Geometry Basics – Quantitative Aptitude
(Coast Guard Program – Navigation & Surveying)
1. Simple Explanation
Geometry is the branch of mathematics that deals with shapes, sizes, positions, and dimensions of objects. In the Coast Guard context, geometry helps you:
- Measure distances between points (ships, buoys, coastlines).
- Determine angles for navigation (bearing, heading).
- Calculate areas of lakes, harbours, or protected zones.
- Work with circles (radial distances, waves).
All these skills are built on a few basic concepts: points, lines, angles, shapes, and the formulas that link their attributes.
2. Key Points
| Concept | Definition | Important Formula(s) |
|---|---|---|
| Point | Exact location with no size. | – |
| Line | Infinite set of points extending in two directions. | – |
| Line Segment | Part of a line between two endpoints. | Length = ` |
| Ray | One endpoint and extends infinitely in one direction. | – |
| Angle | Formed by two rays sharing a common endpoint (vertex). | Measure in degrees or radians |
| Triangle | Three sides, three angles. | Area = ½ × base × height (base × height) |
| Quadrilateral | Four sides. | Area (rectangle) = length × width; parallelogram = base × height |
| Circle | Set of points at the same distance (radius) from center. | Circumference = 2πr |
| Perimeter | Sum of all side lengths. | P = Σ side lengths |
| Coordinate Geometry | Uses (x, y) to locate points. | Distance = √[(x₂–x₁)² + (y₂–y₁)²] |
| Midpoint | Point exactly halfway between two points. | ( (x₁+x₂)/2 , (y₁+y₂)/2 ) |
| Slope | Rate of rise over run. | m = (y₂–y₁)/(x₂–x₁) |
| Angle Sum | Sum of interior angles in a triangle = 180°. | – |
| Pythagoras | In a right‑angled triangle, a² + b² = c². | – |
3. Worked Example
Problem:
A rectangular harbour has a length of 240 m and a width of 120 m. A lighthouse is positioned at one corner (point A). A ship is 80 m from the lighthouse along the length side and 60 m from the lighthouse along the width side.
- Find the distance from the ship to the opposite corner (point C).
- Calculate the area of the harbour.
- Determine the bearing (angle) from the ship to the opposite corner.
Solution
| Step | What to do | Calculation | Result |
|---|---|---|---|
| 1. Coordinates | Place A at (0, 0). | – | – |
| Ship S at (80, 60). | – | – | |
| Opposite corner C at (240, 120). | – | – | |
| 2. Distance (SC) | Use distance formula. | √[(240–80)² + (120–60)²] = √[(160)² + (60)²] | √(25 600 + 3 600) = √29 200 = 171 m |
| 3. Harbour area | Rectangle area = length × width. | 240 m × 120 m | 28 800 m² |
| 4. Bearing from S to C | Find slope, then angle to horizontal. | m = (120–60)/(240–80) = 60/160 = 0.375 | θ = arctan(0.375) ≈ 20.6° above east |
| Bearing from east counter‑clockwise = 20.6° N of E. | – | 20.6° N E |
Answers
- Distance (SC) ≈ 171 m
- Harbour area ≈ 28 800 m²
- Bearing ≈ 20.6° N E
4. Quick Revision Summary
- Perimeter = sum of all sides.
- Area of:
- Rectangle = length × width.
- Triangle = ½ × base × height.
- Circle = π r².
- Parallelogram = base × height.
- Circle Circumference = 2πr.
- Distance (Coordinate) = √[(Δx)² + (Δy)²].
- Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2).
- Slope = Δy/Δx; angle = arctan(slope).
- Pythagoras: a² + b² = c² (right‑angled triangles).
- Angle Sum: Triangle ≈ 180°.
Tip for the Coast Guard – Always convert bearings to angles relative to the cardinal directions (N, E, S, W) when plotting courses or interpreting maps.
Use these fundamentals to solve navigation, surveying, and safety‑related geometry problems quickly and accurately.