Question

the perimeter of a rectangle is 103 m and its area is 640 m sq. find its length and breadth.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{be}\mathrm{the}\mathrm{length}\mathrm{and}\mathrm{breadth}\mathrm{of}\mathrm{rectangle}. \\ \mathrm{SInce},\mathrm{perimieter}=103\mathrm{m} \\ 2\mathrm{x}+2\mathrm{y}=103--------\left(1\right) \\ \mathrm{Since},\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{rectangle}=\mathrm{xy} \\ \mathrm{xy}=640 \\ \mathrm{x}\left(\frac{103-2x}{2}\right)=640\left\{\mathrm{from}\left(1\right)\right\} \\ 103x-2x^2=1280 \\ 2{\mathrm{x}}^2-103\mathrm{x}+1280=0 \\ x=\frac{103\pm \sqrt{{\left(103\right)}^2-8\left(1280\right)}}{4}=\frac{103\pm 19.2}{4} \\ x=30.55,x=20.95 \\ implies,y=20.94andy=30.55 \\ \mathrm{Thus},\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{sets}\mathrm{of}\mathrm{possible}\mathrm{values}\mathrm{of}\mathrm{length}\mathrm{and}\mathrm{breadth}. \end{gathered}$$