Question

The length of a rectangle is eight more than twice its width. The perimeter is 88 feet. Find the dimensions of the rectangle.

Answer 1

length: 31

width: 12

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

Answer 2

`l=32` and
`w=12`

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*12+8=32`

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