You choose these perimeter Word problems often in mathematics. Many of them, it will require familiarity with basic math, algebra skills or a combination of the two for the problems

As I solve these perimeter Word problems, I will make an attempt to give a few problems to solve their problems and show you that the problems could be solved with basic, algebra math, or both.

**Word problem # 1:**

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

**Important concept**: hexagon. This means 6 equal parts.

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. This means that 3 equal parts.

There are many ways to approach this problem.

**Use of basic mathematical skills:**

Since the triangle have 3 equal parts, you can say to yourself, “what number even add three times to get 6?”

Since 2 + 2 + 2 = 6, then the length of one side is 2

**Use of algebra:**

Let x either side you are looking for

x + x + x = 6

x 3 = 6

x 3 / 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 duration?

**Use of basic mathematical skills:**

**Important concept**: a rectangle has four sides. Parallel and opposite sides are equal.

Since opposite sides are equal, there are two sides (width) measure 8-8

Therefore, the addition of both sides give 8 + 8 = 16

The length of the other two sides total to 42-16 = 26

Given that 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 of 13

**Use of algebra:**

P = 2 ? L + ? 2 W

Replace all values in the formula.

42 = 2 ? 2 8 L + ?

42 = 2 L + 16 ?

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 mathematical skills:**

**Important concept**:

Regular polygon. A polygon with equal parts and parts equal.Divided by 5 to obtain the length of a side. It is the Pentagon as it has 5 sides.

So p = 5 s ?

To obtain 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 mathematical skills:

Crystalize ican help solving you perimeter Word problems sometimes.

Pretend wide = 1, length = 6 (1 + 5)

? 2 ? 1 + 2 7 = 2 + 14 = 16. Note that 16 is a perimeter of 34

Try a much larger number.

What if we…

Pretend width = 4, while length = 9 (4 + 5)

? 2 4 + 2 ? 9 = 6 + 18 = 24. We are 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. It is superior to a perimeter of 34. If the width should be greater than 5 and less than 7. May be that a width of 6 will work.

Pretend width = 6 and length = 11 (6 + 5)

? 2 6 + 2 ? 11 = 12 + 22 = 34.

**Use of algebra:**

Leave the width = x

Let the length = x + 5

P = 2 ? L + ? 2 W

34 = 2 ? (x + 5) + ? 2 x

34 = 2 x + 10 + 2 x

34 = 4 x + 10

34 – 10 = 4 x + 10-10.

24 = 4 x

24/4 = x 4 / 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 the shows in the past problems. And the use of basic mathematical skills is perhaps not the best way to go in solving perimeter Word problems. Algebra is sometimes better!

Have a question about these perimeter Word problems? send me a note.