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 $ABCD$.
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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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Updated on: 10-Oct-2022

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