Mensuration – Quick Study Guide
(For Lakshadweep Teacher Eligibility Test – Quantitative Aptitude)
1. What is Mensuration?
Mensuration is the branch of mathematics that deals with measurement of geometric figures – their area, perimeter, volume, and surface area.
Simple definition – “Finding the size of a shape.”
In TET, you’ll mostly need the standard formulas for common 2‑D and 3‑D figures and the ability to apply them in word problems.
2. Key Points & Formulas
| Figure | 2‑D – Area | 2‑D – Perimeter | 3‑D – Volume | 3‑D – Surface Area |
|---|---|---|---|---|
| Rectangle | (A = l \times w) | (P = 2(l+w)) | – | – |
| Square | (A = a^2) | (P = 4a) | – | – |
| Triangle | (A = \frac{1}{2} b \times h) | Sum of sides | – | – |
| Right‑angled Triangle | (A = \frac{1}{2} a \times b) | (a+b+c) (c = hypotenuse) | – | – |
| Circle | (A = \pi r^2) | (C = 2\pi r) | – | – |
| Semi‑circle | (A = \frac{1}{2}\pi r^2) | (C = \pi r + \pi r) (half circumference + diameter) | – | – |
| Parallelogram | (A = b \times h) | (P = 2(a+b)) | – | – |
| Trapezium (Trapezoid) | (A = \frac{1}{2}(a+b)h) | – | – | – |
| 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) is slant height |
| Cuboid / Rectangular Prism | – | – | (V = lwh) | (SA = 2(lw+wh+lh)) |
| Cube | – | – | (V = a^3) | (SA = 6a^2) |
Important Tips
- π is usually taken as 22/7 in competitive exams.
- When a problem involves “half” a figure, remember to divide the area or surface area by 2 but keep the perimeter (or circumference) as is for semi‑circles.
- For a sphere, the volume has the factor (\frac{4}{3}); for a cone, the factor is (\frac{1}{3}).
3. Worked Example
Problem
*A cylindrical water tank has a radius of 5 m and height of 12 m.
- What is the volume of water it can hold?
- What is the total surface area (including top and bottom)?*
Solution
-
Volume
[ V = \pi r^2 h = \frac{22}{7} \times 5^2 \times 12 = \frac{22}{7} \times 25 \times 12 ] [ V = \frac{22 \times 300}{7} = \frac{6600}{7} \approx 942.86 \text{ m}^3 ] -
Surface Area
[ SA = 2\pi r (h + r) = 2 \times \frac{22}{7} \times 5 \times (12 + 5) ] [ SA = \frac{44}{7} \times 5 \times 17 = \frac{220}{7} \times 17 \approx 534.29 \text{ m}^2 ]
Answer
- Volume ≈ 942.9 m³
- Surface Area ≈ 534.3 m²
4. Quick Revision Summary
| Topic | Formula | Quick Check |
|---|---|---|
| Perimeter of rectangle | (2(l+w)) | Add opposite sides and double |
| Area of triangle | (\frac{1}{2}bh) | Base × height ÷ 2 |
| Circumference | (2\pi r) | Radius × 2 × π |
| Area of circle | (\pi r^2) | Radius squared times π |
| Volume of cylinder | (\pi r^2 h) | Base area × height |
| Surface area of cylinder | (2\pi r(h+r)) | Two circles + side rectangle |
| Volume of sphere | (\frac{4}{3}\pi r^3) | 4/3 × π × radius cubed |
| Surface area of sphere | (4\pi r^2) | 4 × π × radius squared |
| Volume of cone | (\frac{1}{3}\pi r^2 h) | 1/3 × base area × height |
| Surface area of cone | (\pi r(r+l)) | π × radius × (radius + slant height) |
Mnemonic for 3‑D volumes
- Cylinder: (\pi r^2 h)
- Sphere: