# The width of a rectangle is one-fifth as much as the length and the perimeter is 120 cm, how do you find the length and width?

Jul 21, 2015

$L = 50$
$W = 10$

#### Explanation:

Questions of these type are often called Worded Problems.

This is because the key equations that you will use, are already described for you in the question.

Here's how it goes:

Let the length be $L$ and the width be $W$ and the perimeter is $P$

The question says: "the width$\left(W\right)$ is one-fifth$\left(\frac{1}{5}\right)$ as much as the length$\left(L\right)$"

In mathematical terms this means,

$W = \frac{1}{5} L$

$\implies L = 5 W \ldots \ldots \ldots \textcolor{red}{\left(1\right)}$

Secondly: "the perimeter$\left(P\right)$ is $120$"

Meaning, $P = 120$

The Perimeter of any rectangle $= 2 \times \left(L + W\right)$

Hence, $2 \left(L + W\right) = 120$

$\implies L + W = 60 \ldots \ldots \ldots \textcolor{red}{\left(2\right)}$

Now, all that is left to do is to solve $\textcolor{red}{\left(1\right)}$ and $\textcolor{red}{\left(2\right)}$ simultaneously.

$5 W + W = 60 \implies W = 10$

$\implies L = 5 \cdot \left(10\right) = 50$

And You're Done.