Mensuration – Quantitative Aptitude (Meghalaya Forest Guard)
Mensuration is the branch of mathematics dealing with the measurement of geometric figures – perimeter, area, volume, and surface area.
In the Forest Guard exam, you’ll need to solve problems on basic shapes (rectangles, triangles, circles, cylinders, etc.) and sometimes on compound shapes.
1. Simple Explanation
- Perimeter – “The length of the boundary” of a 2‑D figure.
- Area – “The amount of space inside” a 2‑D figure.
- Volume – “The amount of space inside” a 3‑D figure.
- Surface Area – “Total area of all outer surfaces” of a 3‑D figure.
The key to quick calculations is memorizing a few core formulas and being able to break a complex shape into simpler ones.
2. Key Points & Formulas
| Shape | Perimeter | Area | Volume | Surface Area |
|---|---|---|---|---|
| Rectangle | P = 2(l + w) |
A = l × w |
– | – |
| Square | P = 4a |
A = a² |
– | – |
| Triangle | P = a + b + c |
A = ½ × base × height |
– | – |
| Isosceles Triangle | – | A = ½ × base × √(s(s‑a)(s‑b)(s‑c)) (Heron) |
– | – |
| Circle | C = 2πr |
A = πr² |
– | – |
| Ellipse | Approx. π[(3(a+b) – √((3a+b)(a+3b))] |
A = πab |
– | – |
| Parallelogram | P = 2(l + b) |
A = base × height |
– | – |
| Trapezium (trapezoid) | P = a + b + c + d |
A = ½ × (a + c) × height |
– | – |
| Right‑Angled Triangle | P = a + b + √(a²+b²) |
A = ½ab |
– | – |
| Cylinder | – | – | V = πr²h |
SA = 2πr(h + r) |
| Cone | – | – | V = ⅓πr²h |
SA = πr(r + s) (where s = √(r²+h²)) |
| Sphere | – | – | V = ⅔πr³ |
SA = 4πr² |
| Right‑Circular Prism | – | – | V = A_base × h |
SA = 2A_base + perimeter_base × h |
| Right‑Circular Pyramid | – | – | V = ⅓ A_base × h |
SA = A_base + ½ × perimeter_base × slant_height |
Common Acronyms
π (pi) ≈ 3.1416
s– slant height (cone, pyramid)
h– height (any figure)
r– radius (circle, cylinder, cone, sphere)
Quick Tips
- Identify the shape first – then look for the applicable formula.
- Break compound shapes into rectangles, triangles, circles, etc.
- Use Heron’s formula only when you know all three sides.
- Remember: Surface area = area of all faces; volume = area of base × height (prism, cylinder) or
⅓ × base area × height(pyramid, cone). - Convert units if needed (cm ↔ m, mm ↔ cm).
- Check for symmetry – often simplifies the problem.
3. Worked Example
Problem
A rectangular plot of land measures 30 m × 20 m. A semicircular pond of radius 5 m is cut out from one corner of the plot. Find the remaining area of the plot.
Step‑by‑Step
-
Area of the rectangle
[ A_{\text{rect}} = 30 \times 20 = 600\ \text{m}^2 ] -
Area of the semicircle
Full circle area:
[ A_{\text{circ}} = \pi r^2 = 3.1416 \times 5^2 = 78.54\ \text{m}^2 ]
Semicircle area:
[ A_{\text{semi}} = \tfrac{1}{2} \times 78.54 \approx 39.27\ \text{m}^2 ] -
Remaining area
[ A_{\text{remaining}} = A_{\text{rect}} - A_{\text{semi}} = 600 - 39.27 \approx