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(B) Let the length and breadth of the rectangle be $a$ and $b$ respectively.Given,diagonal $d = 25\, cm$ and area $A = 168\, cm^2$.We know that $a^2 +

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$24$

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(B) Let the length and breadth of the rectangle be $a$ and $b$ respectively.Given,diagonal $d = 25\, cm$ and area $A = 168\, cm^2$.We know that $a^2 + b^2 = d^2 = 25^2 = 625$.Also,$ab = 168$.Using the identity $(a+b)^2 = a^2 + b^2 + 2ab$,we get $(a+b)^2 = 625 + 2(168) = 625 + 336 = 961$.Therefore,$a+b = \sqrt{961} = 31\, cm$ (Equation $1$).Using the identity $(a-b)^2 = a^2 + b^2 - 2ab$,we get $(a-b)^2 = 625 - 336 = 289$.Therefore,$a-b = \sqrt{289} = 17\, cm$ (Equation $2$).Adding Equation $1$ and Equation $2$: $2a = 31 + 17 = 48$,which gives $a = 24\, cm$.Subtracting Equation $2$ from Equation $1$: $2b = 31 - 17 = 14$,which gives $b = 7\, cm$.Thus,the length of the rectangle is $24\, cm$.

If the measures of a diagonal and the area of a rectangle are $25\, cm$ and $168\, cm^2$ respectively,what is the length of the rectangle? (in $cm$)

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