D, E and F are the mid point of sides BC , CA,AB of a triangle ABC
1. BDEF is a parallelogram
2. $a r (D E F) = \frac{1}{4} a r (A B C)$
3. $(B D E F) = \frac{1}{2} a r (A B C)$

Q. D, E and F are respectively the mid-points of the sides BC, CA, and AB of a ABC
(i) BDEF is a parallelogram
(ii) ar (DEF) =

\frac{1}{4}

ar (ABC)
(iii) ar BDEF)

\frac{1}{2}

ar (ABC)

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D,E,F are mid points of the sides BC,CA& AB respectively of △ABC, then the area of the parallelogram BDEF is equal to

The correct option is D 12ar(△ABC)Using the mid point theorem for triangles show that ADEF is a parallelogram.Draw AQ⊥BC so that AQ is the height of the triangle ABC.PQ will be the height of BDEF.

$D, E, F$ are mid points of the sides $B C, C A & A B$ respectively of $△ A B C$ , then the area of the parallelogram $B D E F$ is equal to

The correct option is D $\frac{1}{2} a r (△ A B C)$
Using the mid point theorem for triangles show that ADEF is a parallelogram.
Draw $A Q ⊥ B C$ so that AQ is the height of the triangle ABC.
PQ will be the height of BDEF.