In the figure, $ABCD$ is a trapezium in which $AB \parallel DC$ and $DC = 40\ cm$ and $AB = 60\ cm$. If $X$ and $Y$ are respectively the mid-points of $AD$ and $BC$, prove that $XY = 50\ cm$.
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Given:

$ABCD$ is a trapezium in which $AB \parallel DC$ and $DC = 40\ cm$ and $AB = 60\ cm$.

$X$ and $Y$ are respectively the mid-points of $AD$ and $BC$.

To do:

We have to prove that $XY = 50\ cm$.

Solution:

Join $DY$ and produce it to meet $AB$ produced at $P$.

$\mathrm{DC}=40 \mathrm{~cm}, \mathrm{AB}=60 \mathrm{~cm}$

This implies,

$X Y=\frac{1}{2}(A B+D C)$

$=\frac{1}{2}(60+40) \mathrm{cm}$

$=\frac{100}{2}$

$=50 \mathrm{~cm}$

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Updated on: 10-Oct-2022

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