ABCD is a trapezium in which AB ll DC and 2AB=3DC, find the ratio of areas of∆ ABC and∆ COD
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the height of both triangles woulde be same as they lie between same pair of parallel lines
Height of triangle abc=Height of triangle cod
h=h
2ab=3dc
2ab/3=dc
ar (abc)/ar (cod)=(1/2×ab×h)/(1/2×dc×h)
=ab/(2ab/3)
=1/(2/3)
=3/2
So, the ratio of there areas = 3:2
Height of triangle abc=Height of triangle cod
h=h
2ab=3dc
2ab/3=dc
ar (abc)/ar (cod)=(1/2×ab×h)/(1/2×dc×h)
=ab/(2ab/3)
=1/(2/3)
=3/2
So, the ratio of there areas = 3:2
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