2.1: The Shapes of Things
 Page ID
 7127
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In order to transport water from the source to treatment, to the distribution system, and, eventually, to the customer, it needs to flow through geometric shapes. An aqueduct brings water from Northern California to Southern California. Reservoirs and tanks store water before it enters the treatment process. Pipes flow water throughout the treatment plant and through the distribution system. Above ground storage tanks and elevated storage tanks hold water and provide pressure to the distribution system. This is a crude description of the path water takes, but it illustrates the point of different structures and shapes that water must transfer through.
Areas
Calculating areas is the first step in working with geometric shapes. Areas are used to determine how much paint to buy, how much water can flow through a pipe, and many other things. A circle, a rectangle, and a trapezoid are probably the most common shapes you will encounter in the water industry. However, a sphere, a triangle, a half circle with a rectangle can also be found. These are the structures we will focus on in this chapter.
\[0.785 \mathrm{D}^{2} \quad \mathrm{LW}\left(\frac{b_{1}+b_{2}}{2}\right) H\]
\[\frac{\left(0.785 \mathrm{D}^{2}\right)}{2}+\mathrm{LW} \quad 4\left(0.785 \mathrm{D}^{2}\right) \quad \pi \mathrm{DH}\]
Circles
To calculate the area of a circle, multiply 0.785 by the diameter squared. This means to multiply the diameter times 0.785. If you recall from the 030 course, we use 0.785 in the “area” formula.
\[A = 3.14 \times r^2\]
or
\[A = 0.785 \times d^2\]
0.785 replaces \(\pi\) and diameter replaces radius. Diameter squared is four times greater than radius squared and 0.785 is onefourth of \(\pi\).
Take special note of the units for the diameter. Many times (especially when talking about pipes) the diameter will be given on some other unit besides feet (e.g. inches). Converting the diameter to feet as your first step will avoid ending up with squared units other than square feet. Sometimes the diameter of a pipe might be given in metric units. This is common when working with the California Department of Transportation.
Examples
What is the area for each of the following diameters listed below?
Given Diameter 
Conversion 
Formula 
Answer 

24 in 
24 in / 12 in = 2 ft 
0.785 x (2 ft)^{2} 
3.14 ft^{2} 
130 in 
130 in / 12 in = 10.8 ft 
0.785 x (10.8 ft)^{2} 
91.5 ft^{2} 
813 mm 
813 mm / 304.5 mm = 2.67 ft 
0.785 x (2.67 ft)^{2} 
5.6 ft^{2} 
Rectangles
Calculating the area of a rectangle or a square simply involves multiplying the length by the width. If you are painting the walls, ceiling, or floors of a room the perspective changes slightly. For example, the dimensions of a wall might look like a width and height when you are standing looking at it. A floor might look like a width and a length. Regardless of the perspective, the area formula is the same.
Examples
What is the area for each of the following rectangles listed below?
 Length = 30 ft, width = 10 ft 30 ft x 10 ft = 300 ft^{2}
 Height = 15 ft, width = 7 ft 15 ft x 7 ft = 105 ft^{2}
Trapezoids
Trapezoids are most commonly the shape of an aqueduct. Aqueducts are typically miles and miles of trapezoidal shaped concrete channels. They have flat narrow bottoms that slope up to wider distances at the top. In order to calculate the varying distances across a trapezoid, add the distance (width, b2) across the bottom to the distance (width, b1) across the top and divide by 2. This gives the average width. Then multiply the average width by the height or depth of the trapezoid to calculate the area.
Examples
 Widths = 5 ft and 7 ft, height = 6 ft
 (5 ft + 7 ft) / 2 x 6 ft = 36 ft^{2}
 Widths = 8 ft and 12 ft, depth = 10 ft
 (8 ft + 12 ft) / 2 x 10 ft = 100 ft^{2}
Spheres and Other Shapes
As previously stated, circles, rectangles, and trapezoids are the most common shapes in the water industry. However, large standpipes shaped like a cylinder with a sphere on top or an elevated storage tank shaped like a sphere can be very common in the midwest. Half circles and rectangles can also be found as reservoirs or sedimentation basins. Therefore, understanding how to calculate the area for these types of structures is also important.
The distance around the cylinder is calculated as Pi (3.14) multiplied by the diameter. Pi is a unitless constant. It can also be looked at as the “length” around a cylinder. Once the “length” is calculated multiply this number by the height or depth to get the area.
Examples
 D = 100 ft, height = 20 ft
 3.14 x 100 ft = 314 ft (which is “L”)
 314 ft x 20 ft = 6,280 ft^{2}
When the area is calculated for a sphere, it is the entire surface area of a “ball.” Spheres can be commonplace in the midwest as elevated storage structures. The formula for the area of a sphere is 4 times 0.785 times the diameter squared.
Examples
D = 50 ft 4 x 0.785 x (50 ft)^{2} = 7,850 ft^{2}
D = 35 ft 4 x 0.785 x (35 ft)^{2} = 3,845 ft^{2}
Exercises

What is the area of the opening of a 10” diameter pipe?

A rectangular channel flows millions of gallons of water through it and dumps into a storage reservoir. The channel is 2 miles long 3 feet wide and 2 feet deep. What is the area of the channel opening?

A channel is 3 feet wide at the bottom and 5 feet wide at the top and the water is 4 feet deep when the channel is full. What is the area of the channel?

An elevated storage tank is shaped like a sphere and needs to be recoated. If the diameter of the tank is 65 feet, what is the surface area?

A standpipe needs to be painted. What is the surface area of the entire standpipe? It has a diameter of 30 feet and is 80 feet tall.

What is the surface area of a spherical structure that has a 42foot diameter?

A box structure needs to be painted. It is 20 feet wide, 30 feet long and 10 feet tall. What is the total area of all six surfaces (inside only)?

What is the inside surface area of a 32” diameter pipe that is 1,000 feet long and is capped with half a sphere at the end? (Assume the sphere diameter is not included in the length)
Volumes
In order to calculate the volume inside a structure, a third dimension needs to be included in the “area” formula. For example, if a circle is given a length or height, it becomes a cylinder and a volume can be calculated. If a trapezoid or a rectangle has a length it becomes a threedimensional structure with a volume that can be calculated.
Exercises

What is the volume of a 30foot diameter sphere?

What is the volume in 2,000 feet of 18inch diameter pipe?

A 5mile long aqueduct is 5 feet wide at the bottom and 8 feet wide at the water line. If the water depth is 6 ½ feet, how many acrefeet of water are in the aqueduct?

A sedimentation basin is 100 feet long, 30 feet wide, and 20 feet deep. How many gallons can it hold?

A standpipe is 80 feet tall and has a 25foot diameter. Assuming the top of the standpipe is a halfsphere shape, how many gallons will it hold?