Question

the length of rectangle exceeds its breadth by 14 cm. Length is decreased by 8 cm and breadth is increased by 6 cm, the area of the both same. Find the length and breadth

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{original}\mathrm{breadth}\mathrm{of}\mathrm{the}\mathrm{rectangle}=\mathrm{x}\mathrm{cm} \\ \mathrm{Then},\mathrm{original}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rectangle}=\left(\mathrm{x}+14\right)\mathrm{cm} \\ \mathrm{Original}\mathrm{area}=\mathrm{length}\times \mathrm{breadth} \\ =\left(\mathrm{x}+14\right)\times \mathrm{x} \\ =\left({\mathrm{x}}^2+14\mathrm{x}\right){\mathrm{cm}}^2 \\ \mathrm{New}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rectangle}=\left(\mathrm{x}+14-8\right)\mathrm{cm}=\left(\mathrm{x}+6\right)\mathrm{cm} \\ \mathrm{new}\mathrm{breadth}\mathrm{of}\mathrm{the}\mathrm{rectangle}=\left(\mathrm{x}+\right)\mathrm{cm} \\ \mathrm{New}\mathrm{area}=\left(\mathrm{x}+6\right)\left(\mathrm{x}+6\right)=\left({\mathrm{x}}^2+36+12\mathrm{x}\right){\mathrm{cm}}^2 \\ \mathrm{According}\mathrm{to}\mathrm{question}: \\ {\mathrm{x}}^2+14\mathrm{x}={\mathrm{x}}^2+12\mathrm{x}+36 \\ \Rightarrow 14\mathrm{x}-12\mathrm{x}=36 \\ \Rightarrow 2\mathrm{x}=36 \\ \Rightarrow \mathrm{x}=\frac{36}{2}=18 \\ \mathrm{So},\mathrm{original}\mathrm{breadth}=18\mathrm{cm} \\ \mathrm{original}\mathrm{length}=\left(18+14\right)=32\mathrm{cm} \end{gathered}$$