11.3 Area Formulas & Area of Composite Figures

11.3 Area Formulas & Area of Composite Figures

Area calculations represent one of the most heavily emphasized topics on the TABE 13&14 Mathematics assessment. Real-world adult math problems—from estimating building materials like drywall, flooring, and siding to calculating agricultural acreage and commercial paint requirements—rely on applying core 2D area formulas and decomposing complex composite plans into manageable geometric shapes.

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The Concept of 2D Area & Square Units

Area ($A$) is the amount of two-dimensional surface space enclosed within the interior boundary of a flat plane figure. While perimeter measures linear length along an edge, area measures square units—the number of unit squares (such as $1\text{ ft} \times 1\text{ ft}$) required to pave the surface completely without gaps or overlaps.

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Master Reference: Core 2D Area Formulas

2D Geometric Shapes & Area Dimensions:
Square:            Rectangle:           Parallelogram:       Triangle:           Trapezoid:
  +-----+            +---------+          /---------/          /\                  +-------+
  |     | s          |         | w       /    |h   /          /  \ h              /    |h   \
  +-----+            +---------+        /_____|___/          /____\              /_____|_____\
     s                    l                  b                  b                  b₂
  A = s²               A = l · w          A = b · h          A = ½ b · h         A = ½(b₁ + b₂)h

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Critical Geometric Rules & Traps to Avoid

1. The Perpendicular Height Rule (Parallelograms & Triangles)

On standardized tests, figures often display both the perpendicular height ($h$) and the slant side length ($s$). You must always multiply the base by the line forming a $90^\circ$ angle with the base. The slant side is a distractor for area calculations.

Worked Example: A parallelogram has a base of $16\text{ cm}$, a slant side length of $10\text{ cm}$, and a vertical height of $8\text{ cm}$ perpendicular to the base. $A = b \times h = 16\text{ cm} \times 8\text{ cm} = \mathbf{128\text{ cm}^2} \quad (\text{ignoring the } 10\text{ cm slant side})$

2. Triangle Height Variations

• Right Triangle: The two legs form a $90^\circ$ angle and serve directly as base and height ($A = \frac{1}{2} a b$).
• Obtuse Triangle: The perpendicular altitude is often dropped outside the triangle's base to meet an extended base line.

3. The Diameter Squaring Trap (Circles)

When a problem provides a diameter ($d = 12\text{ ft}$), you must divide by $2$ to find the radius ($r = 6\text{ ft}$) before squaring: $\text{Correct: } A = \pi (6)^2 = 36\pi \approx 113.04\text{ sq ft} \qquad | \qquad \text{Trap: } \pi (12)^2 = 144\pi \approx 452.16\text{ sq ft } (4\times \text{ too large!})$

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Strategies for Composite 2D Figures

Real-world architectural and technical drawings rarely consist of single basic shapes. Instead, they are composite figures formed by combining two or more standard polygons and curves. Two core strategies solve all composite area problems:

Composite Area Strategies:
1. Decomposition Method (Additive):         2. Subtraction Method (Cutout / Negative Space):
   +------------+                           +-----------------------+
   | Area 1     |                           | Total Bounding Area   |
   | (Triangle) |                           |       +-------+       |
   +------------+                           |       | Hole  |       |
   | Area 2     |                           |       +-------+       |
   | (Rectangle)|                           +-----------------------+
   Total = Area 1 + Area 2                  Net Area = Bounding Area - Cutout Area

Strategy 1: The Decomposition Method (Additive Strategy)

1. Draw dashed construction lines to divide the irregular shape into non-overlapping standard geometric figures (rectangles, triangles, trapezoids, semicircles).
2. Calculate missing dimensions for each sub-shape using parallel opposite side geometry.
3. Compute the area of each individual sub-shape.
4. Sum all individual areas: $A_{\text{total}} = A_1 + A_2 + A_3 + \dots + A_k$

Strategy 2: The Subtraction Method (Cutout / Negative Space Strategy)

1. Calculate the gross area of the entire outer bounding geometric figure (e.g., the complete rectangular wall or floor).
2. Calculate the area of all interior empty openings or cutouts (e.g., doors, windows, skylights, unpaved center courtyards, or pipe holes).
3. Subtract the cutout areas from the gross bounding area: $A_{\text{net}} = A_{\text{gross}} - \sum A_{\text{cutouts}}$

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Step-by-Step Worked Examples

Worked Example 1: Additive Decomposition (Gym Key / Athletic Area)

Problem: A basketball court key consists of a rectangle measuring $12\text{ ft}$ wide by $19\text{ ft}$ long, capped at the top by a semicircular free-throw area whose diameter matches the $12\text{ ft}$ width of the rectangle. Using $\pi \approx 3.14$, find the total area of the key.

1. Area of Rectangle: $A_{\text{rect}} = l \times w = 19\text{ ft} \times 12\text{ ft} = 228\text{ sq ft}$
2. Radius of Semicircle: $r = \frac{d}{2} = \frac{12}{2} = 6\text{ ft}$
3. Area of Semicircle (Half of Circle Area): $A_{\text{semi}} = \frac{1}{2}\pi r^2 = \frac{1}{2}(3.14)(6^2) = \frac{1}{2}(3.14)(36) = 18 \times 3.14 = 56.52\text{ sq ft}$
4. Total Combined Area: $A_{\text{total}} = 228 + 56.52 = \mathbf{284.52\text{ sq ft}}$

Worked Example 2: Negative Space Subtraction (Commercial Wall Painting)

Problem: A painter must paint an exterior commercial building wall measuring $28\text{ ft}$ wide by $12\text{ ft}$ high. The wall contains two identical display windows measuring $5\text{ ft} \times 4\text{ ft}$ each and one double-entry door measuring $6\text{ ft} \times 7\text{ ft}$. One gallon of exterior primer covers $300\text{ sq ft}$. How many square feet of wall must be painted, and how many whole gallons of primer must be purchased?

1. Gross Wall Area: $A_{\text{gross}} = 28\text{ ft} \times 12\text{ ft} = 336\text{ sq ft}$
2. Area of Openings (Cutouts): $\text{Windows Area} = 2 \times (5\text{ ft} \times 4\text{ ft}) = 2 \times 20 = 40\text{ sq ft}$ $\text{Door Area} = 6\text{ ft} \times 7\text{ ft} = 42\text{ sq ft}$ $\text{Total Cutout Area} = 40 + 42 = 82\text{ sq ft}$
3. Net Paintable Surface Area: $A_{\text{net}} = 336 - 82 = \mathbf{254\text{ sq ft}}$
4. Gallons of Primer Needed: $\text{Gallons} = \frac{254\text{ sq ft}}{300\text{ sq ft/gal}} = 0.8467\text{ gal} \implies \mathbf{1\text{ whole gallon}}$

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Practical Trade & Agricultural Applications

Flooring & Tile Waste Allowance

When purchasing ceramic tile, hardwood planks, or laminate flooring, contractors always add a $10%$ waste allowance to cover cuts, trimmings, and breakage: $\text{Total Material Ordered} = A_{\text{floor}} \times 1.10$ Example: If a room area is $450\text{ sq ft}$, order $450 \times 1.10 = 495\text{ sq ft}$ of flooring.

Agricultural Seed & Turf Coverage per Acre

For agricultural landscaping problems involving acres, convert total square footage using $1\text{ acre} = 43,560\text{ sq ft}$: $\text{Number of Acres} = \frac{\text{Total Square Feet}}{43,560}$ Example: A rectangular sod farm parcel measures $660\text{ ft} \times 330\text{ ft} = 217,800\text{ sq ft}$. The acreage is $\frac{217,800}{43,560} = \mathbf{5\text{ acres}}$.