If E,F,G and $ \mathrm{H} $ are respectively the mid-points of the sides of a parallelogram $ \mathrm{ABCD} $, show that $ \operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) $
Given:
$E,F,G$ and \( \mathrm{H} \) are respectively the mid-points of the sides of a parallelogram \( \mathrm{ABCD} \)
To do:
We have to show that \( \operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \).
Solution:
Join $EF, FG, GH, HE$ and $FH$.
We know that,
Opposite sides of a parallelogram are equal and parallel.
This implies,
$A D \| B C$
$A D=B C$
$\Rightarrow \frac{1}{2} A D=\frac{1}{2}B C$
$\mathrm{AH} \| \mathrm{BF}$
$\mathrm{DH} \| \mathrm{CF}$
$\Rightarrow \mathrm{AH}=\mathrm{BF}$
$\mathrm{DH}=\mathrm{CF}$
$\mathrm{H}$ and $\mathrm{F}$ are mid points
Therefore, $\mathrm{ABFH}$ and $\mathrm{HFCD}$ are parallelograms.
$\triangle \mathrm{EFH}$ and parallelogram $\mathrm{ABFH}$ both lie on the same base $\mathrm{FH}$ and between the same parallels $\mathrm{AB}$ and $\mathrm{HF}$.
This implies,
Area of $\triangle \mathrm{EFH}=\frac{1}{2}\times$ area of $\mathrm{ABFH}$.......(i)
Area of $\triangle \mathrm{GHF}=\frac{1}{2}\times$ area of $\mathrm{HFCD}$..........(ii)
Adding (i) and (ii), we get,
Area of $\triangle \mathrm{EFH}+$ area of $\triangle \mathrm{GHF}=\frac{1}{2}\times$ area of $\mathrm{ABFH}+\frac{1}{2}\times$ area of $\mathrm{HFCD}$
Area of $\mathrm{EFGH}=$ area of $\mathrm{ABFH}$
Therefore,
$\operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$
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