4. the length of a rectangle is eight more
than twice its width. the perimeter is 96
feet. find the dimensions of the rectangle.

width 32 length 32

Step-by-step explanation:

length: 31

width: 12

Step-by-step explanation:

l=8+2w

p=2l+2w=88

(8+2w)+(8+2w)+w+w

16+6w=88

6w=72

w=12

l=8+2(12)

l=32

sorry if this was confusing

and

Step-by-step explanation:

Given: The length of a rectangle is eight more than twice its width. The perimeter is 88 feet.

To find: The dimensions of the rectangle.

Solution:

It is given that the length of a rectangle is eight more than twice its width.

Let the width of the rectangle be w.

So, the length of the rectangle

We know that the perimeter of a rectangle is

Here, perimeter of rectangle

So, we have

Therefore, width of the rectangle is 12 feet

length

Hence, length of the rectangle is 32 feet and width of the rectangle is 12 feet.

48 = x + x + x + 8 + x + 8

48 = 4x + 16

32 = 4x

x = 8

length = 16

width = 8

24 and 8

Step-by-step explanation:

The formula for the perimeter of the rect. is P = 2W + 2L.

Here, L = 2W + 8, so the area formula becomes:

P = 2W + 2(2W + 8) = 64

Dividing all terms by 2 reduces this to:

         W + 2W + 8 = 32, or 3W = 24, or W = 8.

Then:  L = 2W + 8 = 2(8) + 8 = 24

Check:  does 2W + 2L = 64 when these values for W and L are substituted?

              Is 2(8) + 2(24) = 64   true?  Yes

The length and width are 24 and 8 respectively.