Mensuration – Study Notes
(Quantitative Aptitude – NABARD Development Assistant)
1. Simple Explanation
Mensuration is the branch of mathematics that deals with finding the measurements (length, area, volume, surface area, etc.) of geometric figures.
- Perimeter / Circumference – total boundary length.
- Area – space covered inside the boundary.
- Volume – space occupied by a 3‑D object.
- Surface Area – total area of all the faces of a 3‑D object.
All formulas are derived from basic geometry (triangles, rectangles, circles, etc.) and can be applied to composite figures as well.
2. Key Points & Formulae
| Figure | Perimeter / Circumference | Area | Volume (if 3‑D) | Surface Area (if 3‑D) |
|---|---|---|---|---|
| Line Segment | (P = 0) | – | – | – |
| Square (a) | (4a) | (a^2) | – | – |
| Rectangle (l \times w) | (2(l+w)) | (lw) | – | – |
| Parallelogram (base (b), height (h)) | – | (b h) | – | – |
| Rhombus (diagonals (d_1, d_2)) | – | (\frac{d_1 d_2}{2}) | – | – |
| Trapezium (bases (a,b), height (h)) | – | (\frac{(a+b)h}{2}) | – | – |
| Circle (radius (r)) | (2\pi r) | (\pi r^2) | – | – |
| Ellipse (semi‑axes (a,b)) | – | (\pi a b) | – | – |
| Triangle (base (b), height (h)) | – | (\frac{b h}{2}) | – | – |
| Right‑Angled Triangle | – | – | – | – |
| Sphere (radius (r)) | – | – | (\frac{4}{3}\pi r^3) | (4\pi r^2) |
| Cylinder (radius (r), height (h)) | – | – | (\pi r^2 h) | (2\pi r h + 2\pi r^2) |
| Cone (radius (r), height (h), slant (l)) | – | – | (\frac{1}{3}\pi r^2 h) | (\pi r(l+r)) |
| Right Circular Prism (base area (B), height (h)) | – | – | (B h) | (2B + P h) |
| Right Circular Pyramid (base area (B), slant height (l), height (h)) | – | – | (\frac{1}{3} B h) | (B + \frac{1}{2} P l) |
Tip:
- For composite figures, break them into basic shapes, compute each area/volume, then sum.
- For irregular shapes, approximate using known shapes or use calculus (integral) if required.
3. Worked Example
Problem:
A composite figure consists of a rectangle of dimensions (12\text{ cm} \times 5\text{ cm}) and a right‑angled triangle attached to one of its longer sides. The triangle has a base of (5\text{ cm}) (same as the rectangle’s width) and a height of (8\text{ cm}).
Find the total area of the composite figure.
Step‑by‑Step Solution
| Step | Explanation | Calculation |
|---|---|---|
| 1 | Area of rectangle | (A_{\text{rect}} = l \times w = 12 \times 5 = 60\ \text{cm}^2) |
| 2 | Area of right‑angled triangle | (A_{\text{tri}} = \frac{1}{2}\times \text{base}\times \text{height} = \frac{1}{2}\times 5 \times 8 = 20\ \text{cm}^2) |
| 3 | Total area | (A_{\text{total}} = A_{\text{rect}} + A_{\text{tri}} = 60 + 20 = 80\ \text{cm}^2) |
Answer: The area of the composite figure is 80 cm².
4. Quick Revision Summary
| Topic | Formula | Typical Question |
|---|---|---|
| Perimeter | (P = 2(l+w)) (rectangle) | “Find the perimeter of a rectangle 8 cm × 6 cm.” |
| Area – Rectangle | (A = lw) | “Area of a 9 cm × 4 cm rectangle.” |
| Area – Triangle | (A = \frac{bh}{2}) | “Area of a triangle with base 10 cm and height 7 cm.” |
| Area – Circle | (A = \pi r^2) | “Area of a circle of radius 5 cm.” |
| Area – Trapezium | (A = \frac{(a+b)h}{2}) | “Area of trapez |