10.1 Perimeter, Area & Composite 2D Figures

Perimeter, Area & Composite 2D Figures

Core Concept: In the Edutest Mathematics subtest, 2D measurement questions test far more than basic formula recall. With only about 30 to 45 seconds per question and no calculator, candidates must instantly identify perpendicular altitudes, execute fraction-based circular approximations, account for straight radial boundaries in curved perimeters, and rapidly dissect complex shaded composites using additive and subtractive framing.

Measurement problems in scholarship and selective entry examinations frequently combine multiple geometric figures into irregular shapes or shaded regions. The primary trap is not computational complexity, but visual deception: examiners intentionally present oblique side lengths to bait students into using slanted heights instead of perpendicular altitudes, or omit straight boundary edges when calculating perimeters of circular sectors. Achieving a superior score requires systematic structural analysis and rapid mental calculation tricks.

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The 2D Geometric Formula Architecture

Every 2D area formula is derived from the foundational area of a rectangle ($A = l \times w$). Understanding how triangles, parallelograms, trapeziums, and kites relate to this rectangular baseline prevents formula misapplication under exam stress.

[Fundamental Baseline: Rectangle (Area = base × perpendicular height)]
  ├── Parallelogram: Base sheared horizontally  ──► Area = b × h_perp
  ├── Triangle: Exactly half of a parallelogram ──► Area = ½ × b × h_perp
  ├── Trapezium: Average of two parallel bases  ──► Area = ½(a + b) × h_perp
  └── Rhombus / Kite: Halved bounding rectangle ──► Area = ½ × d₁ × d₂

Comprehensive 2D Measurement Reference Table

The Perpendicular Height Invariant

In any triangle, parallelogram, or trapezium, the height ($h$) must be perpendicular ($90^\circ$) to the base. Exam diagrams frequently label a slanted side (e.g., $10\text{ cm}$) and a perpendicular dashed altitude (e.g., $8\text{ cm}$). Multiplying by the slanted side is one of the most common geometry errors. If the perpendicular height is not directly provided, use Pythagoras' theorem ($a^2 + b^2 = c^2$) on an internal right-angled triangle to deduce $h$ before calculating area.

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Circles, Semicircles & Annuli

In a calculator-free exam, circular geometry requires sharp mental arithmetic strategies for the irrational constant $\pi$.

Radius, Diameter, Circumference & Area

• Diameter ($d$) vs Radius ($r$): $d = 2r$ and $r = \frac{d}{2}$. Always double-check whether the problem provides the radius or the diameter.
• Circumference: $C = 2\pi r = \pi d$.
• Area: $A = \pi r^2 = \frac{\pi d^2}{4}$.

Mental Approximation Tactics for $\pi$

1. The $\frac{22}{7}$ Fraction Rule: Whenever the radius or diameter is a multiple of $7$ (e.g., $7, 14, 21, 28, 35$) or a fraction of $7$ (e.g., $3.5, 10.5$), unconditionally substitute $\pi \approx \frac{22}{7}$. This allows immediate cross-cancellation: $A = \pi r^2 = \frac{22}{7} \times 14^2 = \frac{22}{7} \times 196 = 22 \times 28 = 616\text{ cm}^2$
2. The $3.14$ Decimal Benchmark: When dimensions are multiples of 10 or decimals, use $3.14$. Remember key benchmark products:
   • $3.14 \times 2 = 6.28$
   • $3.14 \times 4 = 12.56$
   • $3.14 \times 5 = 15.7$
   • $3.14 \times 8 = 25.12$
   • $3.14 \times 9 = 28.26$
   • $3.14 \times 16 = 50.24$
   • $3.14 \times 25 = 78.5$
3. Symbolic $\pi$ Retention: Inspect the multiple-choice options before calculating! Some multiple-choice questions leave answers in terms of $\pi$ (e.g., $25\pi - 50$). If options contain $\pi$, never convert $\pi$ into decimals.

The Annulus (Concentric Rings)

An annulus is the flat region bounded between two concentric circles with radii $R$ (outer) and $r$ (inner):

$A_{\text{annulus}} = \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = \pi (R - r)(R + r)$

Applying the difference of two squares identity ($(R - r)(R + r)$) turns a tedious two-step squaring calculation into a rapid mental multiplication.

Worked Example: What is the area of a circular track with outer radius $13\text{ m}$ and inner radius $7\text{ m}$? (Leave in terms of $\pi$). $A = \pi (13^2 - 7^2) = \pi (13 - 7)(13 + 7) = \pi (6)(20) = 120\pi\text{ m}^2$

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Circular Sectors & The Boundary Perimeter Trap

A sector is a pie-slice fraction of a circle defined by a central angle $\theta$ (in degrees):

• Fraction of circle: $f = \frac{\theta}{360^\circ}$
• Arc Length ($l$): $l = \frac{\theta}{360^\circ} \times 2\pi r$
• Sector Area ($A$): $A = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{2} l r$

          Circular Sector Boundary Trap
                  ╭───────╮  ◄── Curved Arc: l = (θ/360) × 2πr
                 ╱         ╲
      Radius: r ╱     θ     ╲ Radius: r
               ╱      •      ╲
              ─────────┴───────
    Total Perimeter = Curved Arc + 2 × Radius  (P = l + 2r)
    DO NOT forget the two straight radial edges!

The Universal Perimeter Trap

When asked to find the perimeter of a semicircle, quadrant, or sector, students instinctively calculate only the curved arc length and forget to add the straight bounding edges:

• Perimeter of Semicircle: $P = \pi r + 2r = \pi r + d$ (Curved arc $+ 1$ diameter).
• Perimeter of Quadrant ($\theta = 90^\circ$): $P = \frac{1}{2}\pi r + 2r$ (Curved arc $+ 2$ radii).
• Perimeter of Sector: $P = \frac{\theta}{360^\circ}(2\pi r) + 2r$.

Worked Example: A sector of a circle has radius $14\text{ cm}$ and central angle $45^\circ$. What is the total perimeter of the sector? (Use $\pi \approx \frac{22}{7}$).

• Step 1 (Circle fraction): $\frac{45^\circ}{360^\circ} = \frac{1}{8}$.
• Step 2 (Curved arc length): $l = \frac{1}{8} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{8} \times 88 = 11\text{ cm}$.
• Step 3 (Add straight radial edges): $P = l + 2r = 11 + 2(14) = 11 + 28 = 39\text{ cm}$.
• Distractor Alert: Expect $11\text{ cm}$ to appear as an option for candidates who forget the straight edges.

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Composite Shape Dissection: Additive vs Subtractive

Irregular 2D shapes are resolved through one of two distinct structural paradigms:

1. Additive Decomposition

Partition the shape into non-overlapping elementary polygons whose dimensions are explicitly stated or deducible. Sum the individual areas:

$A_{\text{total}} = A_1 + A_2 + A_3 + \dots$

• Common in L-shaped rooms, T-shaped cross-sections, and penta-gables (rectangle topped by a triangle).
• Rule: Always draw the partition line that minimizes fractional arithmetic. For an L-shape, cut vertically or horizontally to create two whole-number rectangles.

2. Subtractive Framing (The Bounding Box Method)

Enclose the irregular shape inside a minimal bounding rectangle (or circle). Calculate the area of the entire bounding frame, then subtract the empty voids or cutouts:

$A_{\text{figure}} = A_{\text{bounding box}} - \sum A_{\text{voids}}$

• Extremely fast for shapes with missing corners, notched borders, or irregular polygons plotted on coordinate grids.

      Additive Decomposition                 Subtractive Framing
      ┌──────────┬──────────┐               ┌──────────┬ - - - - -┐
      │  Area 1  │  Area 2  │               │          │   Void   │
      │(Triangle)│(Rectangle│               │  Shape   │ (Cutout) │
      ├──────────┴──────────┤               │          │ - - - - -┤
      │       Area 3        │               │          │          │
      │     (Trapezium)     │               └──────────┴──────────┘
      └─────────────────────┘               Total = Bounding Box - Voids

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Shaded Region Canonical Models & Shortcuts

Scholarship exams reuse three classic geometric relationships where circles and squares intersect:

Model 1: Circle Inscribed in a Square

A circle of radius $r$ is inscribed in a square of side length $s = 2r$. The four corners outside the circle are shaded:

• Area of square: $A_{\text{square}} = s^2 = 4r^2$
• Area of circle: $A_{\text{circle}} = \pi r^2 = \frac{\pi}{4} s^2 \approx 0.7854 s^2$
• Shaded Corner Area: $A_{\text{shaded}} = s^2 - \frac{\pi}{4}s^2 = s^2\left(1 - \frac{\pi}{4}\right) \approx 0.2146 s^2$
• Mental Shortcut: The inscribed circle occupies roughly $78.5\%$ of the square; the four corners comprise roughly $21.5\%$.

Model 2: Square Inscribed in a Circle

A square is inscribed in a circle of radius $r$. The square's diagonal equals the circle's diameter ($d = 2r$):

• Area of circle: $A_{\text{circle}} = \pi r^2$
• Area of square: $A_{\text{square}} = \frac{1}{2} d^2 = \frac{1}{2}(2r)^2 = 2r^2$
• Shaded Segment Area (outside square, inside circle): $A_{\text{shaded}} = \pi r^2 - 2r^2 = r^2(\pi - 2) \approx 1.1416 r^2$
• Mental Shortcut: The inscribed square occupies exactly $\frac{2}{\pi} \approx 63.7\%$ of the circle.

Model 3: Overlapping Circular Quadrants (The Lens / Leaf)

Two quarter-circles of radius $r$ are drawn inside a square of side $r$ from opposite corners. The overlapping central lens is shaded:

• Area of each quadrant: $\frac{1}{4}\pi r^2$
• Combined area of two quadrants: $2 \times \frac{1}{4}\pi r^2 = \frac{1}{2}\pi r^2$
• By inclusion-exclusion, the overlapping region equals the sum of the two quadrants minus the square: $A_{\text{lens}} = \frac{1}{2}\pi r^2 - r^2 = r^2\left(\frac{\pi}{2} - 1\right) = \frac{\pi - 2}{2} r^2 \approx 0.5708 r^2$
• Unshaded Corners: $r^2 - A_{\text{lens}} = r^2\left(2 - \frac{\pi}{2}\right) \approx 0.4292 r^2$.

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Measurement Conversions: Length, Mass, Capacity & Time

Edutest lists measurement as one of the five strands of its Mathematics test. Many measurement errors come from units rather than formulas.

24-hour time and elapsed time

• In 24-hour time, 1:45 pm is 13:45 and 12:30 am is 00:30.
• To find elapsed time, count up in friendly steps. From 10:50 am to 2:15 pm: 10:50 to 11:00 is 10 minutes, 11:00 to 2:00 pm is 3 hours, and 2:00 to 2:15 is 15 minutes, so the total is 3 hours 25 minutes.

Australian time zones (standard time)

Australian Eastern Standard Time is UTC+10, Australian Central Standard Time is UTC+9:30 and Australian Western Standard Time is UTC+8. In standard time, Perth is 2 hours behind Sydney, so a 9:00 am Sydney meeting starts at 7:00 am in Perth. Daylight saving changes these gaps in summer, so read each question's stated times carefully.

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High-Yield Traps in 2D Measurement

1. The Slant Height Trap

In a parallelogram with base $12\text{ cm}$, side $8\text{ cm}$, and perpendicular height $6\text{ cm}$, candidates multiply $12 \times 8 = 96\text{ cm}^2$. The correct area is $12 \times 6 = 72\text{ cm}^2$. The side length $8\text{ cm}$ is required only for perimeter ($2(12 + 8) = 40\text{ cm}$).

2. The Linear vs Area Scaling Illusion

If all linear dimensions of a 2D shape are scaled by a factor of $k$:

• The perimeter increases by a factor of $k$.
• The area increases by a factor of $k^2$. If the radius of a circle is tripled ($k = 3$), its circumference triples ($3\times$), but its area increases by $3^2 = 9\times$!

3. Metric Area Unit Conversions

Never convert square units using linear factors:

• $1\text{ m} = 100\text{ cm} \implies 1\text{ m}^2 = 100\text{ cm} \times 100\text{ cm} = 10,000\text{ cm}^2$.
• $1\text{ cm} = 10\text{ mm} \implies 1\text{ cm}^2 = 10\text{ mm} \times 10\text{ mm} = 100\text{ mm}^2$.
• $1\text{ hectare (ha)} = 10,000\text{ m}^2$ (a square of $100\text{ m} \times 100\text{ m}$).