Question

$E$ is the mid point of diagonal $BD$ of a parallelogram $ABCD$. If the point $E$ is joined to a point $F$ on $DA$ such that $DF=\frac{1}{3}DA$, then the ratio of $A\left(\Delta DFE\right)$ to the $A\left(\square ABEF\right)$ is:

• A. $1:3$
• B. $1:4$
• C. $1:5$
• D. $2:5$

Answer 1

The correct option is C $1:5$

Given,$DF=\frac{1}{3}DA$

From the fig, $BP||EQ$

In $△BPD$,

$E$ is the midpoint of $BD$ and $BP||EQ$

$\therefore Q$ is the midpoint of $PD$. ---(converse of midpoint theorem)

$EQ=\frac{1}{2}BP$

$Areaof△ABD=\frac{1}{2}\times AD\times BP$

$Areaof△EFD=\frac{1}{2}\times DF\times EQ$

$\therefore \frac{Areaof△EFD}{Areaof△ABD}=\frac{\frac{1}{2}\times DF\times EQ}{\frac{1}{2}\times AD\times BP}$

$=\frac{\frac{1}{2}\times \frac{1}{3}AD\times \frac{1}{2}\times BP}{\frac{1}{2}\times AD\times BP}$

$=\frac{1}{6}$

$\therefore ,Areaof△EFD=\frac{1}{6}Areaof△ABD$

$AreaofQuadrilateralABEF=Areaof△ABD-Areaof△EFD$

$AreaofQuadrilateralABEF=Areaof△ABD-\frac{1}{6}Areaof△ABD$

$AreaofQuadrilateralABEF=\frac{5}{6}Areaof△ABD$

$\frac{Areaof△EFD}{AreaofQuadrilateralABEF}=\frac{\frac{1}{6}Areaof△ABD}{\frac{5}{6}Areaof△ABD}=\frac{1}{5}$