Geometry Basics – Azure Certification Quantitative Aptitude
(Study Notes in Markdown)
1️⃣ Simple Explanation
Geometry is the branch of mathematics that deals with shapes, sizes, positions, and properties of space. For quantitative aptitude, you’ll mainly need to know:
- Basic elements (points, lines, planes).
- Common shapes (triangles, quadrilaterals, circles).
- Key measurements – angles, lengths, area, perimeter, and volume.
- Simple formulas and how to apply them in word‑problems.
2️⃣ Key Points & Formulas
| Concept | Symbol | Formula | Notes |
|---|---|---|---|
| Point | – | – | An exact location, no size. |
| Line | – | – | Infinite in both directions. |
| Line segment | – | – | Finite part of a line with two endpoints. |
| Ray | – | – | Starts at a point and extends infinitely in one direction. |
| Plane | – | – | Flat, 2‑D surface. |
| Angle | ∠ | – | Measured in degrees (°). |
| Right angle | – | 90° | Perpendicular lines. |
| Straight angle | – | 180° | Opposite rays. |
| Complementary angles | – | a + b = 90° | Sum to 90°. |
| Supplementary angles | – | a + b = 180° | Sum to 180°. |
| Triangle | – | – | 3 sides, 3 angles. |
| Triangle perimeter | P | P = a + b + c | Sum of side lengths. |
| Triangle area | A | A = ½ × base × height | Height is perpendicular to base. |
| Pythagoras | – | a² + b² = c² | For right triangles. |
| Quadrilateral | – | – | 4 sides. |
| Rectangle | – | – | Opposite sides equal; all angles 90°. |
| Square | – | – | All sides equal; all angles 90°. |
| Parallelogram | – | – | Opposite sides equal & parallel; opposite angles equal. |
| Rhombus | – | – | All sides equal; diagonals bisect each other at right angles. |
| Parallelogram area | A | A = base × height | Height is perpendicular to base. |
| Trapezium (US) / Trapezoid (UK) | – | – | One pair of parallel sides. |
| Trapezium area | A | A = ½ × (a + b) × h | a,b = lengths of parallel sides; h = height. |
| Circle | – | – | All points equidistant from centre. |
| Radius | r | – | Distance from centre to any point on circle. |
| Diameter | d | d = 2r | Longest chord through centre. |
| Circumference | C | C = πd = 2πr | 3.1416 ≈ π. |
| Circle area | A | A = πr² | |
| Sphere | – | – | 3‑D analogue of a circle. |
| Sphere volume | V | V = ⅔πd³ = 4/3πr³ | |
| Cylinder | – | – | 2 circles + 2 parallel rectangles. |
| Cylinder volume | V | V = πr²h | h = height. |
| Cone | – | – | Circular base + apex. |
| Cone volume | V | V = ⅓πr²h |
Angle Rules
| Property | Statement |
|---|---|
| Angle sum of triangle | a + b + c = 180° |
| Exterior angle | Exterior = sum of two remote interior angles |
| Adjacent angles on a line | a + b = 180° |
| Vertically opposite angles | Equal |
| Corresponding angles | Equal when two lines cut by a transversal are parallel |
3️⃣ Worked Example
Problem
A right‑angled triangle has one leg of length 6 cm and a hypotenuse of length 10 cm.
- Find the length of the other leg.
- Find the area and perimeter.
- If the triangle is doubled in size (each side multiplied by 2), what are the new area and perimeter?
Solution
1. Find the missing leg
Use Pythagoras:
(a^2 + b^2 = c^2)
Let (a = 6) cm, (c = 10) cm, solve for (b):
[ b^2 = c^2 - a^2 = 10^2 - 6^2 = 100 - 36 = 64 \ b = \sqrt{64} = 8 \text{ cm} ]
2. Area & Perimeter
- Area:
[ A = \frac{1}{2} a b = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2 ] - Perimeter:
[ P = a + b + c = 6 + 8 + 10 = 24 \text{ cm} ]
3. Scaled triangle (scale factor 2)
All side lengths double:
-
New legs: (6 \times 2 = 12) cm, (8 \times 2 = 16) cm
-
New hypotenuse: (10 \times 2 = 20) cm
-
New area (scales as (k^2)):
[ A' = k^2 A = 2^2 \times 24 = 4 \times 24 = 96 \text{ cm}^2 ] -
New perimeter (scales as (k)):
[ P' = k P = 2 \times 24 = 48 \text