Mensuration – Quick Study Guide
(For Meghalaya Police Constable – Quantitative Aptitude)
1. Simple Explanation
Mensuration is the branch of mathematics that deals with finding the area, perimeter, volume, and surface area of geometric shapes. In the Police Constable exam, you will mainly encounter problems involving:
| Shape | 2‑D (Area & Perimeter) | 3‑D (Volume & Surface Area) |
|---|---|---|
| Square, Rectangle, Triangle, Trapezium | ✔ | – |
| Circle | ✔ | – |
| Cylinder, Cone, Sphere, Prism, Pyramid | – | ✔ |
The key to success is memorising a few core formulas and practicing them quickly.
2. Key Points & Formula Cheat‑Sheet
| Quantity | Formula | Notes |
|---|---|---|
| Perimeter | P = 2×(l + w) for rectangle | For square, l = w |
| Area of Rectangle / Square | A = l × w | For square, A = s² |
| Area of Triangle | A = ½ × base × height | Height is perpendicular to base |
| Area of Parallelogram | A = base × height | Same as triangle but no ½ |
| Area of Trapezium | A = ½ × (a + b) × h | a & b are parallel sides |
| Area of Circle | A = π r² | π ≈ 3.14 |
| Circumference of Circle | C = 2πr | |
| Surface Area of Cylinder | SA = 2πr(h + r) | 2πr² + 2πrh |
| Volume of Cylinder | V = πr²h | |
| Surface Area of Cone | SA = πr(l + r) | l = slant height |
| Volume of Cone | V = (1/3)πr²h | |
| Surface Area of Sphere | SA = 4πr² | |
| Volume of Sphere | V = (4/3)πr³ | |
| Surface Area of Cube | SA = 6s² | |
| Volume of Cube | V = s³ | |
| Surface Area of Rectangular Prism | SA = 2(lw + lh + wh) | |
| Volume of Rectangular Prism | V = lwh | |
| Surface Area & Volume of Pyramid | SA = base area + ½ × base perimeter × slant height V = (1/3) × base area × height |
For right square pyramid, base = s², height = h |
Tip: Always identify the shape first; then pick the corresponding formula.
3. Worked Example
Problem (Typical for M.P.C. Exam)
A cylindrical water tank has a radius of 4 m and a height of 12 m. The tank is filled with water to a depth of 8 m. What is the volume of water inside the tank? Also, find the surface area of the water surface.
Step‑by‑Step Solution
- Identify the shape: Cylinder.
- Given data:
- Radius, r = 4 m
- Height of water, h = 8 m (only the depth of water matters for volume)
- Volume of water
[ V = \pi r^2 h = \pi (4^2)(8) = \pi \times 16 \times 8 = 128\pi \text{ m}^3 ] Numerical value (π ≈ 3.14):
[ V \approx 128 \times 3.14 = 401.92 \text{ m}^3 ] - Surface area of the water surface
This is just the area of the top circular face:
[ A_{\text{top}} = \pi r^2 = \pi (4^2) = 16\pi \text{ m}^2 \approx 50.24 \text{ m}^2 ]
Answer:
- Volume of water ≈ 401.92 m³
- Surface area of water surface ≈ 50.24 m²
4. Quick Revision Summary
| Concept | Formula | Quick Mnemonic |
|---|---|---|
| Perimeter of rectangle | 2(l+w) | “Two‑Times length plus width” |
| Area of triangle | ½×b×h | “Half the base times height” |
| Area of circle | πr² | “Pi times radius squared” |
| Circumference | 2πr | “Two times pi times radius” |
| Volume of cylinder | πr²h | “Pi times base area times height” |
| Surface area of sphere | 4πr² | “Four times pi times radius squared” |
| Volume of cone | 1/3 πr²h | “One‑third the base area times height” |
- Always check units (m, cm, km, etc.) and convert if necessary.
- For prisms and pyramids, remember the ¾‑formula for volume.
- In trapezium problems, use the average of parallel sides.
- When a problem gives “height” that is actually a slant height, use the right‑triangle relationship:
( l = \sqrt{h^2 + r^2