Question

The length of a rectangle is six times its width.

If the perimeter of a rectangle is $112$ ft, find its area.

Answer 1

Step-1: Finding the length and width of the rectangle:

Let the width $\left(\mathrm{W}\right)$ of the rectangle be $x$.

So, the length $\left(\mathrm{L}\right)$of the rectangle is $6x$.

The perimeter $\left(\mathrm{P}\right)$ of the rectangle is $112$ ft

$\mathrm{P}=2(\mathrm{L}+\mathrm{W})$

Put the values of $\mathrm{W}$,$\mathrm{L}$ and $\mathrm{P}$ as $x$, $6x$ and $112$ in the above formula:

$\begin{array}{rcl}& \Rightarrow & 112=2(x+6x)\\ & \Rightarrow & 112=2\left(7x\right)\\ & \Rightarrow & 112=14x\\ & \Rightarrow & x=\frac{112}{14}\\ & \Rightarrow & x=8\end{array}$

Hence, the width $\left(W\right)$ of the rectangle is $8$ ft and the length $\left(L\right)$ of the rectangle is $48$ ft.

Step-2: Find the area $\left(A\right)$ of the rectangle.

$\mathrm{A}=\mathrm{L}\times \mathrm{W}$

Put the values of $L$ and $W$ as $6$ and $48$ in the above formula:

$$\begin{gathered} \Rightarrow A=6\times 48 \\ \Rightarrow A=288 \end{gathered}$$

Hence, the area $\left(A\right)$ of the rectangle is $288$ square ft.