DIMENSIONAL ANALYSIS #2
Area is measured in square units, such as square feet or square centimeters. These units can be
abbreviated as ft2 (square feet) and cm2 (square centimeters). For example, we might say that the
area of a room is 328 ft2. Generally, when you see a unit of measure with an exponent of 2, it is a
measurement of area.
Similarly, volume is measured in cubic units. The volume of a tank might be written as 200 cubic
feet or 200 ft3. Generally, when you see a unit of measure with an exponent of 3, it is a
measurement of volume.
The technique of dimensional analysis can be used to convert from one unit of area or volume to
another unit of area or volume. However, it is important to remember that more than one dimension
is being considered.
EXAMPLE 1: Determine the number of square inches in 1 square foot.
SOLUTION: This problem is equivalent to solving 1 ft2 = ? in.2 To solve this problem,
remember that 1 square foot means the area of a square that measures 1 foot on each side. The area
of the square is A = lw = (1 ft)(1 ft) = 1 ft2.
1 ft
1 ft
2
1 ft
To convert from square feet to square inches, we must consider that each of the two dimensions of
the square is 1 foot, and each foot is equivalent to 12 inches.
We can perform the conversion using unit fractions, but we must use the conversion factor of
12 inches
twice, once for each dimension of the square.
1 ft
(1 ft)(1 ft)
(12 in.)(12 in.)
(1)(1)(12)(12) in.2
1 ft 2
=
×
=
= 144 in 2
(1 ft)(1 ft)
1
1
1(1)(1)
So there are 144 square inches in 1 square foot. Notice that the unit fraction we used to perform the
conversion had 1 foot twice in the denominator (equivalent to 1 ft2) and 12 inches twice in the
numerator.
REMEMBER: To convert from one unit of area to another unit of area, always use the
conversion factor two times. The exponent of 2 on the unit is a reminder that there are two
dimensions to consider.
printed 11/16/2004 ©Copyright Linn-Benton Community College Mathematics Department. Used with permission.
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Dimensional Analysis #2, Continued
Performing a conversion between two different units of volume is very similar. Volume involves
three dimensions and is expressed in cubic units. For example, 1 cubic meter means the volume of a
cube that measures 1 meter by 1 meter by 1 meter. The volume of that cube is given by
V = lwh = (1 m)(1 m)(1 m) = 1 m3.
1m
1m
1m
3
1m
When we perform a conversion using cubic meters, we must remember that there are three
dimensions to the cube, each of which measures 1 meter. So the conversion factor must be used
three times, as shown in the next example.
EXAMPLE 2: Convert 15,800,000 cubic millimeters to cubic meters.
SOLUTION: This problem is equivalent to solving 15,800,000 mm3 = ? m3. We begin by
15,800, 000 mm3
3
writing 15,800,000 mm as
. Now we multiply by a unit fraction that relates mm3
1
to m3 and that has mm3 in the denominator. From the measurement and conversion table, we know
that 1 m = 1000 mm, so we write a unit fraction using the conversion factor of 1000 three times,
once for each dimension.
(1 m)(1 m)(1 m)
(15,800,000)(1)(1)(1) m 3
15,800, 000 mm3
×
=
= 0.0158 m 3
1
(1000 mm)(1000 mm)(1000 mm)
1(1000)(1000)(1000)
So 15,800,000 cubic millimeters is equivalent to 0.0158 cubic meters.
The conversion we wrote above with the unit fraction relating mm3 and m3 can be written more
briefly as shown below.
15,800, 000 mm3
1 m3
15,800,000(1) m3
×
=
= 0.0158 m 3
1
1000(1000)(1000) mm3
1(1000)(1000)(1000)
or even as
1 m3
15,800,000(1) m 3
15,800, 000 mm3
×
=
= 0.0158 m 3
1
(1000)3 mm3
1(1000)(1000)(1000)
The important thing is to be sure to use the conversion factor three times.
REMEMBER: To convert from one unit of volume to another unit of volume, always use the
conversion factor three times. The exponent of 3 is a reminder that there are three dimensions
to consider.
printed 11/16/2004 ©Copyright Linn-Benton Community College Mathematics Department. Used with permission.
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Dimensional Analysis #2, Continued
We can “chain” unit fractions to perform conversions with square and cubic units as we did in the
earlier problems. The next example illustrates this process.
EXAMPLE 3: The displacement of an engine is often measured in liters or in cubic inches.
Suppose that a compact car has a 1.8 L engine. What is the displacement of this engine in cubic
inches?
SOLUTION: This problem is equivalent to the conversion: 1.8 L = ? in.3 Notice only the unit
on the right-hand side has an exponent of 3. We can compare liters and cubic inches, however, since
a liter is a measure of capacity. Remember that 1 mL = 1 cm3. We will use this relationship to
perform the conversion.
First we will convert from liters to cubic centimeters by multiplying by two unit fractions.
1000 mL
1 cm3
1.8 L
×
×
1 mL
1L
1
If we multiplied these fractions, liters and milliliters would “cancel” and we would be left with cubic
centimeters. Now we must convert from cubic centimeters to cubic inches, so we multiply by a unit
fraction with cubic inches in the numerator and cubic centimeters in the denominator.
1000 mL
1 cm3
1 in.3
1.8(1000)(1)(1) in.3
1.8 L
×
×
×
=
≈ 110 in.3
3
3
3
1L
1 mL
(2.54) cm
1(1)(1)(2.54)
1
Multiplying these fractions together gives us the equivalent displacement in cubic inches (rounded to
the nearest whole number). Therefore, the displacement of the engine in the compact car is
approximately 110 cubic inches.
printed 11/16/2004 ©Copyright Linn-Benton Community College Mathematics Department. Used with permission.
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Dimensional Analysis #2, Continued
PROBLEM SET 2
Answers to the odd-numbered problems are given at the end of the problem set.
WARM-UP EXERCISES
Use dimensional analysis to perform the following conversions. Show the procedure that you used,
including all of your unit fractions. If an answer is not exact, round to two decimal places.
1. 10 ft2 to in.2
2. 16 in.2 to cm2
3. 7.2 mm3 to cm3
4. 3 m3 to L
5. 6 mi2 to km2
6. 150 gal to yd3
7. 2500 cm3 to qt
8. 2.5 acres to yd2
PROBLEMS
Use dimensional analysis to solve each of the following problems. Show the procedure that you
used, including all of your unit fractions. Answer the question in a complete sentence.
9. Frank just moved to the U.S. and wants to buy a 40-hectare farm. How many acres should he
tell the real estate agent he wants to buy? Round your answer to the nearest acre.
10. A room measures 16 ft by 12 ft.
a. Find the area of the room in square yards. Round your answer up to the nearest
whole number.
b. What would it cost to carpet the room if the carpet sells for $21.99 per square yard?
Round your answer to the nearest cent.
11. A basement for a 30 ft by 40 ft house is to be dug at a depth of 7 feet. How many cubic
yards of earth need to be hauled away? Assume that the surface of the ground is level.
Round your answer to the nearest cubic yard.
12. A cylindrical tank is 8 meters long and 4 meters in diameter. How many liters of liquid will
the tank hold? Round your answer to the nearest thousand.
EXTRA PROBLEMS
Use dimensional analysis to perform the following conversions. Show the procedure that you used,
including all of your unit fractions. If an answer is not exact, round to two decimal places.
13. 150 cm2 to m2
14. 6 yd 3 to ft3
15. 0.009 m3 to in.3
16. 350 in.3 to L
Use dimensional analysis to solve each of the following problems. Show the procedure that you
used, including all of your unit fractions. Answer the question in a complete sentence. Round your
answers to the nearest whole number.
17. Ann’s back yard is 60 feet by 70 feet. What is the area of her backyard in square meters?
18. A fish tank measures 25 inches by 16 inches by 12 inches. How many cubic feet of water
will it hold? How many gallons will it hold?
printed 11/16/2004 ©Copyright Linn-Benton Community College Mathematics Department. Used with permission.
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Dimensional Analysis #2, Continued
ANSWERS TO ODD-NUMBERED PROBLEMS:
(NOTE: For some of these problems there are several ways to set up the problem. Therefore, your
unit fractions may look different from the sequence of unit fractions shown here. Your final answer,
however, should be approximately the same as the one given below.)
1.
10 ft 2
(12)2 in.2
×
= 1440 in.2
2
1
1 ft
3.
7.2 mm 3
1 cm3
3
×
3
3 = 0.0072 cm
1
(10) mm
6 mi2
(1.609)2 km 2
5.
×
≈ 15.53 km 2
2
1
1 mi
2500 cm3
1 mL
1L
1.057 qt
7.
×
×
≈ 2.64 qt
3 ×
1
1 cm
1000 mL
1L
40 ha
10,000 m 2
(39.37) 2 in.2
1 ft 2
1 acre
9.
×
×
×
×
2
2
2
2 ≈ 99 acres
1 ha
1m
(12) in.
1
43,560 ft
Frank needs a farm of approximately 99 acres.
11. Volume = (30 ft)(40 ft)(7 ft) = 8400 ft3
8400 ft 3
1 yd 3
3
×
3
3 ≈ 311 yd
1
(3) ft
Approximately 311 cubic yards of dirt must be hauled away.
13.
150 cm2
1 m2
2
×
2
2 = 0.015 m
1
(100) cm
15.
0.009 m3
(100)3 cm3
1 in.3
3
×
×
3
3
3 ≈ 549.21 in.
1
1m
(2.54) cm
17. Area = (60 ft)(70 ft) = 4200 square feet.
4200 ft 2
(12)2 in.2
1 m2
2
×
×
2
2
2 ≈ 390 m
1
1 ft
(39.37) in.
The area of Ann’s back yard is approximately 390 square meters.
printed 11/16/2004 ©Copyright Linn-Benton Community College Mathematics Department. Used with permission.
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