# 2.1: The Shapes of Things

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

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 one-fourth 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

24 in

24 in / 12 in = 2 ft

0.785 x (2 ft)2

3.14 ft2

130 in

130 in / 12 in = 10.8 ft

0.785 x (10.8 ft)2

91.5 ft2

813 mm

813 mm / 304.5 mm = 2.67 ft

0.785 x (2.67 ft)2

5.6 ft2

### 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 ft2
• Height = 15 ft, width = 7 ft 15 ft x 7 ft = 105 ft2

### 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 ft2
• Widths = 8 ft and 12 ft, depth = 10 ft
• (8 ft + 12 ft) / 2 x 10 ft = 100 ft2

### 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 mid-west. 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 ft2

When the area is calculated for a sphere, it is the entire surface area of a “ball.” Spheres can be commonplace in the mid-west 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 ft2

D = 35 ft 4 x 0.785 x (35 ft)2 = 3,845 ft2

## 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 three-dimensional structure with a volume that can be calculated.

## Exercises

This page titled 2.1: The Shapes of Things is shared under a CC BY license and was authored, remixed, and/or curated by Mike Alvord (ZTC Textbooks) .