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Area of parallelogram $ABCD$ is $x \space cm^2.$ If $E, F, G,$ and $H$ are mid-points of the sides, find the area of $EFGH.$

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Answer:

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

Step by Step Explanation:

1. 
   It is given that$, E, F, G$ and $H$ are respectively the mid-points of the sides of the parallelogram $ABCD.$
2. Let's join the midpoints $E, F, G$ and $H$ and join $HF, EG.$
   
3. Line $HF$ drawn from the midpoints of the parallelogram$ABCD$ devides the parallelogram into two equal parts.
   $$\therefore \text{ Area of the parallelogram } HFCD = \dfrac { \text{ Area of the parallelogram } ABCD } { 2 } = \dfrac { x } { 2 }$$
4. Lines $HF$ and $EG$ drawn from the midpoints of the parallelogram $ABCD$ bisect each other other at the $O.$ $$\therefore \text{ Area of the parallelogram } HOGD = \dfrac{ \text { Area of the parallelogram } HFCD } { 2 } = \dfrac { x }{ 4 }$$
5. Diagonal $GH$ of the parallelogram $HOGD$ devides the parallelogram into two equal parts. $$\therefore \text{ Area of HOG } = \dfrac { \text{ Area of the parallelogram HOGD } } { 2 } = \dfrac { x }{ 8 }$$
6. Area of $FOG =$ Area of $FOE =$ Area of $EOH = \dfrac { x } { 8 }$
7. Thus the area of the $EFGH =$ = Area of $HOG$ + Area of $FOG +$ Area of $FOE +$ Area of $EOH = 4 \times \dfrac { x } { 8 } = \dfrac { x }{ 2 }$

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