The perimeter of a square is double the perim | vedclass

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(D) Let the side of the square be $a$ and the length and breadth of the rectangle be $l$ and $b$ respectively.Perimeter of the square = $4a$.Perimeter

Vedclass · Accepted Answer

Cannot be determined

Vedclass · Answer

(D) Let the side of the square be $a$ and the length and breadth of the rectangle be $l$ and $b$ respectively.Perimeter of the square = $4a$.Perimeter of the rectangle = $2(l + b)$.According to the problem,$4a = 2 	imes [2(l + b)]$,which simplifies to $4a = 4(l + b)$,so $a = l + b$.The area of the rectangle is given as $l 	imes b = 240 \, cm^2$.We need to find the area of the square,which is $a^2 = (l + b)^2 = l^2 + b^2 + 2lb$.Since we only know the value of $lb = 240$,we cannot determine the value of $l^2 + b^2$ without knowing the individual values of $l$ and $b$ or their sum. Thus,the area of the square cannot be uniquely determined.

The perimeter of a square is double the perimeter of a rectangle. The area of the rectangle is $240 \, cm^2$. What is the area of the square? (in $cm^2$)

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