📑 Mensuration – Study Notes for Madhya Pradesh Patwari / Lekhpal Exam
Topic: Quantitative Aptitude – Mensuration
Exam: Madhya Pradesh Patwari / Lekhpal (Quantitative Aptitude)
Format: Markdown (easy to read & print)
1️⃣ Simple Explanation
Mensuration is the branch of mathematics that deals with measuring the area, perimeter, volume, and surface area of geometric figures.
In the Patwari exam, you’ll mainly encounter:
- 2‑D shapes – rectangles, squares, triangles, circles, trapeziums, parallelograms, etc.
- 3‑D shapes – cubes, cuboids, cylinders, spheres, cones, pyramids, etc.
Knowing the right formulas and being able to apply them quickly is the key.
2️⃣ Key Points & Formulas
| 2‑D Shape | Area Formula | Perimeter (or Circumference) | Notes |
|---|---|---|---|
| Rectangle | (A = l \times w) | (P = 2(l + w)) | Simple |
| Square | (A = a^2) | (P = 4a) | (l = w = a) |
| Triangle | (A = \frac{1}{2} \times b \times h) | (P = a + b + c) | If right‑angled: (c = \sqrt{a^2 + b^2}) |
| Circle | (A = \pi r^2) | (C = 2\pi r) | (π ≈ 3.14) |
| Parallelogram | (A = b \times h) | (P = 2(a + b)) | Base & side, not base & height |
| Trapezium | (A = \frac{1}{2}(a + b) \times h) | (P = a + b + c + d) | (a, b) are parallel sides |
| 3‑D Shape | Volume | Surface Area (SA) | Notes |
|---|---|---|---|
| Cube | (V = a^3) | (SA = 6a^2) | |
| Cuboid | (V = lwh) | (SA = 2(lw + wh + lh)) | |
| Cylinder | (V = \pi r^2 h) | (SA = 2\pi r(r + h)) | Lateral + top + bottom |
| 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)) | (l) = slant height |
| Pyramid | (V = \frac{1}{3} B h) | (SA = B + \frac{1}{2} P \times s) | (B) = base area, (P) = perimeter of base, (s) = slant height |
Quick Mnemonics
| Quantity | Mnemonic |
|---|---|
| Area of Circle | “π r²” – “π” stands for “pi” (3.14), “r²” is “radius squared.” |
| Volume of Cylinder | “π r² h” – think “π r²” is the base area, then multiply by height. |
| Surface Area of Cylinder | “2π r (r + h)” – 2πr = circumference of base, times (r + h) = radius + height. |
| Area of Triangle | “½ × base × height” – half of rectangle with same base & height. |
| Pyramid Volume | “⅓ × Base Area × Height” – similar to cone but no π. |
3️⃣ Worked Example (Patwari‑style)
Problem:
A farmer has an irregular plot that can be approximated as a trapezium.
The parallel sides are 12 m and 20 m.
The height (distance between them) is 8 m.
The plot is fenced on all four sides.
Q1: Find the area of the plot.
Q2: Find the total length of the fence (perimeter).
Q3: After selling the plot, the farmer wants to build a rectangular house on the plot. The house will occupy 60 % of the plot’s area. What will be the perimeter of the house if it is a square?
Step‑by‑Step Solution
Q1 – Area of the trapezium
[ A = \frac{1}{2} \times (a + b) \times h ] where (a = 12,\text{m}), (b = 20,\text{m}), (h = 8,\text{m}).
[ A = \frac{1}{2} (12 + 20) \times 8 = \frac{1}{2} \times 32 \times 8 = 16 \times 8 = 128 \text{ m}^2 ]
Q2 – Perimeter of the trapezium
We need the lengths of the two non‑parallel sides.
Assume the trapezium is isosceles, so the two slant sides are equal.
Let (x) be the length of each slant side.
Using the right triangle formed by half the difference of the bases:
[ x = \sqrt{h^2 + \left(\frac{b - a}{2}\right)^2} = \sqrt{8^2 + \left(\frac{20 - 12}{2}\right)^2}