The length of a rectangle exceeds its breadth | vedclass

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Question

(C) Let the breadth of the rectangle be $x \, cm$. Then,the length of the rectangle is $(x + 3) \, cm$. According to the problem,the numerical value o

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let the breadth of the rectangle be $x \, cm$. Then,the length of the rectangle is $(x + 3) \, cm$. According to the problem,the numerical value of the area is equal to the numerical value of the perimeter. Area $= 	ext{Length} 	imes 	ext{Breadth} = x(x + 3) \, cm^2$. Perimeter $= 2(	ext{Length} + 	ext{Breadth}) = 2(x + 3 + x) = 2(2x + 3) = (4x + 6) \, cm$. Equating the two: $x(x + 3) = 4x + 6$. $x^2 + 3x = 4x + 6$. $x^2 - x - 6 = 0$. Factoring the quadratic equation: $(x - 3)(x + 2) = 0$. This gives $x = 3$ or $x = -2$. Since the breadth cannot be negative,the breadth is $3 \, cm$.

The length of a rectangle exceeds its breadth by $3 \, cm$. If the numerical values of the area and the perimeter of the rectangle are equal,then the breadth is ...... $cm$.

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