11.2 Perimeter of Polygons, Circle Anatomy, & Circumference Calculations

11.2 Perimeter of Polygons, Circle Anatomy, & Circumference Calculations

Perimeter and circumference are fundamental linear measurement concepts on the TABE 13&14 Mathematics assessment. Whether calculating the linear footage of security fencing around an irregular warehouse lot, sizing baseboard molding for a residential remodel, or determining the travel distance of a circular gear, solving these problems requires identifying outer boundaries, determining missing segment lengths, and applying circle formulas accurately.

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The Concept of Perimeter & Linear Distance

Perimeter ($P$) is the continuous total distance around the exterior boundary of a closed two-dimensional figure. Perimeter is strictly a one-dimensional (linear) measurement expressed in units of length:

• U.S. Customary Units: Inches ($\text{in}$), feet ($\text{ft}$), yards ($\text{yd}$), miles ($\text{mi}$)
• Metric Units: Millimeters ($\text{mm}$), centimeters ($\text{cm}$), meters ($\text{m}$), kilometers ($\text{km}$)

Standard Polygon Perimeter Formulas

Regular Polygon Example: A regular octagon has side lengths of $7.25\text{ inches}$. Its perimeter is:

$$P = 8 \times 7.25\text{ in} = \mathbf{58\text{ in}}$$

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Perimeter of Composite & Stepped Shapes

On the TABE, many perimeter questions feature composite "stepped" or L-shaped floor plans where one or more side lengths are not labeled. You must use the geometric principle of opposing parallel segments to deduce the missing lengths before adding the boundary.

Stepped L-Shaped Floor Plan Boundary:

         <-------- Top = 24 ft -------->
       +--------------------------------+
       |                                |
       |                                | Right = 14 ft
Left   |                 +--------------+
= 20 ft|                 | Inset = 10 ft
       |                 |
       +-----------------+
         <-- Bottom = ? ->

Rules for Solving Missing Dimensions

1. Horizontal Balance: The total horizontal distance across the top must equal the sum of all horizontal bottom segments: $$\text{Top Side} = \text{Bottom Segment 1} + \text{Bottom Segment 2}$$
2. Vertical Balance: The total vertical distance along the left must equal the sum of all vertical right segments: $$\text{Left Side} = \text{Right Segment 1} + \text{Right Segment 2}$$

Worked Example: L-Shaped Perimeter

Problem: In the diagram above, the total top width is $24\text{ ft}$, the full left height is $20\text{ ft}$, the outer right vertical edge is $14\text{ ft}$, and the inner horizontal inset is $10\text{ ft}$. Find the perimeter of the floor plan.

1. Find Missing Bottom Width: The top ($24\text{ ft}$) equals the bottom plus the horizontal inset ($10\text{ ft}$): $$\text{Bottom Width} = 24 - 10 = 14\text{ ft}$$
2. Find Missing Inner Vertical Height: The left edge ($20\text{ ft}$) equals the right edge ($14\text{ ft}$) plus the inner vertical drop: $$\text{Inner Vertical Drop} = 20 - 14 = 6\text{ ft}$$
3. Sum All Outer Boundary Sides (6 sides total): $$P = 24\text{ (top)} + 14\text{ (right)} + 10\text{ (inset)} + 6\text{ (inner drop)} + 14\text{ (bottom)} + 20\text{ (left)}$$ $$P = 24 + 14 + 10 + 6 + 14 + 20 = \mathbf{88\text{ ft}}$$

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Anatomy of a Circle

A circle is the set of all points in a plane equidistant from a fixed point called the center ($O$).

Circle Anatomy Components:
                 Arc (Curved Boundary)
                /-----\ 
              /         \ 
             |     r     |
             |  •----->  |  <--- Circumference (C = 2πr)
             |  Center   |
              \    d    /
                \-----/ 
             <-- Diameter -->

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Circumference Calculations & The Value of $\pi$

The ratio of any circle's circumference to its diameter is the constant irrational number $\pi$ (Pi):

$$\pi = \frac{C}{d} \approx 3.14159265...$$

Circumference Formulas

$$\mathbf{C = \pi d} \qquad \Longleftrightarrow \qquad \mathbf{C = 2\pi r}$$

Selecting the Best Value for $\pi$

Depending on the problem specifications on the TABE:

• Decimal Approximation: Use $\pi \approx 3.14$ for general decimal calculations.
• Fraction Approximation: Use $\pi \approx \frac{22}{7}$ when the radius or diameter is a multiple of $7$.
• Exact Form: Leave $\pi$ as a symbol in the answer (e.g., $18\pi\text{ cm}$).

Worked Example 1: Basic Circumference with Fraction $\pi$

Problem: A circular decorative garden pool has a radius of $14\text{ feet}$. Using $\pi \approx \frac{22}{7}$, what is the circumference of the pool?

$$C = 2\pi r = 2 \times \frac{22}{7} \times 14 = 2 \times 22 \times 2 = \mathbf{88\text{ ft}}$$

Worked Example 2: Finding Diameter from Circumference

Problem: A commercial storage silo has a measured circumference of $94.2\text{ feet}$. Using $\pi \approx 3.14$, find its diameter and radius.

$$d = \frac{C}{\pi} = \frac{94.2}{3.14} = \mathbf{30\text{ ft}} \qquad | \qquad r = \frac{d}{2} = \frac{30}{2} = \mathbf{15\text{ ft}}$$

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Arc Length Formula

Arc length ($L$) represents the linear distance along a curved portion of a circle's circumference intercepted by a central angle $\theta$ (in degrees):

$$\mathbf{L = \left(\frac{\theta}{360^\circ}\right) \times 2\pi r = \left(\frac{\theta}{360^\circ}\right) \times \pi d}$$

Worked Example: A central angle of $45^\circ$ intercepts an arc on a circular wheel with a radius of $16\text{ inches}$. Using $\pi \approx 3.14$:

$$L = \left(\frac{45^\circ}{360^\circ}\right) \times 2(3.14)(16) = \frac{1}{8} \times 100.48 = \mathbf{12.56\text{ in}}$$

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Real-World Composite Perimeters: Stadium Tracks & Enclosures

Trade and vocational problems frequently test composite perimeters combining straight polygon segments with semicircular curved ends (e.g., running tracks, athletic courts, and conveyor belts).

Athletic Running Track Geometry:
          <-------- Straightaway (L = 100 m) -------->
        +---------------------------------------------+
       /|                                             |\
      / |                                             | \
 (r) /  |                                             |  \ (r)
    | • | Diameter (d = 60 m)                         | • | 
 (r) \  |                                             |  / (r)
      \ |                                             | /
       \|                                             |/
        +---------------------------------------------+
          <-------- Straightaway (L = 100 m) -------->

The Track Perimeter Formula

Notice that the two semicircular ends combine to make one full circle of circumference $\pi d$:

$$P_{\text{track}} = 2l + \pi d = 2l + 2\pi r$$

Worked Example: An asphalt running track consists of two straightaways measuring $100\text{ meters}$ each and two semicircular ends with a diameter of $60\text{ meters}$ ($r = 30\text{ m}$). Using $\pi \approx 3.14$, calculate the total perimeter distance around the track.

1. Length of Straightaways: $2 \times 100\text{ m} = 200\text{ m}$.
2. Circumference of Two Semicircles (1 Full Circle): $$C = \pi d = 3.14 \times 60\text{ m} = 188.4\text{ m}$$
3. Total Track Perimeter: $$P = 200 + 188.4 = \mathbf{388.4\text{ meters}}$$