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

The area of a trapezium is $475c{m}^{2}$ and the height is 19 cm .Find the lengths of its two paallel sides if one side is 4 cm greater than the other.
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Step 1
We know that the area of a trapezium is
A $\frac{1}{2}×\left(\text{sum of parallel lines}\right)×\left(\text{distance between parallel lines}\right)$
Step 2
$A=475c{m}^{2}$
$A=19$
Let one parallel side length x
Another parallel side length $=x+4$
$A=\frac{1}{2}×\left(x+x+4\right)×19$
$475=\frac{1}{2}×\left(2x+4\right)×19$
$2x+4=\frac{2×475}{19}$
$2x+4=50$
$2x=46$
$x=23$
Second parallel side $x+4$
$=23+4$
$=27cm$
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delirija7z
Consider x cm as the smaller side of the trapeziumSo the larger parallel can be written as $\left(x+4\right)cm$We know that the area of trapezium $\frac{1}{2}×$ sum of parallel sides x heightBy substituting the values$475=\frac{1}{2}×\left(x+\left(x+4\right)\right)×19$On further calculation$25=\frac{1}{2}×\left(2x+4\right)$So we get$50=2x+4$By subtraction$2x=50-42$$x=46$
By division
$x=23cm$
So we get Larger parallel side $\left(x+4\right)=23+4=27cm$
Therefore, the length of two parallel sides is 23cm and 27cm.