Mensuration – Quantitative Aptitude (Karnataka Police Constable)
Audience: Candidates preparing for the Karnataka Police Constable (PCC) exam
Depth: MIT‑level (first principles, rigorous proofs, edge cases)
Format: Markdown study notes with derivations, theory, examples, and a challenging problem
1. Introduction
Mensuration is the branch of mathematics that deals with finding the measure of geometric figures. In quantitative aptitude it is used to solve problems involving areas, perimeters, volumes, surface areas, and related quantities.
Why “first principles”?
In aptitude tests you are rarely given a formula without justification. Knowing why the formula works saves time, helps avoid mistakes, and gives you confidence to handle unfamiliar shapes.
2. Fundamental Definitions
| Symbol | Geometric Quantity | Units |
|---|---|---|
| (A) | Area | (\text{units}^2) |
| (P) | Perimeter / Circumference | (\text{units}) |
| (V) | Volume | (\text{units}^3) |
| (S) | Surface Area | (\text{units}^2) |
| (r) | Radius | (\text{units}) |
| (h) | Height / Altitude | (\text{units}) |
| (\theta) | Angle (in radians) | rad |
3. 2‑D Geometry
3.1. Triangle
Area via base × height
[ A = \frac{1}{2},b,h ]
Derivation (first‑principle)
Take a right‑angled triangle. By definition, a rectangle of sides (b) and (h) has area (b,h). The triangle is exactly half of that rectangle (by dropping the diagonal). ∎
Heron’s Formula (general triangle)
[ A = \sqrt{s(s-a)(s-b)(s-c)}, \qquad s=\frac{a+b+c}{2} ]
Derivation (using semiperimeter and Pythagoras)
Drop an altitude (h) from vertex (C) to side (AB). Using the Pythagorean theorem on the two right triangles, express (h) in terms of the sides, substitute into (\frac12bh), and simplify.
3.2. Circle
Area
[ A = \pi r^2 ]
Derivation (calculus)
Circle can be seen as a union of infinitesimal concentric rings of radius (x) and width (dx).
Area of a ring = circumference × width = (2\pi x,dx).
Integrate from (0) to (r):
[ A = \int_{0}^{r} 2\pi x,dx = \pi r^2 ]
Perimeter (Circumference)
[ P = 2\pi r ]
3.3. Regular Polygon
If a regular polygon has (n) sides of length (s) and circumradius (R):
[ A = \frac{1}{2}n,s,R ]
Derivation
Split into (n) isosceles triangles with vertex at the center. Each has area (\frac12,s,R\sin(\frac{2\pi}{n})). Summation gives the above formula. For a regular polygon inscribed in a circle, (s = 2R\sin(\frac{\pi}{n})).
3.4. Special Quadrilaterals
| Shape | Area Formula | Notes |
|---|---|---|
| Rectangle | (A = l\cdot w) | |
| Parallelogram | (A = b,h) | Height perpendicular to base |
| Trapezium | (A = \frac{1}{2}(a+b)h) | (a,b) are parallel sides |
| Rhombus | (A = \frac{1}{2}d_1d_2) | (d_1,d_2) diagonals |
4. 3‑D Geometry
4.1. Prisms & Pyramids
| Type | Volume | Surface Area |
|---|---|---|
| Prism (any cross‑section) | (V = A_{\text{base}}\cdot h) | (S = 2A_{\text{base}} + \sum_{\text{lateral}} \text{area}) |
| Pyramid | (V = \frac{1}{3}A_{\text{base}}\cdot h) | (S = A_{\text{base}} + \sum_{\text{lateral faces}}) |
Derivation (prism)
Imagine the prism as (h) copies of the base area stacked. Each slice has area (A_{\text{base}}). Integrate over (