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

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 $\mathrm{ar}\left(\mathrm{AFDE}\right)=\mathrm{ar}\left(△\mathrm{AFE}\right)+\mathrm{ar}\left(△\mathrm{FED}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{ar}\left(△\mathrm{AFE}\right)\phantom{\rule{0ex}{0ex}}=2×\frac{1}{4}\mathrm{ar}\left(△\mathrm{ABC}\right)=\frac{1}{2}\mathrm{ar}\left(△\mathrm{ABC}\right)$ Hence, the correct answer is option (a). MathematicsRS Aggarwal (2020, 2021)All

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