6.3 Perimeter, Area & Circumference

Perimeter, Area & Circumference

Perimeter and area are essential spatial measurement concepts taught throughout elementary grade levels. On the Praxis 5003 exam, questions evaluate your computational fluency, conceptual understanding of 2D formulas, and ability to break down complex composite shapes into manageable components.

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1. Perimeter of Polygons & Circumference of Circles

Perimeter ($P$)

Perimeter is the total linear distance around the outer boundary of a two-dimensional closed shape. It is expressed in one-dimensional linear units (e.g., inches, feet, centimeters, meters).

For any polygon, perimeter is calculated by adding the lengths of all outer sides:

$$P = s_1 + s_2 + s_3 + \dots + s_n$$

• Rectangle Perimeter: $P = 2l + 2w = 2(l + w)$
• Square Perimeter: $P = 4s$
• Regular $n$-gon Perimeter: $P = n \times s$

Circumference ($C$)

Circumference is the specific term used for the perimeter of a circle (the linear distance around the circular edge).

$$\text{Circumference Formula: } C = 2\pi r \quad \text{or} \quad C = \pi d$$

Where:

• $r$ is the radius (distance from center to edge).
• $d$ is the diameter ($d = 2r$).
• $\pi$ (pi) is the mathematical constant representing the ratio of a circle's circumference to its diameter ($\pi = \frac{C}{d} \approx 3.14159 \dots \approx 3.14$ or $\frac{22}{7}$).

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2. Area Formulas & Geometric Derivations

Area ($A$) is the measure of 2D surface coverage enclosed within a closed figure. Area is expressed in two-dimensional square units (e.g., $\text{in}^2, \text{ft}^2, \text{cm}^2, \text{m}^2$).

Comprehensive Area Formula Table

RECTANGLE                     PARALLELOGRAM                        TRIANGLE
┌───────────────┐             ┌───────────────┐                    /││               │ w           │ /            /│ h                 / │ \ h
│               │             │/             / │                 /  │  └───────────────┘             └───────────────┘                 /───┴───        l                             b                             b
   A = l × w                     A = b × h                     A = 1/2 × b × h

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3. Step-by-Step Worked Calculation Examples

Example 1: Area of a Trapezoid

Problem: Find the area of a trapezoid with parallel bases measuring $12\text{ cm}$ and $18\text{ cm}$, and a perpendicular height of $8\text{ cm}$.

Step-by-Step Solution:

1. Identify given values: $b_1 = 12\text{ cm}$, $b_2 = 18\text{ cm}$, $h = 8\text{ cm}$.
2. Write the formula: $A = \frac{1}{2} (b_1 + b_2) h$.
3. Add the bases inside parentheses: $b_1 + b_2 = 12 + 18 = 30\text{ cm}$.
4. Multiply by height: $30 \times 8 = 240$.
5. Multiply by $\frac{1}{2}$: $A = \frac{1}{2} \times 240 = 120\text{ cm}^2$.

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Example 2: Circle Area vs. Circumference

Problem: A circular table has a diameter of $14\text{ feet}$. Calculate its exact circumference and exact area in terms of $\pi$.

Step-by-Step Solution:

1. Determine radius from diameter: $d = 14\text{ ft} \implies r = \frac{14}{2} = 7\text{ ft}$.
2. Circumference calculation: $C = \pi d = 14\pi\text{ ft}$.
3. Area calculation: $A = \pi r^2 = \pi (7)^2 = 49\pi\text{ ft}^2$.

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4. Decomposing Composite Figures

A composite figure is a 2D shape composed of two or more basic geometric shapes (such as rectangles, triangles, and semi-circles). Praxis 5003 frequently presents composite figure problems.

Core Strategies for Composite Area

1. Additive Method: Divide the irregular shape into separate non-overlapping simple shapes, calculate the area of each individual component, and add the areas together.

$$\text{Total Area} = A_1 + A_2 + A_3 + \dots$$

1. Subtractive Method: Enclose the composite figure inside a larger standard shape (like a bounding rectangle), calculate the total area, and subtract the unshaded or cut-out regions.

$$\text{Total Area} = A_{\text{outer}} - A_{\text{removed}}$$

ADDITIVE METHOD (SPLIT INTO SHAPES)          SUBTRACTIVE METHOD (OUTER MINUS INNER)
┌───────────┐                                ┌────────────────────────┐
│  Area 1   │                                │                        │
├───────────┼───────────┐                    │   ┌──────────────┐     │
│  Area 2   │  Area 3   │                    │   │ Cutout Region│     │
└───────────┴───────────┘                    │   └──────────────┘     │
Total = Area 1 + Area 2 + Area 3             └────────────────────────┘
                                             Total = Outer Area - Cutout Area

Detailed Worked Composite Example

Problem: An L-shaped floor plan is shown below with dimensions: top horizontal side = $6\text{ m}$, far left vertical side = $10\text{ m}$, bottom horizontal side = $14\text{ m}$, and far right vertical side = $4\text{ m}$. Calculate the total floor area.

Step-by-Step Solution:

1. Split into two rectangles:
   • Draw a vertical line down from the inner corner.
   • Rectangle A (left section): Width = $6\text{ m}$, Height = $10\text{ m}$. $$A_{\text{Rectangle A}} = 6 \times 10 = 60\text{ m}^2$$
   • Rectangle B (right section):
     • Base = Total bottom - top left width = $14 - 6 = 8\text{ m}$.
     • Height = $4\text{ m}$.
     $$A_{\text{Rectangle B}} = 8 \times 4 = 32\text{ m}^2$$
2. Sum the component areas: $$\text{Total Floor Area} = 60\text{ m}^2 + 32\text{ m}^2 = 92\text{ m}^2$$

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5. Key Candidate Traps on Praxis 5003

Fractional Side Lengths

Perimeter and area formulas still apply when side lengths are fractions. Example: a rectangle that is $\frac{5}{2}$ cm by $\frac{4}{3}$ cm has

$$P = 2\left(\frac{5}{2}+\frac{4}{3}\right) = 2\left(\frac{15}{6}+\frac{8}{6}\right) = 2\cdot\frac{23}{6} = \frac{23}{3}\ \text{cm},$$ $$A = \frac{5}{2}\times\frac{4}{3} = \frac{10}{3}\ \text{square cm}.$$

Composite-area problems may also use fractional lengths when decomposing into rectangles and triangles.