Question

A rectangle is 7 times as long as it is wide. The perimeter is

64 feet. Find the dimensions. The length is ???feet where the width is ???feet.

Please show a step by step on how to do this.

Answer 1

The length is 28ft and the width is 4ft.

Explanation:

Let x equals to the width. Since the length is 7 times as long as the width, we know that it has to be 7x.

The perimeter equation of a rectangle would be P=2(l+w)

Here are the givens:

l= 7x

w= x

P=32 ft

P=2(l+w)

All you need to do is substitute x and 7x into width and length in the perimeter equation

P=2(7x+x)

P=2(8x)

Substitute 64ft into "P"

64=16x

x=4

After finding x=2, we would substitute x into the length and the width equations

w=4

l=4(7)=28

so the length is 28ft and the width is 4ft.

Answer 2

Answer: Length = 28 feet

Width = 4 feet

Explanation:

Let L represent the length of the rectangle.

Let W represent the width of the rectangle.

The formula for determining the perimeter of a rectangle is expressed as

Perimeter = 2(L + W)

The perimeter of the rectangle is 64 feet. This means that

2(L + W) = 64

Dividing through by 2, it becomes

L + W = 32 - - - - - - - - - - - -1

The rectangle is 7 times as long as it is wide. This means that

L = 7W

Substituting L = 7W into equation 1, it becomes

7W + W = 32

8W = 32

W = 32/8

W = 4

L = 7W = 7 × 4

L = 28