HFCC Math Lab
Beginning Algebra - 15
PERIMETER WORD PROBLEMS
The perimeter of a plane geometric figure is the sum of the lengths of its sides. In this handout, we
will deal with perimeter problems involving rectangles, squares, and triangles.
We will use L and W to denote the length and width, respectively, of a rectangle; we'll use S to
represent the length of a side of a square; and we'll use A, B, and C to represent the lengths of the three
sides of a triangle. Formulas for the perimeter P of each are shown below.
Rectangle
Square
Triangle
B
W
C
S
L
S
P = 2W + 2 L
P = 4S
A
P = A+ B+C
Use the following procedure to set up perimeter problems:
1.
Read the problem carefully several times.
2.
Draw a figure to help visualize the problem.
3.
Identify the unknown measures.
4.
Represent one of the unknown measures by a suitable letter, such as x. Then represent the
other unknown measures in terms of x.
5.
Translate the problem into a linear equation using the appropriate perimeter formula or
formulas.
6.
Solve the equations
7.
State the answer or answers to the problem.
8.
Check the answer or answers in the original statement of the problem, not the equation that you
set up.
Now, let’s look at some example problems involving the perimeter of rectangles, square, and triangles.
Revised 03/09
1
Example l: The length of a rectangle is nine inches more than its width. If the perimeter of the
rectangle is 62 inches, find its length and width.
Let x = the width W.
Then x + 9 = the length L since it is given to be 9 inches more than the width.
x
x+9
Since the perimeter P of a rectangle is given by P = 2W + 2 L , and the perimeter is 62,
62 = 2 x + 2( x + 9)
62 = 2 x + 2 x + 18
62 = 4 x + 18
44 = 4 x
x = 11
x + 9 = 20
The width of the rectangle is 11 inches.
The length of the rectangle is 20 inches.
Example 2: The longest side of a triangle is one foot more than twice the shortest side. The other side
is eight feet less than the longest side. If the perimeter of the triangle is 54 feet, find the length of each
side.
2x-7
x
Let x = the length of the shortest side.
Then 2 x + 1 = the length of the longest side
and (2 x + 1) − 8 or 2 x − 7 = the length of the other side.
2x+1
Since the perimeter P of a triangle is given by P = A + B + C , and the perimeter is 54,
54 = x + (2 x + 1) + (2 x − 7)
54 = 5 x − 6
60 = 5 x
x = 12
2 x + 1 = 25
2 x − 7 = 17
Revised 03/09
The length of the shortest side is 12 feet.
The length of the longest side is 25 feet.
The length of the other side is 17 feet.
2
Example 3: A rectangle is formed by placing two identical squares side by side. If the perimeter of
the rectangle is 90 centimeters, what is the length of a side of each square?
Let x = the length of a side of each square.
The rectangle that is formed has a width of x and a length of 2x.
x
x
x
Since the perimeter of a rectangle is given by P = 2 W + 2 L and the perimeter is 90,
90 = 2 x + 2 ⋅ 2 x
90 = 6 x
x = 15
The length of a side of each square is 15 centimeters.
Example 4: In a certain rectangle, the length is three inches more than the width. If the length is
increased by four inches and the width is increased by two inches, the perimeter of the new rectangle
will be 78 inches. Find the dimensions of the original rectangle.
Let x = the width of the original rectangle.
Then x + 3 = the length of the original rectangle.
Since the width of the original rectangle is increased by 2, the new width is x + 2; and since the
length of the original rectangle is increased by 4, the new length is ( x + 3) + 4 or x + 7 .
Original Rectangle
New Rectangle
x
x+2
x+3
x+7
Using the dimensions of the new rectangle and substituting into the perimeter formula of a rectangle,
P = 2W + 2 L
78 = 2( x + 2) + 2( x + 7)
78 = 2 x + 4 + 2 x + 14
78 = 4 x + 18
4 x = 60
x = 15
x + 3 = 18
Revised 03/09
The width of the original rectangle is 15 inches.
The length of the original rectangle is 18 inches
3
Example 5: An 8 by 12 inch picture has a frame of uniform width. If the outer edge of the frame has a
perimeter of 52 inches, how wide is the frame?
Let x = the width of the frame.
It can be observed in the figure at the right that a frame of uniform width x around the picture
increases the dimensions of the 8 by 12 inch picture by 2x inches in each direction. The outer
dimensions are (8 + 2 x) by (12 + 2 x ) inches.
x
Given the perimeter of the outer edge is 52, we form the equation
2(8 + 2 x ) + 2(12 + 2 x) = 52
12
16 + 4 x + 24 + 4 x = 52
8 x + 40 = 52
8
8 x = 12
x = 1.5
The width of the frame is 1.5 inches.
Example 6: The length of a side of a square is twice the width of a rectangle, and the length of the
rectangle is 21 feet. If the perimeter of the square is the same as the perimeter of the rectangle, find the
dimensions of the square and the rectangle.
Let x = the width of the rectangle.
Since the length of a side of the square is twice the width of the rectangle, 2x = the length of a
side of the square.
x
2x
21
2x
The perimeter of the square is 4 ⋅ 2x or 8x .
The perimeter of the rectangle is 2 x + 2 ⋅ 21 or 2 x + 42 .
Since these perimeters are given to be the same,
8 x = 2 x + 42
6 x = 42
x=7
2 x = 14
Revised 03/09
The dimensions of the rectangle (in feet) are 7 by 21.
The length of a side of the square is 14 feet.
4
PERIMETER PROBLEM EXERCISES
1.
The perimeter of a rectangle is 52 feet. The length of the rectangle is 10 feet longer than the
width. Find the dimensions of the rectangle.
2.
The length of a rectangle is three times its width, and its perimeter is 88 inches. Find its length
and width.
3.
The width of a rectangle is 12 centimeters less than its length, and its perimeter is 48
centimeters. Find the dimensions.
4.
The length of a rectangle is four feet more than twice the width. If the perimeter is 110 feet,
find the length and the width.
5.
The length of a rectangle is three feet less than two times the width. The perimeter is 27 feet.
Find the dimensions.
6.
One side of a triangle is twice another side. The third side is eight centimeters and the
perimeter is 26 centimeters. How long is each side?
7.
The longest side of a triangle is five meters more than the second side. The shortest side is two
meters less than the second side. If the perimeter of the triangle is 24 meters, find the length of
the shortest side.
8.
The perimeter of a triangle is 58 meters. The longest side is three times the shortest side. The
other side is five meters shorter than the longest side. How long is each side?
9.
A rectangle is formed by placing two equal squares side by side. If the perimeter of the
rectangle is 30 inches, how long is a side of each square?
10.
The perimeter of an equilateral triangle (a triangle whose sides have the same length) is 28
inches more than the length of a side. Find the length of a side.
11.
One side of a triangle is six meters less than the second side and twice the length of the third
side. If the perimeter is 106 meters, find the length of the three sides.
12.
The longest side of a triangle is 12 inches longer than the shortest side. The other side is five
inches shorter than the longest side. If the perimeter is 61 inches, find the length of each side.
13.
One side of a triangle is three feet longer than another, and the third side is four feet less than
the sum of the first two. If the perimeter is 54 feet, find the lengths of the sides.
14.
The length of a rectangle is three feet more than twice the width. If the width is increased by
four feet, the perimeter of the new rectangle is 68 feet. Find the dimensions of the original
rectangle.
Revised 03/09
5
15.
In a certain rectangle, the length is six yards more than the width. If the length is increased by
25 yards and the width decreased by one yard, the perimeter of the new rectangle will be twice
that of the original. Find the length and width of the original rectangle.
16.
A 5 by 7 inch rectangular picture has a frame of uniform width. If the outer edge of the frame
is to have a perimeter of 32 inches, how wide should the frame be?
17.
A man plans to build a fence along the four sides of his house. The house is a rectangle that is
40 feet long and 32 feet wide. How far should he uniformly build the fence from the house if
he has enough material for 200 feet of fence?
18.
A 10 by 12 inch rectangular picture is framed with a frame of uniform width. How wide is the
frame if the perimeter of its outer edges is 58 inches?
19.
The side of a square is the same length as the shortest side of a triangle. The other two sides of
the triangle have equal lengths and are four inches longer than the shortest side of the triangle.
Find the lengths of the sides of the triangle and the square if the perimeters of the square and
triangle are the same.
20.
The length of the longest side of a triangle is ten centimeters more than that of the shortest side.
The other side is five centimeters less than the longest side. The perimeter of the triangle is the
same as the perimeter of a square with sides the same length as the shortest side of the triangle.
Find the lengths of the sides of the triangle and the square.
21.
The lengths of the three sides of a triangle are 16, 19, and 24 inches. The shortest side is to be
decreased by a certain amount, the next longer side is decreased by twice this amount, and the
longest side is decreased by three times this amount to form a triangle with a perimeter of 44
inches. Find the lengths of the sides of the new triangle.
22.
A triangle has sides of lengths of 8, 12, and 16 inches. Each side is to be decreased by the
same amount to form a new triangle that will have a perimeter one-half of the original triangle.
How much should each side be decreased?
23.
The length of a rectangle is four times its width. The perimeter of the rectangle is 84 feet more
than the perimeter of a square with a side the same length as the width of the rectangle. Find
the dimensions of the rectangle.
24.
The length of a side of a square is twice the width of a rectangle. If the perimeter of the square
is the same as the perimeter of the rectangle, and the length of the rectangle is 18 feet, find the
dimensions of the square and the rectangle.
25.
CALCULATOR PROBLEM - The length of a rectangle is 3.46 meters more than the width. If
the perimeter is 29.20 meters, find the dimensions of the rectangle.
26.
CALCULATOR PROBLEM - The perimeter of a rectangle is 98.88 inches, and the length is
three times as long as the width. Find the length and width of the rectangle.
Revised 03/09
6
Solutions to odd-numbered problems and answers to even-numbered problems:
1.
Let x = the width and x + 10 = the length
2 x + 2( x + 10) = 52
4 x + 20 = 52
4 x = 32
x=8
x + 10 = 18
The width is 8 ft. and the length is 18 ft.
2.
11 inches by 33 inches
3.
Let x = the length and x − 12 = the width
2 x + 2( x − 12) = 48
4 x − 24 = 48
4 x = 72
x = 18
x − 12 = 6
The length is 18 cm and width is 6cm.
4.
17 feet by 38 feet
5.
Let x = the width and 2 x − 3 = the length
2 x + 2(2 x − 3) = 27
6 x − 6 = 27
6 x = 33
x = 5.5
2x − 3 = 8
The width is 5.5 ft and the length is 8 ft.
6.
6, 8, and 12 centimeters
7.
Let x = the length of the 2nd side, x + 5 = the length of the longest side, and x − 2 = the length
of the shortest side.
x + ( x + 5) + ( x − 2) = 24
3 x + 3 = 24
3 x = 21
x=7
x−2=5
The length of the shortest side is 5 meters.
8.
9, 22, and 27 meters
Revised 03/09
7
9.
Let x = the length of the side of the square. Then the width of the rectangle is x and the length
is 2x.
2 x + 2(2 x) = 30
6 x = 30
x=5
The length of the side of the square is 5 inches.
10.
14 inches
11.
Let x = the length of the 3rd side, 2x = the length of the 1st side and 2 x + 6 = the length of the
2nd side.
x + 2 x + (2 x + 6) = 106
5 x + 6 = 106
5 x = 100
x = 20
2 x = 40
2 x + 6 = 46
The lengths of the three sides are 20, 40, and 46 meters.
12.
14, 21, and 26 inches
13.
Let x = the length of the 2nd side, x + 3 = the length of the 1st side, and x + ( x + 3) − 4 or
2 x − 1 = the length of the 3rd side.
x + ( x + 3) + (2 x − 1) = 54
4 x + 2 = 54
4 x = 52
x = 13
x + 3 = 16
2 x − 1 = 25
The lengths of the three sides are 13, 16, and 25 feet.
14.
9 by 21 feet
15.
Let x = the width of the original rectangle. so that x + 6 = the length of the original rectangle.
Then x + 31 = the length of the new rectangle, and x − 1 = the width of the new rectangle.
2( x + 31) + 2( x − 1) = 2[2 x + 2( x + 6)]
4 x + 60 = 8 x + 24
36 = 4 x
x=9
x + 6 = 15
The width is 9 yards and the length is 15 yards.
Revised 03/09
8
16.
1 inch
17.
Let x = the distance from fence to house.
2(40 + 2 x ) + 2(32 + 2 x) = 200
80 + 4 x + 64 + 4 x = 200
8 x + 144 = 200
8 x = 56
x=7
The distance is 7 feet.
18.
1.75 inches
19.
Let x = the length of the side of the square. Then, also, x = the length of the shortest side of the
triangle. x + 4 = the length of each of the other two sides of the triangle.
4 x = x + ( x + 4) + ( x + 4)
4 x = 3x + 8
x=8
x + 4 = 12
The length of a side of the square is 8 inches and the lengths of the three sides of the triangle
are 8, 12, and 12 inches.
20.
Square: 15 centimeters and Triangle: 15, 20, and 25 centimeters
21.
Let x = the amount. The sides of the new triangle have lengths 16 − x ,19 − 2x and 24 − 3x .
(16 − x ) + (19 − 2 x) + (24 x − 3 x) = 44
−6 x + 59 = 44
−6 x = −15
x = 2.5
16 − x = 13.5
19 − 2 x = 14
24 x − 3 x = 16.5
The lengths of the three sides of the new triangle are 13.5, 14, and 16.5 inches.
22.
6 inches
23.
Let x = the width. Then 4x = the length.
2 x + 2(4 x) = 4 x + 84
10 x = 4 x + 84
6 x = 84
x = 14
4 x = 56
The width is 14 ft. and the length is 56 ft.
Revised 03/09
9
24.
Square: 12 feet and Rectangle: 6 by 18 feet
25.
Let x = the width. Then x + 3.46 = the length.
2 x + 2( x + 3.46) = 29.20
4 x + 6.92 = 29.20
4 x = 22.28
x = 5.57
x + 3.46 = 9.03
The width is 5.57 meters and the length is 9.03 meters
26.
12.36 by 37.08 inches
NOTE: You can get additional instruction and practice by going to the following
web sites:
http://www.purplemath.com/modules/translat.htm This website gives general
help in translating word problems into algebraic equations.
http://www.edhelper.com/geometry.htm This website contains many printable
worksheets on perimeter and area problems.
Revised 03/09
10