The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to

Question:

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to

(a) $\frac{1}{2}(\operatorname{ar} \Delta \mathrm{ABC})$

(b) $\frac{1}{3}($ ar $\Delta \mathrm{ABC})$

(c) $\frac{1}{4}($ ar $\Delta \mathrm{ABC})$

 

Solution:

D, E and F are the midpoints of sides BC, AC and AB respectively. 
On joining FE, we divide ABC into 4 triangles of equal area.
Also, median of a triangle divides it into two triangles with equal area

$\operatorname{ar}(\mathrm{AFDE})=\operatorname{ar}(\triangle \mathrm{AFE})+\operatorname{ar}(\triangle \mathrm{FED})$

$=2 \operatorname{ar}(\triangle \mathrm{AFE})$

$=2 \times \frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABC})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$

Hence, the correct answer is option (a). 

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