Paul Murphys flag has a perimeter of 28 feet. Find the dimensions of the flag if the length is 4 feet less than twice its width.

Answer 1

Here, we can consider the flag as a rectangle.

Below is the steps to solve this problem.

Let width = x, length = (2*x - 4) according to the information from the problem. (x = unknown)

Perimeter = 2*width + 2*length

Expressing perimeter using x

2*x + 2*(2x-4) = 28

2x + 4x - 8 = 28

6x = 36

x = 6

Therefore, width = 6 ft & length = 2*6 - 4 = 8 ft.

Answer 2

Consider the flag as a rectangle, with length L and width W. Then 2L + 2W = 28, and L = 2W - 4. We can plug the second equation into the first, giving: 2(2W - 4) + 2W = 28. Now we can solve for W!

Distributing gives us 4W - 8 + 2W = 28, so 6W = 36, and W = 6. Now we can plug this value in to the second equation!

If W = 6 and L = 2W - 4, then L = 2(6) - 4 = 8.

Thus, our dimensions are length = 8, and width = 6.

Answer 3

A flag is the shape of a rectangle.

The perimeter of a rectangle is twice the sum of the length and the width or 2 times the length plus 2 times the width

P = 2L + 2W or P=2(L+W)

We're told the perimeter is 28 ft and the length is 4 ft less than 2 times the width.

P=28

L=2W-4

Substitute these into the perimeter equation above.

28 = 2(2W-4) + 2W

Distribute and collect like terms

28 = 4W - 8 + 2W

28 = 6W - 8

And solve for W

add 8 to both sides

36 = 6W

divide both sides by 6

W = 6

Now that we have the width, we can find the length.

L = 2W-4

L = 2*6-4 = 12-4=8

Width = 6; Length = 8

You can check this by using the perimeter formula to see if you get a perimeter of 28

2(8) + 2(6) = 16+12 = 28. √