1. Basic Math Area Formulas
Math for Water Technology
MTH 082
Lecture 2
Chapter 9

2. Temperature Conversions
Celsius to Fahrenheit
1. Begin by multiplying the Celsius temperature by 9.
2. Divide the answer by 5.
3. Now add 32.
Convert 17oC to Fahrenheit
oF= (9 * oC) 5
oF= (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.
Convert 451oF to degrees Celsius
oC= 5 * (oF – 32) 9
oC= 5* (oF -32)=232.7oC= 233oC 9

3. 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

4. RULES FOR AREA PROBLEMS
IDENTIFY THE OBJECT
LABEL /DRAW THE OBJECT
LOCATE THE FORMULA
ISOLATE THE PARAMETERS NECESSARY
CARRY OUT CONVERSIONS
USE YOUR UNITS TO GUIDE YOU
SOLVE THE PROBLEM

5. What is area? The amount of space that a figure encloses
It is two-dimensional
It is always answered in square units

6. Water/Wastewater Shapes

7. Source Intake Process JWC built in1976
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

8. Area of a Circle A= π (r2) A= 1/4 π (d2)
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 (d2)
Area of ½ circle= π (r2) 2 m f u c r
D=diameter e r
r=radius n i c c e
A= π (r2)
A= 1/4 π (d2)

9. A circular clarifier has a diameter of 40 ft
A circular clarifier has a diameter of 40 ft. What is the surface area (ft2) of the clarifier??
DRAW:
Given:
Formula:
Solve:
40 ft
D=40 ft
A=1/4 π (d2)
A= (40 ft)2
A=0.785 (1600 ft2)
A= 1256 ft2
31.4 ft
15.7 ft3
1256 ft2
628 ft2

10. The radius of a water tank is 35 ft
The radius of a water tank is 35 ft. What is the circumference (ft) of the tank?
DRAW:
Given:
Formula:
Solve:
35 ft
R= 35 ft; D=70 ft
C= π (d)
C= π(70 ft)
C=3.14 (70ft)
A= ft
15386 ft
220 ft
ft2
55 ft2

11. Area of a triangle A= ½ x b x h
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
H=height
B=base
A= ½ x b x h

12. What is the area (cm2) of this figure?
DRAW:
Given:
Formula:
Solve:
Base= 13 cm, Side= 4 cm
A= (B XH)/2
A=(13 cm X 4 cm)/2)
A= 26 cm2
13 cm
4 cm
102 cm
102 cm3
26 cm2
52 cm2

13. A triangular portion of the water treatment grounds is not being used
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)?
DRAW:
Given:
Formula:
Solve: h
A=17,000 ft2
200 ft
Base= 200 ft, Area= 17,000 ft2, Side= ?
A= (B XH)/2
Solve for unknown H:
2A/B=H
2(17,000ft2)/200 ft=170 ft
85 ft
170 ft2
170 ft
3,400,000 ft2

14. Area of a Square
A square is a 4-sides 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
A=S2

15. An Acti-flo square settling basin is 8ft on a side, what is its area?
DRAW:
Given:
Formula:
Solve:
s= 3ft
A= S X S or S2
A= (8ft)2
A= (8 ft)(8ft)
A= 64 ft2
8 ft
8 ft
6 ft3
16 ft2
64 ft2

16. Area of a rectangle A= b x h or (l x w)
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)

17. What is the area (ft2) of this rectangular up-flow clarifier figure?
DRAW:
Given:
Formula:
Solve:
Length = 8 ft, Width = 3ft
A= (L X W)
A= (8 ft)(3ft)
A= 24 ft2
8 ft.
3 ft
3 ft.
24 ft2
3456 in2
2.23 m2
All of the above

18. 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:
Given:
Formula:
Solve:
L= 50 ft, W=15 ft
A= (L X W)
A= (50 ft)(15ft)
A= 750 ft2
15 ft
50 ft
375 ft2
65 ft2
750 ft2

19. The surface area of a tank is 2,000 sq 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:
Solve:
A=2000ft2, L= ? ft, W=25 ft
A= (L X W)
A/W=L
2000 ft2/(25 ft)= X ft
L= 80 ft
25 ft
L= ? ft
A= 2000 ft2
50000 ft ft
80 ft

20. Area of a parallelogram
A parallelogram is any 4-sided figure 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
H=height or l=length
B=base or w=width
A= b x h or (l x w)

21. B= 40 ft, h=30 ft A= (b X h) A= (30 ft)(40 ft) A= 1200 ft2 1200 ft2
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:
Formula:
Solve:
B= 40 ft, h=30 ft
A= (b X h)
A= (30 ft)(40 ft)
A= 1200 ft2
40 ft
30 ft
1200 ft2
1200 cm2
1200 m2
None of the above

22. Area of a rhombus A= 1/2 d1 x d2
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 d2 d1
d1 is the length of the first diagonal
d2 is the length of the second diagonal.
A= 1/2 d1 x d2

23. 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

24. h=height of the trapezoid
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
b2=base 2
Leg 1
Leg 2
h=height
h=height of the trapezoid
b1= 1st base
b2= second base.
A= (1/2) h (b1 + b2)

25. 3,750,000 ft2 3.75 X 106 ft2 All of the above DRAW: Given: Formula:
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:
Given:
Formula:
Solve:
b1=2500 ft
b2=5000 ft
Leg 1
Leg 2
h=height 1000 ft
b1=2,500 ft, b2=5,000 ft and h=1000 ft
A= (b1 +b2) * (h) 2
A= (2500 ft)+(5000ft) * 1000 ft
A= 7500 ft * 1000 ft
A=3750 ft * 1000 ft
A= 3,750,000 ft2
3,750,000 ft2
3.75 X 106 ft2
All of the above

26. 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

27. Area of a Right Circular Cone
B = π r2 s = sqrt[r2+h2] S = πrs
T = π r(r+s)
V = π r2h/3
H=height
s=slant height
Diameter given
A= π(r2)+π(d)(s)
S=lateral surface area
Radius given
A= π(r2)+π(r)(s)
R=radius

28. 51032 ft2 3533 ft2 1649 ft2 DRAW: Given: Formula: Solve:
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 cone) must be painted?
DRAW:
Given:
Formula:
Solve:
D= 50 ft, s= 20 ft, I know r= 25 ft!
A= π(r2)+π(d)(s) 2
A= π(25 ft)2 + π(50ft)(20ft)
A= π 625 ft2 + π 1000 ft2
A= 1963 ft ft2
A= 1963 ft ft2
A= 3533 ft2
D= 50 ft
r= 25 ft
s= 20 ft
51032 ft2
3533 ft2
1649 ft2

29. 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)
S = π(r+R)s
T = π(r[r+s]+R[R+s])        
V = π(R2+rR+r2)h/3
r=radius of base 2
s=slant height
h=height
S=lateral surface area
R=radius of base 1

30. Review Problems!

31. A circular sedimentation tank has an area of 1962 ft2
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
D=? Ft; AREA = 1962 ft2
A=1/4 π (d2)
1962 ft2= (X ft)2
1962 ft2= (X ft)2
0.785
2499 ft2= (X ft)2
√2499 ft2= X ft
50 ft = X
2499 ft
1250 ft
50 ft

32. 535 ft 14 ft 17500 ft DRAW: Given: Formula:
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 ?
DRAW:
Given:
Formula:
Solve:
Area= 500 ft2
35 ft.
X ft.?
A= 500 ft2, Length = X ft, Width = 35ft
A= (L X W)
A/W=L
500 ft2/35 ft= X ft
14 ft = Length
535 ft
14 ft
17500 ft

33. 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?

34. 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

35. Today’s objective: to become proficient with the concept of area as it pertains to water and wastewater operation has been met
Strongly Agree
Agree
Disagree
Strongly Disagree