Perimeter word problemsYou will accounter these perimeter word problems often in math. Many of them will require familiarity with basic math, algebra skills, or a combination of both to solve the problems As I solve these perimeter word problems, I will make an attempt to give you some problems solving skills and show you that the problems could be solved with basic math, algebra, or both. Word problem #1: The length of a side of a hexagon is 2 inches. What is the perimeter? Important concept: Hexagon. It means 6 equal sides. p = 2 + 2 + 2 + 2 + 2 + 2 = 4 + 4 + 4 = 8 + 4 = 12 inches Word problem #2: The perimeter of an equilateral triangle is 6 inches. What is the length of a side? Important concept: Equilateral. It means 3 equal sides. There are many ways to approach this problem. Use of basic math skills: Since the triangle has 3 equal sides, you can just say to yourself, " What same number do I add three times to get 6?" Since 2 + 2 + 2 = 6, then the length of one side is 2 Use of algebra: Let x be the side you are looking for x + x + x = 6 3x = 6 3x/3 = 6/3 ( Divide both sides by 3) x = 2 Word problem #2: The perimeter of a rectangle is 42 inches. If the width is 8, what is the length? Use of basic math skills: Important concept: A rectangle has four sides. Parallel and opposites sides are equal. Since opposite sides are equal, there are two sides (widths) measuring 8 and 8 Therefore, adding two sides give 8 + 8 = 16 The length of the two remaining sides totals to 42  16= 26 Since these two sides are equal, just divide by 2 to get the measure of the length of the rectangle 26/2 = 13, so the length is 13 Use of algebra: P = 2 × L + 2 × W Replace all known values into the formula. 42 = 2 × L + 2 × 8 42 = 2 × L + 16 Solve the resulting equation: 42  16 = 2 × L + 16  16 26 = 2 × L 26/2 = (2 × L)/2 13 = L Word problem #3: When the perimeter of a regular polygon is divided by 5, the length of a side is 25. What is the name of the polygon? What is the perimeter? Use of basic math skills: Important concept: Regular polygon. A polygon with equal sides and equal sides. Divided by 5 to get the length of a side. It is the pentagon since it has 5 sides. So p = 5 × s To get the perimeter, just multiply a side by 5. Since 25 × 5 = 125, the perimeter is 125. Word problem #4: The length of a rectangle is 5 more than the width. What are the dimensions of the rectangle if the perimeter is 34? Use of basic math skills: Trial and error ican help you solve perimeter word problems sometimes. Pretend width = 1, then length = 6 ( 1 + 5) 2 × 1 + 2 × 7 = 2 + 14 = 16. Notice that 16 is far from a perimeter of 34 Try much bigger number. How about if we... Pretend width = 4, then length = 9 ( 4 + 5) 2 × 4 + 2 × 9 = 6 + 18 = 24. We are getting closer to a perimeter of 34 Pretend width = 5, then length = 10 ( 5 + 5) 2 × 5 + 2 × 10 = 10 + 30 = 30. Pretend width = 7, then length = 12 ( 7 + 5) 2 × 7 + 2 × 12 = 14 + 24 = 38. This is higher than a perimeter of 34. So width should be higher than 5 and smaller than 7. May be a width of 6 will work. Pretend width = 6, then length = 11 ( 6 + 5) 2 × 6 + 2 × 11 = 12 + 22 = 34. Use of algebra: Let width = x Let length = x + 5 P = 2 × L + 2 × W 34 = 2 × ( x + 5) + 2 × x 34 = 2x + 10 + 2x 34 = 4x + 10 34  10 = 4x + 10  10 24 = 4x 24/4 = 4x/4 6 = x Therefore, width = 6 and length = x + 5 = 6 + 5 = 11 As you can see perimeter word problems can become very complicated as shows in the last problems. And the use of basic math skills may not be the best way to go to solve perimeter word problems. Sometimes algebra is better! Have a question about these perimeter word problems? send me a note. 




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