D, E, F are midpoints of the sides BC, CA, AB of a ∆ABC. If area of ∆ABC is 20 cm², then find area of the trapezium FBCE.
Area of $$\triangle\ $$ABC=20 $$cm^2.$$
But, D,E and F are mid points of side BC,AC and AB.
So, $$Area\ of\ \triangle AEF=\triangle CED=\triangle DFB=\triangle DEF.$$
So, $$Area\ of\ \triangle AEF=\frac{20}{4}\ cm^2=5\ cm^2.$$
So,$$Area\ of\ trapizeum\ FBCE=\left(5+5+5\right)=15\ cm^2.$$
B is correct choice.
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