# SOLUTION: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.

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 Question 73092: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.Found 2 solutions by checkley75, rmromero:Answer by checkley75(3666)   (Show Source): You can put this solution on YOUR website!L=2W+2 2L+2W=PERIMETER 2(2W+2)+2W=52 4W+4+2W=52 6W=52-4 6W=48 W=48/6 W=8 ANSWER FOR THE WIDTH. THUS: L=2*8+2 L=16+2 L=18 ANSWER FOR THE LENGTH PROOF 2*18+2*8=52 36+16=52 52=52 Answer by rmromero(383)   (Show Source): You can put this solution on YOUR website!``` The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle. What is asked in the problem? find the dimensions of the rectangle. Given: The length of a rectangle is 2cm more than twice its width The perimeter of the rectangle is 52cm Representation: Let x = the width of the rectangle 2x+2 = the length of the rectangle Equation 2l + 2w = P 2(2x+2) + 2x = 52 Multiply (Distributive Property) 4x + 4 + 2x = 52 Combine like terms 6x + 4 = 52 6x + 4 - 4 = 52 - 4 subtract 4 both sides 6x = 48 divide both sides by 6 x = 8 cm -----> Width length = 2x + 2 = 2(8) + 2 = 16 + 2 = 18 cm ----- length Checking 2L + 2W = P 2(18) + 2(8) = 52 36 + 16 = 52 52 = 52 ```