Question

The sum of the areas of two squares is 157 m². If the sum of their perimeters is 68 m, find the sides of the two squares.

Answer 1

Let the length of the side of one square be x m and the length of the side of another square be y m.

According to the question,
$$\begin{gathered} x^2+y^2=157...\left(1\right) \\ 4x+4y=68...\left(2\right) \\ 4x+4y=68 \\ \Rightarrow 4\left(x+y\right)=68 \\ \Rightarrow x+y=17 \\ \Rightarrow x=17-y...\left(3\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ {\left(17-y\right)}^2+y^2=157 \\ \Rightarrow 289+y^2-34y+y^2=157 \\ \Rightarrow 2y^2-34y+289-157=0 \\ \Rightarrow 2y^2-34y+132=0 \\ \Rightarrow 2\left(y^2-17y+66\right)=0 \\ \Rightarrow y^2-17y+66=0 \\ \Rightarrow y^2-11y-6y+66=0 \\ \Rightarrow y\left(y-11\right)-6\left(y-11\right)=0 \\ \Rightarrow \left(y-6\right)\left(y-11\right)=0 \\ \Rightarrow y=6,11...\left(4\right)\left(\because y>0\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get} \\ x=17-6\mathrm{and}x=17-11 \\ \Rightarrow x=11,6 \\ \mathrm{Hence},\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{squares}\mathrm{are}11\mathrm{m}\mathrm{and}6\mathrm{m}. \end{gathered}$$