IN $△ A B C, D, E, F$ are the mid points of sides $B C, C A$ and $A B$ respectively. If $a r (A B C) = 16 c m$ the $a r (t r a p e z i u m F B C E) =$

4 c m^{2}

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8 c m^{2}

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12 c m^{2}

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10 c m^{2}

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Solution

The correct option is C $12 c m^{2}$
area $(△ A B C) = 16 c m^{2}$
area $(F B C E) = △ A B C - △ A F E$
$\frac{1}{4}$ area $(△ A B C) =$ area $(△ A F E)$
area $(△ F E) = \frac{1}{4} \times 16 = 4 c m^{2}$
area of trapezium $= 16 - 4 = 12$

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