Question

In $||gm$ $ABCD,E$ and $F$ are mid-points of sides $AB$ and $BC,$ if $ar\left(BEF\right)=10c{m}^2$ then $ar\left(||gmABCD\right)=$

Answer 1

Given that$:$ $ABCD$ ia a $||gm.$ $X$ and $Y$ are the mid points of $BC$ and $CD$

Construction: Join $BD$

Since $X$ and $Y$ are the mid points of sides $BC$ and $CD$ respectively$,$

therefore in triangle $BCD,XY//BDandXY=1/2BD$

implies area of triangle $CYX=1/4$ area of triangle $DBC$

In triangle $BCD,$ if $X$ is the mid point of $BC$ and $Y$ is the mid pt of $CD$ then area triangle $CYX=1/4$ area triangle $DBC$

Implies area triangle $CYX=1/8area\left(||gmABCD\right)$

$[$Area of$||gm$ is twice the area of triangle made by the diagonal$]$

Since $||gmABCD$ and triangle $ABX$ are between same $||$ lines $AB$ and $BC$ and $BX=1/2BC$

Therefore$,$ area triangle $ABX=1/4area//gmABCD$

Similarly,$area△$AYD= 1/4 area //gm ABCD$

Now,$area△AXY=area\left(||gmABCD\right)-ar△ABX\$+arAYD+arCYX$

$=ar\left(||gmABCD\right)-\left(1/4+1/4+1/8\right)area\left(||gmABCD\right)$

$=area\left(||gmABCD\right)-5/8area\left(||gmABCD\right)$

$=3/8area\left(||gmABCD\right).$