4.2 Plane Geometry, Perimeter, Area & Mensuration

4.2 Plane Geometry, Perimeter, Area & Mensuration

Spatial awareness and geometric mensuration are essential core capabilities for military recruits. Service personnel regularly confront spatial tasks in garrison and field deployments: setting out perimeter fencing, surveying obstacle courses, establishing defensive camp footprints, estimating excavation volumes, and computing liquid fuel or water capacities. Working by hand under time pressure, candidates must rapidly deploy angle relationships, two-dimensional perimeter and area formulas, three-dimensional volume equations, and Pythagorean shortcuts.

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Fundamental Geometric Angle Rules

Angle theorems allow recruits to determine bearings, line-of-sight angles, and spatial positions from given geometric data.

Intersecting Lines and Straight Lines

• Complementary Angles: Sum to $90^\circ$ ($a + b = 90^\circ$).
• Supplementary Angles: Sum to $180^\circ$ ($a + b = 180^\circ$).
• Angles on a Straight Line: Adjacent angles forming a straight line sum to $180^\circ$.
• Angles at a Point: Angles surrounding a common vertex sum to $360^\circ$.
• Vertically Opposite Angles: Non-adjacent opposite angles formed by intersecting lines are equal.

Parallel Lines and Transversals

When a transversal intersects two parallel lines:

1. Alternate Interior Angles (Z-angles): Angles on opposite sides of the transversal between the parallels are equal ($a = b$).
2. Corresponding Angles (F-angles): Angles occupying matching positions at each intersection are equal ($c = d$).
3. Co-Interior Angles (C-angles): Interior angles on the same side of the transversal sum to $180^\circ$:

$$\alpha + \beta = 180^\circ$$

Polygon Interior Angle Sums

• Triangles: Interior angles always sum to $180^\circ$ ($\angle A + \angle B + \angle C = 180^\circ$). The exterior angle equals the sum of the two opposite interior angles ($\angle \text{ext} = \angle A + \angle B$).
• Quadrilaterals: Interior angles sum to $360^\circ$.
• General $n$-sided Polygons: Interior angles sum to $(n - 2) \times 180^\circ$.

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Two-Dimensional Perimeter and Area Formulas

Perimeter ($P$) is the linear boundary distance around a figure; area ($A$) is the two-dimensional surface enclosed within.

Standard Formulas

Hand-Calculation Strategies for Circles

Exam problems avoid tedious decimal arithmetic by choosing convenient dimensions:

• If radius or diameter is a multiple of $7$ ($7$, $14$, $21$, $28$), use $\pi \approx \frac{22}{7}$:

$$A = \frac{22}{7} \times 14^2 = \frac{22}{7} \times 196 = 22 \times 28 = 616\text{ m}^2$$

• If dimensions involve powers of $10$ or decimals, use $\pi \approx 3.14$:

$$C = 2 \times 3.14 \times 10 = 62.8\text{ m}$$

• Semicircles: Area is $\frac{1}{2}\pi r^2$. The perimeter of a closed semicircle includes the straight diameter: $P = \pi r + 2r$.

Composite Shapes

Break complex ground footprints into elementary rectangles and triangles. Sum individual areas to find total area. For perimeter, sum only exposed external boundary lines.

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Pythagoras' Theorem and Pythagorean Triples

In any right-angled triangle, the square of the hypotenuse ($c$) equals the sum of the squares of the perpendicular legs ($a$ and $b$):

$$a^2 + b^2 = c^2$$

Primitive Triples and Common Multiples

Memorizing primitive integer triples eliminates manual square-root calculations:

• $(3, 4, 5)$: Common multiples include $(6, 8, 10)$, $(9, 12, 15)$, and $(30, 40, 50)$.
• $(5, 12, 13)$: Common multiples include $(10, 24, 26)$ and $(25, 60, 65)$.
• $(8, 15, 17)$: Common multiple: $(16, 30, 34)$.
• $(7, 24, 25)$: Common multiple: $(14, 48, 50)$.

Tactical Navigation Application

A patrol unit marches $15\text{ km}$ North, turns $90^\circ$, and marches $20\text{ km}$ East.

• Identify the ratio: $15 : 20 = 3 : 4$ (scaling factor $k = 5$).
• Hypotenuse is $5 \times 5 = 25\text{ km}$. The direct return distance is $25\text{ km}$ without extracting square roots.

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Three-Dimensional Mensuration: Volume and Surface Area

Solid geometry governs physical storage capacities, ammunition packing, and bulk fuel storage.

Rectangular Prisms (Cuboids)

• Volume: Space enclosed: $V = l \times w \times h$
• Total Surface Area (TSA): Sum of six faces: $TSA = 2(lw + lh + wh)$

Circular Cylinders

A cylinder of radius $r$ and height $h$ has:

• Volume: $V = \pi r^2 h$
• Curved Surface Area (CSA): $CSA = 2\pi r h$
• Total Surface Area (TSA): $TSA = 2\pi r h + 2\pi r^2 = 2\pi r(h + r)$

Capacity Conversions

• $1\text{ cubic meter } (\text{m}^3) = 1,000\text{ litres}$
• $1\text{ litre} = 1,000\text{ cubic centimeters } (\text{cm}^3)$
• $1\text{ cm}^3 = 1\text{ milliliter } (\text{mL})$

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Worked Examples with Step-by-Step Solutions

Example 1: Parallel Line Angle Problem

Problem: In a security grid, two parallel perimeter lines are cut by an access route. An interior angle is $(4x - 10)^\circ$ and its alternate interior angle is $(2x + 30)^\circ$. Find $x$ and the adjacent co-interior angle.

Solution:

1. Alternate interior angles are equal: $4x - 10 = 2x + 30$.
2. Transpose terms: $2x = 40 \implies x = 20$.
3. Substitute: Angle is $2(20) + 30 = 70^\circ$.
4. Co-interior angles are supplementary: $180^\circ - 70^\circ = 110^\circ$. Final Answer: $x = 20$; the co-interior angle is $110^\circ$.

Example 2: Outpost Cylindrical Water Tank Capacity

Problem: An outpost at Shai Hills installs a cylindrical water reservoir with base diameter $1.4\text{ m}$ and height $3\text{ m}$. Using $\pi = \frac{22}{7}$, calculate liquid capacity in litres. If $44$ soldiers consume $15\text{ litres}$ each per day, how many days will the water last?

Solution:

1. Radius: $r = \frac{1.4}{2} = 0.7\text{ m} = \frac{7}{10}\text{ m}$.
2. Volume: $V = \frac{22}{7} \times \frac{7}{10} \times \frac{7}{10} \times 3 = \frac{462}{100} = 4.62\text{ m}^3$.
3. Litres: $4.62 \times 1,000 = 4,620\text{ litres}$.
4. Daily use: $44 \times 15 = 660\text{ litres/day}$.
5. Duration: $\frac{4,620}{660} = 7\text{ days}$. Final Answer: Holds $4,620\text{ litres}$; lasts $7$ days.