Question

The mid point of the sides of a triangle ABC along with any one of the vertices as the fourth point makes a parallelogram whose area is equal to :

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

Answer 1

Answer: (D)

$\frac{1}{2}ar\left(ABC\right)$

let area of $\Delta ABC=A$

=> Area of $\Delta AEF=\frac{A}{4}$ (Hero's Formula)

=> Area of $\Delta GFC=\frac{A}{4}$

=> Area of parallelogram BEFG = Total Area of $\Delta ABC$ - Area of $\Delta AEF$ - Area of $\Delta GFC$

=> Area of parallelogram BEFG = $\frac{A}{2}$