The rectangle length is two times the width, of the perimeter is 54 inches what is the dimensions of the rectangle?

Answer 1

For all of these equations:P is Perimeter, W is width, and L is length

First, we have to find the relationship between length and width which in this case it states:

L=2W

Now we know that a rectangle has 4 sides and the perimeter is all of the sides of the rectangle added up we know the equation would be:

P=W+W+L+L

Then we can plug in the equation from the first step into the second step because we know what L is :

P=W+W+2W+2W

Now we plug in the perimeter:

54 in=W+W+2W+2W

Now we can combine the W's in the equation:

54 in=6W

Now we can divide 6 by both sides to get the width equals 9 in. To get the length you plug it into the first equation L=2W

Then we figure out L=2(9in)

L=18 in

Finally, the dimensions are

Length=18 in

Width=9 in

Now if we want to check the answer we can just plug in all the numbers into the perimeter equation where:

P=W+W+L+L

P= 9 in +9in +18 in +18 in

P=54 in

Answer 2

Let w= width then the length is 2w

The some of the 4 sides is w + w + 2w +2w or 6w

6w=54 w=9

the length is twice the width so it is 18

Answer 3

Since perimeter of a rectangle is equal to twice the length plus twice the width, we can write the equation: P = 2L + 2W. Substituting L=2W and the known perimeter into the equation, we have (54in) = 2(2W) + 2W. Solving for W gives a width of 9in, and doubling this gives us a length of 18 in. Therefore, the dimensions of the rectangles are L=9in and W=18in, or 9"x18".