The area of a trapezium is 475 , cm^2 and the | vedclass

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(A) The area of a trapezium is given by the formula: $	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.Let the le

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$23 \, cm, 27 \, cm$

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(A) The area of a trapezium is given by the formula: $	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.Let the length of one parallel side be $x \, cm$.Then,the length of the other parallel side is $(x + 4) \, cm$.Given,$	ext{Area} = 475 \, cm^2$ and $	ext{height} = 19 \, cm$.Substituting these values into the formula:$475 = \frac{1}{2} 	imes (x + x + 4) 	imes 19$$475 = \frac{1}{2} 	imes (2x + 4) 	imes 19$$475 = (x + 2) 	imes 19$$x + 2 = \frac{475}{19} = 25$$x = 25 - 2 = 23 \, cm$.Therefore,the two parallel sides are $23 \, cm$ and $(23 + 4) \, cm$,which are $23 \, cm$ and $27 \, cm$.

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.

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