The length of a rectangle is twice its breadt | vedclass

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(C) Let the breadth of the rectangle be $x\, cm$. Then,the length is $2x\, cm$.The initial area of the rectangle is $A_1 = 	ext{length} 	imes 	ext{

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Let the breadth of the rectangle be $x\, cm$. Then,the length is $2x\, cm$.The initial area of the rectangle is $A_1 = 	ext{length} 	imes 	ext{breadth} = (2x)(x) = 2x^2\, cm^2$.According to the problem,the new length is $(2x - 5)\, cm$ and the new breadth is $(x + 5)\, cm$.The new area is $A_2 = (2x - 5)(x + 5) = 2x^2 + 10x - 5x - 25 = 2x^2 + 5x - 25$.Given that the area increases by $75\, cm^2$,we have $A_2 = A_1 + 75$.Substituting the expressions: $2x^2 + 5x - 25 = 2x^2 + 75$.Subtracting $2x^2$ from both sides: $5x - 25 = 75$.$5x = 100 \Rightarrow x = 20\, cm$.The length of the rectangle is $2x = 2 	imes 20 = 40\, cm$.

The length of a rectangle is twice its breadth. If its length is decreased by $5\, cm$ and breadth is increased by $5\, cm$,the area of the rectangle is increased by $75\, cm^2$. Then the length of the rectangle is......$cm$.

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