Mensuration – IIFT Quantitative Ability Study Notes
Mensuration is the branch of mathematics that deals with the measurement of the area, perimeter, volume, and surface area of geometric figures.
In competitive exams, problems often involve combinations of shapes, right‑triangle relationships, and simple algebraic manipulation.
1. Simple Explanation
- Area – Size of a 2‑D figure (units²).
- Perimeter / Circumference – Length of the boundary (units).
- Volume – Capacity of a 3‑D figure (units³).
- Surface Area – Total area of all outer faces (units²).
Key idea: Use the correct formula for each figure and simplify carefully.
For composite shapes, break them into simple shapes, compute individually, then add/subtract.
2. Key Points & Formulas
| Figure | Area | Perimeter / Circumference | Volume | Surface Area |
|---|---|---|---|---|
| Rectangle | (A = l \times w) | (P = 2(l+w)) | – | – |
| Square | (A = s^2) | (P = 4s) | – | – |
| Triangle | (A = \frac{1}{2} b h) | (P = a+b+c) | – | – |
| Right‑triangle | (A = \frac{1}{2} a b) | – | – | – |
| Parallelogram | (A = b \times h) | – | – | – |
| Circle | (A = \pi r^2) | (C = 2\pi r) | – | – |
| Ellipse | (A = \pi a b) | – | – | – |
| Cylinder | – | – | (V = \pi r^2 h) | (S = 2\pi r(h + r)) |
| Cone | – | – | (V = \frac{1}{3}\pi r^2 h) | (S = \pi r (r + l)) , (l = \sqrt{r^2+h^2}) |
| Sphere | – | – | (V = \frac{4}{3}\pi r^3) | (S = 4\pi r^2) |
| Regular Pyramid | – | – | (V = \frac{1}{3} B h) | – |
| Regular Prism | – | – | (V = B h) | – |
Useful Relationships
- Pythagoras: (a^2 + b^2 = c^2) (right‑triangle).
- Area of a sector: (\displaystyle A_{\text{sector}} = \frac{\theta}{360} \pi r^2) where (\theta) is in degrees.
- Volume of a right circular cone: (V = \frac{1}{3}) volume of a cylinder of same base and height.
Common Pitfalls
- Confusing radius and diameter – Always verify if the problem gives radius or diameter.
- For composite shapes – Remember to subtract overlapping areas, not add them.
- Units – Keep track of units; they cancel in ratios but not in final numeric answers.
- Exact vs. approximate – Use (\pi = 3.14) unless specified; for exact answers, keep (\pi) as a symbol.
3. Worked Example
Problem
A right circular cylinder has a height of 10 cm and a base radius of 3 cm.
Find (i) its volume and (ii) the total surface area.
Then, a cone is placed on top of the cylinder with the same base radius and height 10 cm.
Find the combined volume of the cylinder and the cone.
Step‑by‑step
-
Cylinder Volume
[ V_{\text{cyl}} = \pi r^2 h = \pi (3)^2 (10) = 90\pi\ \text{cm}^3 ] -
Cylinder Surface Area
[ S_{\text{cyl}} = 2\pi r(h + r) = 2\pi(3)(10 + 3) = 2\pi(3)(13) = 78\pi\ \text{cm}^2 ] -
Cone Volume
[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (10) = 30\pi\ \text{cm}^3 ] -
Combined Volume
[ V_{\text{total}} = V_{\text{cyl}} + V_{\text{cone}} = 90\pi + 30\pi = 120\pi\ \text{cm}^3 ] If a numerical value is required, use (\pi \approx 3.14):
[ V_{\text{total}} \approx 120 \times 3.14 = 376.8\ \text{cm}^3 ]
Answer
- Cylinder volume: (90\pi) cm³ (≈ 282.6 cm³)
- Cylinder surface area: (78\pi) cm² (≈ 245.2 cm²)
- Combined volume: (120\pi) cm³ (≈ 376.8 cm³)
4. Quick Revision Summary
| Concept | Quick Formula | Quick Check |
|---|---|---|
| Rectangle area | (l \times w) | Multiply length by width |
| Circle circumference | (2\pi r) | Double the radius times π |
| Cylinder volume | (\pi r^2 h) | Cylinder ≈ sphere volume × height / diameter |
| Cone volume | (\dfrac{1}{3}\pi r^2 h) | One‑third of cylinder with same base |
| Sphere volume | (\dfrac{4}{3}\pi r^3) | 4/3 × π × |