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

Answer:

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

Answer 2

Answer:

$l=32$ 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.