Question

Given a parallelogram $ABCD$ where $X$ and $Y$ are the mid-points of the sides $BC$ and $CD$ respectively. Prove that :
$ar\left(\Delta AXY\right)=\frac{3}{8}\times ar\left(//gmABCD\right)$

Answer 1

Let the area of a parallelogram be $a$

We know that the diagonal of a parallelogram divides it into two triangles of equal areas.

$\therefore$

Area of $△$ ACD=Area of $△$ABC=Area of $△$ BCD=$\frac{1}{2}a$

In $△$ ACD,AY is the medium

$\therefore$ Area of $△AYD=\frac{1}{2}Areaof△ADC$[Medium of a triangle divides it into two triangles of equal areas ]

$=\frac{1}{2}\times \frac{1}{2}a$

$=\frac{1}{4}a$

In $△ABX=\frac{1}{2}Areaof△ABC$

$=\frac{1}{2}\times \frac{1}{2}a$

$=\frac{1}{4}a$

In $△$ BCD, X and Y are the midpoints of sides BC and CD respectively.

$\therefore CY=\frac{1}{2}BC$

$CY=\frac{1}{2}CD$

$XY=\frac{1}{2}BD$

So, sides of $△$ CXY are half of the sides of the $△$ CBD.

Area of $△$ CXY=$\frac{1}{4}$ area of CBD

=$\frac{1}{4}\times \frac{1}{2}a$

=$\frac{1}{8}a$

Now, area of $△$ AXY=Area of ABCD-[Area of \triangle ADY+Area of \triangle ABX+\triangle CXY ]

=$a-[\frac{1}{4}a+\frac{1}{4}a+\frac{1}{8}a]$

=$a-\frac{5}{8}a$

=$\frac{3}{8}a$

=$\frac{3}{8}$ Area of a parallelogram ABCD