Geometry Basics – District Judge Exam (Quantitative Aptitude)
Goal:
Build a solid foundation in the basic concepts of geometry that are frequently tested in the District Judge examination.
These notes are concise, yet comprehensive enough to give you confidence for quick revision.
1. Simple Explanation
Geometry is the branch of mathematics that deals with the size, shape, proportion, and relative position of figures. In the exam context, you’ll encounter:
| Concept | Typical Question Type |
|---|---|
| Points, lines, angles | Identify types of angles, calculate missing angles |
| Triangles & Quadrilaterals | Find side lengths, area, perimeter |
| Circles | Radius, diameter, circumference, area |
| Pythagoras & Heron | Right‑triangle length relations, area of scalene triangle |
| Similarity & Congruence | Ratio of sides, angles, area comparison |
2. Key Points
2.1 Basic Elements
| Element | Definition | Notation |
|---|---|---|
| Point | Exact location, no size | (P, Q, R) |
| Line | Infinite set of points, no thickness | (\overline{AB}) |
| Line Segment | Part of a line between two points | (\overline{AB}) |
| Ray | One endpoint, extends infinitely | (\overrightarrow{AB}) |
| Angle | Formed by two rays sharing a vertex | (\angle ABC) |
| Straight Angle | (180^\circ) | |
| Right Angle | (90^\circ) | |
| Complementary Angles | Sum to (90^\circ) | |
| Supplementary Angles | Sum to (180^\circ) |
2.2 Angles in Polygons
- Triangle: Sum of interior angles = (180^\circ)
- Quadrilateral: Sum = (360^\circ)
- Circle: Full revolution = (360^\circ)
2.3 Triangle Types
| Type | Conditions | Key Properties |
|---|---|---|
| Equilateral | All sides equal | All angles (60^\circ) |
| Isosceles | Two sides equal | Base angles equal |
| Scalene | All sides different | No equal angles |
| Right | One (90^\circ) angle | Pythagoras applies |
2.4 Quadrilaterals of Interest
| Quadrilateral | Key Properties |
|---|---|
| Rectangle | Opposite sides equal, all angles (90^\circ) |
| Square | All sides equal, all angles (90^\circ) |
| Rhombus | All sides equal, opposite angles equal |
| Parallelogram | Opposite sides equal & parallel, opposite angles equal |
| Trapezium (US) / Trapezoid (UK) | One pair of parallel sides |
2.5 Circle Basics
| Term | Formula |
|---|---|
| Diameter | (2r) |
| Circumference | (C = 2\pi r = \pi d) |
| Area | (A = \pi r^2) |
2.6 Perimeter & Area
| Shape | Perimeter | Area |
|---|---|---|
| Rectangle | (2(l+w)) | (l \times w) |
| Square | (4a) | (a^2) |
| Triangle | (a+b+c) | (\frac{1}{2} \times \text{base} \times \text{height}) |
| Circle | (2\pi r) | (\pi r^2) |
| Parallelogram | (2(a+b)) | (\text{base} \times \text{height}) |
2.7 Pythagoras Theorem
- For right‑angled triangle (ABC) with right angle at (C):
[ a^2 + b^2 = c^2 ] where (c) is the hypotenuse.
2.8 Heron’s Formula
- For any triangle with sides (a, b, c) and semiperimeter (s = \frac{a+b+c}{2}):
[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} ]
2.9 Similarity & Congruence
| Criterion | Condition | Implication |
|---|---|---|
| Similar | Corresponding angles equal, sides proportional | (\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}) |
| Congruent | All corresponding sides equal, all angles equal | Same shape & size |
3. Worked Example
Problem (Typical of the Exam)
*A rectangle has a length of 12 cm and a width that is 3 cm less than the length.
- What is its perimeter?
- What is its area?
- If the rectangle is inscribed in a circle, find the circle’s radius.*
Solution
-
Find the width
[ \text{Width} = \text{Length} - 3 = 12 - 3 = 9\ \text{cm} ] -
Perimeter of rectangle
[ P = 2(l + w) = 2(12 + 9) = 2(21) = 42\ \text{cm} ] -
Area of rectangle
[ A = l \times w = 12 \times 9 = 108\ \text{cm}^2 ] -
Inscribed circle
- The diameter of the circle equals the rectangle’s diagonal.
- Use Pyth