11.2 Perimeter, Circumference, Area & Scale Drawings

Perimeter, Circumference, Area & Scale Drawings

Quick Answer: WEST-B Mathematics Objective 0014 requires fluency in calculating linear boundaries (perimeter and circumference) and two-dimensional surface coverage (area) for standard polygons, circles, and composite geometric figures, as well as interpreting scale drawings. Essential formulas include rectangle perimeter ($P = 2l + 2w$) and area ($A = lw$), triangle area ($A = \frac{1}{2}bh$), parallelogram area ($A = bh$), trapezoid area ($A = \frac{1}{2}(b_1 + b_2)h$), rhombus/kite area ($A = \frac{1}{2}d_1 d_2$), circle circumference ($C = 2\pi r = \pi d$), circle area ($A = \pi r^2$), arc length ($s = \frac{\theta}{360^\circ} 2\pi r$), and sector area ($A = \frac{\theta}{360^\circ} \pi r^2$). A foundational exam principle for scale drawings: when linear dimensions scale by a factor of $k$, perimeter scales by $k$, but area scales quadratically by $k^2$.

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1. Perimeter of Polygons & Boundary Analysis

Perimeter ($P$) is a one-dimensional linear measure representing the total distance around the exterior boundary of a closed two-dimensional polygon. It is measured in linear units ($\text{in}$, $\text{ft}$, $\text{cm}$, $\text{m}$).

+---------------------------------------------------------------------------------------------------+
|                                 POLYGON PERIMETER FORMULA MATRIX                                  |
|                                                                                                   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
|   | GEOMETRIC SHAPE       | PERIMETER FORMULA     | VARIABLE DEFINITIONS & GEOMETRIC NOTES    |   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
|   | General Polygon       | P = s_1 + s_2 + ...   | Sum of all exterior boundary side lengths |   |
|   | Rectangle             | P = 2l + 2w = 2(l + w)| l = length, w = width                     |   |
|   | Square                | P = 4s                | s = length of one side (all 4 equal)      |   |
|   | Equilateral Triangle  | P = 3s                | s = length of one side (all 3 equal)      |   |
|   | Isosceles Triangle    | P = 2a + b            | a = congruent legs, b = base              |   |
|   | Regular n-gon         | P = n * s             | n = number of sides, s = side length      |   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
+---------------------------------------------------------------------------------------------------+

The Interior Seam Trap in Composite Figures

When calculating the perimeter of composite figures (such as a rectangle attached to a triangle or semicircle), never sum internal shared edges. Perimeter includes only the exterior border exposed to the outside.

Worked Example: A classroom banner consists of a rectangle measuring $8\text{ ft}$ long and $4\text{ ft}$ wide, with an equilateral triangle of side length $4\text{ ft}$ attached to one of the $4\text{ ft}$ ends.

• Exterior Edges: Two length sides ($8\text{ ft} + 8\text{ ft} = 16\text{ ft}$), one width end ($4\text{ ft}$), and two triangle sides ($4\text{ ft} + 4\text{ ft} = 8\text{ ft}$).
• Interior Seam: The shared $4\text{ ft}$ line between rectangle and triangle is inside the figure.
• Correct Perimeter: $P = 8 + 8 + 4 + 4 + 4 = 28\text{ ft}$ (NOT $28 + 4 = 32\text{ ft}$).

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2. Circles: Circumference, Area, Arc Length & Sector Area

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

+---------------------------------------------------------------------------------------------------+
|                                    CIRCLE FORMULA MASTER TABLE                                    |
|                                                                                                   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
|   | MEASUREMENT TYPE      | EXACT FORMULA         | APPROXIMATE CONSTANTS / NOTES             |   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
|   | Circumference (C)     | C = 2πr = πd          | π ≈ 3.14159... ≈ 3.14 or 22/7             |   |
|   | Area (A)              | A = πr^2              | Measured in square units (r must be squared)   |
|   | Arc Length (s)        | s = (θ / 360°) * 2πr  | θ = central angle in degrees              |   |
|   | Sector Area (A_sect)  | A = (θ / 360°) * πr^2 | Fraction of total circular disk area      |   |
|   | Annulus (Ring) Area   | A = π(R^2 - r^2)      | R = outer radius, r = inner radius        |   |
|   +-----------------------+-----------------------+-------------------------------------------+   |
+---------------------------------------------------------------------------------------------------+

Fractional Sector Area & Arc Length Proportions

Any sector of a circle is simply a fractional slice of the complete $360^\circ$ disk. The fraction is always defined by the central angle $\theta$ divided by $360^\circ$: $\text{Fraction of Circle} = \frac{\theta}{360^\circ}$

1. Arc Length ($s$): The linear distance along the curved edge of the slice: $s = \left(\frac{\theta}{360^\circ}\right) \times 2\pi r$
2. Sector Area ($A_{\text{sector}}$): The two-dimensional surface area inside the slice: $A_{\text{sector}} = \left(\frac{\theta}{360^\circ}\right) \times \pi r^2$

Worked Example: Sector and Arc Calculations

Problem: A circular lawn sprinkler rotates through a central angle of $120^\circ$ with a spray reach (radius) of $12\text{ feet}$. Find the exact area watered and the outer curved boundary length.

• Step 1: Determine the circle fraction: $\frac{\theta}{360^\circ} = \frac{120^\circ}{360^\circ} = \frac{1}{3}$
• Step 2: Calculate Sector Area: $A_{\text{sector}} = \frac{1}{3} \times \pi (12)^2 = \frac{1}{3} \times 144\pi = 48\pi\text{ sq ft} \approx 48 \times 3.1416 = 150.8\text{ sq ft}$
• Step 3: Calculate Arc Length: $s = \frac{1}{3} \times 2\pi(12) = \frac{1}{3} \times 24\pi = 8\pi\text{ ft} \approx 25.13\text{ ft}$

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3. Two-Dimensional Polygon Area Formulas

Area ($A$) is a two-dimensional measure of the surface region enclosed by a geometric figure, expressed in square units ($\text{in}^2$, $\text{ft}^2$, $\text{cm}^2$, $\text{m}^2$, acres).

+---------------------------------------------------------------------------------------------------+
|                                 POLYGON AREA FORMULAS REFERENCE                                   |
|                                                                                                   |
|   +-------------------+-----------------------+-----------------------------------------------+   |
|   | POLYGON TYPE      | FORMULA               | CRITICAL APPLICATION RULES                    |   |
|   +-------------------+-----------------------+-----------------------------------------------+   |
|   | Rectangle         | A = l * w             | Length * Width; perpendicular sides.          |   |
|   | Square            | A = s^2 = (1/2)d^2    | Side squared, or half the diagonal squared.   |   |
|   | Parallelogram     | A = b * h             | Base * PERPENDICULAR height (never slant side)|   |
|   | Triangle          | A = (1/2) * b * h     | Half base * perpendicular altitude.           |   |
|   | Trapezoid         | A = (1/2)(b_1 + b_2)h | Average of parallel bases * perpendicular h.  |   |
|   | Rhombus / Kite    | A = (1/2) * d_1 * d_2 | Half product of perpendicular diagonals.      |   |
|   | Equilateral Tri.  | A = (√3 / 4) * s^2    | Special case where all sides = s.             |   |
|   +-------------------+-----------------------+-----------------------------------------------+   |
+---------------------------------------------------------------------------------------------------+

The Perpendicular Height Requirement

A frequent error on geometry exams is multiplying the base by the slant side length instead of the perpendicular altitude ($h$). In both triangles and parallelograms, $h$ must form a $90^\circ$ right angle with the base line.

• If a parallelogram has base $b = 10\text{ cm}$, side slant length $s = 6\text{ cm}$, and perpendicular height $h = 5\text{ cm}$:
  • Correct Area: $A = b \times h = 10 \times 5 = 50\text{ cm}^2$.
  • Incorrect Error: $10 \times 6 = 60\text{ cm}^2$.

The Trapezoid Area Derivation

A trapezoid has two parallel bases, $b_1$ and $b_2$, and a perpendicular height $h$. The formula $A = \frac{1}{2}(b_1 + b_2)h$ can be conceptualized as multiplying the average base width by the height: $A = \left(\frac{b_1 + b_2}{2}\right) \times h$

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4. Complex Composite Figures: Additive & Subtraction Methods

Real-world architectural and classroom diagrams often feature irregular composite shapes. Two reliable methods resolve these figures:

Method 1: Additive Decomposition (Partitioning)

Divide the irregular shape into non-overlapping standard sub-shapes (rectangles, triangles, semicircles), compute each individual area, and sum them.

          <------- 14 m ------->
         +----------------------+      
         |                      | 4 m  
         |     Rectangle A      |      
         +------------+---------+      
         |            | 6 m            
         |Rectangle B |                
         |            |                
         +------------+                
            6 m                        

Worked Calculation:

• Rectangle A (Top): $\text{Area}_A = 14\text{ m} \times 4\text{ m} = 56\text{ m}^2$.
• Rectangle B (Bottom): $\text{Area}_B = 6\text{ m} \times 6\text{ m} = 36\text{ m}^2$.
• Total Composite Area: $56 + 36 = 92\text{ m}^2$.

Method 2: Subtraction Method (Negative Space / Shaded Regions)

Calculate the area of a large enclosing bounding shape, then subtract the unshaded cutouts or interior voids.

Worked Example (Shaded Border / Annulus): A rectangular school courtyard measures $30\text{ ft}$ by $20\text{ ft}$. In the center, a concrete patio measuring $18\text{ ft}$ by $12\text{ ft}$ is installed. The remaining perimeter area is planted with grass. What is the grass area?

• Total Enclosing Area: $A_{\text{total}} = 30 \times 20 = 600\text{ sq ft}$.
• Interior Concrete Patio Area: $A_{\text{patio}} = 18 \times 12 = 216\text{ sq ft}$.
• Grass (Shaded) Area: $A_{\text{grass}} = 600 - 216 = 384\text{ sq ft}$.

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5. Scale Drawings, Blueprints & The Area-Scaling Theorem

A scale drawing represents a real-world object proportionally reduced or enlarged. The scale ratio expresses the relationship between drawing distance and actual physical distance (e.g., $1\text{ in} : 5\text{ ft}$ or $1 : 500$).

Linear Scaling vs. Area Scaling Theorem

This is one of the most heavily tested geometric reasoning concepts on the WEST-B:

+---------------------------------------------------------------------------------------------------+
|                                 THE DIMENSIONAL SCALING PRINCIPLE                                 |
|                                                                                                   |
|   If every linear dimension of an object is multiplied by a linear scale factor of k:             |
|                                                                                                   |
|   1. LINEAR MEASURES (Length, Width, Height, Perimeter, Circumference) scale by:                 |
|      --> Linear Dimension_new = k * Linear Dimension_old                                          |
|      --> Perimeter_new = k * Perimeter_old                                                        |
|                                                                                                   |
|   2. TWO-DIMENSIONAL MEASURES (Surface Area, Base Area, Cross-Section Area) scale by:             |
|      --> Area_new = k^2 * Area_old                                                                |
|                                                                                                   |
|   3. THREE-DIMENSIONAL MEASURES (Volume, Liquid Capacity) scale by:                               |
|      --> Volume_new = k^3 * Volume_old                                                            |
+---------------------------------------------------------------------------------------------------+

Step-by-Step Blueprint Calculation

Problem: On an architectural blueprint, the scale is $1\text{ inch} = 6\text{ feet}$. A rectangular science laboratory on the blueprint measures $4\text{ inches}$ wide by $7\text{ inches}$ long.

1. Find the actual perimeter of the laboratory in feet.
2. Find the actual floor area of the laboratory in square feet.

Solution:

• Method A: Convert Linear Dimensions First (Recommended):

  • $\text{Actual Width} = 4\text{ in} \times 6\text{ ft/in} = 24\text{ ft}$
  • $\text{Actual Length} = 7\text{ in} \times 6\text{ ft/in} = 42\text{ ft}$
  • $\text{Actual Perimeter} = 2(24 + 42) = 2(66) = 132\text{ ft}$
  • $\text{Actual Area} = 24\text{ ft} \times 42\text{ ft} = 1,008\text{ sq ft}$

• Method B: Apply the Quadratic Area-Scaling Factor ($k^2$):

  • Drawing area $= 4\text{ in} \times 7\text{ in} = 28\text{ sq in}$
  • Since $1\text{ in} = 6\text{ ft}$, $1\text{ sq in} = (6\text{ ft})^2 = 36\text{ sq ft}$
  • $\text{Actual Area} = 28 \times 36 = 1,008\text{ sq ft}$

Critical Warning: Never multiply the drawing area ($28\text{ sq in}$) by the linear scale factor ($6$). Multiplying $28 \times 6 = 168\text{ sq ft}$ is a classic exam distractor error!