Question

The perimeter of a rectangle is 64 feet. The length is two more than double the

width. Find the dimensions of the rectangle.

(Remember Perimeter of rectangle = 2L+ 2w)

Answer 1

Width = 10
Length = 22

Explanation

1. Write the equation L= 2w + 2 for the given statement length = two more than doubled word.

2. 2(2w+2) represents 2L in the perimeter equation. Solve 2(2w+2) + 2w = 64 to find the value of the width.

3. Distribute the 2 into the parentheses first, which should give you 4w + 4 + 2w = 64.

4. Combine like terms 4w and 2w leaving you 6w + 4 = 64.

5. Subtract 4 on both sides of the equation to get 6w alone. 6w = 60.

6. Divide by 6 on both sides to solve for w. w = 10.

7. Input x = 10 into 2w + 2 to find the length of the rectangle. 2(10) + 2 = l

8, multiply 2 by 10, which is 20. Add that to 2, which is 22.

Answer 2

To solve this problem, we can set up the following equation:

2L + 2W = 64

We are told that the length of the rectangle is two more than double the width, so we can write the following equation:

L = 2W + 2

Substituting this equation into the first equation, we get:

2(2W + 2) + 2W = 64

Simplifying this equation, we get:

6W + 4 = 64

Subtracting 4 from both sides, we get:

6W = 60

Dividing both sides by 6, we get:

W = 10

The width of the rectangle is 10 feet.

To find the length of the rectangle, we can substitute this value back into the equation L = 2W + 2 to get:

L = 2(10) + 2

= 20 + 2

= 22

The length of the rectangle is 22 feet.

Therefore, the dimensions of the rectangle are 10 feet by 22 feet.

Explanation: