Question

$ABCD$ is a parallelogram $X$ and $Y$ are the mid points of $BC$ and $CD$ respectively. Then, ar(parallelogram $ABCD$) is

• A. $4\times ar\left(△AXY\right)$
• B. $2\times ar\left(△AXY\right)$
• C. $\frac{8}{3}\times ar\left(△AXY\right)$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{8}{3}\times ar\left(△AXY\right)$

$X$ and $Y$ are the mid-points of sides $BC$ and $CD$.

In $△BCD,$

$XY\parallel BD$ and $XY=\frac{1}{2}BD$ {From the mid point theorem}

$\Rightarrow$ $ar\left(△CYX\right)=\frac{1}{4}ar\left(△DBC\right)$ {property of triangle having mid points}

$\Rightarrow$ $ar\left(△CYX\right)=\frac{1}{8}ar\left({\parallel }^{gm}ABCD\right)$ [ Area of parallelogram is twice the area of triangle made by diagona ] --- ( 1 )

Since, parallelogram $ABCD$ and $△ABX$ are between the same parallel lines $AD$ and $BC$ and $BX=\frac{1}{2}BC.$

$\Rightarrow$ $ar\left(△ABX\right)=\frac{1}{4}ar\left({\parallel }^{gm}ABCD\right)$ ----- ( 2 )

Similarly, $ar\left(△AYD\right)=\frac{1}{4}ar\left({\parallel }^{gm}ABCD\right)$ ----- ( 3 )

Now, $ar\left(△AXY\right)=ar\left({\parallel }^{gm}ABCD\right)-\left[ar\right(△ABX)+ar(△AYD)+ar(△CYX\left)\right]$

$=ar\left({\parallel }^{gn}ABCD\right)-\left[\frac{1}{4}ar\right({\parallel }^{gm}ABCD+\frac{1}{4}ar\left({\parallel }^{gm}ABCD\right)+\frac{1}{8}ar\left({\parallel }^{gm}ABCD\right)]$ [ From ( 1 ) ( 2 ) and ( 3 ) ]

$=ar\left({\parallel }^{gm}ABCD\right)-\frac{5}{8}\left({\parallel }^{gm}ABCD\right)$

$\Rightarrow$ $ar\left(△AXY\right)=\frac{3}{8}ar\left({\parallel }^{gm}ABCD\right)$

$\therefore$ $ar\left({\parallel }^{gm}ABCD\right)=\frac{8}{3}\times ar\left(△AXY\right)$