Question

The width of a rectangle is 5 less than twice the length. if the perimeter of the rectangle is 110 inches, find the width and lenght

Answer 1

Width is 35inches and Length is 20inches

Explanation:

Let's use W to represent the rectangle's width, L to represent its length, and P to represent its perimeter.

We're told the width of a rectangle is 5 less than twice the length. Let's break this down and represent it algebraically.

Twice the length: 2L

5 less than twice the length: 2L-5

The width is 5 less than twice the length: W=2L-5

Next we're told that the perimieter of the rectangle is 110in: P=110

We also know the perimeter of a shape is the sum of its sides, and that a rectangle ahs two long sides and two wide sides. So: P=2L+2W

If P=110 and P=2L+2W, then 2L+2W=110

Since W=2L-5, let's replace W with 2L-5 in 2L+2W=110 to find out L:

2L+2W=110

2L+2(2L-5)=110

2L+4L-10=110

6L-10=110

6L=120

L=20, Length is 20inches

Now that we know L=20, let's sub 20 into W=2L-5 to find W:

W=2L-5

W=2(20)-5

W=40-5

W=35, Width is 35inches

Answer 2

The length of the rectangle is 20 inches

The width is 35 inches.

Explanation:

Let:

L = length

W = width

According to the problem

W = 2L - 5

The formula for the perimeter of a rectangle is:

Perimeter = 2(L + W)

If the Perimeter is 110 inches, we have:

P = 2(L+W)

110 = 2(L + (2L - 5)) (Because W = 2L - 5)

110 / 2 = 2(L + (2L - 5)) / 2

55 = L + (2L - 5)

55 = 1L + 2L - 5

55 = 3L - 5

55 + 5 = 3L - 5 + 5

60 = 3L

60/3 = 3L/3

20 = L

L = 20

So, the length will be 20 inches

Now, use the first equation to find the width:

W = 2L - 5

W = 2(20) - 5

W = 40 - 5

W = 35

The width of the rectangle is 35 inches.