A rectangular carpet has an area of 60 , m^{2 | vedclass

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(D) Let the length of the carpet be $l$ and the breadth be $b$. Given,$l 	imes b = 60 \, m^{2}$.By the Pythagorean theorem,the diagonal $d = \sqrt{l^

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$12$

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(D) Let the length of the carpet be $l$ and the breadth be $b$. Given,$l 	imes b = 60 \, m^{2}$.By the Pythagorean theorem,the diagonal $d = \sqrt{l^{2} + b^{2}}$,so $d^{2} = l^{2} + b^{2}$.According to the problem,$d + l = 5b$,which implies $d = 5b - l$.Squaring both sides,$d^{2} = (5b - l)^{2} = 25b^{2} + l^{2} - 10bl$.Substituting $d^{2} = l^{2} + b^{2}$,we get $l^{2} + b^{2} = 25b^{2} + l^{2} - 10bl$.Simplifying,$b^{2} = 25b^{2} - 10(60)$,which gives $24b^{2} = 600$.Thus,$b^{2} = 25$,so $b = 5 \, m$.Since $l 	imes b = 60$,$l = 60 / 5 = 12 \, m$.

$A$ rectangular carpet has an area of $60 \, m^{2}$. Its diagonal and longer side together equal $5$ times the shorter side. The length of the carpet is.........$m$

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