Basic Math Area Formulas
Math for Water Technology
MTH 082
Lecture 2
Chapter 9
Temperature Conversions
Celsius to Fahrenheit
1. Begin by multiplying the Celsius temperature by 9.
2. Divide the answer by 5.
3. Now add 32.
oF=
oC)
(9 *
5
+ 32
oF=
Convert 17oC to Fahrenheit
(9 *17)+32=62.6oF=
63oF
5
Fahrenheit to Celsius
1. Begin by subtracting 32 from the Fahrenheit #.
2. Divide the answer by 9.
4. Then multiply that answer by 5.
oC=
5 * (oF – 32)
9
Convert 451oF to degrees Celsius
oC=
5* (oF -32)=232.7oC= 233oC
9
Objectives
•
•
Become proficient with the concept of
area as it pertains to common geometric
shapes.
Solve waterworks math problems
equivalent to those on State of Oregon
Level I and Washington OIT Certification
Exams
RULES FOR AREA PROBLEMS
1. IDENTIFY THE OBJECT
2. LABEL /DRAW THE OBJECT
3. LOCATE THE FORMULA
4. ISOLATE THE PARAMETERS NECESSARY
5. CARRY OUT CONVERSIONS
6. USE YOUR UNITS TO GUIDE YOU
7. SOLVE THE PROBLEM
What is area?
• The amount of space that a figure
encloses
• It is two-dimensional
• It is always answered in square units
Water/Wastewater Shapes
Source Intake Process
• Water is collected from the bottom of the Tualatin River through
intake screens that bring in water at a very low velocity.
Rapid Mix
Sedimentation Basins
Settling Tanks
Filter beds
Baffles
JWC built in1976
Area of a Circle
• Distance around circle is
circumference
• C=π (d)
• Distance through circle
is diameter
• radius is 1/2 as large as
diameter
• Area= π (r2) or 1/4 π (d2)
or 0.785 (d2)
• Area of ½ circle= π (r2)
2
m
f
u
c
r
D=diameter
r
e
r=radius
n
i
c
c
e
A= π (r2)
A= 1/4 π (d2)
A circular clarifier has a diameter
of 40 ft. What is the surface area
(ft2) of the clarifier??
DRAW:
• Given:
• Formula:
• Solve:
1.
2.
3.
4.
31.4 ft
15.7 ft3
1256 ft2
628 ft2
D=40 ft
40 ft
A=1/4 π (d2)
A= 0.785 (40 ft)2
A=0.785 (1600 ft2)
A= 1256 ft2
The radius of a water tank is 35
ft. What is the circumference (ft)
of the tank?
35 ft
DRAW:
.5
46
38
22
25%
ft2
0
ft2
ft
ft
25% 25%
55
25%
C= π(70 ft)
C=3.14 (70ft)
A= 219.8 ft
6
15386 ft
220 ft
3846.5 ft2
55 ft2
C= π (d)
38
1.
2.
3.
4.
R= 35 ft; D=70 ft
15
• Given:
• Formula:
• Solve:
Area of a triangle
H=height
• If two triangles look
different but have the
same base and height,
they will have the same
areas.
• B is the base
• H is the height
• Area= ½ x base x height
• A= b x h
2
B=base
A= ½ x b x h
2
(cm )
What is the area
figure?
of this
DRAW:
Base= 13 cm, Side= 4 cm
• Given:
• Formula: A= (B XH)/2
4 cm
• Solve:
A=(13 cm X 4 cm)/2)
A= 26 cm2
1.
2.
3.
4.
102 cm
102 cm3
26 cm2
52 cm2
13 cm
A triangular portion of the water treatment grounds is
not being used. If the area is 17,000 sq ft and the base
of the area is 200 ft, what is the height of the area (ft)?
ft
00
0
3,
40
0,
0
17
ft2
ft2
25% 25% 25% 25%
0
2. 170 ft2
3. 170 ft
4. 3,400,000 ft2
200 ft
17
• Formula:
• Solve:
Solve for unknown H:
2A/B=H
1. 85 ft 2(17,000ft2)/200 ft=170 ft
ft
A= (B XH)/2
A=17,000 ft2
85
• Given:
DRAW:
2, Side= ?
Base= 200 ft, Area= 17,000
ft
h
Area of a Square
• A square is a 4sides figure that has
4 right angles and 4
congruent sides.
• S is the side
• Area= side x side
• A= s x s
• A=s2
S=side
S=side
2
A=S
An Acti-flo square settling basin is
8ft on a side, what is its area?
DRAW:
s= 3ft
8 ft
8 ft
ft2
25%
64
ft2
25%
16
ft
8 ft
6 ft3
16 ft2
64 ft2
8
1.
2.
3.
4.
25%
ft3
25%
A= (8ft)
A= (8 ft)(8ft)
A= 64 ft2
6
• Given:
• Formula: A= S X S or S2
• Solve:
2
Area of a rectangle
• A rectangle is a 4-sides figure that
has 4 right angles.
• A rectangle is also a parallelogram
since it has 2 pairs of opposite
sides that are parallel.
• b is the base (w is the width)
• h is the height (l is the height)
• Area= base x height (length x
width)
• A= b x h or (l x w)
H=height or l=length
B=base or w=width
A= b x h or (l x w)
What is the area (ft2) of this
rectangular up-flow DRAW:
clarifier figure?
• Given: Length = 8 ft, Width = 3ft
• Formula:
8 ft.
A= (L X W)
• Solve:
3 ft
A= (8 ft)(3ft)
A= 24 ft2
1.
2.
3.
4.
24 ft2
3456 in2
2.23 m2
All of the above
A treatment plant has three drying beds,
each of which is 50 ft long and 15 ft wide.
How many ft2 does one drying bed occupy?
DRAW:
L= 50 ft, W=15 ft
• Given:
• Formula:
A= (L X W)
• Solve:
A= (50 ft)(15ft)
A= 750 ft2
1. 375 ft2
2. 65 ft2
3. 750 ft2
15 ft
50 ft
The surface area of a tank is 2,000 sq ft. If
the width of the tank is 25 ft, what is the
length of the tank (ft)?
DRAW:
• Given:
• Formula:
A= (L X W)
• Solve:
A/W=L
25 ft
A=2000ft2, L= ? ft, W=25 ft
A= 2000 ft2
L= ? ft
2000 ft2/(25 ft)= X ft
L= 80 ft
1. 50000 ft
2. 0.0125 ft
3. 80 ft
Area of a parallelogram
• A parallelogram is
any 4-sided figure H=height or l=length
that has 2 pairs of
opposite sides that
are parallel.
• b is the base
• h is the height
• Area= base x height
• A= b x h
A= b x h or (l x w)
A settling pond has a shape like a parallelogram.
If the base is 40 ft and the height is 30 ft what is
the area (ft2)?
DRAW:
• Given:
B= 40 ft, h=30 ft
30 ft
• Formula: A= (b X h)
• Solve:
A= (30 ft)(40 ft)
A= 1200 ft2
1.
2.
3.
4.
1200 ft2
1200 cm2
1200 m2
None of the above
40 ft
Area of a rhombus
• A rhombus is a quadrilateral
with four sides of the same
length.
• Every rhombus is a
parallelogram (a 4-sides figure
with opposite sides parallel)
• The diagonal of a rhombus is
the segment that joins 2
vertices that are nonadjacent
• The area of a rhombus is
• A= (1/2) d1 X d2
d1
d2
•d1 is the length of
the first diagonal
•d2 is the length of
the second diagonal.
A= 1/2 d1 x d2
Area of a Trapezoid
•
•
•
•
A trapezoid is a
quadrilateral (a 4-sided
figure) with exactly one
pair of parallel sides.
The parallel sides are
called the bases of the
trapezoid and the other two
sides are called the legs
An isosceles trapezoid is a
trapezoid with congruent
legs.
The median of a trapezoid
is the line segment that
connects the midpoints of
the 2 nonparallel sides of
the trapezoid (the legs).
b1=base 1
height
Leg 1
Leg 2
median
b2=base 2
Area of a Trapezoid
•
•
•
•
•
A trapezoid is 4-sided
figure with exactly one
pair of parallel sides.
The two parallel sides
are the bases; we will
call them b1(base one)
and b2(base two).
The other 2 sides are
called the legs. THEY
ARE NOT PARALLEL.
The area of a
trapezoid is
A= (1/2) h (b1 + b2)
b1=base 1
Leg 1
h=height
Leg 2
b2=base 2
•h=height of the trapezoid
• b1= 1st base
• b2= second base.
A= (1/2) h (b1 + b2)
A water treatment plant is being built on a trapezoidal parcel of land. An
aerial view from Google Earth revealed that the large boundary (property
line) or base is 5,000 ft and the small base (property line) is 2500 ft and
the height is 1000 ft. What is the area (ft2)?
DRAW:
b1=2500 ft
• Given:
b1=2,500 ft, b2=5,000 ft and h=1000 ft
• Formula:
Leg 1
A= (b1 +b2) * (h)
2
• Solve:
A= (2500 ft)+(5000ft) * 1000 ft
2
A= 7500 ft * 1000 ft
2
A=3750 ft * 1000 ft
A= 3,750,000 ft2
1. 3,750,000 ft2
2. 3.75 X 106 ft2
h=height 1000 ft
b2=5000 ft
Leg 2
Area of a Right Circular Cone
• In a right circular cone, the
axis is perpendicular to the
base.
• The length of any line
segment connecting the
vertex to the directrix is
called the slant height of
the cone.
•
Height: h
Radius of base: r
Slant height: s
Lateral surface area: S
Total surface area: T
Volume: V
s=slant height
H=height
S=lateral
surface
area
R=radius
Area of a Right Circular Cone
• B = π r2
s = sqrt[r2+h2]
S = πrs
T = π r(r+s)
V = π r2h/3
Diameter given
A= π(r2)+π(d)(s)
2
Radius given
A= π(r2)+π(r)(s)
H=height
s=slant height
S=lateral
surface area
R=radius
The cone portion of a upflow clarifier tank must be
painted. If the diameter of the cone is 50 ft and the
slant height is 20 ft, how many sq ft. (total area of the
DRAW:
cone) must be painted?
D= 50 ft, s= 20 ft, I know r= 25 ft!
• Given:
A= π(r2)+π(d)(s)
s= 20 ft
r= 25 ft
• Formula:
2
A= π(25 ft)2 + π(50ft)(20ft)
• Solve:
D= 50 ft
2
A= π 625 ft2 + π 1000 ft2
2
A= 1963 ft2 + 3140 ft2
2
A= 1963 ft2 + 1570 ft2
1. 51032 ft2
2. 3533 ft2
3. 1649 ft2
A= 3533 ft2
Frustum of a Right Circular Cone
• The part of a right circular cone
between the base and a plane
parallel to the base whose
distance from the base is less
than the height of the cone.
Height: h
Radius of bases: r, R
Slant height: s
Lateral surface area: S
Total surface area: T
Volume: V
s = sqrt([R-r]2+h2)
r=radius of base 2
h=height
s=slant height
S=lateral
surface
area
S = π(r+R)s
T = π(r[r+s]+R[R+s])
V = π(R2+rR+r2)h/3
R=radius of base 1
Review Problems!
A circular sedimentation tank
has an area of 1962 ft2. What is
the diameter of the tank??
DRAW:
• Given:
• Formula:
• Solve:
D=? Ft; AREA = 1962 ft2
AREA=1962 ft2
A=1/4 π (d2)
1962 ft2= 0.785 (X ft)2
1962 ft2= (X ft)2
0.785
1. 2499 ft
2. 1250 ft
3. 50 ft
D= ? Ft
2499 ft2= (X ft)2
√2499 ft2= X ft
50 ft = X
The top of a Monterey Bay aquarium tank has a surface
area of 500 ft2. If the width of the tank is 35 ft, what is
the length of the tank ?
• Solve: 500
ft= X ft
14 ft = Length
ft2/35
1. 535 ft
2. 14 ft
Area= 500 ft2
35 ft.
DRAW:
• Given: A= 500 ft2, Length = X ft, Width = 35ft
• Formula:
A= (L X W)
A/W=L
X ft.?
What did you learn?
•
•
•
•
What is area?
How are the units of area usually expressed?
How many dimensions is area?
What does the “S” mean in the formula for the area
of a square?
• What is the formula for the area of a square?
• What is the formula for the area of a rectangle?
• What does the “h” stand for in the formula for the
area of a triangle?
Review Area Formulas!
•
•
•
•
•
•
•
•
Circle= A= (π) r2
Square: A = s x s
Rectangle : A = l x w
Triangle: A = 1/2 b x h
Parallelogram: A= b x h
Rhombus: A= (1/2) d1 X d2
Trapezoid: A= (1/2) h (b1 + b2)
Cone = A= π(r2)+πdhs or A= π(r2)+πrs
2
Today’s objective: to become proficient with
the concept of area as it pertains to water
and wastewater operation has been met
1.
2.
3.
4.
Strongly Agree
Agree
Disagree
Strongly Disagree