📚 Delhi PSC – Quantitative Aptitude
Geometry Basics – Study Notes
Why Geometry?
In the Delhi PSC exam, geometry questions test your ability to apply simple formulas, recognise patterns, and perform quick mental calculations. Mastering the “basics” will help you tackle most of the geometry problems in the test.
1️⃣ Simple Explanation
Geometry is the branch of mathematics that deals with points, lines, angles, shapes, and the space they occupy. In the PSC exam you’ll mainly encounter:
- Basic shapes (triangles, quadrilaterals, circles, polygons).
- Angle relationships (acute, obtuse, right, supplementary, complementary, vertical).
- Perimeter / Area formulas for common shapes.
- Coordinate geometry (distance, mid‑point, slope).
All these concepts are usually combined in simple, “plug‑and‑play” problems.
2️⃣ Key Points
| Concept | What to Know | Formula / Property |
|---|---|---|
| Points, Lines & Planes | Points have no dimension. Lines are infinite in one dimension. Planes are two‑dimensional. | – |
| Angles | Sum of angles in a triangle = 180°. In a circle, a central angle subtends an arc equal to its measure (in degrees). | – |
| Types of Angles | <90° = acute, 90° = right, >90° = obtuse, 180° = straight. | – |
| Triangles | - Equilateral (all sides & angles 60°). - Isosceles (two equal sides, base angles equal). - Scalene (all sides & angles different). - Right triangle (one 90° angle). |
Pythagoras: a² + b² = c² (right triangles) |
| Quadrilaterals | - Rectangle (opposite sides equal, all angles 90°). - Square (all sides equal, all angles 90°). - Parallelogram (opposite sides equal & parallel). - Rhombus (all sides equal, opposite angles equal). |
– |
| Circle | Radius (r), Diameter (d = 2r), Circumference (C = πd), Area (A = πr²). | – |
| Perimeter | Sum of all side lengths of a closed shape. | – |
| Area | Depends on shape (see below). | – |
| Coordinate Geometry (Basics) | Distance between (x₁, y₁) & (x₂, y₂): √[(x₂‑x₁)² + (y₂‑y₁)²]. Mid‑point: ((x₁+x₂)/2, (y₁+y₂)/2). |
– |
Tip: Most PSC geometry questions require only a single formula. Memorise the few most common ones (triangle area, rectangle area, circle area, distance formula).
3️⃣ Worked Example
Question (PSC style):
A rectangular plot has a length of 30 m and a width of 12 m. A square garden of side x m is cut out from one corner of the rectangle. The remaining area of the plot is 240 m². Find the side length of the square garden.
Step‑by‑Step
| Step | Explanation | Calculation |
|---|---|---|
| 1. | Compute total area of the rectangle. | 30 × 12 = 360 m² |
| 2. | Set up equation: remaining area = total area – area of square. | 360 – x² = 240 |
| 3. | Solve for x². |
x² = 360 – 240 = 120 |
| 4. | Take square root. | x = √120 ≈ 10.95 m |
Answer: The square garden’s side ≈ 10.95 m (or 11 m if rounding to nearest whole number).
Why it works: We used the area formula for a rectangle and a square, and basic algebra to isolate the unknown.
4️⃣ Quick Revision Summary
| Topic | Formula / Key Idea | Quick Check |
|---|---|---|
| Triangle Area | A = ½ × base × height |
Remember: “½” factor! |
| Rectangle Area | A = length × width |
Same as “Perimeter/Area” in many problems. |
| Square Area | A = side² |
Easy to compute quickly. |
| Parallelogram Area | A = base × height |
Height may be given or deduced. |
| Circle Area | A = πr² |
Use π = 3.14 (or 22/7 for exact). |
| Circle Circumference | C = 2πr or πd |
Often needed for “distance around a circle”. |
| Pythagoras | a² + b² = c² |
Right‑triangle check. |
| Distance (Coordinate) | √[(x₂‑x₁)² + (y₂‑y₁)²] |
Common in “find distance between two points” problems. |
| Mid‑point | ((x₁+x₂)/2 , (y₁+y₂)/2) |
Handy for symmetry questions. |
| Angle Sum in Triangle | 180° |
Quick sanity check on any triangle problem. |
| Supplementary Angles | `θ + |