Philippines
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.^@
Answer:
^@ \dfrac{ x }{ 2 } cm^2 ^@
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 }^@
