x + \left( \sqrt{3} - 1 \right) y = 0$ and $\left( \sqrt{3} - 1 \right) x - \left( \sqrt{3} + 1 \right) y + 8\sqrt{3} = 0.$ Then $ a^2 $ is equal to

• A. 24
• B. 32
• C. 48
• D. 16

Hint:

- Diagonals of square intersect at 90° and bisect each other - Distance from center to vertex gives half-diagonal length - \( \text{Diagonal} = a\sqrt{2} \) for square of side \( a \) - Verify perpendicularity by checking \( m_1m_2 = -1 \)

Answer 1

The Correct Option is C

Step 1: Find the angle between diagonals The diagonals of a square intersect at 90°.
Let's verify: \[ m_1 = -\frac{\sqrt{3}+1}{\sqrt{3}-1}, \quad m_2 = \frac{\sqrt{3}-1}{\sqrt{3}+1} \] \[ m_1 \times m_2 = -1 \quad \text{(perpendicular)} \]
Step 2: Find intersection point (center) Solve the diagonal equations simultaneously: \[ x = \frac{8\sqrt{3}(\sqrt{3}-1)}{8} = 3 - \sqrt{3} \] \[ y = \frac{8\sqrt{3}(\sqrt{3}+1)}{8} = 3 + \sqrt{3} \]
Step 3: Calculate side length Distance from origin to center: \[ \sqrt{(3-\sqrt{3})^2 + (3+\sqrt{3})^2} = \sqrt{24} = 2\sqrt{6} \] For a square, diagonal \( d = a\sqrt{2} \), and distance to center is \( d/2 \): \[ 2\sqrt{6} = \frac{a\sqrt{2}}{2} \Rightarrow a = 4\sqrt{3} \] Thus: \[ a^2 = (4\sqrt{3})^2 = 48 \]