IN A TRIANGLE

Asked by hemant2020 | 25th Aug, 2010, 10:08: PM

Expert Answer:

angle B = 1350From the point A draw perpendicular on BC to meet the extended line BC at DSo AD is height of triangle ABC.angle ABC= 1350 so angle ABD = 450In right angled triangle ABD,  AD=BDConsider,AB2+BC2+4 area of triangle ABC= (AD2 + DB2) + (DC-BD)2+4 (1/2) BC.AD (by pythagoras thm)=(AD2 + DB2) + (DC2+BD2-2DC.BD)+2BC.BD        (AD=BD)=(AD2+DC2)+2BD2-2BD(DC-BC)                   (by pythagoras thm)= AC2+2BD2-2BD(BD)=AC2">

Answered by  | 26th Aug, 2010, 10:44: AM

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