Question

The sum of the areas of the two squares is $468{\mathrm{m}}^2$. If the difference of their perimeters is $24\mathrm{m}$, find the sides of the two squares.

Answer 1

Step 1: Find the sides of the two squares:

Given: The sum of the areas of the two squares is $468{\mathrm{m}}^2$.

The difference of their perimeters is $24\mathrm{m}$.

Let the side of the first square be $\mathrm{x}\mathrm{m}$ and that of the second square be $\mathrm{y}\mathrm{m}$.

Step 2: Forming a linear equation in variable $\mathrm{x}$ and $\mathrm{y}$:

The perimeter of the first square $=4\mathrm{x}$ (Perimeter of a square $=4\times \mathrm{side}\mathrm{of}\mathrm{square}$)

The perimeter of the second square $=4\mathrm{y}$

Area of the first square $={\mathrm{x}}^2$ (Area of a square $={\left(\mathrm{side}\right)}^2$)

Area of the second square $={\mathrm{y}}^2$

According to the question,

Difference of their perimeters $=24\mathrm{m}$

$$\begin{gathered} 4\mathrm{x}-4\mathrm{y}=24 \\ 4\left(\mathrm{x}-\mathrm{y}\right)=24 \\ \mathrm{x}-\mathrm{y}=6 \\ \mathrm{x}=\mathrm{y}+6............\left(1\right) \end{gathered}$$

Step 3: Forming a quadratic equation and solve it:

Also given, the sum of the areas of two squares $=468{\mathrm{m}}^2$

${\mathrm{x}}^2+{\mathrm{y}}^2=468............\left(2\right)$

Substituting the value of $\mathrm{x}$ from equation $\left(1\right)$ in the equation $\left(2\right)$, We get;

$\begin{array}{rcl}{\left(\mathrm{y}+6\right)}^2+{\mathrm{y}}^2& =& 468\end{array}$

$\begin{array}{rcl}& \Rightarrow & {\mathrm{y}}^2+12\mathrm{y}+36+{\mathrm{y}}^2=468\end{array}$ (By using the algebraic identity $\begin{array}{rcl}{\left(\mathrm{a}+\mathrm{b}\right)}^2& =& {\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2\end{array}$)

$\begin{array}{rcl}& \Rightarrow & 2{\mathrm{y}}^2+12\mathrm{y}+36=468\end{array}$

$\begin{array}{rcl}& \Rightarrow & {\mathrm{y}}^2+6\mathrm{y}+18=234\end{array}$ (Dividing by the common factor $2$)

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & {\mathrm{y}}^2+6\mathrm{y}+18-234=0\end{array} \\ \Rightarrow \begin{array}{rcl}& & \mathrm{y}\end{array}\begin{array}{rcl}& & \end{array} \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & {\mathrm{y}}^2+18\mathrm{y}-12\mathrm{y}-216=0\end{array}$ (By splitting the middle term)

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & \mathrm{y}\left(\mathrm{y}+18\right)-12\left(\mathrm{y}+18\right)=0\end{array} \\ \Rightarrow \left(\mathrm{y}+18\right)\left(\mathrm{y}-12\right)=0 \end{gathered}$$

Thus $\mathrm{y}=-18,12$

Since the side of a square cannot be negative, therefore $\mathrm{y}=-18$ cannot be possible.

Hence $\mathrm{y}=12$

Substitute $\mathrm{y}=12$ in equation $\left(1\right)$

$$\begin{gathered} \Rightarrow \mathrm{x}=12+6 \\ \Rightarrow \mathrm{x}=18 \end{gathered}$$

Hence $\mathrm{y}=12\mathrm{m}$ and $\mathrm{x}=18\mathrm{m}$

Therefore the sides of two squares are $12\mathrm{m}$and $18\mathrm{m}$.