8.3 Perimeter, Circumference & Area of 2D Figures

Key Takeaways

  • Perimeter measures the total outer boundary distance of a polygon (linear units), while area measures the 2D surface enclosed inside a shape (square units).
  • Circle circumference is $C = 2\pi r = \pi d$, and circle area is $A = \pi r^2$. On test day, use $\pi \approx 3.14$, $\frac{22}{7}$, or exact $\pi$ notation as indicated.
  • Standard area formulas: Rectangle ($A = lw$), Square ($A = s^2$), Triangle ($A = \frac{1}{2}bh$), Parallelogram ($A = bh$), and Trapezoid ($A = \frac{1}{2}(b_1 + b_2)h$).
  • Altitude/height ($h$) in triangles, parallelograms, and trapezoids must always be measured perpendicular ($90^\circ$) to the base, never along a slanted side.
  • Composite figures are solved by decomposing into simpler geometric shapes (additive method) or by subtracting inner unshaded regions from an outer bounding area (subtractive method).
Last updated: September 2026

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.


Perimeter & Circumference Fundamentals

  • Perimeter ($P$): The total distance around the exterior boundary of a polygon. Calculated by summing the lengths of all outer sides. Units: inches (in), feet (ft), meters (m), centimeters (cm)\text{Units: inches (in), feet (ft), meters (m), centimeters (cm)}
  • Circumference ($C$): The perimeter of a circle. C=2πr=πdC = 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
              •──────────•       └───────────────┘

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

Geometric FigurePerimeter / Circumference FormulaArea FormulaKey Measurement Rules
Square$P = 4s$$A = s^2$All 4 sides $s$ are congruent and meet at $90^\circ$.
Rectangle$P = 2l + 2w = 2(l + w)$$A = l \cdot w$Opposite sides are equal; angles are $90^\circ$.
Triangle$P = s_1 + s_2 + s_3$$A = \frac{1}{2} b h$Height $h$ is perpendicular (at $90^\circ$) to base $b$.
Parallelogram$P = 2a + 2b$$A = b \cdot h$Never use the slant side as height; use perpendicular altitude $h$.
Trapezoid$P = s_1 + s_2 + b_1 + b_2$$A = \frac{1}{2}(b_1 + b_2)h$$b_1$ and $b_2$ are parallel bases; $h$ is perpendicular distance between them.
Circle$C = 2\pi r = \pi d$$A = \pi r^2$If given diameter $d$, divide by $2$ to get radius $r$ before squaring!
SemicircleBoundary $= \pi r + 2r$$A = \frac{1}{2}\pi r^2$Half the area of a full circle.
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2D Measurement & Composite Figure Strategy Pathway

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: Ashaded=AtotalAunshadedA_{\text{shaded}} = A_{\text{total}} - A_{\text{unshaded}}

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=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)h A=12(14+22)(9)=12(36)(9)=18×9=162 sq ftA = \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 $162\text{ square feet}$.

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: Asquare=s2=202=400 sq ftA_{\text{square}} = s^2 = 20^2 = 400\text{ sq ft}
  2. Area of circular pool:
    • Diameter $d = 12\text{ ft} \implies \text{radius } r = \frac{12}{2} = 6\text{ ft}$. Acircle=πr23.14×(6)2=3.14×36=113.04 sq ftA_{\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: Aexposed deck=400113.04=286.96 sq ftA_{\text{exposed deck}} = 400 - 113.04 = 286.96\text{ sq ft}
  4. Conclusion: The exposed deck area is $286.96\text{ square feet}$.

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 ($h$).
Test Your Knowledge

A concrete patio is shaped like a trapezoid with parallel bases measuring 14 feet and 22 feet, and a perpendicular height of 9 feet. What is the total surface area of the patio?

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Test Your Knowledge

A circular pool with a diameter of 12 feet is installed in the center of a square wooden deck measuring 20 feet on each side. What is the area of the remaining exposed wooden deck in square feet? (Use pi = 3.14)

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Test Your Knowledge

A decorative garden plot is in the shape of a rectangle surmounted by a semicircle on one of its shorter ends. The rectangle is 12 meters long and 6 meters wide. The diameter of the attached semicircle equals the 6-meter width of the rectangle. What is the total perimeter (outer boundary length) of the garden plot? (Use pi = 3.14)

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Test Your Knowledge

A triangle has a base that is 4 inches longer than its perpendicular height. If the total area of the triangle is 48 square inches, what is the length of the base?

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