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  • The area of a trapezium is 475 cm^{2} and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

The area of a trapezium is $475 \mathrm{~cm}^2$ and the height is 19 cm . Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

Answers (1)

Solution

Let the smaller parallel side be $CD = x$ cm

                  

Then other parallel side $A B=(x+4) \mathrm{cm}$
Given, area of trapezium $=475 \mathrm{~cm}^2$
Height DE $=19 \mathrm{~cm}$
We know that, Area oftrapezium $=\frac{1}{2} \times$ height $\times($ sum of parallel sides $)$

$
\begin{aligned}
& 475=\frac{1}{2} \times D E \times(D C+A B) \\
& 475 \times 2=19 \times(x+x+4)
\end{aligned}
$

$\begin{aligned} & \frac{475 \times 2}{19}=2 x+4 \\ & 25 \times 2=2 x+4 \\ & 50=2 x+4 \\ & 50-4=2 x \\ & 46=2 x \\ & x=\frac{46}{2} \\ & x=23 \mathrm{~cm}\end{aligned}$

So the smaller side CD is $23$ cm and the other parallel side AB is $(23 + 4)$ cm = $27$ cm

 

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infoexpert21

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