in triangle abc , de is parallel to bc and ad/db=2/3 and bcde is a trapezium find the ratio of the area of triangle with the trapezium

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If your question is , in triangle abc , de is parallel to bc and ad/db=2/3 and bcde is a trapezium find the ratio of the area of triangle  ADE with the trapezium BCED.


Given: In ABC, D divides the side AB such that AD:DB=2:3, E is a point onAC such that DEBC.To find: The ratio of the areas of ADE and the trapezium BDEC.In ADE and ABCADE=B        [Corresponding angles]DAE=BAC      [Common]So, ADE~ABC     [AA similarity]We know that the ratio of areas of two similar triangles is equal to the ratio of squares of theircorresponding sides.arADEarABC=AD2AB2arADEarABC=4x22x+3x2arADEarABC=4x225x2arADEarABC=425Let Area of ADE=4 square units and Area of ABC=25 sq unitsartrap BCED=arABC-arADE=25x2-4x2=21x2 sq unitsNow, arADEartrap BCED=4x221x2=421
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