Mensuration – Quick Study Guide
(Maharashtra TET – Quantitative Aptitude)
Mensuration is the study of the measurement of geometric quantities like area, perimeter, surface area, and volume.
In TET, you’ll encounter both 2‑D (area, perimeter) and 3‑D (volume, surface area) problems.
1. Simple Explanation
| Dimension | What you measure | Typical Formula(s) | Example |
|---|---|---|---|
| 2‑D shape | Area (square units) | Depends on shape | Area of a rectangle = length × breadth |
| 2‑D shape | Perimeter / Circumference (linear units) | Sum of all sides or 2πr for circles | Perimeter of a square = 4 × side |
| 3‑D solid | Volume (cubic units) | Depends on solid | Volume of a cube = a³ |
| 3‑D solid | Surface Area (square units) | Sum of areas of all faces | Surface area of a cube = 6a² |
Tip: Always write down the formula before doing any calculation.
Keep a separate page for common formulas, as they will save time.
2. Key Points & Common Formulas
2‑D Shapes
| Shape | Area Formula | Perimeter / Circumference |
|---|---|---|
| Square | (A = a^2) | (P = 4a) |
| Rectangle | (A = l \times b) | (P = 2(l+b)) |
| Triangle | (\displaystyle A = \frac{1}{2} \times \text{base} \times \text{height}) | (P = a+b+c) |
| Parallelogram | (\displaystyle A = \text{base} \times \text{height}) | (P = 2(a+b)) |
| Rhombus | (\displaystyle A = \frac{1}{2} \times d_1 \times d_2) | (P = 4a) |
| Trapezium | (\displaystyle A = \frac{1}{2} \times (a+b) \times h) | (P = a+b+c+d) |
| Circle | (\displaystyle A = \pi r^2) | (C = 2\pi r) |
| Ellipse | (\displaystyle A = \pi a b) | No simple closed form for perimeter (use approximation) |
3‑D Solids
| Solid | Volume Formula | Surface Area Formula |
|---|---|---|
| Cube | (V = a^3) | (S = 6a^2) |
| Cuboid | (V = l \times b \times h) | (S = 2(lb + bh + lh)) |
| Sphere | (V = \frac{4}{3}\pi r^3) | (S = 4\pi r^2) |
| 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)) (where (l) = slant height) |
| Right Pyramid (square base) | (V = \frac{1}{3} \times \text{Base Area} \times \text{Height}) | (S = \text{Base Area} + \frac{1}{2} \times \text{Perimeter of Base} \times \text{Slant Height}) |
Common Mistake: Mixing up height and slant height in cones/pyramids.
Check: In a cone, height is perpendicular from apex to base centre; slant height is along the side.
3. Worked Example
Problem:
A right circular cylinder has a radius of ( 6 , \text{cm} ) and a height of ( 9 , \text{cm} ).
Find its (i) volume and (ii) total surface area.
Solution
-
Volume
[ V = \pi r^2 h = \pi (6)^2 \times 9 = \pi \times 36 \times 9 = 324\pi , \text{cm}^3 ] Approximate: (324 \times 3.14 = 1017.36 , \text{cm}^3) -
Surface Area
[ S = 2\pi r (h + r) = 2\pi (6)(9+6) = 12\pi \times 15 = 180\pi , \text{cm}^2 ] Approximate: (180 \times 3.14 = 565.2 , \text{cm}^2)
Answer:
- Volume = (324\pi , \text{cm}^3) (≈ 1017.36 cm³)
- Surface Area = (180\pi , \text{cm}^2) (≈ 565.2 cm²)
Tip: Always check units:
- Area → square units (cm², m²)
- Volume → cubic units (cm³, m³)
4. Quick Revision Summary
- Area of circle: (A = \pi r^2)
- Circumference of circle: (C = 2\pi r)
- Volume of cylinder: (V = \pi r^2 h)