The areas of a square and a rectangle are equ | vedclass

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(D) Let the side of the square be $a 	ext{ cm}$.Therefore,the length of the rectangle $l = (a + 5) 	ext{ cm}$ and the breadth of the rectangle $b =

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$34$

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(D) Let the side of the square be $a 	ext{ cm}$.Therefore,the length of the rectangle $l = (a + 5) 	ext{ cm}$ and the breadth of the rectangle $b = (a - 3) 	ext{ cm}$.Given that the area of the square is equal to the area of the rectangle:$a^2 = (a + 5)(a - 3)$Expanding the right side:$a^2 = a^2 - 3a + 5a - 15$$a^2 = a^2 + 2a - 15$Subtracting $a^2$ from both sides:$0 = 2a - 15$$2a = 15$$a = 7.5 	ext{ cm}$Now,calculate the dimensions of the rectangle:Length $l = 7.5 + 5 = 12.5 	ext{ cm}$Breadth $b = 7.5 - 3 = 4.5 	ext{ cm}$The perimeter of the rectangle is $P = 2(l + b)$:$P = 2(12.5 + 4.5) = 2(17) = 34 	ext{ cm}$.

The areas of a square and a rectangle are equal. The length of the rectangle is greater than the side of the square by $5 \text{ cm}$ and the breadth is less than the side of the square by $3 \text{ cm}$. Find the perimeter of the rectangle (in $\text{cm}$).

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