Question

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Answer 1

Let the length of the side of the first and the second square be $x$ and y, respectively.

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}: \\ {x}^2+y^2=640...\left(\text{i}\right) \\ \mathrm{Also}, \\ 4x-4y=64 \\ \Rightarrow x-y=16 \\ \Rightarrow x=16+y \end{gathered}$$

Putting the value of $x$ in (i), we get:

$$\begin{gathered} x^2+y^2=640 \\ \Rightarrow {(16+y)}^2+y^2=640 \\ \Rightarrow 256+32y+y^2+y^2=640 \\ \Rightarrow 2y^2+32y-384=0 \\ \Rightarrow y^2+16y-192=0 \\ \Rightarrow y^2+(24-8)y-192=0 \\ \Rightarrow y^2+24y-8y-192=0 \\ \Rightarrow y(y+24)-8(y+24)=0 \\ \Rightarrow (y+24)(y-8)=0 \\ \Rightarrow y=-24\text{or}y=8 \\ \therefore y=8(\because \mathrm{S}\text{ide cannot be negative}) \\ \therefore x=16+y=16+8=24\text{m} \\ \text{Thus, the sides of the squares are 8 m and 24 m.} \end{gathered}$$