Draw a rectangle that meets both of the following conditions and label the rectangle’s length and width. The perimeter of the rectangle is $36units$ and the length of the rectangle is $5$ times its width.

Hint: Here, in the given question, we need to draw and label a rectangle that meets both of the given conditions. We are given that the perimeter of the rectangle is $36units$ and the length of the rectangle is $5$ times its width. So, first we need to find the width of the rectangle. As we know, the perimeter is the total length of the boundary of any closed 2-dimensional shape. Perimeter of the rectangle is $2\left( {l + w} \right)$ where $l $ is the length and $w$ is the width of the rectangle. As here we are given that the length of the rectangle is $5$ times its width, so we will substitute the value of length in terms of width in the perimeter formula.

Complete step by step answer:
As it is given that the length of the rectangle is $5$ times its width.
$ \Rightarrow l = 5w$
As we know, the perimeter of a rectangle = $2\left( {l + w} \right)$.
On substituting the value of perimeter = $36units$ and $l = 5w$, we get
$ \Rightarrow 36 = 2\left( {5w + w} \right)$
In addition to terms, we get
$ \Rightarrow 36 = 2\left( {6w} \right)$
On multiplication of terms, we get
$ \Rightarrow 36 = 12w$
$ \Rightarrow w = \dfrac{{36}}{{12}}$
On division, we get
$ \Rightarrow w = 3units$
As we know $l = 5w$. Therefore on substituting the value of $w$, get
$ \Rightarrow l = 5 \times 3$
$ \Rightarrow l = 15units$


Note: The key concept to solve this type of question is to learn the formulas of different shapes. Don’t get confused between perimeter and area. Perimeter is the distance around the outside of a shape and area measures the space inside a shape. We should take care of the calculations so as to be sure of our final answer. Remember to write units.