District Judge Exam /Quantitative Aptitude

Geometry basics

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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

  1. Find the width
    [ \text{Width} = \text{Length} - 3 = 12 - 3 = 9\ \text{cm} ]

  2. Perimeter of rectangle
    [ P = 2(l + w) = 2(12 + 9) = 2(21) = 42\ \text{cm} ]

  3. Area of rectangle
    [ A = l \times w = 12 \times 9 = 108\ \text{cm}^2 ]

  4. Inscribed circle

    • The diameter of the circle equals the rectangle’s diagonal.
    • Use Pyth

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