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(A) Let $l$ and $b$ be the length and breadth of the rectangular paper.When folded along the breadth,the new dimensions are $l/2$ and $b$. The perimet

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(A) Let $l$ and $b$ be the length and breadth of the rectangular paper.When folded along the breadth,the new dimensions are $l/2$ and $b$. The perimeter is $2(l/2 + b) = 34$,which simplifies to $l + 2b = 34$ $....(1)$When folded along the length,the new dimensions are $l$ and $b/2$. The perimeter is $2(l + b/2) = 38$,which simplifies to $2l + b = 38$ $....(2)$Multiplying equation $(1)$ by $2$,we get $2l + 4b = 68$ $....(3)$Subtracting equation $(2)$ from equation $(3)$,we get $(2l + 4b) - (2l + b) = 68 - 38$,which gives $3b = 30$,so $b = 10\, cm$.Substituting $b = 10$ into equation $(1)$,we get $l + 2(10) = 34$,so $l + 20 = 34$,which gives $l = 14\, cm$.The area of the paper is $l 	imes b = 14 	imes 10 = 140\, cm^2$.

$A$ rectangular paper,when folded into two congruent parts,has a perimeter of $34\, cm$ for each part when folded along one set of sides and a perimeter of $38\, cm$ when folded along the other set of sides. What is the area of the original paper in $cm^2$?

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