MathematicsMedium19×since 2002Q5042If in a triangle ABC, AB = 5 units, ∠B=cos−1(35)\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)∠B=cos−1(53) and radius of circumcircle of Δ\DeltaΔABC is 5 units, then the area (in sq. units) of Δ\DeltaΔABC is :A10+6210 + 6\sqrt 210+62B8+228 + 2\sqrt 28+22C6+836 + 8\sqrt 36+83D4+234 + 2\sqrt 34+23Check answerSkip