In the figure, $ABCD$ is a trapezium in which $AB = 7\ cm, AD = BC = 5\ cm, DC = x\ cm$, and distance between $AB$ and $DC$ is $4\ cm$. Find the value of $x$ and area of trapezium $ABCD$. "
Given:
$ABCD$ is a trapezium in which $AB = 7\ cm, AD = BC = 5\ cm, DC = x\ cm$, and distance between $AB$ and $DC$ is $4\ cm$.
To do:
We have to find the value of $x$ and the area of trapezium $ABCD$.
Solution:
In $\triangle \mathrm{ADL}$,
$\mathrm{AD}^{2}=\mathrm{AL}^{2}+\mathrm{LD}^{2}$
$5^{2}=4^{2}+\mathrm{LD}^{2}$
$25=16+\mathrm{LD}^{2}$
$\mathrm{LD}^{2}=25-16$
$=9$
$=(3)^{2}$
$\Rightarrow \mathrm{LD}=3 \mathrm{~cm}$
Similarly,
$\mathrm{AL}$ and $\mathrm{BM}$ are perpendiculars on $\mathrm{CD}$$\mathrm{MC}=\mathrm{LD}=3 \mathrm{~cm}$
$\mathrm{LM}=\mathrm{AB}=7 \mathrm{~cm}$
Therefore,
$x=\mathrm{DL}+\mathrm{LM}+\mathrm{MC}$
$=3+7+3$
$=13 \mathrm{~cm}$
Area of trapezium $ABCD=\frac{1}{2}(a+b) \times h$
$=\frac{1}{2}(7+13) \times 4$
$=\frac{1}{2} \times 20 \times 4$
$=40 \mathrm{~cm}^{2}$
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