abc is a right angled triangle with angle b is equal to 90 degree . m is the mid point of ac and bm =under root 117 . sum of the lenght of other two sides ab and bc is 30 degree .the the value of 1\12 (area of triangle abc)is

At first the line bm is perpendicular to AC and let the two sides x and y and therefore (x^2+y^2)^0.5=AC AM=CM=(1/2)×ACIN TRIANGLE AMB ANGLE AMB IS 90°ThereforeAM^2 +BM^2=AB^2AM=1/2×ACAM ={(x^2+y^2)^0.5}×1/2Am ^2 =(x^2+y^2)×1/4BM^2=117AB=xx^2+y^2+468=4×x^2Similarlyx^2+y^2+468=4×x^2Add this two equationx^2+y^2+936=3×y^2+3×x^2936=2×x^2+2×x^2468=x^2+y^2We know that x+y=30We have to find 1/12×1/2×xyAt first we have to find xy so (X+y)^2-(x^2+y^2)=2xyX+y=30(X+y)^2=900X^2 +y^2=468900-468=2xy432/2=xy216=xyWe have to find 1/12×1/2×xy1/24×216=9Therefore ans is 9