6.2 Mensuration — Perimeter, Area, Surface Area & Volume

6.2 Mensuration — Perimeter, Area, Surface Area & Volume

Mensuration provides the quantitative mathematical framework for physical measurement. In military service, combat engineers, logistics quartermasters, and artillery officers continuously apply mensuration: determining the earthwork volume to be displaced for anti-tank ditches, calculating the capacity of bulk fuel tankers at forward operating bases, computing the canvas area required for tactical field hospitals, and projecting supply storage constraints. Precision in calculating perimeters, surface areas, and volumetric capacities is an operational necessity.

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2D Plane Figures: Perimeter & Area Formulations

Two-dimensional mensuration deals with boundaries (perimeter) and surface enclosures (area) across plane figures.

1. Triangles

• Standard Base-Height Formula: $A = \frac{1}{2} b h$
• Right-Angled Triangle: With perpendicular legs $a$ and $b$: $A = \frac{1}{2} a b$
• Equilateral Triangle: With side length $a$: $h = \frac{\sqrt{3}}{2} a, \quad A = \frac{\sqrt{3}}{4} a^2$
• Trigonometric Area Formula: When two sides $a$ and $b$ and their included angle $\theta$ are known: $A = \frac{1}{2} a b \sin\theta$
• Heron's Formula (Scalene Triangles): When all three side lengths $a$, $b$, and $c$ are known without a measured altitude: $s = \frac{a + b + c}{2} \quad (\text{semi-perimeter})$ $A = \sqrt{s(s - a)(s - b)(s - c)}$

Worked Example — Heron's Formula:
Problem: Calculate the area of a triangular helicopter landing zone with sides measuring 13 m, 14 m, and 15 m.
Step 1: Calculate semi-perimeter s:
s = (13 + 14 + 15) / 2 = 42 / 2 = 21 m.
Step 2: Calculate differences (s - a), (s - b), (s - c):
s - a = 21 - 13 = 8 m
s - b = 21 - 14 = 7 m
s - c = 21 - 15 = 6 m
Step 3: Evaluate product under square root:
A = √[21 × 8 × 7 × 6] = √[(3 × 7) × (8) × (7) × (6)]
  = √[7² × (3 × 8 × 6)] = √[49 × 144] = 7 × 12 = 84 m².
Conclusion: The landing zone surface area is exactly 84 square meters.

2. Quadrilaterals

3. Circles, Annuli, Sectors & Segments

• Circumference and Area of a Circle: $C = 2\pi r = \pi d, \quad A = \pi r^2 = \frac{\pi d^2}{4}$
• Circular Annulus (Ring / Protective Earth Berm): Area between concentric circles of outer radius $R$ and inner radius $r$: $A = \pi(R^2 - r^2) = \pi(R - r)(R + r)$
• Circular Sector: A wedge bounded by two radii and an arc with central angle $\theta$ (in degrees): $\text{Arc Length } L = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\theta \pi r}{180^\circ}$ $\text{Sector Area } A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{2} L r$ $\text{Perimeter of Sector} = L + 2r$
• Circular Segment: The region bounded by a chord and its intercepted arc: $A_{\text{segment}} = A_{\text{sector}} - A_{\Delta} = \left(\frac{\theta}{360^\circ}\pi r^2\right) - \left(\frac{1}{2} r^2 \sin\theta\right)$

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3D Solid Geometry: Surface Area & Volume

Solid geometry deals with the spatial capacity (volume) and boundary enclosure (total surface area and curved/lateral surface area) of three-dimensional objects.

1. Cubes and Rectangular Cuboids

• Cube (Side length $a$): $V = a^3, \quad \text{TSA} = 6a^2, \quad \text{LSA} = 4a^2, \quad \text{Internal Diagonal } d = a\sqrt{3}$
• Rectangular Cuboid (Length $l$, Width $w$, Height $h$): $V = l \times w \times h$ $\text{TSA} = 2(lw + lh + wh)$ $\text{LSA (Area of 4 walls)} = 2h(l + w)$ $\text{Space Diagonal } d = \sqrt{l^2 + w^2 + h^2}$

2. Right Circular Cylinders

A cylinder consists of two parallel circular bases of radius $r$ separated by height $h$:

• Volume ($V$): $V = \pi r^2 h$
• Curved Surface Area (CSA): $\text{CSA} = 2\pi r h$
• Total Surface Area (TSA): $\text{TSA} = 2\pi r h + 2\pi r^2 = 2\pi r(r + h)$
• Hollow Cylinder (Pipes and Gun Barrels): With external radius $R$ and internal bore radius $r$: $V_{\text{material}} = \pi(R^2 - r^2)h, \quad \text{Internal Capacity} = \pi r^2 h$

3. Right Circular Cones

A cone has a circular base of radius $r$, vertical altitude $h$, and slant height $l$:

• Slant Height ($l$): $l = \sqrt{r^2 + h^2}$
• Volume ($V$): $V = \frac{1}{3} \pi r^2 h$
• Curved Surface Area (CSA): $\text{CSA} = \pi r l$
• Total Surface Area (TSA): $\text{TSA} = \pi r l + \pi r^2 = \pi r(l + r)$

4. Spheres and Hemispheres

• Solid Sphere (Radius $r$): $V = \frac{4}{3} \pi r^3, \quad A = 4\pi r^2$
• Solid Hemisphere: $V = \frac{2}{3} \pi r^3$ $\text{Curved Surface Area (CSA)} = 2\pi r^2$ $\text{Total Surface Area (TSA, including flat base)} = 2\pi r^2 + \pi r^2 = 3\pi r^2$

5. Prisms and Regular Pyramids

• Right Prism: A solid with uniform polygonal cross-section of area $A_{\text{base}}$ and length $h$: $V = A_{\text{base}} \times h, \quad \text{LSA} = P_{\text{base}} \times h, \quad \text{TSA} = \text{LSA} + 2A_{\text{base}}$
• Regular Pyramid: A solid with polygonal base and triangular lateral faces meeting at an apex: $V = \frac{1}{3} A_{\text{base}} \times h, \quad \text{LSA} = \frac{1}{2} P_{\text{base}} \times l_{\text{slant}}$

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Composite Solids, Recasting & Liquid Displacement

Many physical engineering structures combine multiple geometric solids.

1. Composite Structures

• Ammunition Bunker / Silo: Composed of a cylinder of radius $r$ and height $h_1$, capped with a conical roof of slant height $l$ and height $h_2$: $V_{\text{total}} = \pi r^2 h_1 + \frac{1}{3} \pi r^2 h_2 = \pi r^2 \left(h_1 + \frac{1}{3}h_2\right)$ $\text{Exposed Surface Area} = 2\pi r h_1 + \pi r l = \pi r(2h_1 + l)$
• Fuel Storage Capsule: A central cylinder of length $h$ with hemispherical ends of radius $r$: $V_{\text{total}} = \pi r^2 h + \frac{4}{3} \pi r^3 = \pi r^2 \left(h + \frac{4}{3}r\right)$ $\text{Surface Area} = 2\pi r h + 4\pi r^2 = 2\pi r(h + 2r)$

2. Recasting and Melting (Conservation of Volume)

When a metal ingot or projectile is melted down and recast into smaller units, total material volume remains strictly invariant: $V_{\text{initial}} = n \times V_{\text{single unit}}$

3. Liquid Displacement (Archimedes' Principle)

When an irregular solid object is submerged into a liquid-filled container (such as a cylindrical testing vessel of radius $r$), the volume of the submerged object equals the volume of the displaced liquid column: $V_{\text{object}} = \Delta V_{\text{liquid}} = A_{\text{container base}} \times \Delta h = \pi r^2 \times \Delta h$

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Dimensional Scaling Principles (Similitude Laws)

In military modeling, cartography, and ballistic blast wave calculations, scaling principles dictate how physical dimensions change relative to one another.

If all linear dimensions of an object scale by a uniform ratio factor $k = \frac{L_2}{L_1}$:

1. Linear Measures (Perimeter, Circumference, Height, Radius): $\frac{P_2}{P_1} = k$
2. Surface Area Measures (Base Area, TSA, CSA, Cross-Section): $\frac{A_2}{A_1} = k^2$
3. Volumetric Measures (Volume, Capacity, Mass at constant density): $\frac{V_2}{V_1} = k^3$

Scaling Rule in Practice:
If the radius of an artillery storage tank is tripled (k = 3) while height remains constant:
Volume scales by 3² = 9 (since V = πr²h depends on r²).
If both radius and height are tripled (uniform 3D scaling, k = 3):
Volume scales by 3³ = 27 times its original capacity!

Map Scales & Physical Area Equivalents

On a military survey map with a Representative Fraction (RF) of $1 : M$ (e.g., $1 : 50,000$):

• Linear Ground Distance $= \text{Map Distance} \times M$
• Ground Area $= \text{Map Area} \times M^2$ $\text{If a cantonment measures } 4\text{ cm by } 5\text{ cm on a } 1 : 50,000\text{ map:}$ $\text{Map Area} = 20\text{ cm}^2$ $\text{Ground Area} = 20 \times (50,000)^2\text{ cm}^2 = 20 \times (2.5 \times 10^9)\text{ cm}^2 = 50 \times 10^9\text{ cm}^2 = 5,000,000\text{ m}^2 = 5.0\text{ km}^2$

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Military Field Engineering Calculations

1. Earthwork Trench Excavation

Infantry anti-tank and defensive ditch profiles are constructed with trapezoidal cross-sections to prevent soil collapse. Given top width $w_{\text{top}}$, bottom width $w_{\text{bottom}}$, depth $d$, and total trench length $L$: $A_{\text{cross}} = \frac{w_{\text{top}} + w_{\text{bottom}}}{2} \times d$ $V_{\text{excavation}} = A_{\text{cross}} \times L = \left(\frac{w_{\text{top}} + w_{\text{bottom}}}{2} \times d\right) \times L$

2. Fuel Depot Capacity & Conversion

Fuel storage requires volumetric conversion between cubic meters and liquid volumes: $1\text{ m}^3 = 1,000\text{ liters}, \quad 1\text{ liter} = 1,000\text{ cm}^3, \quad 1\text{ cm}^3 = 1\text{ mL}$ A forward tank measuring $r = 2\text{ m}$ and $h = 5\text{ m}$ holds: $V = \pi (2)^2 (5) = 20\pi \approx 62.83\text{ m}^3 = 62,830\text{ liters}$ At an assumed consumption rate of 2,500 liters per operational day, this single tank provides $62,830 / 2,500 \approx 25.1$ days of endurance.