Mensuration – Quick & Easy Study Notes
(For the Arunachal Pradesh Patwari / Lekhpal Exam – Quantitative Aptitude)
Mensuration is all about measuring the size of shapes – how big they are (area), how long the edges are (perimeter), and how much space they occupy (volume).
Think of it as the same way you’d measure a field, a pizza, or a box.
1. Why It Matters
- Everyday life: figuring out how much paint you need, how many tiles will cover a floor, or how much rice fits in a bag.
- Exam: quick mental math once you know the basic formulas.
2. Basic Terms (No Jargon)
| Term | What It Means | Everyday Analogy |
|---|---|---|
| Perimeter | Total length around the shape. | The fence that goes around a garden. |
| Area | Amount of flat surface inside the shape. | How many square‑meter tiles you need to cover your floor. |
| Volume | How much 3‑D space the shape takes up. | How many 1‑liter bottles fit in a box. |
| Surface Area | Total area of all sides of a 3‑D shape. | How much paint you need to cover every side of a box. |
3. 2‑D Shapes – Simple Formulas
Tip: Keep the units the same (cm, m, etc.) when you plug numbers in.
| Shape | Perimeter Formula | Area Formula | Tiny Example |
|---|---|---|---|
| Rectangle | 2 × (length + breadth) |
length × breadth |
A room 4 m × 3 m → Perimeter = 14 m, Area = 12 m² |
| Square | 4 × side |
side² |
A 5 m side → P = 20 m, A = 25 m² |
| Triangle | a + b + c (add all sides) |
½ × base × height |
A 3‑m‑by‑4‑m right triangle → A = 6 m² |
| Parallelogram | 2 × (base + side) |
base × height |
Base 6 m, side 4 m, height 3 m → A = 18 m² |
| Trapezium (trapezoid) | a + b + c + d |
½ × (sum of two bases) × height |
Bases 10 m & 6 m, height 4 m → A = 32 m² |
| Circle | 2 × π × radius (≈ 6.28 × radius) |
π × radius² |
Radius 2 m → P ≈ 12.57 m, A ≈ 12.57 m² |
Quick note on π (pi): Use 3.14 or 22/7. For exam shortcuts, you can use
π ≈ 3.14.
4. 3‑D Shapes – Quick Rules
| Shape | Volume Formula | Surface Area Formula | Tiny Example |
|---|---|---|---|
| Cube | side³ |
6 × side² |
Side 2 m → V = 8 m³, SA = 24 m² |
| Rectangular Prism (Box) | length × breadth × height |
2 × (lb + bh + hl) |
3 m × 2 m × 1 m → V = 6 m³, SA = 22 m² |
| Cylinder | π × radius² × height |
2 × π × radius × (radius + height) |
Radius 1 m, Height 3 m → V ≈ 9.42 m³, SA ≈ 28.27 m² |
| Cone | ⅓ × π × radius² × height |
π × radius × (radius + slant height) |
Radius 1 m, Height 2 m → V ≈ 2.09 m³ |
| Sphere | (4/3) × π × radius³ |
4 × π × radius² |
Radius 1 m → V ≈ 4.19 m³, SA ≈ 12.57 m² |
Slant height (cone/cylinder): the slanted side you’d see if you cut the cone or cylinder straight across.
(For exam, you’ll usually be given the slant height directly.)
5. Sample Problems (with One‑Line Solutions)
-
Perimeter of a rectangle
A garden is 8 m long and 5 m wide. How long is the fence?
P = 2 × (8 + 5) = 26 m -
Area of a triangle
A triangular plot has base 9 m and height 4 m.
A = ½ × 9 × 4 = 18 m² -
Volume of a cube
A storage box has side 3 m.
V = 3³ = 27 m³ -
Surface area of a cylinder
A can has radius 0.5 m and height 2 m.
`SA = 2 × π