Geometry Basics – Meghalaya TET (Quantitative Aptitude)
Targeted for: Meghalaya Teacher Eligibility Test (TET) – Quantitative Aptitude Section
Time to study: ~30 minutes
Key objective: Master the fundamental concepts of geometry that appear frequently in the TET exam.
1. Simple Explanation
Geometry is the branch of mathematics that deals with the properties and relations of points, lines, angles, surfaces, and solids. In TET, you’ll mainly encounter plane geometry (2‑D shapes) and a few basic solid geometry concepts. The exam focuses on:
- Recognizing shapes and their properties
- Calculating perimeter, area, and volume
- Solving problems involving angles, triangles, circles, and quadrilaterals
- Applying simple algebraic relationships (e.g., Sum of angles of polygons)
2. Key Points
| Concept | Definition | Formula / Property | Typical TET Question Type |
|---|---|---|---|
| Point, Line, Plane | Fundamental building blocks | — | Identify or classify (e.g., “Which is a line?”) |
| Angle | Space between two intersecting lines | Acute (< 90°), Right (= 90°), Obtuse (> 90°) | Find missing angle in a figure |
| Triangle | Three‑sided polygon | Types: Equilateral, Isosceles, Scalene; Right, Acute, Obtuse | Name type, find missing sides/angles, area |
| Sum of Interior Angles | ((n-2)×180°) where (n) = vertices | — | Check consistency of a polygon |
| Quadrilateral | Four‑sided polygon | Parallelogram, Rectangle, Rhombus, Square, Trapezium | Perimeter/area, property checks |
| Circle | Set of points equidistant from a centre | Radius (r), Diameter (d=2r), Circumference (C = πd = 2πr), Area (A = πr^2) | Find missing radius or area |
| Perimeter | Total length of a shape’s boundary | (P_{\text{rect}} = 2(l+w)), (P_{\text{tri}} = a+b+c), (P_{\text{circ}} = πd) | Compute perimeter of given shapes |
| Area | Surface covered | (A_{\text{rect}} = l×w), (A_{\text{tri}} = \frac{1}{2}bh), (A_{\text{circ}} = πr^2) | Compute area of standard shapes |
| Volume (Basic) | 3‑D space occupied | (V_{\text{cuboid}} = lwh), (V_{\text{cube}} = a^3), (V_{\text{sphere}} = \frac{4}{3}πr^3) | Simple volume calculations (often in optional sections) |
| Pythagoras’ Theorem | Relation in right‑angled triangles | (a^2 + b^2 = c^2) | Find missing side in right triangle |
| Circle Theorems | Basic angle properties | Inscribed angle = ½ central angle, Tangents to a circle are equal in length from a point | Find missing angles, chord lengths |
| Coordinate Geometry (Basic) | Points in the plane | Distance formula (d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}) | Find distance between two points |
3. Worked Example
Question (Typical TET Style):
A right‑angled triangle has one acute angle of (30^\circ). The hypotenuse is (10) cm.
i) Find the lengths of the other two sides.
ii) Compute the area of the triangle.
Step‑by‑Step Solution
i) Finding the sides
-
Identify the sides relative to the angles
- Opposite the (30^\circ) angle → shorter leg (a)
- Opposite the (60^\circ) angle → longer leg (b)
- Hypotenuse (c = 10) cm
-
Use the 30‑60‑90 triangle ratio
- (a : b : c = 1 : \sqrt{3} : 2)
-
Scale the ratio to match the hypotenuse
[ \text{Scale factor} = \frac{c}{2} = \frac{10}{2} = 5 ]- Shorter leg: (a = 1 × 5 = 5) cm
- Longer leg: (b = \sqrt{3} × 5 = 5\sqrt{3}) cm ≈ 8.66 cm
ii) Computing the area
- Area formula for a triangle: (A = \frac{1}{2} × \text{base} × \text{height})
Here, base and height are the two legs.
[ A = \frac{1}{2} × a × b = \frac{1}{2} × 5 × 5\sqrt{3} = \frac{25\sqrt{3}}{2} \text{ cm}^2 \approx 21.65 \text{ cm}^2 ]
Answer:
- Shorter leg = 5 cm, longer leg = (5\sqrt{3}) cm
- Area = (\dfrac{25\sqrt{3}}{2}) cm² (≈ 21.65 cm²)
4. Quick Revision Summary
| Topic | Formula / Property | Quick Check |
|---|---|---|
| **Angle sum of triangle |