Let a triangle ABC be inscribed in the circle ${x^2} - \sqrt 2 (x + y) + {y^2} = 0$ such that $\angle BAC = {\pi \over 2}$. If the length of side AB is $\sqrt 2 $, then the area of the $\Delta$ABC is equal to :

Question

$\left( {\sqrt 6 + \sqrt 3 } \right)/2$

$\left( {3 + \sqrt 3 } \right)/4$

$\left( {\sqrt 6 + 2\sqrt 3 } \right)/4$

Solution

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