The length of a rectangle is one more than four times its width. if the perimeter of the rectangle is 62 meters, how do you find the dimensions of the rectangle?

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Answer 1 · 2017-01-08T23:38:55

See full process for how to solve this problem below in the Explanation:

First, let's define the length of the rectangle as $l$ $l$ and the width of the rectangle as $w$ $w$ .

Next, we can write the relationship between the length and width as:

$l = 4 w + 1$ $l = 4w + 1$

We also know the formula for the perimeter of a rectangle is:
$p = 2 l + 2 w$ $p = 2l + 2w$

Where:

$p$ $p$ is the perimeter
$l$ $l$ is the length
$w$ $w$ is the width

We can now substitute $4 w + 1$ $color(red)(4w + 1)$ for $l$ $l$ in this equation and 62 for $p$ $p$ and solve for $w$ $w$ :

$62 = 2 (4 w + 1) + 2 w$ $62 = 2(color(red)(4w + 1)) + 2w$

$62 = 8 w + 2 + 2 w$ $62 = 8w + 2 + 2w$

$62 = 8 w + 2 w + 2$ $62 = 8w + 2w + 2$

$62 = 10 w + 2$ $62 = 10w + 2$

$62 - 2 = 10 w + 2 - 2$ $62 - color(red)(2) = 10w + 2 - color(red)(2)$

$60 = 10 w + 0$ $60 = 10w + 0$

$60 = 10 w$ $60 = 10w$

$\frac{60}{10} = \frac{10 w}{10}$ $60/color(red)(10) = (10w)/color(red)(10)$

$6 = (color(red)(cancel(color(black)(10)))w)/cancel(color(red)(10))$

$6 =$ or $w = 6$

We can now substitute $w$ into our formula for the relationship between $l$ and $w$ and calculate $l$ :

$l = (4 xx 6) + 1$

$l = 24 + 1$

$l =25$

The length of the rectangle is 25 meters and the width of the rectangle is 6 meters.