Mensuration – Study Notes for Haryana Police Constable (Quantitative Aptitude)
Mensuration is the branch of mathematics that deals with finding the perimeter, area and volume of various 2‑D and 3‑D figures.
In the Police Constable exam, questions usually involve simple shapes (rectangles, circles, triangles) and a few common solids (cylinder, sphere, cone).
1. Simple Explanation
- Perimeter – The total length of the boundary of a figure.
- Area – The amount of surface the figure covers (in square units).
- Surface Area (SA) – The total area of all the faces of a 3‑D figure (in square units).
- Volume (V) – The amount of space a 3‑D figure occupies (in cubic units).
Tip: Memorise the most common formulas first; the rest can be derived quickly.
2. Key Points & Formulas
| Figure | Perimeter / SA | Area | Volume |
|---|---|---|---|
| Rectangle | (P = 2(l + w)) | (A = l \times w) | – |
| Square | (P = 4a) | (A = a^2) | – |
| Triangle | (P = a+b+c) | (A = \frac{1}{2} \times base \times height) | – |
| Circle | (C = 2\pi r) | (A = \pi r^2) | – |
| Trapezium | (P = a+b+c+d) | (A = \frac{1}{2} \times (a + c) \times h) | – |
| Cube | (SA = 6a^2) | – | (V = a^3) |
| Cuboid | (SA = 2(lw + lh + wh)) | – | (V = l \times w \times h) |
| Sphere | – | – | (V = \frac{4}{3}\pi r^3) |
| Cylinder | – | – | (V = \pi r^2 h) |
| Cone | – | – | (V = \frac{1}{3}\pi r^2 h) |
| Right Circular Cylinder SA | – | – | (SA = 2\pi r(h + r)) |
| Right Circular Cone SA | – | – | (SA = \pi r(r + l)) where (l = \sqrt{r^2 + h^2}) (slant height) |
| Pyramid (square base) | – | – | (V = \frac{1}{3} \times base \times height) |
Common Mistakes to Avoid
- Confusing π with 3.14 – Use π in symbolic form for exact answers; use 3.14 only for approximations.
- Mixing up height and slant height – In cones/pyramids,
h= perpendicular height,l= slant height. - For circles: radius vs diameter – Remember
r = d/2. - Neglecting units – Always keep track of units (sq. cm, cubic cm, etc.).
Quick Mental Calculation Hacks
- Area of circle: ( \pi r^2 \approx 3.14 \times r^2).
- Volume of cylinder: (πr^2h) – multiply (r^2) first, then with π and h.
- Surface area of cube: 6 × side² – easy to remember because all faces equal.
3. Worked Example
Question (Sample):
A cylindrical water tank has a radius of 5 m and a height of 12 m. Find:
a) The volume of the tank (in cubic metres).
b) The total surface area (including top and bottom).
Solution
Given:
(r = 5,\text{m})
(h = 12,\text{m})
a) Volume
(V = πr^2h)
- Compute (r^2 = 5^2 = 25)
- Multiply by (h): (25 \times 12 = 300)
- Multiply by π: (V = 300π \approx 300 \times 3.14 = 942) cubic metres.
Answer: (V = 300π \text{ m}^3 ;(\approx 942 \text{ m}^3))
b) Surface Area
(SA = 2πr(h + r))
- Inside parentheses: (h + r = 12 + 5 = 17)
- Multiply by (r): (5 \times 17 = 85)
- Multiply by (2π): (SA = 2π \times 85 = 170π)
- Approximate: (170 × 3.14 = 533.8) square metres.
Answer: (SA = 170π \text{ m}^2 ;(\approx 534 \text{ m}^2))
4. Quick Revision Summary
| Concept | Key Formula | Quick Mnemonic |
|---|---|---|
| Perimeter of Rectangle | (2(l+w)) | “Double the sum of sides.” |
| Area of Triangle | (\frac{1}{2} \times base \times height) | “Half the base‑height rectangle.” |
| Area of Circle | (\pi r^2 |