Area of parallelogram ^@ABCD^@ is ^@x \space cm^2.^@ If ^@E, F, | Math Question and Answer

Answer:

$\dfrac{ x }{ 2 } cm^2$

Step by Step Explanation:

It is given that $, E, F, G$ and $H$ are respectively the mid-points of the sides of the parallelogram $ABCD.$
Let's join the midpoints $E, F, G$ and $H$ and join $HF, EG.$
Line $HF$ drawn from the midpoints of the parallelogram $ABCD$ devides the parallelogram into two equal parts.
$\therefore \text{ Area of the parallelogram } HFCD = \dfrac { \text{ Area of the parallelogram } ABCD } { 2 } = \dfrac { x } { 2 }$
Lines $HF$ and $EG$ drawn from the midpoints of the parallelogram $ABCD$ bisect each other other at the $O.$ $\therefore \text{ Area of the parallelogram } HOGD = \dfrac{ \text { Area of the parallelogram } HFCD } { 2 } = \dfrac { x }{ 4 }$
Diagonal $GH$ of the parallelogram $HOGD$ devides the parallelogram into two equal parts. $\therefore \text{ Area of HOG } = \dfrac { \text{ Area of the parallelogram HOGD } } { 2 } = \dfrac { x }{ 8 }$
Area of $FOG =$ Area of $FOE =$ Area of $EOH = \dfrac { x } { 8 }$
Thus the area of the $EFGH =$ = Area of $HOG$ + Area of $FOG +$ Area of $FOE +$ Area of $EOH = 4 \times \dfrac { x } { 8 } = \dfrac { x }{ 2 }$