Mizoram TET (Teacher Eligibility) /Quantitative Aptitude

Geometry basics

Speed & Geometry · free study notes · Updated

Geometry Basics – Mizoram TET – Quantitative Aptitude

(For 10‑year‑old teachers)


1️⃣ Simple Explanation

Geometry is the branch of mathematics that deals with the shape, size, position, and relative properties of objects in space.

  • Points – a position in space, no size.
  • Lines – a set of points extending infinitely in two directions.
  • Line segments – part of a line between two points.
  • Angles – formed by two rays sharing a common endpoint.
  • Shapes – collections of lines and angles (triangles, quadrilaterals, circles, etc.).

In the TET, you’ll need to solve problems that involve:

  • Measuring lengths and angles.
  • Computing area and perimeter of basic shapes.
  • Using simple theorems (Pythagoras, congruence, similarity).

2️⃣ Key Points (Bullet‑style)

Concept Definition Formula / Important Notes
Point Exact location, no dimension —
Line Infinite length, no thickness —
Line Segment Finite part of a line Length = distance between endpoints
Ray One endpoint, extends infinitely in one direction —
Angle Measure between two rays • Acute < 90°, obtuse > 90°, right = 90°
Straight Angle 180°
Full Angle 360°
Triangle 3 sides, 3 angles • Sum of interior angles = 180°
Quadrilateral 4 sides • Sum of interior angles = 360°
Rectangle Opposite sides equal, all angles 90° • Area = l × w, Perimeter = 2(l + w)
Rhombus All sides equal, opposite angles equal • Area = (diagonal1 × diagonal2)/2
Parallelogram Opposite sides parallel & equal • Area = base × height
Trapezium (US: Trapezoid) One pair of parallel sides • Area = ½ × (sum of parallel sides) × height
Circle All points equidistant from center • Circumference = 2πr, Area = πr²
Pythagoras (Right Triangle) (a^2 + b^2 = c^2) c = hypotenuse
Area of Triangle (\frac{1}{2} × \text{base} × \text{height}) –
Area of Regular Polygon (\frac{1}{4} n s^2 \cot(\frac{\pi}{n})) n = number of sides, s = side length
360° around a point Sum of all angles around a point –

3️⃣ Worked Example

Problem
A rectangle has a length of 12 cm and a width of 5 cm.
a) Find its perimeter.
b) Find its area.
c) If the length of the diagonal is required, compute it.

Solution

Step Calculation Result
Perimeter (P = 2(l + w)) (2(12 + 5) = 2(17) = 34) cm
Area (A = l \times w) (12 \times 5 = 60) cm²
Diagonal (d) Use Pythagoras: (d = \sqrt{l^2 + w^2}) (\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13) cm

Answers

  • Perimeter = 34 cm
  • Area = 60 cm²
  • Diagonal = 13 cm

4️⃣ Quick Revision Summary

Topic Formula Quick Tip
Sum of angles in triangle (180°) Always check if angles add up to 180°
Sum of angles in quadrilateral (360°) Use for checking correctness
Area of triangle (\frac{1}{2} \times \text{base} \times \text{height}) Remember “half”
Area of rectangle (l \times w) Simple multiplication
Perimeter of rectangle (2(l + w)) Double the sum of sides
Circle circumference (2\pi r) Use (π ≈ 3.14)
Circle area (\pi r^2) Square the radius first
Pythagoras (a^2 + b^2 = c^2) Right‑triangle helper
Diagonal of rectangle (\sqrt{l^2 + w^2}) Use Pythagoras on rectangle sides

Remember:

  • Always label known quantities.
  • Check units (cm, m, etc.).
  • Verify that computed angles or perimeters are logical (e.g., perimeter cannot be less than the sum of two sides).

Good luck with your TET preparation!

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