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

Answers

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

l=32 and w=12

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=2w+8

We know that the perimeter of a rectangle is 2(l+w)

Here, perimeter of rectangle=88

So, we have

2(2w+8+w)=88

\implies2w+w+8=44

\implies3w=44-8

\implies3w=36

\implies w=12

Therefore, width of the rectangle is 12 feet

length=2\times12+8=32

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.



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