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) $\frac{1}{2}\left(\mathrm{ar}∆\mathrm{ABC}\right)$
(b) $\frac{{\displaystyle 1}}{{\displaystyle 3}}\left(\mathrm{ar}∆\mathrm{ABC}\right)$
(c) $\frac{{\displaystyle 1}}{{\displaystyle 4}}\left(\mathrm{ar}∆\mathrm{ABC}\right)$

Answer 1

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
$$\begin{gathered} \mathrm{ar}\left(\mathrm{AFDE}\right)=\mathrm{ar}\left(△\mathrm{AFE}\right)+\mathrm{ar}\left(△\mathrm{FED}\right) \\ =2\mathrm{ar}\left(△\mathrm{AFE}\right) \\ =2\times \frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right) \end{gathered}$$
Hence, the correct answer is option (a).