Question

the length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the are of the rectangle would have increased by 140 square metres. Find its dimensions

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rectangle}=x\mathrm{m} \\ \mathrm{Let}\mathrm{the}\mathrm{breadth}\mathrm{of}\mathrm{the}\mathrm{rectangle}=y\mathrm{m} \\ \mathrm{area}\mathrm{of}\mathrm{rectangle}=xy{\mathrm{m}}^2 \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{condition}, \\ x=5+y.......\left(1\right) \\ \mathrm{Now},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{when}\mathrm{the}\mathrm{breadth}\mathrm{is}\mathrm{doubled}\mathrm{and}\mathrm{length}\mathrm{is}\mathrm{decreased}\mathrm{by}9\mathrm{m},\mathrm{then}\mathrm{area}\mathrm{is}\mathrm{increased}\mathrm{by}140{\mathrm{m}}^2. \\ \mathrm{so},\left(x-9\right)2y=xy+140 \\ \Rightarrow 2xy-18y-xy-140=0 \\ \Rightarrow xy-18y-140=0 \\ \Rightarrow y\left(5+y\right)-18y-140=0\left[\mathrm{using}\left(1\right)\right] \\ \Rightarrow y^2+5y-18y-140=0 \\ \Rightarrow y^2-13y-140=0 \\ \Rightarrow y^2-20y+7y-140=0 \\ \Rightarrow y\left(y-20\right)+7\left(y-20\right)=0 \\ \Rightarrow \left(y-20\right)\left(y+7\right)=0 \\ \Rightarrow y-20=0ory+7=0 \\ \Rightarrow y=20ory=-7\left[\mathrm{rejected}\right] \\ \mathrm{so},\mathrm{breadth}\mathrm{of}\mathrm{rectangle}=y=20\mathrm{m} \\ \mathrm{length}\mathrm{of}\mathrm{rectangle}=x=5+y=25\mathrm{m} \end{gathered}$$