CHAPTER TWO
ESTIMATING MATH
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CHAPTER TWO
A. ESTIMATING MATH
a. An estimator needs to determine lengths, areas, and volumes in order to
determine the quantities of materials needed to complete the project.
i. In many cases the estimator needs to convert the quantities from one set
of units to another set of units
B. LENGTHS
a. In most cases lengths are determined by calculating the distances based on
dimensions shown on the plans or by measuring the distance using a scale.
b. Example #1: From the above drawing determine the distance (A) between the
corner of the building and the center of the door.
i. First, convert the fractions of an inch to a common denominator
1. Distance = 5’5½” + 3’11¾” = 5’5 2 2/4” + 3’11¾” = 8’16 5/4”
ii. Reduce the fraction of inches to less than 1 inch
1. Distance = 8’ 16 5/4” = 8’ 17¼”
iii. Reduce the inches portion of the dimension to less than 12
1. Distance = 8’ 17¼” = 9’ 5¼”
2. The distance is 9 feet 5 ¼ inches
c. Example #2: From the above drawing determine the distance (B) between the
center of the closet wall and the center of the window
i. Make sure the fraction of an inch have the same denominator
1. Distance = 5’ 5½” – 2’ 6¾” = 5’ 5 2/4” – 2’ 6¾”
ii. Make sure that the fraction of an inch in the larger dimension is greater
than the fraction of an inch in the smaller dimension
1. Distance = 5’ 5 2/4” – 2’ 6¾” = 4’ 16 6/4” – 2’ 6¾”
iii. Subtract the feet from the feet, the inches from the inches, and the
fractions of an inch from the fractions of an inch
1. Distance = 4’ 16 6/4” – 2’ 6¾” = 2’ 10¾”
iv. The distance is 2 feet 10¾ inches
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C. SCALING
a. Plans are prepared at a reduced scale.
b. The length of components on the plans may be determine by scaling the length
using a scale
i. When scaling drawings it is important to make sure that the correct scale is
used and that the drawings are to scale
ARCHITECT’S SCALE
ENGINEER’S SCALE
c. Example #3: Determine the length of the wall shown using the Architect’s scale
above. The figure is drawn at a scale of 1/8” = 1’
i. The scale is lined up with a foot mark at the right side of the drawing, and
the left side of the drawing is lined up within the inch marks at the left side
of the scale.
ii. The length of the wall is 28 feet 3 inches
d. Example #4: Determine the length of the building shown using the Engineer’s
scale. The figure is drawn at a scale of 1” = 20’
i. The scale is lined up with the zero; therefore, the feet can be read off of
the right side of the scale
ii. The length of the wall is between 68 and 69 feet
D. PYTHAGOREAN THEOREM
a. The Pythagorean Theorem is useful when two dimensions at RIGHT angles are
used to describe a construction component
i. This is commonly found in roof-framing, where the run of the roof joists is
shown and there is a rise in the joists that is calculated from the slope of
the roof
ii. The Pythagorean Theorem states that the square of the hypotenuse equals
the sun of the squares of the other two sides of the triangle
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1. C2 = A2 + B2
8.33’
900
20’ 0”
40’ 0”
2’ OVERHANG
2. Example #5: Determine the length of the roof rafters as shown in the
drawing above.
a. C2 = A2 + B2
b. C2 = (8.33)2 + (20)2
c. C2 = (69.39) + (400)
d. C2 = 469.39
e. C = 469.39
f. C = 21.67 or 22’ 0” + the overhang
g. 22’ + 2’ = 24’ 0” material for rafters
E. AREAS
a. Estimators are often required to calculate areas in order to complete the quantity
takeoff.
i. Geometric Method: The geometric method breaks complex shapes down
into six simpler geometric shapes: the circle, the square, the rectangle, the
triangle, the trapezoid and the parallelogram
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1. Area of a Circle = π R2 (see drawing above)
a. or from the diameter using: A = π D2/4
b. where: π = 3.14139
R = radius of the circle
D = diameter of the circle
c. Example #6: Find the area of a circle whose radius is 12”
i. A = π R2 = 3.14(12”)2 = 452 in2
2. Area of a Rectangle = LW (see drawing above)
L = length
W = width
a. Example #7: Find the area of a rectangle whose length is 24”
and width is 12”
i. A = LW = (24”)(12”) = 288 in2
3. Area of a Square = L2 (see drawing above)
a. Example #8: Find the area of a square whose length is 12”
i. A = L2 = (12”)2 = 144 in2
4. Area of a Triangle = BH/2 (see drawing above)
B = base
H = height
a. The height of a triangle must be measured perpendicular to
the base (see drawing above)
b. Example #9: Find the area of a triangle whose base is 12” and
height is 24”
i. A = BH/2 = (12”)(24”)/2 = 144 in2
5. Area of a Trapezoid = (B + C)/2 (D) (see drawing above)
B and C = lengths of the parallel sides
D = perpendicular distance between the parallel sides
a. A trapezoid is a four sided figure with two parallel sides
b. Example #10: Find the area of a trapezoid whose parallel sides
are 12 and 16 inches with a distance between the parallel sides
of 24”
i. A = (B + C)/2 (D) = (12” + 16”)/2 (24”) = 336 in2
6. Area of a Parallelogram = BD (see drawing above)
B = length of each of a given pair of parallel sides
D = perpendicular distance between the two sides that have
length B
a. A parallelogram is a four sided plane with opposite sides
parallel.
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b. Because both pairs of opposite sides are parallel, the opposite
sides have the same length
c. Example #11: Find the area of a parallelogram whose parallel
sides are 12” long with a distance between the parallel sides of
24”
i. A = BD = (12”)(24”) = 288 in2
7. Area of a complex shape
a. The area of a complex shape is determined by breaking the
shape down into its geometric shapes, determining the area of
the basic geometric shapes, and adding their areas
Figure A
Figure B
b. Example #12: Determine the area of the shape in figure A
i. Divide the complex shape into two rectangles and a
square as shown in figure B
ii. Area of the left rectangle = LW = (30’)(20’) = 600 ft2
iii. Area of the right rectangle = LW = (20’)(10’) = 200 ft2
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iv. Area of the square = L2 = (30’)2 = 900 ft2
v. The area of the complex shape is determined by
summing the areas of the smaller shapes
1. A = 600 ft2 + 200 ft2 + 900 ft2 = 1,700 ft2
c. Sometimes it is advantageous to make the shape a part of a
larger shape and subtract the missing pieces (see figure
A)
i. Area of the large rectangle = LW = (60’)(30’) = 1,800 ft2
ii. Area of the square = L2 = (10’)2 = 100 ft2
iii. Area of complex shape = 1,800 ft2 – 100 ft2 = 1,700 ft2
d. Another common example of creating a large shape and
subcontracting the missing pieces arises in the case of an
outside radius that creates a complex shape.
The area of this shape is calculated by taking the area of the
square of length R and subtracting the area of a quarter of a
circle
Area = R2(1- π/4)
Where:
Π = 3.14159
R = radius of the circle
e. Example #13: Determine the area of the complex shape below
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f. Divide the shape into two rectangles and a square less a
quarter circle as shown in the drawing below
g. The area of the left rectangle = LW = (70’)(30’) = 2,100 ft2
The area of the right rectangle = LW = (40’)(30’) = 1,200 ft2
The area of the square less a quarter circle
Area = R2(1- π/4) = (20’)2(1- π/4) = 86 ft2
The area of the complex shape is determined by summing the
area of the smaller shapes
Area = 2,100 ft2 + 1,200 ft2 + 86 ft2 = 3,386 ft2
THE AREA OF THE COMPLEX SHAPE IS 3,386 SQUARE
FEET
F. VOLUMES
a. Most volumes are created by extending a two-dimensional shape along a line
perpendicular to the plane it occupies.
i. Volumetric shapes
ii. Cylinder
1. The volume of a cylinder is calculated by multiplying the area of a
circle by the height of the cylinder
2. Volume = πR2H
or
2
Volume = πD H/4
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Where: π = 3.14159
R = Radius of circle
D = Diameter of circle
H = Height measured perpendicular to the area of the
circle
3. Example #14: Find the volume of a cylinder whose radius is 1’ and
height is 10’
a. Volume = πR2H = 3.14(1’)(10’) = 31.4 ft3
iii. Column
1. The volume of a column is calculated by multiplying the area of a
rectangle by the height of the column, where the height of the
column is measured perpendicular to the area of the rectangle
2. Volume = LWH
Where: L = length
W = width
H = height measured perpendicular to the area of
the rectangle
3. Example #15: Find the volume of the a column whose height is 10’,
length is 2’ and width is 1’
a. Volume = LWH = (2’)(1’)(10’) = 20 ft3
iv. Prism
1. The volume of a prism is calculated by multiplying the area of a
triangle by the length of the prism where the length of a prism is
measured perpendicular to the area of the triangle
2. Volume = BHL/2
Where: H = height measured perpendicular to the base of the
triangle
B = base
L = length measured perpendicular to the area of the
triangle
3. Example #16: Find the volume of a prism whose height is 5’. Length is
20’ and base is 3’
a. Volume = BHL/2 = (3’)(5’)(20’)/2 = 150 ft3
v. Pyramids and Cones
1. For the volumetric shapes created by extending the shape to a point,
the volume is one-third of the area of the base of the object times
the height of the object, where the height is measured perpendicular
to the area
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2. For a pyramid:
Volume = LWH/3
Where: L = length
W = width
H = height measured perpendicular to the area of
the rectangle
3. Example #17: Find the volume of a pyramid whose height is 12’.
Length is 2’ and width is 1’
a. Volume = LWH/3 = (2’)(1’)(12’)/3 = 8 ft3
4. For a cone
Volume = πR2H/3
From the diameter: Volume = πD2H/12
Where: π = 3.14159
R = radius of the circle
D = diameter of the circle
H = height measured perpendicular to the area of the
circle
5. Example #18: Find the volume of a cone whose radius is 1’ and height
is 10’
a. Volume = πR2/3 = 3.14(1’)(10’)/3 = 10.5 ft3
vi. Volume of a shape with a constant area
1. The volume of any shape with a constant area over a given length is
calculated by multiplying the cross-sectional area of the shape by the
length of the object over which the cross-sectional is representative
of the shape of the object.
2. Volume = AL
Where: A = Cross-sectional area of the shape measured
perpendicular to the length
L = length
3. The length should be measured along the centerline of the shape, and
the cross-sectional area should be measured perpendicular to the
centerline of the shape. When the centerline of the shape forms a
curved line, the volume is an approximation
4. Example #19: Find the volume of a curb and gutter whose crosssectional area is 1.7 ft2 and whose length is 200’
a. Volume = AL = (1.7 ft2)(200’) = 340 ft3
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vii. Complex volume
1. The volume of complex three-dimensional bodies can be calculated
by breaking the complex body into simpler shapes for which the
volume can be calculated
2. For example, the complex volume created by the building excavation
above can be divided into four pyramids (one at each corner), four
prism (one along each side), and one column (located in the center).
G. The volume of this complex shape is calculated by determining the volumes for the
pyramids, prisms, and column and adding them together
H. BOARD FEET
a. Lumber especially random lengths and finish lumber, are often purchased by the
board feet
b. A board feet is the amount of wood contained in a 12 inch by 12 inch by 1 inch
nominal board.
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c. The nominal dimensions are the dimensions used to describe the board, for
example, 2 X 4 or 2 X 12.
d. Board Feet = (Thickness)(Width)(Lineal Feet)
12in/ft
e. Example #21: Find the number of board feet in a 10 foot long 2 X 4
i. Board Feet = (Thickness)(Width)(Lineal Feet)
12in/ft
= (2 in)(4 in)(10 ft) = 6.67 in-ft2 = 6.67 bft
12in/ft
THERE ARE 6.67 BOARD FEET IN A 10 FOOT-LONG 2 X 4
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The Stretch-of-Length Concept (SOL)
 This may be defined as the length of the centerline of the strips that form the perimeter of a
figure.
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STEP 1: In problem above, the outside dimensions are 43 ft. 0 in. by 46 ft. 6 in., except for
the fact that there are two setbacks, one measuring 5 ft. 3 in. and one measuring 10 ft. 2 in.
In figuring the outside perimeter, P0, add 43 ft. 0 in. to 46 ft. 6 in., and multiply the result by
2 to get 179 ft. But the two setbacks of 5 ft. 3 in. and 10 ft. 2 in. must be taken into account.
Add them together and multiply by 2, for a result of 30 ft. 10 in. When this sum is added to
the 179 ft., the total of 209 ft. 10 in. represents the adjusted perimeter length. Therefore, P0
= 209.82 ft.
STEP 2: Stretch-out-length (SOL = P0 – 4t)
SOL = 209.83 – (4X1) = 205.83 ft.
STEP 3: The area of the top of the strip
Area = SOL X 1
= 205.83 X 1 = 205.83 sq.ft.
STEP 4: Volume of concrete required for the wall

Volume = area of the top of the wall X height of the wall
= 205.83 X 8 = 1646.64 CF = 61 CY.
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