Question

9. A rectangle has a width that is 2 feet longer than four times the length. What is the length of the rectangle
if the perimeter is 64 feet? (HINT: Draw a picture and use algebra to solve)​

Answer 1

Answer:

6 feet

Step-by-step explanation:

Perimeter = 2L + 2W

Since you know the perimeter is 64 feet, you would plug that into the perimeter side.

64 = 2L + 2W

You also know the width is 2 feet longer than 4 times the length.

64 = 2L + 2(2 + 4L)  ----> Multiply 2 * 2 and 2 * 4l

64 = 2L + 4 + 8L  ----> Subtract 4 from each side

-4           -4

60 = 2L + 8l ----> add these two

60 = 10L

/10   /10 -----> divide by 10 on each side

6 = L

Now that we know the length, find the width.

64 = 2(6) + 2(2 + 4(6)) -----> multiply 2 * 6 and 4 * 6

64 = 12 + 2(2 + 24) -------> add 24 and 2

64 = 12 + 2(26) -------> multiply 2 and 26

Before you go any further, you now know that the width is 26 feet. Continue solving.

64 = 12 + 52

64 = 64

Everything checks out.

The length is 6 feet and the width is 26 feet.

Answer 2

The length of the rectangle is 6 feet

Calculation of the length of the rectangle:

We know that

Perimeter = 2L + 2W

So,

64 = 2L + 2W

Since the width is 2 feet longer than 4 times the length.

So,

64 = 2L + 2(2 + 4L)

64 = 2L + 4 + 8L  

60 = 2L + 8l

60 = 10L

L = 6 feet

learn more about the perimeter here: https://brainly.com/question/2468841