Question

Question 6
The midpoint of the sides of the triangle along with any of the vertices as the point make a parallelogram of area equal to:

A) $\frac{1}{2}$ ar(ABC)
B) $\frac{1}{3}$ ar(ABC)
C) $\frac{1}{4}$ ar(ABC)
D) $ar\left(ABC\right)$

Answer 1

The answer is A.

We know that, if D, E and F are respectively the mid-points of the sides BC, CA and AB of a DABC, then all four triangles has equal area i.e.,
ar ($△$AFE) = ar ($△$BFD) = ar ($△$EDC) = ar ($△$DEF) ....(i)
area of $△$DEF = $\frac{1}{4}$ area of $\Delta$ ABC ....(ii)

if we take D as the fourth vertex, then area of parallelogram AFDE
= Area of $△$ AFE + Area of $△$ DEF
= Area of $△$DEF + Area of $△$DEF = 2 Area of $△$ DEF [using Eq. (i)]
= 2 $\times \frac{1}{4}$ Area of $△$ ABC [using Eq. (ii)]
= $\frac{1}{2}$ Area of $△$ ABC