Let a be the length of a side of a square OAB | vedclass

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(A) The slope of the diagonal $OB$ is given by $m_1 = -\frac{\sqrt{3}+1}{\sqrt{3}-1} = -\frac{(\sqrt{3}+1)^2}{3-1} = -\frac{4+2\sqrt{3}}{2} = -(2+\sqr

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) The slope of the diagonal $OB$ is given by $m_1 = -\frac{\sqrt{3}+1}{\sqrt{3}-1} = -\frac{(\sqrt{3}+1)^2}{3-1} = -\frac{4+2\sqrt{3}}{2} = -(2+\sqrt{3}) = 	an(105^{\circ})$.Since $OB$ makes an angle of $105^{\circ}$ with the positive $x$-axis and the diagonal $OB$ bisects the angle $\angle AOC = 90^{\circ}$,the angle $\alpha$ that $OA$ makes with the $x$-axis is $105^{\circ} - 45^{\circ} = 60^{\circ}$.Thus,the coordinates of vertex $A$ are $(a \cos 60^{\circ}, a \sin 60^{\circ}) = (\frac{a}{2}, \frac{\sqrt{3}a}{2})$.Vertex $A$ lies on the other diagonal $(\sqrt{3}-1)x - (\sqrt{3}+1)y + 8\sqrt{3} = 0$.Substituting the coordinates of $A$ into this equation:$(\sqrt{3}-1)(\frac{a}{2}) - (\sqrt{3}+1)(\frac{\sqrt{3}a}{2}) + 8\sqrt{3} = 0$$a(\frac{\sqrt{3}-1 - 3 - \sqrt{3}}{2}) = -8\sqrt{3}$$a(\frac{-4}{2}) = -8\sqrt{3}$$-2a = -8\sqrt{3} \implies a = 4\sqrt{3}$.Therefore,$a^2 = (4\sqrt{3})^2 = 16 	imes 3 = 48$.

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

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