Question

A piece of wire $44\text{cm}$ long is cut into two parts and each part is bent to form a square.

If the total area of the two squares is $65{\text{cm}}^2$, find the perimeter of the two squares?

Answer 1

Step 1. Recall some facts about squares:

Recall that the perimeter of a square is four times the length of its sides.

Since, $\text{Perimeter=4×Side}$, we can write $\frac{\text{Perimeter}}{4}\text{=Side}$.

Also recall that the area of a square is the square of the length of its sides.

Since, $\text{Area of Square =}{\left(\text{Side}\right)}^2$, and $\frac{\text{Perimeter}}{4}\text{=Side}$, we can write $\text{Area of Square =}{\left(\frac{\text{Perimeter}}{4}\right)}^2$.

Step 2. Model the given situation as a quadratic equation:

Let the first part have the length $x$.

Since, the total length of the given piece of wire is $44\text{cm}$, the length of the second part is $44\text{cm}-x$.

It is also given that each of the two parts is bent to form a square.

This means the length of a piece of wire will be the perimeter of the respective square.

Thus the first square has the perimeter $x$ and the second square has the perimeter $44\text{cm}-x$.

Applying $\text{Area of Square =}{\left(\frac{\text{Perimeter}}{4}\right)}^2$, we also get that the area of the first square is ${\left(\frac{x}{4}\right)}^2$ and that of the second square is ${\left(\frac{44\text{cm}-x}{4}\right)}^2$.

The sum of the areas of the squares can be written as ${\left(\frac{x}{4}\right)}^2+{\left(\frac{44\text{cm}-x}{4}\right)}^2$.

It is given that the sum of the areas of the two squares is $65{\text{cm}}^2$.

Thus, write ${\left(\frac{x}{4}\right)}^2+{\left(\frac{44\text{cm}-x}{4}\right)}^2=65{\text{cm}}^2$.

Step 3. Solve the quadratic equation ${\left(\frac{x}{4}\right)}^{\mathbf{2}}\mathbf{+}{\left(\frac{44\text{cm}-x}{4}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{65}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{2}}$.

$\begin{array}{rcl}& \Rightarrow & \frac{x^2}{4^2}\mathrm{+}\frac{{\left(44\text{cm}-x\right)}^2}{4^2}=65{\text{cm}}^2\\ & \Rightarrow & \frac{x^2+{\left(44\text{cm}-x\right)}^2}{4^2}=65{\text{cm}}^2\\ & \Rightarrow & x^2+{\left(44\text{cm}-x\right)}^2=4^2\times 65{\text{cm}}^2\\ & \Rightarrow & x^2+{\left(44\text{cm}\right)}^2+x^2-2\times \left(44\text{cm}\right)x=4^2\times 65{\text{cm}}^2\text{(Applied the identity}{\left(a-b\right)}^2=a^2+b^2-2ab\text{)}\\ & \Rightarrow & 2x^2-\left(88\text{cm}\right)x+1936{\text{cm}}^2=1040{\text{cm}}^2\\ & \Rightarrow & 2x^2-\left(88\text{cm}\right)x+1936{\text{cm}}^2-1040{\text{cm}}^2=0\\ & \Rightarrow & 2x^2-\left(88\text{cm}\right)x+896{\text{cm}}^2=0\\ & \Rightarrow & x^2-\left(44\text{cm}\right)x+448{\text{cm}}^2=0\text{(Divided both sides of the equation by 2)}\\ & \Rightarrow & x^2-\left(16\text{cm}\right)x-\left(28\text{cm}\right)x+448{\text{cm}}^2=0\text{(Split the middle term to factorize.)}\\ & \Rightarrow & x\left(x-16\text{cm}\right)-\left(28\text{cm}\right)\left(x-16\text{cm}\right)=0\\ & \Rightarrow & \left(x-16\text{cm}\right)\left(x-28\text{cm}\right)=0\\ & \Rightarrow & x=16\text{cm},28\text{cm}\\ & & \end{array}$

Thus, $x=16\text{cm}$ and $x=28\text{cm}$ are solutions of ${\left(\frac{x}{4}\right)}^2+{\left(\frac{44\text{cm}-x}{4}\right)}^2=65{\text{cm}}^2$.

Step 4. Interpret the solutions.

The given situation is satisfied if the length of the first part is $x=16\text{cm}$ or $x=28\text{cm}$.

The corresponding lengths of the second parts are $44-16\text{cm}=28\text{cm}$ and $44-28\text{cm}=16\text{cm}$ respectively.

Thus, in both cases, the lengths of the two parts are $16\text{cm}$ and $28\text{cm}$.

Since the lengths of the parts are equal to the perimeters of the squares made by bending the respective part.

The perimeters of the two squares are $16\text{cm}$ and $28\text{cm}$.

Hence, the perimeters of the two squares are $\mathbf{16}\mathbf{}\mathbf{\text{cm}}$ and $\mathbf{28}\mathbf{}\mathbf{\text{cm}}$.