# Lesson Word Problems Involving Perimeter

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 This Lesson (Word Problems Involving Perimeter) was created by by algebrahouse.com(1079)  : View Source, ShowAbout algebrahouse.com: Visit: www.algebrahouse.com to ask questions and for notes, examples, and more When solving word problems involving shapes and perimeter, you need to: 1.) represent the sides of the shape in terms of a variable 2.) set-up an equation to solve what the problem is asking *Hint on how to start representing the sides, in terms of a variable* - A lot of times, the problems use one side of the shape to describe one or more other sides. Usually the word "than" is used. Usually, whatever comes after the word "than" is what you let the variable equal. For example, the problem may say this: "the length of a rectangle is three more than twice the width" ---> notice "width" comes after the word "than", therefore let x equal the width x = width 2x + 3 = length {"length is three more than twice the width"} Here are some examples of word problems involving perimeter, and how to solve them: ------------------------------------------------------------ The length of a rectangle is 12 cm more than the width. The perimeter is 44 cm. Find the length and the width. {The width is being used to describe the length, so let x = width} {Also “width” comes after the word “than”} x = width x + 12 = length {length is 12 more than the width} Perimeter of a rectangle = (2 x width) + (2 x length) 2x + 2(x + 12) = 44 {perimeter = 2(length) + 2(width)} 2x + 2x + 24 = 44 {used distributive property} 4x + 24 = 44 {combined like terms{ 4x = 20 {subtracted 24 from both sides} x = 5 {divided both sides by 4} x + 12 = 17 {substituted 5, in for x, into x + 12} width = 5 cm and length = 17 cm www.algebrahouse.com ------------------------------------------------------------ The perimeter of a rectangle is 104m. The length is 7m more than twice the width. What are the dimensions of the rectangle? x = width {the width is being used to describe the length..."width" comes after "than"} 2x + 7 = length {length is seven more than twice the width} Perimeter of a rectangle = (2 x length) + (2 x width) 2x + 2(2x + 7) = 104 {perimeter = 2(length + 2(width)} 2x + 4x + 14 = 104 {used distributive property} 6x + 14 = 104 {combined like terms} 6x = 90 {subtracted 14 from both sides} x = 15 {divided both sides by 6} 2x + 7 = 37 {substituted 15, in for x, into 2x + 7} width = 15 m and length = 37 m www.algebrahouse.com ------------------------------------------------------------ Side a of a triangle is 4 cm longer than side b. Side c is twice as long as side b. What is the length of each side of the triangle if its perimeter is 28 cm? x = side b x + 4 = side a {side a is 4 longer than side b} 2x = side c {side c is twice as long as side b} x + x + 4 + 2x = 28 {added up all sides of the triangle and set equal to perimeter, 28} 4x + 4 = 28 {combined like terms} 4x = 24 {subtracted 4 from both sides} x = 6 {divided both sides by 4} x + 4 = 10 {substituted 6, in for x, into x + 4} 2x = 12 {substituted 6, in for x, into 2x} side a = 10 cm side b = 6 cm side c = 12 cm www.algebrahouse.com ------------------------------------------------------------ The length of a rectangle is 5 inches less than three times its width. The perimeter is 86 inches. Find the length and width. x = width {the width comes after the word "than" and is being used to describe the length} 3x - 5 = length {length is 5 less than 3 times width} Perimeter of a rectangle = (2 x width) + (2 x length) 2x + 2(3x - 5) = 86 {perimeter = (2 x width) + (2 x length)} 2x + 6x - 10 = 86 {used distributive property} 8x - 10 = 86 {combined like terms} 8x = 96 {added 10 to both sides} x = 12 {divided both sides by 8} 3x - 5 = 31 {substituted 12, in for x, into 3x - 5} width = 12 in. length = 31 in. www.algebrahouse.com ------------------------------------------------------------ The two longest sides of a pentagon are each 3 times as long as the shortest side. The other sides are each 8 m longer than the shortest side. Find the length of each side if the perimeter is 79 m. A pentagon has five sides. x = shortest side {the shortest side is being used to describe the other sides} 3x = long side {the two longest sides are 3 times as long as the shortest side} 3x = long side x + 8 = other side {the other sides are 8 longer than the shortest side} x + 8 = other side x + 3x + 3x + x + 8 + x + 8 = 79 {added up all sides and set equal to perimeter, 79} 9x + 16 = 79 {combined like terms} 9x = 63 {subtracted 16 from both sides} x = 7 {divided both sides by 9} 3x = 21 {substituted 7, in for x, into 3x} x + 8 = 15 {substituted 7, in for x, into x + 8} The five sides are: x = 7 m 3x = 21 m 3x = 21 m x + 8 = 15 m x + 8 = 15 m www.algebrahouse.com ------------------------------------------------------------ This lesson has been accessed 26289 times.