Karnataka Forest Guard – Quantitative Aptitude
Mensuration – Study Notes
Mensuration is the branch of mathematics that deals with the measurement of geometric figures.
In the Forest Guard exam you will encounter problems involving area, perimeter, volume, and surface area of common 2‑D and 3‑D shapes.
1. Simple Explanation
| Concept | What it represents | Typical formula (2‑D) | Typical formula (3‑D) |
|---|---|---|---|
| Perimeter | Sum of all sides of a figure | N/A – depends on figure | N/A |
| Area | Size of region enclosed by a figure | N/A – depends on figure | N/A |
| Volume | Space occupied by a 3‑D figure | N/A | (V = \text{Base Area} \times \text{Height}) |
| Surface Area | Total area of all outer faces | N/A | (SA = 2(\text{Base Area}) + \text{Lateral Area}) |
In tests, you’ll mainly need the formulae for each figure; units (cm², m³, etc.) are usually omitted.
2. Key Points & Formulae
2.1 2‑D Figures
| Figure | Perimeter | Area |
|---|---|---|
| Square | (P = 4a) | (A = a^2) |
| Rectangle | (P = 2(l + w)) | (A = l \times w) |
| Parallelogram | (P = 2(a + b)) | (A = \text{Base} \times \text{Height}) |
| Triangle | (P = a + b + c) | (A = \frac{1}{2}\times \text{Base}\times \text{Height}) |
| Right‑angled Triangle | same as triangle | same as triangle |
| Isosceles / Equilateral Triangle | use side lengths | same formula |
| Circle | (P = 2\pi r) (circumference) | (A = \pi r^2) |
| Trapezium (trapezoid) | (P = a + b + c + d) | (A = \frac{1}{2}(a + b)\times h) (parallel sides (a, b)) |
Pythagoras: (a^2 + b^2 = c^2) – useful for right‑angled triangles.
2.2 3‑D Figures
| Figure | Volume | Surface Area |
|---|---|---|
| Cube | (V = a^3) | (SA = 6a^2) |
| Cuboid (Rectangular Prism) | (V = lwh) | (SA = 2(lw + lh + wh)) |
| Cylinder | (V = \pi r^2 h) | (SA = 2\pi r(h + r)) |
| Sphere | (V = \frac{4}{3}\pi r^3) | (SA = 4\pi r^2) |
| Cone | (V = \frac{1}{3}\pi r^2 h) | (SA = \pi r(r + l)) where (l = \sqrt{r^2 + h^2}) |
| Right Circular Pyramid (square base) | (V = \frac{1}{3}a^2 h) | (SA = a^2 + 2a l) (l = slant height) |
| Regular Pyramid (base of n sides) | (V = \frac{1}{3}B h) (B = base area) | (SA = B + \frac{1}{2}p l) (p = perimeter of base) |
3. Worked Example
Problem:
A cylindrical water tank has a radius of 5 m and a height of 12 m.
(a) Find its volume.
(b) Find its total surface area.
(a) Volume
[ V = \pi r^2 h = \pi (5)^2 (12) = 25\pi \times 12 = 300\pi \ \text{m}^3 ]
Numerical value (π ≈ 3.1416):
[ V \approx 300 \times 3.1416 = 942.48 \ \text{m}^3 ]
(b) Surface Area
[ SA = 2\pi r(h + r) = 2\pi (5)(12 + 5) = 10\pi \times 17 = 170\pi \ \text{m}^2 ]
[ SA \approx 170 \times 3.1416 = 534.01 \ \text{m}^2 ]
Answer:
- Volume ≈ 942.48 m³
- Surface Area ≈ 534.01 m²
4. Quick Revision Summary
| Shape | Area | Volume | Surface Area |
|---|---|---|---|
| Square | (a^2) | N/A | N/A |
| Rectangle | (lw) | N/A | N/A |
| Triangle | (\frac{1}{2}bh) | N/A | N/A |
| Circle | (\pi r^2) | N/A | (2\pi r |