Question

The width of a rectangle is 6 2/3 inches. The length of the is twice it’s width. What so the perimeter of the rectangle?

Answer 1

`\text {Width = } 6(2)/(3) \text { inches}`


The length is twice its width:

`\text {Length = } 2 * 6(2)/(3) \text { inches}`

Change to improper fraction:

`\text {Length = }2 * (20)/(3) \text { inches}`

Combine into single fraction:

`\text {Length = } (40)/(3) \text { inches}`


Find Perimeter :

`\text {Perimeter = Length + Length + Width + Width}`


`\text {Perimeter = } (40)/(3) + (40)/(3) + (20)/(3) + (20)/(3) = (120)/(3) = 40 \text { inches}`



`\bf \text {Answer: Perimeter = 40 inches}`

Answer 2

L = 2*W
L = 2 * 6 2/3 = 12 + 2*2/3 = 12 + 1 1/3 = 13 1/3
P = 2 * L + 2*W
P = 2 * (13 1/3) + 2 * 6 2/3
P = 26 2/3 + 13 1/3
P = 39 + 2/3 + 1/3
P = 40

It might be easier to understand if I used decimals. Let's try that.
L = 2 * 6 2/3 = 2 * 6.666666666 = 13.333333333
W = 6.666666666

P = 2L + 2W = 2*13.33333333 + 2*6.6666666666
P = 26.66666666 + 13.33333333
P = 40