the line segment XYlltoAC of triABC and it divides the tri into two parts of equal areas. Prove that AX/AB=root2-1/root2
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we have XYllAC [GIVEN]
So <BXY=<A AND <BYX=<C [CORRESPONDING ANGLES]
THERE FORE TIANGLE ABC SIMILAR TRIANGLE XBY
SO ar[ABC]byar[XBY]=AB BY XB WHOLE SQUAR
ALSO ar[ABC]=2ar[XBY]
SO ar[ABC] BY ar[XBY]=2by1
AB byXB squar=2 by 1 i;e AB by XB=root 2 by 1
or XB by AB=1 by root 2
or AB-XB by AB=root2-1 by root 2 i;eAX by AB=root2-1 by root 2=2-root2 by 2
So <BXY=<A AND <BYX=<C [CORRESPONDING ANGLES]
THERE FORE TIANGLE ABC SIMILAR TRIANGLE XBY
SO ar[ABC]byar[XBY]=AB BY XB WHOLE SQUAR
ALSO ar[ABC]=2ar[XBY]
SO ar[ABC] BY ar[XBY]=2by1
AB byXB squar=2 by 1 i;e AB by XB=root 2 by 1
or XB by AB=1 by root 2
or AB-XB by AB=root2-1 by root 2 i;eAX by AB=root2-1 by root 2=2-root2 by 2
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