Question

The width of a rectangle is three more than twice the length.

The perimeter is $96$ inches.

Find the length and the width.

Answer 1

Finding the length and width of the rectangle:

Forming the expression for the width

Considering, length of rectangle $=x$

And width of rectangle $=y$

Then according to the question we can write,

$\Rightarrow y=2x+3...\left(i\right)$

Finding values of $x\&y$

$$\begin{gathered} \Rightarrow 2\left(\mathrm{x}+\mathrm{y}\right)=96\left[\because \mathrm{Given},\mathrm{perimeter}=96\right] \\ \Rightarrow \mathrm{x}+\mathrm{y}=48 \\ \Rightarrow \mathrm{x}+\left(2\mathrm{x}+3\right)=48\left[\mathrm{from}\left(\mathrm{i}\right)\right] \\ \Rightarrow 3\mathrm{x}=45 \\ \Rightarrow \mathrm{x}=15\mathrm{inch} \end{gathered}$$

Substituting the value of $x$ into $\left(i\right)$

$$\begin{gathered} \Rightarrow \mathrm{y}=15\times 2+3 \\ \Rightarrow \mathrm{y}=33\mathrm{inch} \end{gathered}$$

Hence, the length and width of rectangle are $15\mathrm{inch}\&33\mathrm{inch}$ respectively.