You 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.

The length of a side of a hexagon is 2 inches. What is the perimeter?

p = 2 + 2 + 2 + 2 + 2 + 2 = 4 + 4 + 4 = 8 + 4 = 12 inches

The perimeter of an equilateral triangle is 6 inches. What is the length of a side?

There are many ways to approach this problem.

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

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

The perimeter of a rectangle is 42 inches. If the width is 8, what is the length?

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

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

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?

Regular polygon. A polygon with equal sides and equal sides.

Since 25 × 5 = 125, the perimeter is 125.

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.

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!

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