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Simple Problem of Areas Of Parallelograms and triangles?

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  • Simple Problem of Areas Of Parallelograms and triangles?


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Answer #1 | 29/12 2013 02:51
What do do want to know? Let DC=x Area of DFBE= DFxEG = DF =x^2/3 Area of DCF = x^2/3 From DCF x(FC/2) = DF So x(FC/2) = x^2/3 FC=2x/3

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