# A rectangle has a perimeter of 40 feet. The length is 4 feet longer than the width. Find the dimensions of the rectangle.

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### 2 Answers

In the problem, the length is being compared to the width. So, let x be the width of the rectangle.

width = x

Then, express in math form the condition "the length is 4 ft longer than the width".

length = x + 4

To set-up the equation, apply the formula of perimeter of rectangle.

`P = 2 * (L + W)`

where L is the length and W is the width.

Plugging in the length and width of the given rectangle, the formula becomes:

`P = 2((x+4)+x)`

Also, plug-in the given perimeter P=40.

Hence, the equation is:

`40=2((x+4)+x)`

Simplifying the right side, it becomes:

`40=2(2x+4)`

`40=4x+8`

Isolating the x result in:

`32=4x`

`8=x`

So, the width is 8. And, its length is:

length` =x + 4 = 8 + 4 =12`

**Therefore, the rectangle has a length of 12 feet and a width of 8 feet.**

A rectangle has a perimeter of 40 feet. The length is 4 feet longer than the width. Find the dimensions of the rectangle. The formula for the perimeter of a rectangle is 2l + 2w= Perimeter. Or you can add up all the sides together.

Let's call the width=w.

Since the length is 4 feet longer than the width, the equation for the length would be:

(4+w). Using the perimeter equation we can solve:

2(4+w) + 2(w)=40

8+2w+2w=40

8+4w=40

w=8. Therefore, the width is 8. Since length is 4+w, it's 4+8 which equals 12. So, the width is 8 feet and length is 12.