Mensuration – Quick Study Notes
(For Jharkhand Forest Guard – Quantitative Aptitude)
1. Simple Explanation
Mensuration is the branch of mathematics that deals with the measurement of geometric figures – finding their area, perimeter, surface area and volume.
- Area – 2‑D “size” of a shape.
- Perimeter (or Circumference) – total length of the boundary.
- Surface Area – total area of all faces of a 3‑D figure.
- Volume – amount of space a 3‑D figure occupies.
All formulas are derived from the basic building blocks: rectangle, triangle, circle, cylinder, cone, prism, pyramid, sphere.
2. Key Points & Formulas
| Shape | Symbol | Perimeter / Circumference | Area | Surface Area (SA) | Volume |
|---|---|---|---|---|---|
| Rectangle | – | (P = 2(l+w)) | (A = l \times w) | – | – |
| Square | – | (P = 4a) | (A = a^2) | – | – |
| Triangle | – | (P = a+b+c) | (A = \frac{1}{2} \times \text{base} \times \text{height}) | – | – |
| Parallelogram | – | (P = 2(a+b)) | (A = \text{base} \times \text{height}) | – | – |
| Trapezium (Trapezoid) | – | (P = a+b+c+d) | (A = \frac{1}{2} \times (a+b) \times h) | – | – |
| Circle | (r) | (C = 2\pi r) | (A = \pi r^2) | – | – |
| Cylinder | – | – | (A_{\text{lateral}} = 2\pi r h) | (SA = 2\pi r(r+h)) | (V = \pi r^2 h) |
| Cone | – | – | (A_{\text{lateral}} = \pi r l) | (SA = \pi r(r+l)) | (V = \frac{1}{3}\pi r^2 h) |
| Sphere | – | – | – | (SA = 4\pi r^2) | (V = \frac{4}{3}\pi r^3) |
| Right Circular Prism | – | – | (A_{\text{lateral}} = P_{\text{base}}\times h) | (SA = 2A_{\text{base}} + A_{\text{lateral}}) | (V = A_{\text{base}}\times h) |
| Right Circular Pyramid | – | – | (A_{\text{lateral}} = \frac{1}{2} P_{\text{base}} \times s) | (SA = A_{\text{base}} + A_{\text{lateral}}) | (V = \frac{1}{3} A_{\text{base}} \times h) |
Important Constants
- (\pi \approx 3.14) (use as per question).
- For a regular polygon, the perimeter is (n \times \text{side}) and area can be found using (A = \frac{1}{4} n s^2 \cot(\pi/n)).
3. Worked Example
Problem
A right circular cone has a slant height of 13 cm and a base radius of 5 cm. Find
- Surface area
- Volume
Solution
-
Surface Area
[ SA = \pi r(r + l) ] [ SA = \pi \times 5(5 + 13) = \pi \times 5 \times 18 = 90\pi ] Using (\pi = 3.14): [ SA \approx 90 \times 3.14 = 282.6 \text{ cm}^2 ] -
Volume
[ V = \frac{1}{3}\pi r^2 h ] First find the height (h) using Pythagoras in the right triangle (radius, height, slant height): [ l^2 = r^2 + h^2 \Rightarrow 13^2 = 5^2 + h^2 \Rightarrow 169 = 25 + h^2 \Rightarrow h^2 = 144 \Rightarrow h = 12 \text{ cm} ] Now plug in: [ V = \frac{1}{3}\pi (5)^2 (12) = \frac{1}{3}\pi \times 25 \times 12 = 100\pi ] [ V \approx 100 \times 3.14 = 314 \text{ cm}^3 ]
Answer
- Surface area ≈ 282.6 cm²
- Volume ≈ 314 cm³
4. Quick Revision Summary
| Topic | Formula | Notes |
|---|---|---|
| Perimeter (P) | (P = \sum \text{all sides}) | For circles: (C = 2\pi r) |
| Area (A) | Depends on shape | Use (A = \frac{1}{2}bh) for triangles |
| Surface Area (SA) | 2 × base area + lateral area | Cylinder: (SA = 2\pi r(r+h)) |
| Volume (V) | Base area × height |