Mensuration – Study Notes
(Tamil Nadu Patwari / Lekhpal – Quantitative Aptitude)
Why Mensuration?
In a Patwari exam, questions on the area of land parcels, volume of water containers, and perimeter of fenced fields are common. A solid grasp of the basic formulae, their derivations, and a few trick techniques will save you time and avoid mistakes.
1. Basics
| Symbol | Meaning | Units (SI) |
|---|---|---|
| (l,w,h) | Length, Width, Height | metres (m) |
| (r) | Radius | metres (m) |
| (\theta) | Angle in degrees | degrees |
| (A) | Area | (m^2) |
| (P) | Perimeter / Circumference | metres (m) |
| (V) | Volume | (m^3) |
| (\pi) | Ratio of circumference to diameter | (3.1415926535…) |
| (\phi) | Euler’s totient function (rarely used in mensuration) | — |
Tip: Always keep track of units—if you mix feet with metres you’ll get the wrong answer.
2. Plane Figures
| Figure | Area Formula | Circumference / Perimeter |
|---|---|---|
| Rectangle | (A = l \times w) | (P = 2(l + w)) |
| Square | (A = a^2) | (P = 4a) |
| Triangle | (general) (A = \frac{1}{2} \times \text{base} \times \text{height}) (right‑angled) (A = \frac{1}{2} l w) (equilateral) (A = \frac{\sqrt{3}}{4} a^2) |
(general) (P = a+b+c) |
| Circle | (A = \pi r^2) | (P = 2\pi r = \pi d) |
| Parallelogram | (A = \text{base} \times \text{height}) | (P = 2(\text{base} + \text{side})) |
| Trapezium (Trapezoid) | (A = \frac{1}{2} (a+b) h) | (P = a+b+c+d) |
Derivation Highlights
-
Circle Area
[ A = \underbrace{\pi r^2}_{\text{limit of regular polygons}} ] Proof Sketch: Inscribe a regular (n)-gon → area (= \frac{1}{2} n r^2 \sin\frac{2\pi}{n}). Let (n \to \infty), (\sin x \approx x) → (A = \pi r^2). -
Parallelogram Area
[ A = \text{base} \times \text{height} = b h ] Proof Sketch: Drop altitudes, the parallelogram becomes a set of (b) rectangles of height (h). -
Trapezium Area
[ A = \frac{1}{2}(a+b)h ] Proof Sketch: Divide trapezium into a rectangle + two right‑angled triangles.
3. Solids (Prisms, Cylinders, Cones, Spheres, Pyramids)
3.1 Cylinders & Pyramids (Common in land & water bodies)
| Solid | Volume (V) | Surface Area (S) |
|---|---|---|
| Right Circular Cylinder | (V = \pi r^2 h) | (S = 2\pi r h + 2\pi r^2) |
| Right Circular Cone | (V = \frac{1}{3}\pi r^2 h) | (S = \pi r l + \pi r^2) (l = slant height) |
| Right Circular Prism (rectangular) | (V = lwh) | (S = 2(lw + lh + wh)) |
| Right Square Pyramid | (V = \frac{1}{3} a^2 h) | (S = a^2 + 2a\ell) (ℓ = slant height) |
Derivations
-
Cone Volume: Take a cone, cut into infinitesimal disks of thickness (dz).
[ dV = \pi r(z)^2 dz ] Since (r = \frac{R}{H} z) (linear relation along height), integrate from (0) to (H).
[ V = \int_0^H \pi \left(\frac{R}{H} z\right)^2 dz = \frac{1}{3}\pi R^2 H ] -
Sphere Surface Area (the "crucial trick"):
The surface area (S) can be derived by considering the spherical shell method or by using the fact that the volume of a sphere with radius (r) is ( \frac{4}{3}\pi r^3).
Differentiating (V) w.r.t. (r) gives: [ \frac{dV}{dr} = 4\pi r^2 = S ] Thus (S = 4\pi r^2).
4. Problem‑Solving Techniques
| Situation | Trick / Shortcut |
|---|---|