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**Basic Math Area Formulas**

Math for Water Technology MTH 082 Lecture 2 Chapter 9

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**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

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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

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**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

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**What is area? The amount of space that a figure encloses**

It is two-dimensional It is always answered in square units

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**Water/Wastewater Shapes**

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**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

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**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)

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**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

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**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

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**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

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**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

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**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

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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

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**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

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**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)

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**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

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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

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**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

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**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)

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**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

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**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

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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

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**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)

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**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

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**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

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**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

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**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

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**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

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Review Problems!

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**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

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**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

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**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?

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**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

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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

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