📚 Mensuration – Quantitative Aptitude
For the Maharashtra Forest Guard (MFG) Examination
1. Simple Explanation
Mensuration is the branch of mathematics that deals with the measurement of geometric figures – in particular, length, area, perimeter, and volume. In the MFG exam, you’ll encounter problems involving:
| Shape | What to find |
|---|---|
| 1‑D (line) | Length, distance |
| 2‑D (plane) | Perimeter, area |
| 3‑D (solid) | Surface area, volume |
Why it matters for MFG
- Many duty‑related tasks involve estimating land area, volume of timber, or distances between forest posts.
- Quick mental calculations using memorized formulas save time in exams and on‑field work.
2. Key Points & Formulas
| Dimension | Formula | Notes |
|---|---|---|
| Perimeter / Circumference | ||
| Rectangle | (P = 2(l + w)) | (l) = length, (w) = width |
| Square | (P = 4a) | (a) = side |
| Triangle | (P = a + b + c) | (a, b, c) = sides |
| Circle | (C = 2\pi r = \pi d) | (\pi \approx 22/7) |
| Area | ||
| Rectangle | (A = l \times w) | |
| Square | (A = a^2) | |
| Triangle | (A = \frac{1}{2} \times \text{base} \times \text{height}) | |
| Parallelogram | (A = \text{base} \times \text{height}) | |
| Trapezium | (A = \frac{1}{2} (b_1 + b_2) \times h) | |
| Circle | (A = \pi r^2) | |
| Volume | ||
| Cube | (V = a^3) | |
| Rectangular Prism | (V = l \times w \times h) | |
| Cuboid (box) | Same as rectangular prism | |
| Cylinder | (V = \pi r^2 h) | |
| Sphere | (V = \frac{4}{3} \pi r^3) | |
| Cone | (V = \frac{1}{3} \pi r^2 h) | |
| Surface Area | ||
| Cube | (SA = 6a^2) | |
| Cylinder | (SA = 2\pi r (r + h)) | |
| Sphere | (SA = 4\pi r^2) | |
| Cone (including base) | (SA = \pi r (r + l)) where (l) is slant height |
Quick Tips for MFG Exams
| Tip | Why it helps |
|---|---|
| Memorise (\pi = 22/7) | Simplifies calculations; often used in exam problems. |
| Convert lengths to common units early | Avoids mistakes when combining lengths (e.g., meters + centimeters). |
| Use “grid method” for irregular shapes | Break into rectangles/triangles; sum areas. |
| Check for “composite shape” hints | Look for “cut” or “add” to see if you need to subtract/ add areas. |
| Always double‑check with dimensional analysis | Units should match (e.g., area in (m^2), volume in (m^3)). |
3. Worked Example
Problem (Typical MFG Question)
A forest clearing is in the shape of a rectangle that is 120 m long and 80 m wide. A circular pond of radius 15 m is cut out from the centre of the clearing. What is the net area of the clearing after the pond is removed?
Step‑by‑step Solution
-
Area of the rectangular clearing
[ A_{\text{rect}} = l \times w = 120,\text{m} \times 80,\text{m} = 9{,}600,\text{m}^2 ] -
Area of the circular pond
Use (\pi = 22/7)
[ A_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 15^2 ] [ = \frac{22}{7} \times 225 = 22 \times 32.142857 \approx 706.28 ,\text{m}^2 ] (Rounded to two decimals; exam‑style may accept 706 m².) -
Net area after pond removal
[ A_{\text{net}} = A_{\text{rect}} - A_{\text{circle}} ] [ = 9{,}600 - 706.28 \approx 8{,}893.72,\text{m}^2 ]If the exam requires an integer, round to 8,894 m².
Quick Check
- Units: All in (m^2). ✔️
- Reasonable: Pond area (~700 m²) is smaller than clearing area (~9,600 m²). ✔️
4. Quick Revision Summary
| Concept | Formula | Quick Mnemonic |
|---|---|---|
| Rectangle Perimeter | (2(l+w)) | “Twice the sum” |
| Square Perimeter | (4a) | “Four sides” |
| Circle Circumference | (2\pi r) | “Two times π times radius” |