# If E,F,G and $\mathrm{H}$ are respectively the mid-points of the sides of a parallelogram $\mathrm{ABCD}$, show that $\operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$

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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,

$\operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$

Updated on 10-Oct-2022 13:41:36

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