Mensuration – Quantitative Aptitude Notes
(Arunachal Pradesh Patwari / Lekhpal – Advanced / MIT‑style treatment)
Mensuration is the branch of mathematics that deals with the measurement of the size of geometric figures: lengths, areas, and volumes.
In the Patwari exam it appears in a variety of contexts – land‐area calculations, block‑and‑tier problems, construction cost estimates, etc.
These notes start from first principles, build up the theory rigorously, and finish with a challenging problem that tests both algebraic manipulation and geometric insight.
1. Foundations
| Concept | Symbol | Definition |
|---|---|---|
| Length (linear measure) | (L) | Distance between two points. |
| Perimeter | (P) | Sum of all side‑lengths. |
| Area | (A) | Measure of the two‑dimensional extent. |
| Volume | (V) | Measure of the three‑dimensional extent. |
| Surface area | (SA) | Total area of all bounding surfaces. |
First Principle – All formulas can be derived by considering an infinitesimal element of the figure (a strip, a ring, a slice, etc.) and summing (integrating) over the entire domain.
1.1 Geometry of the Plane
For a simple shape (S) in the plane, the area can be expressed as
[ A(S)=\iint_{S} 1,dA ]
where (dA) is the infinitesimal area element.
When (S) has a simple Cartesian description, this reduces to a single integral:
[ A=\int_{x_1}^{x_2} \big[,f_{\text{top}}(x)-f_{\text{bottom}}(x),\big],dx ]
2. Basic Two‑Dimensional Shapes
Below we give derivations (not just memorised formulas) for the most common shapes.
All derivations are carried out in Cartesian or polar coordinates; the alternative method (e.g., using symmetry) is also noted.
| Shape | Formula | Derivation Sketch | Key Insight |
|---|---|---|---|
| Rectangle | (A=ab) | (A=\int_0^a b,dx) | Area is product of side lengths. |
| Triangle | (A=\frac{1}{2}ab) | (A=\int_0^a \frac{b}{a}x,dx=\frac{ab}{2}) | Triangle is half a parallelogram. |
| Parallelogram | (A=ab\sin\theta) | (A=\int_0^a b\sin\theta,dx) | Height is (b\sin\theta). |
| Circle | (A=\pi r^2) | (A=\int_0^{2\pi}\int_0^r r',dr'd\theta=\pi r^2) | Polar coordinates simplify. |
| Ellipse | (A=\pi ab) | (A=\int_0^{2\pi}\frac{a b}{2},d\theta) | Use parametric form (x=a\cos\theta, y=b\sin\theta). |
| Regular Polygon | (A=\frac{1}{2}Ps) | (A=\frac{n}{2}ab\sin\frac{2\pi}{n}) | Divide into (n) isosceles triangles. |
| Sector | (A=\frac{1}{2}r^2\theta) | (A=\int_0^\theta \frac{1}{2}r^2,d\phi) | Area proportional to angle. |
Remember: For any shape that can be tiled by congruent elementary units (rectangles, triangles), the total area is the sum of unit areas.
3. Three‑Dimensional Shapes – Volume & Surface Area
3.1 Solids of Revolution
For a function (y=f(x)) rotated about the (x)-axis:
- Disk Method
[ V = \pi \int_{x_1}^{x_2} [f(x)]^2,dx ] - Shell Method (rotated about the (y)-axis)
[ V = 2\pi \int_{y_1}^{y_2} y,x(y),dy ]
These are special cases of the general formula
[ V = \iiint_{\Omega} 1,dV ]
3.2 Classic Solids
| Solid | Volume | Surface Area | Derivation Note |
|---|---|---|---|
| Cylinder | (V=\pi r^2 h) | (SA=2\pi rh+2\pi r^2) | Slice into disks; lateral area from (\int 2\pi r,dz). |
| Cone | (V=\frac{1}{3}\pi r^2 h) | (SA=\pi r (r+s)) with (s=\sqrt{r^2+h^2}) | Cone is a special frustum with one base collapsed. |
| Sphere | (V=\frac{4}{3}\pi r^3) | (SA=4\pi r^ |