Geometry help needed?

Question

The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 52 in., find the width of the rectangle.

Answer 1

Length + Length + Width + Width = Perimeter
2 (2x+2) + 2(x) = 52
SIMPLIFY....
4x + 4 + 2x = 52 Add x's together
6x + 4 = 52 Subtract 4 from both sides
6x = 48 Divide both sides by 6
x = 8
The width is 8 inches. Width = x
The length is 18 inches. Length = 2x+2

Answer 2

The above process is right, but the person misread the question.
It's more of an algebra problem than a geometry problem, by the way =p

Let L = Length.
Let W = Width
Length is 2in more than twice the width. L = 2 + 2W
Perimeter is P. P = 52in. Perimeter is the sum of the 4 sides. Two lengths; two widths.
2L + 2W = P
2[2 + 2W] + 2[W] = 52

[4 + 4W] + 2W = 52
6W + 4 = 52
6W = 48
W = 8


Width = 8in

Answer 3

Given:
L=W+2
2L+2W=52

2(W+2)+2W=52
4W+4=52
4W=48
W=12

Answer 4

let breadth=x
length=2x+2
P=2(L+B)
52=2(2x+2+x)
52/2=3x+2
26=3x+2
3x=26-2
3x=24
x=24/3
x=8
so width of rectangle=8 inch

Answer 5

First add up the sides
x width
x width
x+2in length
x+2in length

then solve for x