Geometry Basics – Quantitative Aptitude
(For Dadra & Nagar Haveli & Daman & Diu Police Constable Exam)
Why Geometry?
Geometry questions test your quick mental calculations and your ability to apply basic formulas. In the police exam, they usually involve area, perimeter, angles, and simple shapes. Master the fundamentals and you’ll solve most problems in seconds.
1. Simple Explanation
| Concept | What It Means | Quick Visual |
|---|---|---|
| Point | A location in space, no size. | • |
| Line | A straight set of points extending infinitely in both directions. | – |
| Line Segment | Part of a line between two points. | –– |
| Ray | A line segment that extends infinitely in one direction. | –→ |
| Plane | A flat, two‑dimensional surface that extends infinitely. | ⎯⎯ |
| Angle | Formed by two rays sharing a common endpoint (vertex). | ∠ |
| Triangle | Three sides, three angles. | |
| Quadrilateral | Four sides. | |
| Circle | Set of points equidistant from a center. | — |
Tip: When you see a geometry problem, first identify the shape (triangle, rectangle, circle, etc.) and note any given measurements. Then look for the formula that ties the knowns to what you need.
2. Key Points & Formulas
2.1 Angles
| Type | Definition | Formula / Property |
|---|---|---|
| Complementary | Two angles adding up to 90°. | θ₁ + θ₂ = 90° |
| Supplementary | Two angles adding up to 180°. | θ₁ + θ₂ = 180° |
| Vertical Angles | Opposite angles formed by two intersecting lines. | Equal |
| Alternate Interior/Exterior | In parallel lines cut by a transversal. | Equal |
| Sum of Interior Angles | Triangle = 180°, Quadrilateral = 360° |
2.2 Triangles
| Property | Formula |
|---|---|
| Area | ½ × base × height |
| Perimeter | Sum of all sides, (P = a + b + c) |
| Inradius (r) | (r = \frac{A}{s}) where (s = \frac{P}{2}) |
| Circumradius (R) | (R = \frac{abc}{4A}) |
2.3 Quadrilaterals
| Shape | Key Properties | Area Formula |
|---|---|---|
| Rectangle | Opposite sides equal, all angles 90° | (A = l \times w) |
| Square | All sides equal, all angles 90° | (A = a^2) |
| Parallelogram | Opposite sides equal, opposite angles equal | (A = \text{base} \times \text{height}) |
| Rhombus | All sides equal, diagonals perpendicular | (A = \frac{d_1 \times d_2}{2}) |
| Trapezium (US)/Trapezoid (UK) | One pair of parallel sides | (A = \frac{(a + b)}{2} \times h) |
2.4 Circles
| Quantity | Formula |
|---|---|
| Circumference | (C = 2\pi r) |
| Area | (A = \pi r^2) |
| Arc length (θ in degrees) | (L = \frac{θ}{360°} \times C) |
| Sector area | (A_{\text{sector}} = \frac{θ}{360°} \times \pi r^2) |
| Chord length (θ in degrees) | (c = 2r \sin\left(\frac{θ}{2}\right)) |
Remember: In exam questions, (\pi) is often approximated as 22/7 or 3.14. Use the value given in the question.
2.5 Common Ratios
| Shape | Side : Height : Hypotenuse (Right‑triangle, 30‑60‑90) | Side : Height : Hypotenuse (45‑45‑90) |
|---|---|---|
| 30‑60‑90 | (1 : \sqrt{3} : 2) | – |
| 45‑45‑90 | (1 : 1 : \sqrt{2}) | – |
3. Worked Example
Question (Typical Exam Style)
A right‑angled triangle has a hypotenuse of 10 cm. One of its acute angles is 30°. Find the length of the other two sides and the area of the triangle.
Step‑by‑Step Solution
-
Identify Triangle Type
It’s a 30‑60‑90 right triangle.
Ratios: (\text{short side : long side : hypotenuse} = 1 : \sqrt{3} : 2). -
Set Up the Ratios
Let short side = (x).
Then:
[ \begin{cases} \text{short side} = x \ \text{long side} = x\sqrt{3} \ \text{hypotenuse} = 2x = 10 \text{ cm} \end{cases} ] -
Solve for (x)
(2x = 10 \Rightarrow x = 5) cm. -
Find the Long Side
(x