120104d
Welder
Geometric Formulas
Trade Math
First Period
Table of Contents
Objective One ............................................................................................................................................... 2 Key Terms ................................................................................................................................................. 2 Order of Operations in Solving Equations ................................................................................................ 3 Solving Equations ..................................................................................................................................... 4 Objective One Exercise................................................................................................................................. 6 Objective Two............................................................................................................................................... 7 Perimeter ................................................................................................................................................... 7 Area ......................................................................................................................................................... 13 Volume .................................................................................................................................................... 23 Objective Two Exercise .............................................................................................................................. 27 Objective Three ........................................................................................................................................... 39 Calculating the Weight of a Solid ........................................................................................................... 39 Objective Three Exercise ............................................................................................................................ 42 Objective Four ............................................................................................................................................ 45 Capacity of an Object .............................................................................................................................. 45 Objective Four Exercise .............................................................................................................................. 47 Self-Test ...................................................................................................................................................... 50 Self-Test Answers ....................................................................................................................................... 54 Objective One Exercise Answers................................................................................................................ 55 Objective Two Exercise Answers ............................................................................................................... 55 Objective Three Exercise Answers ............................................................................................................. 57 Objective Four Exercise Answers ............................................................................................................... 57 NOTES
Geometric Formulas
Rationale
Why is it important for you to learn this skill?
Many of the products you will work on in the shop or on the jobsite are made up of
regular geometric shapes like circles, squares or triangles. Whether you work on an oil
storage tank, a hopper or a truck bed, you will eventually need to calculate perimeter,
area, volume, weight or capacity. You may need these calculations to establish cost or
quantity of materials or to determine the size of an object after you fabricate it.
Outcome
When you have completed this module, you will be able to:
Solve problems involving geometric formulas.
Objectives
1.
2.
3.
4.
Identify terms and concepts used in working with formulas.
Identify formulas and solve problems for perimeter, area and volume.
Calculate the weight of a solid.
Calculate the capacity of a container in gallons and litres.
Introduction
This module addresses the terminology and concepts you will need to identify common
formulas and solve problems for finding perimeter, area and volume. It also addresses
how to use the volume calculation to determine the weight of a solid and to convert to
gallons and litres to find capacity.
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Objective One
When you have completed this objective, you will be able to:
Identify terms and concepts used in working with formulas.
Key Terms
The following terms are used throughout the trade. To work effectively with equations,
you should be familiar with them.
Term
Definition
equation
A statement or mathematical expression that uses an equal sign to
indicate two related quantities have the same value. All equations must be
true.
5  3  4x 2
n 2  64
formula
A law or rule that you express as an equation. A formula expresses some
fundamental fact or truth.
Area  Length x Width
Volume = r2h
constant
A numerical value that forms part of an equation or formula. It is a value
that cannot change. In the following examples, 2 and  are constants.
P  2L  2W
C = d
variable
A symbol or letter that represents an unknown quantity in an equation or
formula. A variable has different values depending on the problem, such
as the radius of a circle (r) or the length of the side of a square (s).
A = r2
P = 4s
term
The parts of an equation or formula separated by addition, subtraction or
equal signs. A term can be a single constant or variable (A). A term can
be two or more constants and variables that you multiply together (2r2)
or it can include division (12bh).
A 
1
bh
2
A = 2r2 + LW
solution
A replacement for a variable or unknown that makes an equation true. To
find the solution is the reason why you do all the calculations.
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Order of Operations in Solving Equations
When you solve for an equation that has two or more mathematical functions, you must
follow a standard order of operations. An acronym to describe the order of operations is
Brackets, Exponents, Division, Multiplication, Addition and Subtraction (BEDMAS).
These letters indicate the order in which you must perform the operations.
1. Brackets
2. Exponents
3. Division
4. Multiplication
5. Addition
6. Subtraction
You can sometimes confuse the multiplication symbol in standard arithmetic with the
symbol x that you use as a variable. You can use brackets or no brackets at all, instead.
For example, when you see the following in a mathematical equation, solve it like you
would a multiplication problem.
3(38)  114
You may also see multiplication problems expressed without brackets, like LW, which
means length times the width.
Example 1
The following example shows how you solve using BEDMAS, with the mathematical .
n  8 – 3  2  20  5
n  8 – 6  20  5
n  8 – 6  4
n  2  4
n  6
Calculators follow this order of operations, so you must take care when you input a series
of numbers and functions. Perform division and multiplication operations as they occur
from left to right. Then perform addition and subtraction operations as they occur from
left to right.
Example 2
The following example shows how you solve using BEDMAS, with the mathematical and brackets.
n  3[5  9 –  4  3]
n  3(5  9 – 7)
n  3 45 – 7 
n  3 38
n  114
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NOTE
When brackets surround any group of terms, you must perform the
functions inside the brackets first. If there is more than one set of
brackets, perform the operation within the inner set first and the
operation within the outer set last.
Example 3
The following example shows you find the area of a circle with a radius of 5 inches,
using BEDMAS and brackets.
A  r 2
A  (5"  5")
A  78.5 in 2
NOTE
You will encounter an exponent when finding the area of a circle
(A = r2). You must find r2 before you multiply by . There is a big
difference between r2 and (r)2 if r is not equal to 1.
Solving Equations
In order to solve a simple equation, isolate the variable or unknown on one side of the
equation. An equation is equal on both sides; therefore, if you add or subtract on one side,
you must add or subtract the same amount on the other side. You can also multiply or
divide every number in an equation by the same number. The objective of these
operations is to isolate the variable so that you have a solution to the equation.
Example 1
Solve for the unknown variable (n) in the following equation.
6n – 7  n / 2  15
To solve for this equation, do the following.
1. Add 7 to both sides.
6n – 7  7  n / 2  15  7
6n  n / 2  22
2. Multiply all terms by 2.
2  6n   2  n / 2  2  22
12n  n  44
3. Subtract n from both sides.
12n – n  n – n  44
11n  44
4. Divide all terms by 11.
11n  11  44  11
n  4
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Example 2
You want to build a welding shop with an area of 2000 square feet. If you must limit it to
a width of 32 feet, what is the length of the shop?
Solution
To solve, consider the following.
 Area = Length x Width
 2000 ft2 = Length x 32 feet
 Divide both sides by 32 feet.
 2000 ft2/32ft = Length x (32 feet/32 feet)
 62.5 feet = Length
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Objective One Exercise
1. Define equation.
2. Define formula.
3. Define constant.
4. Find the value for P in the formula P = L x W, if L = 43.7 and W = 13.2.
5. Find the value for L in the formula P = L x W, if P = 56 and W = 7.
6. Find the value for A in the formula A = r2, if  = 3.14 and r = 3.5.
7. Find the value for A in the formula A = h
8. Find the value for h in the formula A = h
( B 2 b ), if h = 6, B = 10 and b = 5.
( B 2 b ), if A = 210, B = 23 and b = 7.
9. Find the value for H in the formula V = LWH, if V = 3600, L = 40 and W = 20.
10. Find the value for h in the formula V = r2h, if V = 251.2,  = 3.14 and r = 5.
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Objective Two
When you have completed this objective, you will be able to:
Identify formulas and solve problems for perimeter, area and volume.
Perimeter
Perimeter is the distance around any geometric figure, such as a square, rectangle,
triangle, trapezoid, parallelogram, circle or semicircle. Any piece of material in the
welding trade could be a geometric figure or a combination of two or more figures. You
can use perimeter to determine cost or quantity of material that you require. When the
figure is a circle, you call the perimeter the circumference.
Perimeter of a Rectangle
A rectangle is a four-sided figure; its interior angles are all right angles (90°) and its
opposite sides are parallel and equal in length (Figure 1).
Figure 1 - Rectangle.
NOTE
The little squares inside the rectangle in Figure 1 (and similar
illustrations) denote a right angle.
To find the perimeter (P) of a rectangle, add the two lengths and the two widths. You can
express the formula in the following ways.
P  L  L  W  W
P  2L  2W
P  2 L  W
Example
Find the perimeter of the rectangle in Figure 2 in centimeters (cm). Convert to similar
units; for example, 10 mm equal 1 cm; therefore, 216.9 mm equal 21.69 cm.
Figure 2 - Perimeter of a rectangle.
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To solve, do the following.
P  2L  2W
P  2  58.2 cm   2  21.69 cm 
P  159.78 cm
Perimeter of a Square
A square is a rectangle that has four equal sides in length (Figure 3).
Figure 3 - Square.
To find the perimeter of a square, add the length of each side (s). You can express this
formula in the following ways.
P  s  s  s  s
P  4s
Example
Find the perimeter of the square in Figure 4 in feet and inches.
Figure 4 - Perimeter of a square.
To solve, do the following.
P = 4s
P = 4(478")
P = 1912"
P = 1' 712"
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Perimeter of a Triangle
A triangle is a figure that can take several different shapes, but always consists of three
straight lines.
Right Triangle
A right triangle (or right-angle triangle) is a triangle that has one 90° angle (Figure 5).
The sloping line opposite the right angle is the hypotenuse.
Figure 5 - Right triangle.
3-4-5 Triangle
A 3-4-5 triangle is a right triangle that you can use in shop layout for squaring frames or
structural members. The hypotenuse is 5 units of length and the other two sides are 3 and
4 units of length. You can also use multiples of 3-4-5 as dimensions, such as 9-12-15.
Equilateral Triangle
An equilateral triangle includes sides that are the same length and angles that all equal to
60° (Figure 6). An equilateral triangle is one of the strongest shapes in nature, so you
often use them in fabricating.
Figure 6 - Equilateral triangle.
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Isosceles Triangle
An isosceles triangle is a triangle in which two of the three sides are of equal length and
two of the three angles are equal (Figure 7).
Figure 7 - An isosceles triangle.
Scalene Triangle
A scalene triangle is a triangle that has no equal sides or equal angles. All three sides are
different lengths and all three angles are different values (Figure 8).
Figure 8 - Scalene triangle.
To find the perimeter of any triangle, add the lengths of the three sides (A, B and C). You
can express this formula in only one way.
P  A  B  C
Example
Find the perimeter of the triangle in Figure 9 in feet and inches.
Figure 9 - Perimeter of a triangle.
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To solve, do the following.
P  A  B  C
P  231 / 8 "  193 / 8 "  101 / 4 "
P  523 / 4 "
P  4' 43 / 4 "
Circumference of a Circle
A circle is a closed curved line on a flat surface in which every point is the same distance
from a fixed given point called the centre. Figure 10 lists some basic terminology you
must know when dealing with circles.
 The circumference is the distance around a circle.
 The diameter is the straight-line distance across a circle and passes directly
through the centre. If you divide the diameter by 2, you get the radius.
 The radius is the straight-line distance from the centre to the edge of a circle. If
you multiply the radius by 2, you get the diameter.
 The symbol  (pi) is a number approximately equal to 3.1416 and it is the
circumference divided by the diameter.
Figure 10 - Circle.
For most calculations, you can use  = 3.1416; however, it is faster and more accurate to
use the  button on a calculator because it fills the display ( = 3.141592654). You have
to be careful when doing calculations involving  because your answers will vary slightly
depending on the value that you use for  (how much you round it).
To find the circumference (C) of a circle, multiply the diameter (d) by . The diameter is
equal to two radii; therefore, you can write the formula as C = 2r.
C  d
C  2r
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Example
Find the circumference of the circle in Figure 11 (round to two decimal places).
Figure 11 - Circumference of a circle.
To find the circumference of the circle in Figure 11, you can perform either of the
calculations in Table 1.
C  d
C  2r
C    7.2 m 
C  2()  3.6 m 
C  22.62 m
C  22.62 m
Table 1 - Solutions for determining the circumference of a circle.
Perimeter of a Combined Geometric Figure
You may need to find the perimeter of objects that are a combination of two or more
geometric figures. In these cases, divide the object into its geometric component parts.
Example
You have to order enough fencing to enclose a skating rink (Figure 12).
Figure 12 - Skating rink.
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If you remove the two end pieces, each a semicircle (half circle), and place them together
(Figure 13), they form a whole circle. The remaining part will have the shape of a square.
NOTES
Figure 13 - Perimeter of a skating rink.
The distance around the skating rink then becomes the circumference of the circle plus
two straight lengths of 180' each.
C = d
C = (180')
C = 565.487'
Therefore, the total length of the fence you require for the skating rink is as follows.
565.487 + (2 x 180) = 925.487'
Area
Area is the amount of material it takes to cover a surface. Area uses two dimensions that
must always be in the same units. For example, if you use millimetres or centimetres for
distance, the area then becomes square millimetres (mm2) or square centimetres (cm2),
respectively. You must sometimes convert units to make them the same.
Area of a Rectangle
Figure 14 represents a rectangle with a length of 5' and a width of 4'.
Figure 14 - Area of a rectangle
The rectangle divides into small squares, one foot on each side, so each small square
represents 1 ft2. Since there are four rows of squares each containing 5 ft2, the area
becomes 4 x 5 = 20 ft2. This method of calculating area works with any lengths, including
fractions and metric measure. The formula for finding the area (A) of a rectangle is equal
to the product of its length (L) and its width (W), which is A = LW.
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Example
Find the area (mm2) of the rectangle in Figure 15. If 1 cm = 10 mm, then 58.2 cm =
582 mm.
Figure 15 - Area of a rectangle.
To solve, do the following.
A  LW
A  582 mm  216.9 mm
A  126 235.8 mm2
Area of a Square
A square is a special rectangle with all four sides equal in length. If you multiply the
length by the width (which are the same), you find the area (A) of a square. You are
multiplying one side (s) by another side (s) or you are squaring one side (s2), as the
equations show.
A  sxs
A  s2
Example
Find the area of the square in Figure 16.
Figure 16 - Area of a square.
To solve, do the following.
A  s2
A  ( 47 / 8 ") 2
A  2349 / 64 in 2
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Area of a Triangle
You can think of a triangle as a rectangle that you cut exactly in half (Figure 17). Instead
of using length and width (as with a rectangle), you use base and height when calculating
the area of a triangle. The height of a triangle is its altitude.
Figure 17 - Area of a triangle.
Since a triangle is equal to half the area of a rectangle with length equal to base and width
equal to height, the formula for finding the area (A) of a triangle is half (1/2) of the
product of the base (b) and the height (h). Either of the following equations will result in
the correct answer.
A  1 / 2 bh
bh
A 
2
These formulas hold true for any triangle. Figure 18 illustrates the height for triangles
that are not right triangles.
NOTE
The base and height must be perpendicular to each other.
Figure 18 - Height of a triangle.
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Example
Find the area of the triangle in Figure 19.
Figure 19 - Finding the area of a triangle.
To solve, do the following.
A  1 / 2 bh
A  1/ 2  231 / 8 " (81 / 2 ")
A  989 / 32 in 2
Area of a Circle
To find the area of a circle, you must know the radius. If you know the diameter, you can
divide it by 2 to find the radius (Figure 20). The formula to find the area (A) of a circle is
as follows.
A  r  r
A  r 2
Example
Find the area of the circle in Figure 20 (round to two decimal places).
Figure 20 - Finding the area of a circle.
To solve, do the following.
A  r 2
A    3.6m 
16
2
A  40.72 m2
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Area of a Trapezoid
A trapezoid is a figure that has only one set of parallel sides. These two sides are the
bases (large base [B] and small base [b]). The height of a trapezoid is the perpendicular
distance between the bases (Figure 21).
Figure 21 - Bases of a trapezoid.
One way to determine the area of a trapezoid is to remove the small triangles from each
bottom half of the trapezoid and reattach them to the upper corners to form a rectangle
(Figure 22).
Figure 22 - Area of a trapezoid.
In Figure 22, the length of the newly formed rectangle is 11", which is the average length
of the large base and the small base. The area of a rectangle is length times width (LW);
therefore, to find the area (A) of a trapezoid, you find the average of the two bases
(B + b)/2 and multiply by the height (h). The bases are the parallel sides and the height
(B  b)
must be perpendicular to the bases. Use the equation A  h
.
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Example
Find the area of the trapezoid in Figure 23. If 12" is 1', then 7' 9" is equal to 7.75' and 31'
6" is equal to 31.5'.
Figure 23 - Finding the area of a trapezoid.
To solve, do the following.
Ah
B b
2
 31.5'  16' 
A  7.75' 

2


2
A  184.0625ft
Area of a Parallelogram
A parallelogram has four sides. The sides opposite each other are the same length and are
parallel and the angles opposite each other are equal. The height of a parallelogram is the
perpendicular distance between two sides (Figure 24).
Figure 24 - Area of a parallelogram.
If you were to cut the parallelogram along the dotted height line and reassemble it on the
other end, you would create a rectangle. The height of the parallelogram would be equal
to the width of the rectangle and the base of the parallelogram would be equal to the
length of the rectangle. To find the area (A) of a parallelogram, you multiply the base (b)
with the perpendicular height (h). Do not use the slant height for finding area.
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Example
Find the area (cm2) of the parallelogram in Figure 25. If 1 m = 100 cm, then 0.23 m =
23.0 cm.
Figure 25 - Finding the area of a parallelogram.
To solve, do the following.
A  bh
A  23.0 cm  9.4 cm
A  216.2cm2
Lateral Surface Area of a Cylinder
The lateral surface area (LSA) of any solid is the surface area of the sides (walls). It does
not include the area on the top and/or bottom. You should use the following calculations
only for cylinders with uniform cross-sections (same shape from top to bottom) and with
sides that are perpendicular to the bases. Figure 26 is an illustration of a cylindrical tank
with the shell rolled out flat to make it look like a rectangle.
Figure 26 - Lateral surface area of a cylinder.
You can find the area of a rectangle by multiplying the LW.
In Figure 26, the length of the rectangle is actually the circumference of the circle, which
you calculate by multiplying  with the diameter. Therefore, the formula to find the
lateral surface area (LSA) of a cylinder is as follows.
LSA  dh
LSA  2rh
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Example
Find the lateral surface area (ft2) of the cylinder in Figure 27.
Figure 27 - Finding the lateral surface area of a cylinder.
To solve, do the following.
LSA  dh
LSA   1.5' 4'
LSA  18.85 ft 2
Total Surface Area of a Closed Cylinder
The total surface area (TSA) of a closed cylinder includes the lateral surface area and the
area of both ends, which are circles. You must add the area of the ends (both circles) to
the area of the lateral surface.
TSA  dh  2( r 2 )
Example
Find the total surface area (in2) of the cylinder in Figure 28. Use π = 3.14 for this
example.
Figure 28 - Finding the total surface area of a cylinder.
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To solve, do the following.
TSA  dh  2( r 2 )
TSA   3.14 18" 48"  2  3.14  9"
2
TSA  2712.96 in2  508.68 in 2
TSA  3221.64 in 2
Lateral Surface Area of a Right Rectangular Solid
A right rectangular solid has a rectangular base with sides that are perpendicular to it.
You can find the lateral surface area by finding the area of each of the four sides and then
adding these areas together. You could also add the two lengths and the two widths
together and then multiply this sum by the height. It is similar to rolling out the outer
shell to make it one large rectangle. In this case, the lateral surface area is the perimeter
of the base multiplied by the height.
Figure 29 illustrates the pieces of the rectangular shape that make up the lateral surface
area.
Figure 29 - Lateral surface area of a rectangular solid.
To find the area of the two large sides, you multiply length (L) by height (H) of both
sides. To find the area of the two ends, you multiply the width (W) by the height (H) of
both ends. You can state the formula in two different ways.
LSA  2LH  2WH
LSA  2  LH  WH 
Using the perimeter to find the lateral surface area, you can also state the formula in two
other ways.
LSA  H  L  L  W  W 
LSA  H  2L  2W 
In all cases, finding lateral surface area of any solid, regardless of the shape of the base, is
a matter of finding the perimeter of the base and multiplying the perimeter by the height.
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Example
Find the lateral surface area (in2) of the rectangular shape in Figure 30.
Figure 30 - Finding the lateral surface area of a rectangular solid.
To solve, do the following.
LSA  2LH  2WH
LSA   2  32  28   2  14  28
LSA  1792  784
LSA  2576 in 2
Total Surface Area of a Right Rectangular Solid
The TSA of a right rectangular solid includes the lateral surface area plus the area of the
top and bottom, which are rectangles. You have to add the area of the top and bottom to
the area of the lateral surface. To find the area of the top and bottom, multiply length (L)
by width (W), as the equation shows.
TSA  2LH  2WH  2LW
Example
Find the total surface area (in2) of the rectangular shape in Figure 31.
Figure 31 - Finding the total surface area of a rectangular solid.
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To solve, do the following.
TSA  2LH  2WH  2LW
TSA   2  32  28   2  14  28   2  32  14
TSA  1792  784  896
TSA  3472in 2
Volume
Volume is the amount of space that an object occupies. You most often use volume to
mean capacity or the number of cubic units enclosed within an object like a bin or tank.
Before you can calculate the volume of any object, you must ensure that the
measurements are in the same units.
Volume of a Right Rectangular Solid
When calculating the volume of a right rectangular solid, use the length, width and
height. For example, if you have a cube that is 12" by 12" by 12" (Figure 32), you want to
find the number of one-inch cubes that will fit in the large cube. You can also find the
area of the base and multiply that area by the height.
Figure 32 - Volume of a rectangular solid.
In the case of a cube, which has all equal sides, you can find the volume (V) by cubing
the length of one side (s3).
V  s  s  s
V  s3
If you have a right rectangular solid, you find the volume (V) by multiplying the length
(L) by the width (W) by the height (H).
V  LWH
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Example
Find the volume of the rectangular solid in Figure 33 using both in3 and ft3. Ensure that
all dimensions are in the same units. 2 yards = 6 ft = 72".
Figure 33 - Finding the volume of a right rectangular solid.
To solve, use Table 2.
Solution in in3
Solution in ft3
V  in 3   LWH
V  ft 3   LWH
V  in 3   72" x 30" x 6"
V  ft 3   6' x 2.5' x 0.5'
V  in 3   12 960 in 3
V  ft 3   7.5 ft 3
Table 2
NOTE
1 ft3 = 1728 in3.
Volume of a Cylinder
To find the volume of a cylinder, you must first find the area of the base and then
multiply that area by the height. The height is the measurement that is perpendicular to
the round end, no matter which way the tank lies (Figure 34).
Figure 34 - Volume of a cylinder.
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You use the formula A = r2 (where the radius is half the diameter) to find the area of a
circle. To find the volume (V) of the cylinder, multiply the area of the base (r2) by the
height (h).
NOTES
V  r 2 h
Example
Find the volume (ft3) of the cylinder in Figure 35 rounded to two decimals.
V  r 2 h
V    52  16
V  1256.64 ft 3
Figure 35 - Finding the volume of a cylinder.
Volume of any Regular Shaped Object
You calculate volume by finding the area of the base and multiplying it by the height.
You can determine the volume of any object with a uniform cross-section by calculating
the area of the cross-section and then multiplying it by the height. Figure 36 shows some
examples of these shapes.
Figure 36 - Uniform cross-section figures.
For all objects that have a uniform cross-section, you can find the volume by finding the
area of the base and multiplying by the height. You can also break the object into familiar
shapes and find the volume of each familiar shape and then add them together.
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Example
Find the volume of the semicircular tank in Figure 37, rounded to two decimal places.
If you look closely at this tank, you will see that it is actually a tank with two half
cylinders at each end and a solid rectangle in the centre. To find the volume, either:
 find the area of the base and multiply by the height or
 find the volume of the two half cylinders and the volume of the rectangular
centre and add them together.
Figure 37 - Semicircular tank.
To solve, use Table 3.
Area of Base × Height
Cylinder Volume + Rectangle
V  ( r 2  LW )  H
V  r 2 H  LWH
V   r 2  (9  4)   14
V    22  14    9  4  14 
No Addition
V  175.93  504
V  679.93 ft
3
V  679.93 ft 3
Table 3
NOTE
You can treat the two half cylinders like one cylinder.
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NOTES
Objective Two Exercise
Perimeter
1. Define perimeter.
2. You use the formula P = 2L + 2W to find the perimeter of what figure?
3. Which formula do you use to find the circumference of a circle?
4. Find the perimeter of the square in Figure 38
Figure 38
5. Calculate the perimeter (feet and inches) of a square tank bottom that measures
2' 81/4".
6. Calculate the perimeter (metres) of a rectangle with a length of 450 cm and a width
of 1800 mm.
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NOTES
7. Determine the perimeter (centimetres) of the rectangular holding tank bottom in
Figure 39.
Figure 39
8. Calculate the perimeter of a triangle (feet and inches) to the nearest 1/16". It has
dimensions of 50.6', 69' and 103.5".
9. Determine the perimeter (centimetres) of the triangle-shaped neon sign frame in
Figure 40.
Figure 40
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10. Find the circumference of each of the following circles to the nearest 1/64".
(Use π button).
a) d = 14"
NOTES
b) r = 12"
c) r = 3.25"
d) r = 6.5'
e) d = 2' 21/2"
11. What is the circumference of the tank lid in Figure 41 to the nearest 1/16"?
(Use π button).
Figure 41
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NOTES
12. Figure 42 shows the shape of a storage yard for a welding shop. Calculate (feet) the
amount of fencing that you require to enclose the yard completely.
Figure 42
13. You must place a drainage pipe around the foundation of a building (Figure 43).
What is the total length of drainage pipe that you require if you allow 15' for waste
and corners to the nearest 1/16"? (Use π button).
Figure 43
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14. You are fabricating a semicircular-ended tank. The bottom comes from 1/8" steel plate
stock with the dimensions in Figure 44. What length of material (inches) do you
require to form the sides of the tank to the nearest 1/64"? (Use π = 3.1416.)
NOTES
Figure 44
Area
1. Define area.
2. What formula do you use to find the area of a triangle?
3. You use the formula A = s2 to find the area of what figure?
4. The lateral surface area of a rectangular solid includes adding together the area of
how many surfaces?
5. What formula do you use to find the total surface area of a cylinder?
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NOTES
6. Determine the area (in2) of the pieces of sheet metal in Figure 45. Round your
answers to the nearest 116".
Figure 45
2
7. Find the area (ft ) of a square building lot that measures 76 yards on one of its sides.
8. A piece of carpet measures 12' x 15'. Find its total cost at $22.50/yd2.
9. Calculate the area (ft2) of a triangle if the base is 36" long and the height is 8".
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10. Calculate the area (in2) of the triangular-shaped pieces of sheet metal in Figure 46.
NOTES
Figure 46
11. Find the area (cm2) of a circle with a diameter of 650 mm, rounded to two decimal
places. (Use π button).
12. Calculate the area (in2) of the shaded portion of the steel spacer plate in Figure 47,
rounded to three decimals. (Use π button)
Figure 47
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NOTES
13. Calculate the area (ift2) of the piece of sheet metal in Figure 48.
Figure 48
14. Find the area (in2) of the stair stringer support plate in Figure 49.
Figure 49
15. How many square feet of 58" steel plate do you require for a highway bridge surface
if you need 375 sections of the shape in Figure 50 to complete the job?
Figure 50
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16. Find the lateral surface area (ft2) and rounded to two decimal places of a cylinder if
the diameter is 32" and the height is 8.6'. (Use π button).
NOTES
17. Find the total surface area (ft2) and rounded to two decimal places of an enclosed
cylinder if the diameter is 53" and the height is 4.5'. (Use π button).
18. Determine the lateral surface area (in2) of the steel tank in Figure 51.
Figure 51
Volume
1. Define volume.
2. What formula do you use to find the volume of a right rectangular solid?
3. You use the formula V = r2h to find the volume of what figure?
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NOTES
4. Convert each of the following volumes. Round your answers to two decimal places
where applicable.
a) 2 ft3 to in3
b) 123 ft3 to in3
c) 3.65 ft3 to in3
d) 3466 in3 to ft3
e) 18.144 in3 to ft3
f) 2946 in3 to ft3 and yd3
g) 3912 ft3 to yd3
5. What is the volume (in3) of the three steel bars in Figure 52? (Round your answers to
two decimal places where applicable.)
a) A =
b) B =
c) C =
Figure 52
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6. The two tanks in Figure 53 are made from 14 gauge steel plate. Determine the
volume of tank A (in3) and the volume of tank B (ft3). Round your answers to two
decimal places.
NOTES
Figure 53
3
7. Find the volume of a cylinder (ft ), rounded to two decimal places, if the diameter is
24" and the height is 200'. (Use π button).
8. What is the volume (ft3), (rounded to two decimal places, of the hot water tank in
Figure 54? (Use π button).
Figure 54
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9. Three cylindrical coolant tanks are welded together (Figure 55). What is the volume
(ft3), rounded to two decimal places, if you fill the system and connecting pipe
completely? (Use π button).
Figure 55
3
3
10. Calculate the volume (in ) and (ft ) of the heavy equipment fuel tank in Figure 56,
rounded to two decimal places. (Use π button).
Figure 56
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NOTES
Objective Three
When you have completed this objective, you will be able to:
Calculate the weight of a solid.
Calculating the Weight of a Solid
In order to calculate the mass or weight of an object, you must first find its volume. You
then multiply the volume by the known weight of one unit of volume of the material in
question. In the metric system, the weight of one cubic centimetre of steel is 7.849 grams.
In the imperial system, the weight of one cubic inch of steel is 0.2835 lb.
Example 1
Find the weight (grams and kg) of the round bar in Figure 57, rounded to the nearest
tenth.
Figure 57 - Solid round bar.
To solve, follow these steps.
1. Find the volume.
V  r 2 h
V   152  100 
V  70 686.83471 cm3
NOTE
For accuracy, do not round any number until you complete your last
calculation. Leave 70 686.83471 in the calculator display as you
proceed to calculate the weight.
2. Multiply volume by 7.849 g/cm3 to find the weight.
Weight  70 686.83471 cm 3  7.849 g / cm 3
The weight for the solid round bar in Figure 57 in grams is 554 813.1 g (rounded) and in
kilograms is 554.8 kg (rounded).
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NOTES
Example 2
Find the weight (lb) of the rectangular bar in Figure 58, rounded to the nearest tenth.
Figure 58 - Solid rectangular bar.
To solve, follow these steps.
1. Find the volume in cubic inches.
V  LWH
V  72"  3"  2"
V  432 in 3
2. Multiply the volume by 0.2835 lb/in3 to find the weight.
Weight  432 in 3  0.2835 lb / in 3
The weight for the solid rectangular bar in Figure 58 is 122.5 lb.
NOTE
You can calculate the weight of any solid if you know the volume and
if you know the weight of one unit of volume of the material.
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NOTES
Example 3
Calculate the weight (lb) of the solid steel in Figure 59. For this example, use π = 3.14.
The weight of one cubic inch of steel is 0.2835 lb/in3.
Figure 59 - Calculating the weight of a solid.
To solve for Example 3, do the following.
Weight  Area of Base  Height  0.2835 lb / in 3
Weight  (1 / 2 r 2  LW 1 / 2 bh)   H    0.2835
Weight  [(1 / 2 )  3.14   22    6  4   (1 / 2  5  4)]  8  0.2835
Weight   6.28  24  10  8  0.2835
Weight  40.28  8  0.2835
The weight is 91.36 lb.
NOTE
Ensure that you use the correct measurements in your formulas. In this
example, h = 4" is the height when you calculate the area of the
triangle; however, H = 8" is the height of the solid when you calculate
the volume. You may want to use altitude instead of height when
dealing with triangles to avoid confusion.
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Objective Three Exercise
1. Describe the procedure for calculating the weight of a solid.
2. Find the weight (lb) of a 20' long piece of 118" diameter steel shaft, rounded to two
decimal places. (Use the  button).
3. Find the weight (grams and kilograms) of the piece of steel shaft in Figure 60
and round to the nearest tenth. (Use  button)
Figure 60
4. You must field erect an enclosed oil storage tank 60'in diameter and 25' high at an oil
company battery site. The weight of the tank’s 38" sheet steel is 15.3 lbs/ft2. Find the
total weight of the tank (rounded to two decimal places) if you use an extra 756 lb for
bracing material. (Use the  button).
5. Find the weight (kg) of a steel plate if the plate measures 3 m long, 30 cm wide and
30 mm thick. Round your answer to two decimal places.
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6. What is the weight (lb) of the solid bar of aluminum in Figure 61, rounded to two
decimal places? Aluminum weighs 0.093 lb/in3.
NOTES
Figure 61
7. Calculate the weight (lb) of the cast iron overhead crane counterweight in Figure 62,
rounded to two decimal places. Cast iron weighs 475 lb/ft3.
Figure 62
8. Determine the weight (lb) of the iron casting in Figure 63, rounded to two decimal
places. Cast iron weighs 475 lb/ft3.
Figure 63
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NOTES
9. Find the total weight (kg) of the steel column support gussets in Figure 64, if you
have to fabricate 52 gussets. Round your answer to three decimal places.
Figure 64
10. Find the weight (lb) of the steel roller shaft support in Figure 65. Round your answer
to three decimal places. (Use  button).
Figure 65
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NOTES
Objective Four
When you have completed this objective, you will be able to:
Calculate the capacity of a container in gallons and litres.
Capacity of an Object
To find the capacity of an object, you must first find the volume and then convert from
cubic measure into gallons (imperial measurement) or litres (metric measurement).
Some of the basic relationships that you must know are as follows.
 1 imperial gallon = 277 in3 (approximately)
 1 ft3 = 6.239 imperial gallons (approximately)
 1 ft3 = 1728 in3
 1 yd3 = 27 ft3
 1 ml = 1 cm3
 1 litre (L) = 1000 millilitres (mL)
 1 L = 1000 cubic centimetres
 1 cubic metre = 1000 litres
Example 1
What is the capacity (gallons) of the tank in Figure 66? The volume of the tank is 7.5 ft3
or 12 960 in3.
Figure 66 - Finding the capacity of a rectangular tank.
To solve for Example 1, you can either multiply 7.5 ft3 by 6.239 galft3 to equal 46.8
gallons (rounded) or divide 12 960 in3 by 277 in3gal to equal 46.8 gallons (rounded).
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NOTES
Example 2
What is the capacity (gallons) of the tank in Figure 67? The volume of the tank is
679.93 ft3.
Figure 67 - Finding the capacity of a semicircular tank.
To solve for Example 2, you multiply 679.93 ft3 by 6.239 gal/ft3 to equal 4242.08 gallons
(rounded).
Example 3
What is the capacity (litres) of the tank in Figure 68? The volume of the tank is
392.6990817 cubic metres (392 699 081.7 cubic centimetres). Round your answer to the
nearest tenth.
Figure 68 - Finding the capacity of a circular holding tank.
To solve for Example 3, you can multiply 392.6990817 m3 by 1000 L/m3 to equal
392 699.1 L or divide 392699081.7 cm3 by 1000 cm3/L to equal 392 699.1 L.
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NOTES
Objective Four Exercise
1. Describe the procedure for finding the capacity of an object.
2. Find the number of gallons that the hydraulic oil reservoir tank in Figure 69 can hold.
Round your answer to two decimal places.
Figure 69
3. What is the capacity (litres) of a rectangular tank that is 3.25 m long, 92.75 cm wide
and 190 mm high? Round to one decimal place.
4. A saddle tank for a delivery truck has the same dimensions as Figure 70. Determine
the number of gallons that the tank can hold and round to two decimal places.
Figure 70
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5. What is the capacity (gallons) and rounded to one decimal place, of a cylindrical tank
that is 5' in diameter and 5' high? (Use  button).
6. What is the capacity (gallons) of the cylindrical steel tank in Figure 71? Round your
answer to two decimal places. (Use  button)
Figure 71
7. How many gallons are in the semicircular ended tank in Figure 72, if you fill it to a
height of 212' ? Round your answer to two decimal places. (Use  button)
Figure 72
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8. A circular stainless steel tank has a diameter of 3 m and you fill it with gasoline to a
depth of 450 cm. If you lower the depth by 2 m, how many litres of gasoline did you
draw off, rounded to two decimal places? (Use  button)
NOTES
9. The fuel tank in Figure 73 fits onto a piece of heavy equipment. Determine its
capacity (gallons) and round your answer to two decimal places. (Use  button)
Figure 73
10. Calculate the capacity (gallons) of the coolant reservoir tank and manifold system in
Figure 74. Round your answer to two decimal places. (Use  button)
Figure 74
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NOTES
Self-Test
1. Which statement is an equation?
a) 2r2 + dh
b) 2LW + 2LH + 2WH
c) h[(B+b)/2]
d) C = d
2. What is the definition of a formula?
a) a law or rule that you express as an equation
b) a numerical value in a term
c) an unknown quantity in a term
d) a replacement for a variable
3. An unchanging numerical value that forms part of an equation or formula is a(n):
a) variable.
b) constant.
c) solution.
d) unknown.
4. A symbol that represents the unknown quantity in an equation or formula is a:
a) solution.
b) constant.
c) variable.
d) law or rule.
5. What is the solution for the formula n = 9 – 2 x 4 + 12  8?
a) n = 2.5
b) n = 3.5
c) n = 5
d) n = 29.5
6. What is the perimeter of a square if the length of each side is 812"?
a) 72.25"
b) 34"
c) 32"
d) 17"
7. What is the perimeter of a rectangle if the width is 3" and the length is four times the
width?
a) 14"
b) 15"
c) 24"
d) 30"
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8. What is the perimeter of a triangle if the lengths of the sides are 1958", 22916"
and 27132"?
a) 681532"
b) 5' 9732"
c) 6' 9732"
d) 68932"
NOTES
9. Calculate the circumference of a 25' diameter tank, rounded to two decimal places.
a) 39.27'
b) 78.54'
c) 157.08'
d) 490.87'
10. Calculate the area (square inches) of a tank lid that is 2' in diameter. Round to the
nearest tenth.
a) 37.7 in2
b) 75.4 in2
c) 452.4 in2
d) 1809.6 in2
11. What is the area of a square that measures 30" on each side?
a) 6.25 ft2
b) 62.5 ft2
c) 625 ft2
d) 900 ft2
12. What is the area of a rectangle if the length is 28" and the width is 14 of the length?
a) 49 in2
b) 70 in2
c) 112 in2
d) 196 in2
13. You cut a triangular piece of steel off the corner of a large square sheet. What is the
area of the triangular shaped piece if you measured 2' along one side and 39" along
the other side?
a) 3.9 ft2
b) 6.3 ft2
c) 468 in2
d) 936 in2
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NOTES
14. What is the area (mm2) of the trapezoid in Figure 75?
a) 54 mm2
b) 75 mm2
c) 87.5 mm2
d) 112.5 mm2
Figure 75
2
15. What is the area (cm ) of the parallelogram in Figure 76?
a) 377 cm2
b) 435 cm2
c) 812 cm2
d) 841 cm2
Figure 76
16. What is the lateral surface area of a piece of pipe with an outside diameter of 10" and
a length of 10', rounded to the nearest tenth?
a) 314.2 in2
b) 785.4 in2
c) 1885.0 in2
d) 3769.9 in2
17. What is the total surface area of a closed cylindrical tank that measures 3' in diameter
and is 36" high? Round to the nearest tenth.
a) 42.4 ft2
b) 35.3 ft2
c) 28.3 ft2
d) 14.1 ft2
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18. What is the volume of a flat bar that measures 34" by 512" and is 20' long?
a) 82.5 in3
b) 180 in3
c) 990 in3
d) 1320 in3
NOTES
19. Calculate the volume of a piece of 118" diameter cold rolled steel that is 12' long
Round to the nearest tenth.
a) 572.6 in3
b) 143.1 in3
c) 508.9 in3
d) 11.9 in3
20. What is the capacity in gallons (1 gal = 277 cubic inches) of a rectangular tank that
measures 20" by 32" and is 114' high? Round to the nearest tenth.
a) 1538.7 gal
b) 800 gal
c) 128.2 gal
d) 34.7 gal
21. Calculate the capacity in gallons (1 cubic foot = 6.239 gal) of a tank with a diameter
of 8' and a height of 16'. Round to the nearest tenth.
a) 5017.7 gal
b) 3217.0 gal
c) 2508.9 gal
d) 804.2 gal
22. Steel weighs 0.2835 lb/in3. What is the weight of a 12" thick sheet of steel that is 4'
wide and 32' long? (Round to the nearest tenth).
a) 217.7 lb
b) 2612.7 lb
c) 4608 lb
d) 9216 lb
23. Steel weighs 0.2835 lb/in3. What is the weight of a 1" round steel bar that is 20' long?
(Round to the nearest tenth).
a) 53.4 lb
b) 62.8 lb
c) 188.5 lb
d) 213.8 lb
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NOTES
Self-Test Answers
1. d) C = d
2. a) a law or rule that you express as an equation
3. b) constant.
4. c) variable.
5. a) n = 2.5
6. b) 34"
7. d) 30"
8. b) 5' 9732"
9. b) 78.54'
10. c) 452.4 in2
11. a) 6.25 ft2
12. d) 196 in2
13. c) 468 in2
14. b) 75 mm2
15. a) 377 cm2
16. d) 3769.9 in2
17. a) 42.4 ft2
18. c) 990 in2
19. b) 143.1 in3
20. d) 34.7 gal
21. a) 5017.7 gal
22. b) 2612.7 lb
23. a) 53.4 lb
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NOTES
Objective One Exercise Answers
1. An equation is a statement or mathematical expression that uses an equal sign to
indicate that two related quantities have the same value. All equations must be true.
2. A formula is a law or rule that you express as an equation. A formula expresses some
fundamental fact or truth.
3. A constant is a numerical value that forms part of an equation or formula and cannot
change. An example of a constant is .
4. P = 576.84
5. L = 8
6. A = 38.465
7. A = 45
8. h = 14
9. 4.5
10. h = 3.2
Objective Two Exercise Answers
Perimeter
1. Perimeter is the distance around any geometric figure, such as a square, rectangle,
triangle, trapezoid, parallelogram, circle or semicircle. When the figure is a circle, the
perimeter is the circumference.
2. rectangle
3. C = d or C = 2r
4. 152 mm
5. 10' 9".
6. 12.6 m
7. 819.2 cm
8. 128' 211/16"
9. 23.1 cm
10. a)
b)
c)
d)
e)
4363/64"
7525/64"
2027/64"
40' 103/32"
831/4"
11. 36' 31/8"
12. 1032'
13. 1443' 515/16"
14. 11763/64"
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NOTES
Area
1. Area is the amount of material it takes to cover a surface.
2. A = 12bh
3. square
4. 4 surfaces
5. TSA = dh + 2(r2)
6. a)
b)
c)
d)
4058 in2
8218 in2
162916 in2
614 in2
7. 51 984 ft2
8. $450.00
9. 1 ft2
10. a)
b)
c)
d)
32 in2
45 in2
72 in2
66.5in2
11. 3318.31 cm2
12. 1.575 in2
13. 1.91 6 ft2
14. 10.68 in2
15. 78 000 ft2
16. 72.05 ft2
17. 93.08 ft2
18. 608 in2
Volume
1. Volume is the amount of space that an object occupies.
2. V = LWH
3. volume of a cylinder
4. a)
b)
c)
d)
e)
f)
g)
3456 in3
2880 in3
6307.2 in3
2.01 ft3
0.01 ft3
1.7 ft3; 0.06 yd3
1.46 yd3
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NOTES
5. a) 226.88 in3
b) 294 in3
c) 324 in3
6. a) 1199.46 in3
b) 5.36 ft3
7. 628.32 ft3
8. 31.01 ft3
9. 115.56 ft3
10. 58.18 ft3, 100 531.34 in3
Objective Three Exercise Answers
1. In order to calculate the mass or weight of an object, you must first find its volume.
You then multiply the volume by the known weight of one unit of volume of the
material in question.
2. 67.63 lb
3. 389 010.3 grams or389.0 kg
4. 159 375.01 lb
5. 211.92 kg
6. 6.28 lb
7. 396.93 lb
8. 164.93 lb
9. 179.585 kg
10. 48.476 lb
Objective Four Exercise Answers
1. To find the capacity of an object, you must first find the volume and then convert
from cubic measure into gallons (imperial measurement) or litres (metric
measurement).
2. 131.22 gallons
3. 572.7 litres
4. 77.98 gallons
5. 612.5 gallons
6. 640.27 gallons
7. 142.57 gallons
8. 14 137.17 litres
9. 212.39 gallons
10. 61.35 gallons
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NOTES
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120104dp7.0.docx
© 2014, Her Majesty the Queen in right of the Province of Alberta
Module Number 120104d
Version 7.0