Geometry Basics – Study Notes
(For the Puducherry Patwari / Lekhpal Quantitative Aptitude Test)
1. Simple Explanation
Geometry deals with the properties of points, lines, angles, shapes, and space.
In the Patwari exam, you’ll encounter:
| Concept | What it means | Typical use in the exam |
|---|---|---|
| Point | An exact location with no size | Used only to define other items |
| Line | A straight set of points extending infinitely | Helps in describing angles |
| Angle | The space between two rays meeting at a point | Needed for triangles, circles, polygons |
| Triangle | Three-sided polygon | Area, perimeter, Pythagoras, special types |
| Quadrilateral | Four-sided polygon | Perimeter, area (rectangles, squares, parallelograms, trapeziums) |
| Circle | All points equidistant from a centre | Radius, diameter, circumference, area |
| Coordinate Geometry (basic) | Represent points as ((x, y)) | Distance formula, midpoint formula |
2. Key Points & Formulae
| Shape / Concept | Formula(s) | Notes |
|---|---|---|
| Line Segment | Length = ( | x_2-x_1 |
| Distance (between 2 points) | (d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}) | Standard “Pythagoras” in 2‑D |
| Midpoint | (\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)) | Useful in trapezium problems |
| Right‑triangle (Pythagoras) | (a^2+b^2=c^2) | (c) = hypotenuse |
| Area of Triangle | (\frac{1}{2}\times \text{base}\times \text{height}) | Height is perpendicular to base |
| Area of Triangle (Heron) | (\sqrt{s(s-a)(s-b)(s-c)}) where (s=\frac{a+b+c}{2}) | For all triangles |
| Area of Rectangle | (l \times b) | |
| Area of Square | (s^2) | |
| Area of Parallelogram | (\text{base}\times \text{height}) | |
| Area of Trapezium | (\frac{1}{2}\times(\text{sum of parallel sides})\times \text{height}) | |
| Circumference of Circle | (C = 2\pi r) | |
| Area of Circle | (A = \pi r^2) | |
| Diameter | (d = 2r) | |
| Special Triangles | - 30‑60‑90: sides (x:\sqrt{3}x:2x) - 45‑45‑90: sides (x:x:\sqrt{2}x) |
Helpful for quick estimation |
3. Worked Example
Problem (typical of Patwari exam)
A rectangular plot of land has a length of (180,\text{m}) and a breadth of (75,\text{m}).
A straight road of width (4,\text{m}) runs parallel to the breadth and divides the plot into two sub‑plots.
Find the perimeter of the smaller sub‑plot.
Step‑by‑Step Solution
-
Visualise the situation
- The road is parallel to the breadth, so it is a vertical strip (4,\text{m}) wide.
- The plot is divided into two rectangles:
- Smaller sub‑plot: length = (180,\text{m}) (unchanged), breadth = (75,\text{m} - 4,\text{m} = 71,\text{m}).
- Road: length = (180,\text{m}), breadth = (4,\text{m}).
- Larger sub‑plot: length = (180,\text{m}), breadth = (4,\text{m}).
-
Calculate perimeter of the smaller sub‑plot
[ P = 2(\text{length} + \text{breadth}) = 2(180 + 71) = 2 \times 251 = 502,\text{m} ] -
Answer
- Perimeter = 502 m
Quick Check
- If the road width were 0, perimeter would be (2(180+75)=510) m.
- Reducing breadth by 4 m reduces perimeter by (2 \times 4 = 8) m → (510 - 8 = 502) m, confirming the calculation.
4. Quick Revision Summary
| Topic | Quick Formula | When to Use |
|---|---|---|
| Triangle Area | (\frac{1}{2}\times\text{base}\times\text{height}) | When height is known |
| Triangle (Heron) | (\sqrt{s(s-a)(s-b)(s-c)}) | When all sides known |
| Rectangle / Square Area | (l\times b) / (s^2) | Simple rectangle or square |
| Parallelogram Area | (\text{base}\times\text{height}) |