Question

If $ABCD$ is a rectangle, $E,F$ are the mid-points of $BC$ and $AD$ respectively and $G$ is any point on $EF$, then $△GAB$ equals.

• A. $\frac{1}{2}\left(\square ABCD\right)$
• B. $\frac{1}{3}\left(\square ABCD\right)$
• C. $\frac{1}{4}\left(\square ABCD\right)$
• D. $\frac{1}{6}\left(\square ABCD\right)$

Answer 1

Answer: (C)

$\frac{1}{4}\left(\square ABCD\right)$

Given that $E$ and $F$ are mid-points of $BC$ and $AD$ respectively.

$ar\left(\square ABFE\right)=\frac{1}{2}ar\left(\square ABCD\right)[\because$ $E$ and $F$ are mid points of $BC$ and $AD$ respectively]

$ar\left(△GAB\right)=\frac{1}{2}ar\left(\square ABFE\right)[\because$ Area of triangle is half of area of rectangle on the same base and between the same parallels]

$=\frac{1}{2}\times \frac{1}{2}ar\left(\square ABCD\right)$

$=\frac{1}{4}ar\left(\square ABCD\right)$