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The length of a rectangle is 2 ft longer than its width.

If the perimeter of the rectangle is 56 ft, find its length and width.

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Hi! I'm happy to help!

To solve this, we first need to look at the perimeter equation:

P=2L+2W

We don't know our length, so we can represent it with x. Since our width is 2 feet shorter than x, we can represent it with x-2. Now, we plug these values into our equation:

56=2x+(2(x-2))

Let's simplify what the width is by multiplying:

56=2x+2x-4

Now, let's combine our 2xs

56=4x-4

Now, we just need to solve for x in order to find our length and width.

First, we need to isolate x on one side of the equation. We can do this by adding 4 to both sides:

56=4x-4

+4 +4

60=4x

Now, all we have to do is divide both sides by 4 and x will be fully isolated:

60=4x

÷4 ÷4

15=x

Now that we know x, let's plug this into our previous equations:

L=x=15

L=15

W=x-2=15-2=13

W=13

To verify our answers, we can plug this into our perimeter equation:

56=2(15)+2(13)

56=30+36

56=56

After double checking our answers, we know that our length is 15 and our width is 13.

I hope this was helpful, keep learning! :D

User Nasukkin
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