Geometry Basics – Study Notes
(For Maharashtra Patwari / Lekhpal – Quantitative Aptitude)
Purpose: Quick reference to the most common geometry concepts needed for the Patwari exam.
Format: Simple explanation → Key points → Worked example → Quick revision summary
1. Simple Explanation
Geometry deals with the properties of points, lines, angles, shapes, and their relationships.
In the Patwari exam you’ll mainly handle:
- Perimeter (or Circumference) – total boundary length of a shape.
- Area – amount of flat surface inside a shape.
- Basic shapes – rectangle, square, triangle, circle, parallelogram, trapezium, etc.
- Simple transformations – scaling, rotation (rarely needed).
Tip: Remember the “Area = (Base × Height)/2” for triangles and “Area = πr²” for circles.
2. Key Points & Formulae
| Shape | Perimeter | Area | Notes |
|---|---|---|---|
| Rectangle | (P = 2(l + w)) | (A = l \times w) | (l) = length, (w) = width |
| Square | (P = 4a) | (A = a^2) | (a) = side |
| Triangle | Depends on sides | (A = \frac{b \times h}{2}) | (b)=base, (h)=height |
| Isosceles right triangle | (P = a + 2a\sqrt{2}) | (A = \frac{a^2}{2}) | |
| Circle | (C = 2\pi r) | (A = \pi r^2) | Use (\pi \approx 3.14) or (22/7) |
| Parallelogram | (P = 2(a + b)) | (A = \text{base} \times \text{height}) | |
| Trapezium (trapezoid) | (P = a + b + c + d) | (A = \frac{(a + b) \times h}{2}) | (a,b) are parallel sides |
Common Values
- π ≈ 3.14 (or 22/7 for quick mental calculation).
- Square root of 2 ≈ 1.414.
Quick Conversion Rules
- If you know the perimeter of a rectangle and one side, you can find the other side:
[ 2(l + w) = P \implies l + w = \frac{P}{2} ] - For a right triangle with legs (a, b) and hypotenuse (c), use Pythagoras:
[ c = \sqrt{a^2 + b^2} ]
3. Worked Example
Problem:
A rectangular plot measures 48 m in length and 30 m in width. A square garden of side 10 m is cut out from one corner. Find the area of the remaining plot.
Step‑by‑Step
-
Area of the whole rectangle
[ A_{\text{rect}} = l \times w = 48 \times 30 = 1440 \text{ m}^2 ] -
Area of the square garden
[ A_{\text{square}} = a^2 = 10^2 = 100 \text{ m}^2 ] -
Area of the remaining plot
[ A_{\text{remaining}} = A_{\text{rect}} - A_{\text{square}} = 1440 - 100 = 1340 \text{ m}^2 ]
Answer: 1340 m²
Check:
- Perimeter of the rectangle: (2(48 + 30) = 156) m (not needed here).
- The cut‑out is fully inside, so subtraction is valid.
4. Quick Revision Summary
-
Perimeter
- Rectangle: (2(l+w))
- Square: (4a)
- Triangle: sum of sides
- Circle: (2\pi r)
-
Area
- Rectangle: (l \times w)
- Square: (a^2)
- Triangle: (\frac{b \times h}{2})
- Circle: (\pi r^2)
-
π approximations: 3.14 or 22/7.
-
Pythagoras for right triangles: (c^2 = a^2 + b^2).
-
Common mistakes
- Mixing up radius and diameter in circles.
- Forgetting the “/2” in triangle area.
- Using degrees instead of radians for angles in advanced problems (rare here).
Quick Practice Problems (Try before next test)
- A trapezium has parallel sides of 12 m and 20 m, and height 6 m. Find its area.
- A circle has a diameter of 18 cm. What is its circumference?
- A right triangle has legs 9 m and 12 m. Find the hypotenuse.
- A square plot is cut into 3 equal smaller squares. If the side of the large square is 15 m, what is the area of each small square?
Answer key
- (A = \frac{(12+20)\times6}{2} = 84) m²
- (C = 2\pi r = 2\pi \times 9 = 56.52) cm (≈ 56.5 cm)
- (c = \sqrt{9^2 + 12^2} = 15) m
- Each small square side = (15/3 = 5) m → area = (25) m²