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(C) Let the parallel sides of the trapezium be $l_1$ and $l_2$,where $l_1 > l_2$.Given that the difference between the parallel sides is $4\, cm$,so $

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$27\, cm, 23\, cm$

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(C) Let the parallel sides of the trapezium be $l_1$ and $l_2$,where $l_1 > l_2$.Given that the difference between the parallel sides is $4\, cm$,so $l_1 - l_2 = 4$,which implies $l_1 = l_2 + 4$.The perpendicular distance (height) between the parallel sides is $h = 19\, cm$.The area of a trapezium is given by the formula: $	ext{Area} = \frac{1}{2} 	imes (l_1 + l_2) 	imes h$.Substituting the given values: $475 = \frac{1}{2} 	imes (l_2 + 4 + l_2) 	imes 19$.$475 = \frac{19}{2} 	imes (2l_2 + 4)$.$475 = 19 	imes (l_2 + 2)$.$l_2 + 2 = \frac{475}{19} = 25$.$l_2 = 25 - 2 = 23\, cm$.Now,find $l_1$: $l_1 = l_2 + 4 = 23 + 4 = 27\, cm$.Thus,the lengths of the parallel sides are $27\, cm$ and $23\, cm$.

The difference between two parallel sides of a trapezium is $4\, cm$. The perpendicular distance between them is $19\, cm$. If the area of the trapezium is $475\, cm^2$,find the length of the parallel sides.

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