📚 Puducherry TET – Quantitative Aptitude
Mensuration
Mensuration is the branch of mathematics that deals with the measurement of lengths, areas, and volumes of geometric figures.
In the TET exam, you’ll encounter questions on the area/perimeter of 2‑D shapes and the surface area/volume of 3‑D solids.
Tip: Keep the basic formulas in memory and practice quick conversions (square units, cubic units, π ≈ 3.14).
1️⃣ Simple Explanation
- 2‑D Figures – Find area (space inside) and perimeter (boundary length).
- 3‑D Figures – Find volume (space inside) and surface area (total area of all faces).
All formulas are derived from the definition of area/volume and the geometric properties of the shape (e.g., base × height for a rectangle, ½ × base × height for a triangle).
2️⃣ Key Points & Formulas
| Shape | Area | Perimeter | Volume | Surface Area |
|---|---|---|---|---|
| Rectangle | (A = l \times w) | (P = 2(l+w)) | – | – |
| Square | (A = s^2) | (P = 4s) | – | – |
| Triangle (any) | (A = \frac{1}{2} \times \text{base} \times \text{height}) | (P = a+b+c) | – | – |
| Parallelogram | (A = \text{base} \times \text{height}) | (P = 2(\text{base+side})) | – | – |
| Rhombus | (A = \frac{1}{2} \times d_1 \times d_2) | (P = 4a) | – | – |
| Trapezium | (A = \frac{1}{2} \times (a+b) \times h) | (P = a+b+c+d) | – | – |
| Circle | (A = \pi r^2) | (P = 2\pi r) | – | – |
| Sphere | – | – | (V = \frac{4}{3}\pi r^3) | (SA = 4\pi r^2) |
| Cylinder | – | – | (V = \pi r^2 h) | (SA = 2\pi r(r+h)) |
| Cone | – | – | (V = \frac{1}{3}\pi r^2 h) | (SA = \pi r (r + l)) (l = slant height) |
| Right Circular Prism | – | – | (V = \text{Base Area} \times \text{height}) | (SA = 2(\text{Base Area}) + \text{Perimeter of Base} \times \text{height}) |
| Rectangular Prism (Cuboid) | – | – | (V = l \times w \times h) | (SA = 2(lw+wh+lh)) |
Common Constants
- (\pi \approx 3.14) (use 22/7 if exact fraction needed)
- Units:
- Length: cm, m, ft, in
- Area: (cm^2, m^2, ft^2)
- Volume: (cm^3, m^3, ft^3)
Quick Conversion Rules
| Unit | Symbol | 1 unit in |
|---|---|---|
| 1 m | 100 cm | (100^2 = 10,000) (cm^2) |
| 1 m | 100 cm | (100^3 = 1,000,000) (cm^3) |
| 1 ft | 12 in | (12^2 = 144) (in^2) |
| 1 ft | 12 in | (12^3 = 1,728) (in^3) |
3️⃣ Worked Example
Problem:
A cylindrical water tank has a radius of (5 \text{ cm}) and height (20 \text{ cm}).
- Find the volume of the tank.
- Find the total surface area (including top and bottom).
Solution
-
Volume
[ V = \pi r^2 h = 3.14 \times 5^2 \times 20 = 3.14 \times 25 \times 20 ] [ 25 \times 20 = 500 \ V = 3.14 \times 500 = 1570 \text{ cm}^3 ] -
Surface Area
[ SA = 2\pi r (r + h) = 2 \times 3.14 \times 5 \times (5 + 20) ] [ 5 + 20 = 25 \ 2 \times 3.14 \times 5 = 31.4 \ SA = 31.4 \times 25 = 785 \text{ cm}^2 ]
Answer:
- Volume = 1570 cm³
- Surface Area = 785 cm²
4️⃣ Quick Revision Summary
| Concept | Formula | Key Tip |
|---|---|---|
| Rectangle Area | (l \times w) | Multiply sides |
| Square Perimeter | (4s) | All sides equal |
| Triangle Area | (\frac{1}{2} \times \text{base} \times \text{height}) | Always half |
| Circle Area | (\pi r^2) | Square radius first |
| Circle Circumference | (2\pi r) | Often needed for perimeter |
| Parallelogram Area |