SOLUTION: the length of a rectangle is 2 inches longer than the width of the rectangle. If the perimeter of the rectangle is 46 inches, what is the area of the rectangle?

Question 418106: the length of a rectangle is 2 inches longer than the width of the rectangle. If the perimeter of the rectangle is 46 inches, what is the area of the rectangle?
Found 2 solutions by algebrahouse.com, ewatrrr:
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"the length of a rectangle is 2 inches longer than the width of the rectangle. If the perimeter of the rectangle is 46 inches, what is the area of the rectangle?"

x = width
x + 2 = length {length is 2 longer than width}

Perimeter of a rectangle = (2 x width) + (2 x length)

2x + 2(x + 2) = 46 {perimeter = 2(width) + 2(length)}
2x + 2x + 4 = 46 {used distributive property}
4x + 4 = 46 {combined like terms}
4x = 42 {subtracted 4 from both sides}
x = 10.5 {divided both sides by 4}
x + 2 = 12.5 {substituted 10.5, in for x, into x + 2}

width = 10.5in. and length = 12.5in.

Area of a rectangle = length x width
A = 12.5(10.5) {area is length x width}
Area = 131.25in.^2
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Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi
length of a rectangle is 2 inches longer than the width of the rectangle
Let x and (x+2) represent the width and length respectively
2(x+2) + 2x = 46in
4x = 42in
x = 10.5in, the width. The length is 12.5in (10.5 + 2)
CHECKING our Answer***
2*12.5in + 2*10.5in = 25in + 21in = 46in

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