Question

The perimeters of two squares are $40$ and $96$ metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.

Answer 1

Given , Perimeter of first square = $40m.$

Perimeter of second square = $96m.$

Let the side of the first square = $xm$and side of second square = $ym.$

and side of another square whose perimeter we have to find = $zm.$

We know that perimeter of a square = $4\times$ side

So , Perimeter of first square = $4x$ and Perimeter of second square = $4y.$

According to question ,

$$\begin{gathered} 4x=40 \\ orx=\frac{40}{4}=10m \end{gathered}$$

and ,

$$\begin{gathered} 4y=96 \\ \Rightarrow y=\frac{96}{4} \\ \Rightarrow y=24m \end{gathered}$$

According to given question ,

Area of another square = Area of square $1$ + Area of square $2$

$$\begin{gathered} z^2={\left(10\right)}^2+{\left(24\right)}^2 \\ \Rightarrow z^2=100+576 \\ \Rightarrow z^2=676 \\ \Rightarrow z=\sqrt{676} \\ \Rightarrow z=26m \end{gathered}$$

Perimeter of square = $4\times 26m=104m.$

Hence ,the perimeter of another square equal in area to the sum of the first two squares is $104m.$