6.3 Perimeter & Area of 2D Polygons & Circles

6.3 Perimeter & Area of 2D Polygons & Circles

Quick Summary: Calculating boundary lengths and surface coverage is a core spatial competency assessed on the Ontario MPT. This section reviews 2D perimeter and area formulas for standard polygons (rectangles, triangles, parallelograms, trapezoids) and circular shapes (circumference, area, arcs, sectors). It provides explicit strategies for solving multi-step composite 2D figure problems using additive and subtractive methods.

---

Core Concepts: Perimeter vs. Area

When working with two-dimensional geometry, it is vital to distinguish between boundary distance and surface coverage:

• Perimeter ($P$): The total 1D continuous linear distance around the outer boundary of a two-dimensional shape. Measured in linear units (e.g., $\text{mm}$, $\text{cm}$, $\text{m}$, $\text{km}$).
• Area ($A$): The total amount of 2D surface space enclosed within a boundary. Measured in square units (e.g., $\text{mm}^2$, $\text{cm}^2$, $\text{m}^2$, $\text{km}^2$).

---

Formulas for 2D Polygons

1. Rectangles & Squares

• Rectangle: $P = 2l + 2w = 2(l + w)$ $A = l \times w$
• Square (where $l = w = s$): $P = 4s \quad \text{and} \quad A = s^2$

2. Triangles

• Perimeter: Sum of all three side lengths: $P = a + b + c$
• Area: $A = \frac{1}{2} b h = \frac{b \times h}{2}$ Note: The height $h$ must be perpendicular ($90^\circ$) to the chosen base $b$. Never use a slant side length as the height unless the triangle is a right-angled triangle with that side as a leg.

3. Parallelograms

A quadrilateral with two pairs of parallel opposite sides.

• Perimeter: $P = 2a + 2b = 2(a + b)$ (where $a$ and $b$ are adjacent side lengths).
• Area: $A = b \times h$ Note: Like triangles, $h$ is the vertical perpendicular distance between the parallel bases, not the slanted side length $a$.

4. Trapezoids (Trapeziums)

A quadrilateral with at least one pair of parallel sides (called bases $a$ and $b$).

• Perimeter: Sum of all four sides: $P = a + b + c + d$.
• Area: $A = \frac{a + b}{2} \times h = \frac{1}{2}(a + b)h$ Conceptual Meaning: The area of a trapezoid is equal to the average length of its parallel bases ($\frac{a + b}{2}$) multiplied by the perpendicular height $h$.

---

Circles: Circumference, Area, Arcs & Sectors

A circle is defined as the locus of all points equidistant from a central point. The distance from the center to any point on the boundary is the radius ($r$). The distance across the circle through the center is the diameter ($d = 2r$).

1. Circumference ($C$)

The perimeter of a circle is called its circumference: $C = \pi d = 2\pi r$ Where $\pi \approx 3.14159$ (or $\frac{22}{7}$ in non-calculator estimations).

2. Area of a Circle ($A$)

$A = \pi r^2$ Warning: Do not confuse $C = 2\pi r$ with $A = \pi r^2$. Remember that area units are squared, matching $r^2$.

3. Arc Length ($L$) & Sector Area ($A_{\text{sector}}$)

A sector is a pie-shaped fraction of a circle subtended by a central angle $\theta$ (in degrees).

• Arc Length ($L$): The curved boundary portion of the sector: $L = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\theta}{360^\circ} \times \pi d$
• Sector Area ($A_{\text{sector}}$): The surface region enclosed by the sector: $A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$

---

Composite 2D Figures: Additive & Subtractive Strategies

Many MPT exam items test composite figures—shapes created by combining or cutting out basic geometric figures.

graph TD
    COMP["Composite 2D Figure Problem"]
    COMP --> ADD["Additive Strategy<br/>(Sum Sub-Region Areas)"]
    COMP --> SUB["Subtractive Strategy<br/>(Outer Area - Cutout Area)"]
    COMP --> PERIM["Perimeter Rule<br/>Sum ONLY Outer Exposed Edges!<br/>EXCLUDE Internal Boundaries!"]

1. Additive Area Method

Decompose the composite shape into non-overlapping standard shapes (e.g., a rectangle and a triangle), calculate each area separately, and sum them: $A_{\text{total}} = A_1 + A_2 + A_3$

2. Subtractive Area Method

Calculate the area of a larger enclosing standard shape and subtract the area of any unshaded cutouts or holes: $A_{\text{remaining}} = A_{\text{outer}} - A_{\text{cutout}}$

3. Critical Perimeter Rule for Composite Shapes

When calculating the perimeter of a composite figure, DO NOT sum the perimeters of the individual component shapes! Doing so incorrectly counts internal boundary lines where the shapes join. Only add the lengths of edges that lie on the exterior boundary of the final combined figure.

---

Multi-Step Worked Composite Example

Problem Statement: A school playground field design consists of a central rectangular field measuring $40 \text{ metres}$ long by $20 \text{ metres}$ wide, flanked by a semicircular play area attached to one of its $20 \text{ m}$ ends. Inside the rectangular portion, a square sandbox measuring $6 \text{ metres}$ on each side is dug out.

Calculate:

1. The total outer perimeter of the playground field (using $\pi \approx 3.14$).
2. The total usable surface area of the playground field (excluding the sandbox).

Step-by-Step Solution:

1. Step 1: Analyze Dimensions

   • Rectangle: $l = 40 \text{ m}$, $w = 20 \text{ m}$.
   • Semicircle: Attached to $20 \text{ m}$ width $\implies$ diameter $d = 20 \text{ m}$, radius $r = 10 \text{ m}$.
   • Square Sandbox cutout: $s = 6 \text{ m}$.

2. Step 2: Calculate Total Outer Perimeter Identify the exterior boundary components:

   • Bottom long side of rectangle = $40 \text{ m}$.
   • Top long side of rectangle = $40 \text{ m}$.
   • Flat left short side of rectangle = $20 \text{ m}$.
   • Curved right semicircular arc boundary = $\frac{1}{2} C = \frac{1}{2} (\pi d) = \frac{1}{2} (3.14 \times 20) = 31.4 \text{ m}$. (Note: The internal line of $20 \text{ m}$ separating the rectangle and semicircle is internal, so it is excluded!) $\text{Perimeter} = 40 + 40 + 20 + 31.4 = 131.4 \text{ metres}$

3. Step 3: Calculate Total Usable Area Use a combined additive and subtractive area approach:

   • Area of Central Rectangle ($A_1$): $A_1 = l \times w = 40 \times 20 = 800 \text{ m}^2$
   • Area of Semicircle ($A_2$): $A_2 = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times (10)^2 = 0.5 \times 3.14 \times 100 = 157 \text{ m}^2$
   • Area of Square Sandbox Cutout ($A_3$): $A_3 = s^2 = 6^2 = 36 \text{ m}^2$
   • Total Usable Area ($A_{\text{usable}}$): $A_{\text{usable}} = A_1 + A_2 - A_3 = 800 + 157 - 36 = 921 \text{ m}^2$

Final Answer: Total outer perimeter = $131.4 \text{ m}$, Total usable surface area = $921 \text{ m}^2$.