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

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Temperature Conversions o F= (9 * o C) + 32 5 o F= (9 * o C) + 32 5 o C= 5 * ( o F – 32) 9 o C= 5 * ( o F – 32) 9 Convert 17 o C to Fahrenheit Convert 451 o F to degrees Celsius o F= (9 *17)+32=62.6 o F= 63 o F 5 o F= (9 *17)+32=62.6 o F= 63 o F 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 Celsius 1. Begin by subtracting 32 from the Fahrenheit #. 2. Divide the answer by 9. 4. Then multiply that answer by 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. o C= 5* ( o F -32)=232.7 oC = 233 o C 9 o C= 5* ( o F -32)=232.7 oC = 233 o C 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 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 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 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

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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 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 Water is collected from the bottom of the Tualatin River through intake screens that bring in water at a very low velocity. JWC built in1976 Sedimentation Basins Baffles Filter beds Settling Tanks Rapid Mix

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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 (r 2 ) or 1/4 π (d 2 ) or 0.785 (d 2 ) Area of ½ circle= π (r 2 ) 2 Distance around circle is circumference C=π (d) Distance through circle is diameter radius is 1/2 as large as diameter Area= π (r 2 ) or 1/4 π (d 2 ) or 0.785 (d 2 ) Area of ½ circle= π (r 2 ) 2 A= π (r 2 ) D=diameter r=radius c c i i r r c c u u m m f f r r e e n n c c e e A= 1/4 π (d 2 )

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A circular clarifier has a diameter of 40 ft. What is the surface area (ft 2 ) of the clarifier?? DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: D=40 ft A=1/4 π (d 2 ) A= 0.785 (40 ft) 2 A=0.785 (1600 ft 2 ) A= 1256 ft 2 D=40 ft A=1/4 π (d 2 ) A= 0.785 (40 ft) 2 A=0.785 (1600 ft 2 ) A= 1256 ft 2 1.31.4 ft 2.15.7 ft 3 3.1256 ft 2 4.628 ft 2 1.31.4 ft 2.15.7 ft 3 3.1256 ft 2 4.628 ft 2 40 ft

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The radius of a water tank is 35 ft. What is the circumference (ft) of the tank? 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 ft 2 4.55 ft 2 1.15386 ft 2.220 ft 3.3846.5 ft 2 4.55 ft 2

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Area 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=base H=height A= ½ x b x h

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What is the area (cm 2 ) of this figure? 13 cm 4 cm DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: 1.102 cm 2.102 cm 3 3.26 cm 2 4.52 cm 2 1.102 cm 2.102 cm 3 3.26 cm 2 4.52 cm 2 Base= 13 cm, Side= 4 cm A= (B XH)/2 A=(13 cm X 4 cm)/2) A= 26 cm 2 Base= 13 cm, Side= 4 cm A= (B XH)/2 A=(13 cm X 4 cm)/2) A= 26 cm 2

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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: DRAW: Given: Formula: Solve: h h A=17,000 ft 2 200 ft Base= 200 ft, Area= 17,000 ft 2, Side= ? A= (B XH)/2 Solve for unknown H: 2A/B=H 2(17,000ft 2 )/200 ft=170 ft Base= 200 ft, Area= 17,000 ft 2, Side= ? A= (B XH)/2 Solve for unknown H: 2A/B=H 2(17,000ft 2 )/200 ft=170 ft 1.85 ft 2.170 ft 2 3.170 ft 4.3,400,000 ft 2 1.85 ft 2.170 ft 2 3.170 ft 4.3,400,000 ft 2

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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=s 2 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=s 2 Area of a Square S=side A=S 2

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An Acti-flo square settling basin is 8ft on a side, what is its area? DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: s= 3ft A= S X S or S 2 A= (8ft) 2 A= (8 ft)(8ft) A= 64 ft 2 s= 3ft A= S X S or S 2 A= (8ft) 2 A= (8 ft)(8ft) A= 64 ft 2 8 ft 1.8 ft 2.6 ft 3 3.16 ft 2 4.64 ft 2 1.8 ft 2.6 ft 3 3.16 ft 2 4.64 ft 2

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Area of a rectangle B=base or w=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) 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

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What is the area (ft 2 ) 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 ft 2 Length = 8 ft, Width = 3ft A= (L X W) A= (8 ft)(3ft) A= 24 ft 2 DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: 1.24 ft 2 2.3456 in 2 3.2.23 m 2 4.All of the above 1.24 ft 2 2.3456 in 2 3.2.23 m 2 4.All of the above 3 ft

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A treatment plant has three drying beds, each of which is 50 ft long and 15 ft wide. How many ft 2 does one drying bed occupy? L= 50 ft, W=15 ft A= (L X W) A= (50 ft)(15ft) A= 750 ft 2 L= 50 ft, W=15 ft A= (L X W) A= (50 ft)(15ft) A= 750 ft 2 DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: 1.375 ft 2 2.65 ft 2 3.750 ft 2 1.375 ft 2 2.65 ft 2 3.750 ft 2 50 ft 15 ft

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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=2000ft 2, L= ? ft, W=25 ft A= (L X W) A/W=L 2000 ft 2 /(25 ft)= X ft L= 80 ft A=2000ft 2, L= ? ft, W=25 ft A= (L X W) A/W=L 2000 ft 2 /(25 ft)= X ft L= 80 ft DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: 25 ft L= ? ft A= 2000 ft 2 1.50000 ft 2.0.0125 ft 3.80 ft 1.50000 ft 2.0.0125 ft 3.80 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 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=width H=height or l=length A= b x h or (l x w)

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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 (ft 2 )? 40 ft 30 ft B= 40 ft, h=30 ft A= (b X h) A= (30 ft)(40 ft) A= 1200 ft 2 B= 40 ft, h=30 ft A= (b X h) A= (30 ft)(40 ft) A= 1200 ft 2 1.1200 ft 2 2.1200 cm 2 3.1200 m 2 4.None of the above 1.1200 ft 2 2.1200 cm 2 3.1200 m 2 4.None of the above

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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 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 d2 d1 d2 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.

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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). 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 b2=base 2 Leg 1 Leg 2 height median

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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) 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. h=height of the trapezoid b1= 1st base b2= second base.

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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 (ft 2 )? b1=2,500 ft, b2=5,000 ft and h=1000 ft A= (b1 +b2) * (h) 2 A= (2500 ft)+(5000ft) * 1000 ft 2 A= 7500 ft * 1000 ft 2 A=3750 ft * 1000 ft A= 3,750,000 ft 2 b1=2,500 ft, b2=5,000 ft and h=1000 ft A= (b1 +b2) * (h) 2 A= (2500 ft)+(5000ft) * 1000 ft 2 A= 7500 ft * 1000 ft 2 A=3750 ft * 1000 ft A= 3,750,000 ft 2 b1=2500 ft b2=5000 ft Leg 1 Leg 2 h=height 1000 ft 1.3,750,000 ft 2 2.3.75 X 10 6 ft 2 3.All of the above 1.3,750,000 ft 2 2.3.75 X 10 6 ft 2 3.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 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=height R=radius s=slant height S=lateral surface area

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B = π r 2 s = sqrt[r 2 +h 2 ] S = πrs T = π r(r+s) V = π r 2 h/3 B = π r 2 s = sqrt[r 2 +h 2 ] S = πrs T = π r(r+s) V = π r 2 h/3 Diameter given A= π(r 2 )+π(d)(s) 2 Diameter given A= π(r 2 )+π(d)(s) 2 Radius given A= π(r 2 )+π(r)(s) Radius given A= π(r 2 )+π(r)(s) H=height R=radius s=slant height S=lateral surface area Area of a Right Circular Cone

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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? D= 50 ft, s= 20 ft, I know r= 25 ft! A= π(r 2 )+π(d)(s) 2 A= π(25 ft) 2 + π(50ft)(20ft) 2 A= π 625 ft 2 + π 1000 ft 2 2 A= 1963 ft 2 + 3140 ft 2 2 A= 1963 ft 2 + 1570 ft 2 A= 3533 ft 2 D= 50 ft, s= 20 ft, I know r= 25 ft! A= π(r 2 )+π(d)(s) 2 A= π(25 ft) 2 + π(50ft)(20ft) 2 A= π 625 ft 2 + π 1000 ft 2 2 A= 1963 ft 2 + 3140 ft 2 2 A= 1963 ft 2 + 1570 ft 2 A= 3533 ft 2 1.51032 ft 2 2.3533 ft 2 3.1649 ft 2 1.51032 ft 2 2.3533 ft 2 3.1649 ft 2 D= 50 ft r= 25 ft s= 20 ft

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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 +h 2 ) S = π(r+R)s T = π(r[r+s]+R[R+s]) V = π(R 2 +rR+r 2 )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 +h 2 ) S = π(r+R)s T = π(r[r+s]+R[R+s]) V = π(R 2 +rR+r 2 )h/3 Frustum of a Right Circular Cone R=radius of base 1 h=height s=slant height r=radius of base 2 S=lateral surface area

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Review Problems!

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DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: A circular sedimentation tank has an area of 1962 ft 2. What is the diameter of the tank?? D=? Ft; AREA = 1962 ft 2 A=1/4 π (d 2 ) 1962 ft 2 = 0.785 (X ft) 2 1962 ft 2 = (X ft) 2 0.785 2499 ft 2 = (X ft) 2 √2499 ft 2 = X ft 50 ft = X D=? Ft; AREA = 1962 ft 2 A=1/4 π (d 2 ) 1962 ft 2 = 0.785 (X ft) 2 1962 ft 2 = (X ft) 2 0.785 2499 ft 2 = (X ft) 2 √2499 ft 2 = X ft 50 ft = X 1.2499 ft 2.1250 ft 3.50 ft 1.2499 ft 2.1250 ft 3.50 ft D= ? Ft AREA=1962 ft 2 D= ? Ft AREA=1962 ft 2

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DRAW: Given: Formula: Solve: DRAW: Given: Formula: Solve: The top of a Monterey Bay aquarium tank has a surface area of 500 ft 2. If the width of the tank is 35 ft, what is the length of the tank ? Area= 500 ft 2 35 ft. X ft.? A= 500 ft 2, Length = X ft, Width = 35ft A= (L X W) A/W=L 500 ft 2 /35 ft= X ft 14 ft = Length A= 500 ft 2, Length = X ft, Width = 35ft A= (L X W) A/W=L 500 ft 2 /35 ft= X ft 14 ft = Length 1.535 ft 2.14 ft 3.17500 ft 1.535 ft 2.14 ft 3.17500 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? 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) r 2 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(r 2 )+πdhs or A= π(r 2 )+πrs 2 Circle= A= (π) r 2 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(r 2 )+πdhs or A= π(r 2 )+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 1.Strongly Agree 2.Agree 3.Disagree 4.Strongly Disagree 1.Strongly Agree 2.Agree 3.Disagree 4.Strongly Disagree

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