Basic Math Area FormulasBasic Math Area Formulas

Math for Water TechnologyMTH 082Lecture 2Chapter 9

Math for Water TechnologyMTH 082Lecture 2Chapter 9

Temperature ConversionsTemperature Conversions

oF= (9 * oC) + 32 5

oF= (9 * oC) + 32 5

oC= 5 * (oF – 32) 9

oC= 5 * (oF – 32) 9

Convert 17oC to Fahrenheit Convert 17oC to Fahrenheit

Convert 451oF to degrees CelsiusConvert 451oF to degrees Celsius

oF= (9 *17)+32=62.6oF= 63oF 5

oF= (9 *17)+32=62.6oF= 63oF 5

Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32.

Celsius to Fahrenheit 1. Begin by multiplying the Celsius temperature by 9. 2. Divide the answer by 5. 3. Now add 32.

Fahrenheit to Celsius1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5.

Fahrenheit to Celsius1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 5.

oC= 5* (oF -32)=232.7oC= 233oC 9

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

• 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 PROBLEMSRULES FOR AREA PROBLEMS1. 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

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

• 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.• Water is collected from the bottom of the Tualatin River through

intake screens that bring in water at a very low velocity.

JWC built in1976JWC built in1976

Sedimentation BasinsSedimentation Basins

BafflesBaffles

Filter bedsFilter beds

Settling TanksSettling Tanks

Rapid MixRapid Mix

Area of a CircleArea 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

• 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 A= π (r2)A= π (r2)

D=diameterD=diameter

r=radiusr=radius

cc

ii

rr

cc

uumm

ff

rr

ee

nn

ccee

A= 1/4 π (d2) 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:

DRAW:• Given:• Formula:• Solve:

D=40 ft

A=1/4 π (d2)

A= 0.785 (40 ft)2

A=0.785 (1600 ft2)A= 1256 ft2

D=40 ft

A=1/4 π (d2)

A= 0.785 (40 ft)2

A=0.785 (1600 ft2)A= 1256 ft21. 31.4 ft

2. 15.7 ft3

3. 1256 ft2

4. 628 ft2

1. 31.4 ft

2. 15.7 ft3

3. 1256 ft2

4. 628 ft2

40 ft

The radius of a water tank is 35 ft. What is the circumference (ft)

of the tank?

153

86 ft

220

ft

384

6.5

ft2

55

ft2

25% 25%25%25%

DRAW:• Given:• Formula:• Solve:

DRAW:• Given:• Formula:• Solve:

R= 35 ft; D=70 ft

C= π (d)

C= π(70 ft)C=3.14 (70ft)A= 219.8 ft

R= 35 ft; D=70 ft

C= π (d)

C= π(70 ft)C=3.14 (70ft)A= 219.8 ft

35 ft

1. 15386 ft

2. 220 ft

3. 3846.5 ft2

4. 55 ft2

1. 15386 ft

2. 220 ft

3. 3846.5 ft2

4. 55 ft2

Area of a triangleArea of a triangle

• 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

• 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=baseB=base

H=

heig

ht

H=

heig

ht

A= ½ x b x hA= ½ x b x h

What is the area (cm2) of this figure?

What is the area (cm2) of this figure?

13 cm

4 cm

DRAW:• Given:• Formula:• Solve:

DRAW:• Given:• Formula:• Solve:

1. 102 cm

2. 102 cm3

3. 26 cm2

4. 52 cm2

1. 102 cm

2. 102 cm3

3. 26 cm2

4. 52 cm2

Base= 13 cm, Side= 4 cm

A= (B XH)/2

A=(13 cm X 4 cm)/2)A= 26 cm2

Base= 13 cm, Side= 4 cm

A= (B XH)/2

A=(13 cm X 4 cm)/2)A= 26 cm2

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)?

85

ft

170

ft2

170

ft

3,4

00,0

00 ft

2

25% 25%25%25%

DRAW:• Given:

• Formula:• Solve:

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=H2(17,000ft2)/200 ft=170 ft

Base= 200 ft, Area= 17,000 ft2, Side= ?

A= (B XH)/2

Solve for unknown H:2A/B=H2(17,000ft2)/200 ft=170 ft1. 85 ft

2. 170 ft2

3. 170 ft

4. 3,400,000 ft2

1. 85 ft

2. 170 ft2

3. 170 ft

4. 3,400,000 ft2

• 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

• 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

Area of a SquareArea of a Square

S=sideS=side

S=sideS=side

A=S2A=S2

An Acti-flo square settling basin is 8ft on a side, what is its area?

8 ft

6 ft

3

16

ft2

64

ft2

25% 25%25%25%

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

s= 3ft

A= S X S or S2

A= (8ft)2

A= (8 ft)(8ft)A= 64 ft2

s= 3ft

A= S X S or S2

A= (8ft)2

A= (8 ft)(8ft)A= 64 ft2

8 ft

8 ft

1. 8 ft

2. 6 ft3

3. 16 ft2

4. 64 ft2

1. 8 ft

2. 6 ft3

3. 16 ft2

4. 64 ft2

Area of a rectangleArea of a rectangle

B=base or w=widthB=base or w=width

A= b x h or (l x w)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)

• 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=lengthH=height or l=length

What is the area (ft2) of this rectangular up-flow clarifier figure?

3 ft.

8 ft.

Length = 8 ft, Width = 3ft

A= (L X W)

A= (8 ft)(3ft)A= 24 ft2

Length = 8 ft, Width = 3ft

A= (L X W)

A= (8 ft)(3ft)A= 24 ft2

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

1. 24 ft2

2. 3456 in2

3. 2.23 m2

4. All of the above

1. 24 ft2

2. 3456 in2

3. 2.23 m2

4. All of the above

3 ft

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?

L= 50 ft, W=15 ft

A= (L X W)

A= (50 ft)(15ft)A= 750 ft2

L= 50 ft, W=15 ft

A= (L X W)

A= (50 ft)(15ft)A= 750 ft2

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

1. 375 ft2

2. 65 ft2

3. 750 ft2

1. 375 ft2

2. 65 ft2

3. 750 ft2

50 ft

15 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)?

A=2000ft2, L= ? ft, W=25 ft

A= (L X W)A/W=L

2000 ft2/(25 ft)= X ftL= 80 ft

A=2000ft2, L= ? ft, W=25 ft

A= (L X W)A/W=L

2000 ft2/(25 ft)= X ftL= 80 ft

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

25 f

t

L= ? ft

A= 2000 ft2

1. 50000 ft

2. 0.0125 ft

3. 80 ft

1. 50000 ft

2. 0.0125 ft

3. 80 ft

Area of a parallelogramArea 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

• 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

B=base or w=widthB=base or w=width

H=height or l=lengthH=height or l=length

A= b x h or (l x w)A= b x h or (l x w)

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

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)?

40 ft40 ft

30 ft30 ftB= 40 ft, h=30 ftA= (b X h)

A= (30 ft)(40 ft)A= 1200 ft2

B= 40 ft, h=30 ftA= (b X h)

A= (30 ft)(40 ft)A= 1200 ft2

1. 1200 ft2

2. 1200 cm2

3. 1200 m2

4. None of the above

1. 1200 ft2

2. 1200 cm2

3. 1200 m2

4. None of the above

Area of a rhombusArea 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

• 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

A= 1/2 d1 x d2A= 1/2 d1 x d2

d1d1d2d2

•d1 is the length of the first diagonal •d2 is the length of the second diagonal.

•d1 is the length of the first diagonal •d2 is the length of the second diagonal.

Area of a TrapezoidArea 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).

• 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 1b1=base 1

b2=base 2b2=base 2

Leg 1Leg 1 Leg 2Leg 2heightheight

medianmedian

Area of a TrapezoidArea 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)

• 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)

A= (1/2) h (b1 + b2)A= (1/2) h (b1 + b2)

b1=base 1b1=base 1

b2=base 2b2=base 2

Leg 1Leg 1 Leg 2Leg 2h=heighth=height

•h=height of the trapezoid• b1= 1st base

• b2= second base.

•h=height of the trapezoid• b1= 1st base

• b2= second base.

DRAW:•Given:•Formula:

•Solve:

DRAW:•Given:•Formula:

•Solve:

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)?

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)?

b1=2,500 ft, b2=5,000 ft and h=1000 ft

A= (b1 +b2) * (h) 2A= (2500 ft)+(5000ft) * 1000 ft 2A= 7500 ft * 1000 ft 2A=3750 ft * 1000 ftA= 3,750,000 ft2

b1=2,500 ft, b2=5,000 ft and h=1000 ft

A= (b1 +b2) * (h) 2A= (2500 ft)+(5000ft) * 1000 ft 2A= 7500 ft * 1000 ft 2A=3750 ft * 1000 ftA= 3,750,000 ft2

b1=2500 ftb1=2500 ft

b2=5000 ftb2=5000 ft

Leg 1Leg 1 Leg 2Leg 2h=height 1000 fth=height 1000 ft

1. 3,750,000 ft2

2. 3.75 X 106 ft2

3. All of the above

1. 3,750,000 ft2

2. 3.75 X 106 ft2

3. All of the above

Area of a Right Circular ConeArea 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

• 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

H=heightH=height

R=radiusR=radius

s=slant heights=slant height

S=lateral surface area

S=lateral surface area

• B = π r2

s = sqrt[r2+h2]S = πrs

T = π r(r+s)

V = π r2h/3

• B = π r2

s = sqrt[r2+h2]S = πrs

T = π r(r+s)

V = π r2h/3

Diameter givenA= π(r2)+π(d)(s)

2

Diameter givenA= π(r2)+π(d)(s)

2Radius givenA= π(r2)+π(r)(s)

Radius givenA= π(r2)+π(r)(s)

H=heightH=height

R=radiusR=radius

s=slant heights=slant height

S=lateral surface areaS=lateral surface area

Area of a Right Circular ConeArea of a Right Circular Cone

DRAW:•Given:•Formula:•Solve:

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?

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?

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) 2A= π 625 ft2 + π 1000 ft2

2A= 1963 ft2 + 3140 ft2

2A= 1963 ft2 + 1570 ft2

A= 3533 ft2

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) 2A= π 625 ft2 + π 1000 ft2

2A= 1963 ft2 + 3140 ft2

2A= 1963 ft2 + 1570 ft2

A= 3533 ft21. 51032 ft2

2. 3533 ft2

3. 1649 ft2

1. 51032 ft2

2. 3533 ft2

3. 1649 ft2

D= 50 ft

r= 25 fts= 20 ft

• 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

• 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

Frustum of a Right Circular Cone

R=radius of base 1R=radius of base 1

h=heighth=height s=slant heights=slant height

r=radius of base 2r=radius of base 2

S=lateral surface area

S=lateral surface area

Review Problems!Review Problems!

DRAW:• Given:• Formula:• Solve:

DRAW:• Given:• Formula:• Solve:

A circular sedimentation tank has an area of 1962 ft2. What is

the diameter of the tank??

D=? Ft; AREA = 1962 ft2

A=1/4 π (d2)

1962 ft2= 0.785 (X ft)2

1962 ft2= (X ft)2

0.785

2499 ft2= (X ft)2

√2499 ft2= X ft50 ft = X

D=? Ft; AREA = 1962 ft2

A=1/4 π (d2)

1962 ft2= 0.785 (X ft)2

1962 ft2= (X ft)2

0.785

2499 ft2= (X ft)2

√2499 ft2= X ft50 ft = X

1. 2499 ft

2. 1250 ft

3. 50 ft

1. 2499 ft

2. 1250 ft

3. 50 ft

D= ? Ft

AREA=1962 ft2

D= ? Ft

AREA=1962 ft2

DRAW:•Given:•Formula:

•Solve:

DRAW:•Given:•Formula:

•Solve:

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 ?

Are

a= 5

00 f

t2

35 f

t.

X ft.?

A= 500 ft2, Length = X ft, Width = 35ft

A= (L X W)A/W=L

500 ft2/35 ft= X ft14 ft = Length

A= 500 ft2, Length = X ft, Width = 35ft

A= (L X W)A/W=L

500 ft2/35 ft= X ft14 ft = Length

1. 535 ft

2. 14 ft

3. 17500 ft

1. 535 ft

2. 14 ft

3. 17500 ft

What did you learn?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?

• 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

• 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

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

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree