MathematicsHard127×since 2002Q3026Let a triangle ABC be inscribed in the circle x2−2(x+y)+y2=0{x^2} - \sqrt 2 (x + y) + {y^2} = 0x2−2(x+y)+y2=0 such that ∠BAC=π2\angle BAC = {\pi \over 2}∠BAC=2π. If the length of side AB is 2\sqrt 22, then the area of the Δ\DeltaΔABC is equal to :A1B(6+3)/2\left( {\sqrt 6 + \sqrt 3 } \right)/2(6+3)/2C(3+3)/4\left( {3 + \sqrt 3 } \right)/4(3+3)/4D(6+23)/4\left( {\sqrt 6 + 2\sqrt 3 } \right)/4(6+23)/4Check answerSkip