Question

I’m stuck on trying to find the perimeter of the rectangle

Answer 1

We are told that the rectangle is 5 times as long as it is wide. This means that the length of the rectangle is 5 times its width.

Let the length of the rectangle be L and let the width be w, then, it means that

`\begin{gathered} L=5* w \\ L=5w \end{gathered}`

Also we were told the area is 20 square-feet. But area of a rectangle is calculated as

`\begin{gathered} Area=length* width \\ A=L* w \\ We\text{ have L = 5w, so } \\ A=5w* w \\ A=5w^2,\text{ but the area is 20 ft}^2,\text{ so } \\ 20=5w^2 \end{gathered}`

Solving for w in the equation we just made, we have

`\begin{gathered} 20=5w^2 \\ dividing\text{ by 5} \\ (20)/(5)=(5w^2)/(5) \\ 4=w^2 \\ w^2=4 \\ take\text{ the square to 4, it becomes square root, we have } \\ w=√(4) \\ w=2\text{ ft} \end{gathered}`

So the width is 2 ft, the length becomes

`\begin{gathered} L=5w \\ L=5*2 \\ L=10\text{ ft } \end{gathered}`

Perimeter of a rectangle is calculated using the formula

`\begin{gathered} Perimeter=2(L+w) \\ P=2(L+w) \\ P=2(10+2) \\ P=2(12) \\ P=2*12 \\ P=24\text{ ft } \end{gathered}`

Hence the perimeter is 24 ft