Perimeter word problems
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.
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.