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

Question

asked Sep 2, 2020 140k views

2 Answers

Fruitjs · Answer 1 · 2020-09-04T06:36:12+0000

Answer:

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

Trott · Answer 2 · 2020-09-06T07:35:00+0000

Answer:

and

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.