8.3 Perimeter, Circumference & Area of 2D Figures

Perimeter, Circumference & Area of 2D Figures

Quick Summary: Two-dimensional measurement questions on the HiSET focus on perimeter (the linear distance around the outside edge of a figure) and area (the number of square units enclosed within a boundary). Circles require specialized formulas for circumference ($C = 2\pi r$) and area ($A = \pi r^2$). When working with triangles, parallelograms, and trapezoids, the height ($h$) must always be perpendicular to the base. Complex shapes are solved by decomposing them into standard sub-shapes or subtracting unshaded areas.

Familiarity with the official HiSET formula sheet and mastery of composite shape decomposition ensure quick, accurate solutions on test day.

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Perimeter & Circumference Fundamentals

• Perimeter ($P$): The total distance around the exterior boundary of a polygon. Calculated by summing the lengths of all outer sides. $$\text{Units: inches (in), feet (ft), meters (m), centimeters (cm)}$$
• Circumference ($C$): The perimeter of a circle. $$C = 2\pi r = \pi d$$ Where $r$ is the radius (center to edge) and $d$ is the diameter ($d = 2r$, passing through the center).

   Circumference and Diameter of a Circle
   
                 ┌───────────────┐
              •──┼───────► r     │  Radius (r) = Half-distance across
             /   │        \      │  Diameter (d) = Full distance across (2r)
            │ ◄──┴─────────►     │  Circumference (C) = 2πr = πd
             \        d   /      │  Area (A) = πr²
              •──────────•       └───────────────┘

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Comprehensive 2D Area Reference Guide

Below is the complete inventory of 2D area formulas provided on or tested by the HiSET Mathematics subtest:

   Standard 2D Geometry Area Formulas Visualized
   
     Rectangle               Triangle                  Parallelogram             Trapezoid
     ┌─────────────┐             /\                         /─────────────/           ┌─────────┐ b₁
     │             │ w          /  \  h                    /│            /           /│         │\
     │             │           /    \                     / │ h         /           / │ h       │ \
     └─────────────┘          /______\                   /__│__________/           /__│_________│__\
            l                    b                            b                         b₂
        A = l · w           A = 1/2 · b · h                A = b · h              A = 1/2 · (b₁ + b₂) · h

Formula Reference Table

Composite and Irregular Figures

A composite figure is a geometric shape composed of two or more basic shapes (such as rectangles, triangles, and semicircles). There are two fundamental strategies for finding the area of composite figures:

   Two Methods for Composite Figures
   
   1. Additive Method (Decomposition)       2. Subtractive Method (Hollow / Shaded)
      ┌───────┐                                ┌────────────────────────┐
      │   A₁  │ (Triangle)                     │                        │ (Outer Rect)
      ├───────┴────────┐                       │      ┌──────────┐      │
      │                │                       │      │ Unshaded │      │
      │       A₂       │ (Rectangle)           │      │    A₂    │      │
      │                │                       │      └──────────┘      │
      └────────────────┘                       └────────────────────────┘
      Total Area = A₁ + A₂                     Shaded Area = A_outer - A_inner

Method 1: The Additive Method (Summing Sub-Shapes)

1. Draw auxiliary lines to divide the irregular shape into familiar polygons.
2. Determine the missing dimensions of each sub-region using parallel side relationships.
3. Compute the individual area of each component.
4. Add the component areas together.

Method 2: The Subtractive Method (Shaded Regions)

1. Calculate the total area of the entire outer bounding figure.
2. Calculate the area of the unshaded or hollow inner shape(s).
3. Subtract the unshaded area from the total area:

$$A_{\text{shaded}} = A_{\text{total}} - A_{\text{unshaded}}$$

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Worked Examples

Worked Example 1: Area of a Trapezoid

A concrete patio is shaped like a trapezoid with parallel bases measuring $14\text{ feet}$ and $22\text{ feet}$, and a perpendicular distance (height) between them of $9\text{ feet}$. What is the total surface area of the patio?

1. Identify given dimensions: $b_1 = 14\text{ ft}, b_2 = 22\text{ ft}, h = 9\text{ ft}$.
2. Apply the trapezoid area formula: $$A = \frac{1}{2}(b_1 + b_2)h$$ $$A = \frac{1}{2}(14 + 22)(9) = \frac{1}{2}(36)(9) = 18 \times 9 = 162\text{ sq ft}$$
3. Conclusion: The patio surface area is $162\text{ square feet}$.

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Worked Example 2: Shaded Deck Area with Circular Cutout

A circular pool with a diameter of $12\text{ feet}$ is centered inside a square wooden deck measuring $20\text{ feet}$ on each side. What is the area of the exposed wooden deck? (Use $\pi \approx 3.14$).

1. Area of square deck: $$A_{\text{square}} = s^2 = 20^2 = 400\text{ sq ft}$$
2. Area of circular pool:
   • Diameter $d = 12\text{ ft} \implies \text{radius } r = \frac{12}{2} = 6\text{ ft}$.
   $$A_{\text{circle}} = \pi r^2 \approx 3.14 \times (6)^2 = 3.14 \times 36 = 113.04\text{ sq ft}$$
3. Subtract inner circle from outer square: $$A_{\text{exposed deck}} = 400 - 113.04 = 286.96\text{ sq ft}$$
4. Conclusion: The exposed deck area is $286.96\text{ square feet}$.

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High-Frequency HiSET Traps & Exam Tips

1. Diameter vs. Radius Confusion: In circle area ($A = \pi r^2$), using the diameter instead of the radius produces an answer that is $4\times$ too large! Always double-check if the problem states "diameter" or "radius".
2. Perimeter of Composite Figures: When calculating the perimeter of a composite figure, only sum the outer exposed edges. Never include internal dividing seams where two shapes touch.
3. Slant Height Trap: In triangles and parallelograms, never multiply the base by the slanted side length. Always use the perpendicular altitude ($h$).