Mensuration – Quantitative Aptitude for Navy SSR
What it is
Mensuration is the branch of mathematics that deals with the measurement of geometric figures – lengths, areas, volumes, and surface areas of 2‑D and 3‑D shapes.
In the Navy SSR, questions may involve:
- basic geometry (triangles, circles, squares, rectangles)
- composite figures (L‑shapes, rings, trapeziums)
- 3‑D solids (cubes, cuboids, spheres, cylinders, cones, pyramids)
- surface and volume calculations
- area/volume relationships in similar figures
Why it matters
Strong mensuration skills help you solve time‑based questions quickly, a key requirement for the analytical sections of the test.
1. Simple Explanation
| Shape | Key Formulae | What to remember |
|---|---|---|
| Triangle | Area = ½ × base × height | Base × height (or ½ab sin C) |
| Rectangle / Square | Area = length × breadth | Same for square (l²) |
| Parallelogram / Rhombus | Area = base × height | Rhombus: ½ d₁ d₂ |
| Trapezium | Area = ½ × (a + b) × h | a, b – parallel sides |
| Circle | Area = πr², Circumference = 2πr | π ≈ 3.14 |
| Sphere | Surface = 4πr², Volume = 4/3πr³ | |
| Cylinder | Surface = 2πr(r + h), Volume = πr²h | Lateral = 2πrh |
| Cone | Surface = πr(r + l), Volume = 1/3πr²h | l = slant height |
| Pyramid | Surface = base area + ½ × perimeter × slant height |
Tip: For most SSR questions you only need the basic forms (no slant height unless stated).
2. Key Points
| # | Point | Example |
|---|---|---|
| 1 | Always check units – length, area, volume. A typo can change the answer drastically. | If radius = 5 cm, area = 78.5 cm² (not 78.5 cm). |
| 2 | Use 22/7 for π when time is limited; 3.14 is also acceptable. | 22/7 gives exact rational results. |
| 3 | Similar figures: Area ratio = (scale factor)², Volume ratio = (scale factor)³. | Two cubes with side ratio 2:1 → volume ratio 8:1. |
| 4 | Composite figures: Break into simple shapes, find individual areas, then add/subtract. | L‑shaped board = area of larger rectangle – area of cut‑out rectangle. |
| 5 | Avoid unnecessary algebra. Plug numbers directly when possible. | If a rectangle has area 24 m² and one side = 4 m → other side = 6 m. |
| 6 | Common pitfalls: confusing base & height in a triangle, using diameter instead of radius, missing the ½ factor in trapezium. | |
| 7 | Quick mental shortcuts: | |
| • 0.5 × 2 = 1 (for ½ × 2 = 1) | ||
| • 2πr ≈ 6.28r (use 6.28 for quick circumference). | ||
| • 4/3πr³ ≈ 4.19r³ (volume of sphere). |
3. Worked Example
Problem (typical for Navy SSR)
A right circular cone has a slant height of 12 cm and a radius of 5 cm. Find the volume of the cone. Then, a cylinder is formed by cutting the cone along a plane parallel to its base such that the resulting cone has 1/4 of the volume of the original. Find the height of the cylinder.
Step‑by‑Step
-
Volume of the original cone
[ V_{\text{cone}} = \frac{1}{3}\pi r^{2}h ] We need the height (h). From Pythagoras in the right triangle (slant height, radius, height): [ l^{2} = r^{2} + h^{2} \Rightarrow h = \sqrt{l^{2} - r^{2}} = \sqrt{12^{2} - 5^{2}} = \sqrt{144 - 25} = \sqrt{119}\approx 10.91\text{ cm} ]Now compute the volume: [ V_{\text{cone}} = \frac{1}{3}\pi (5^{2})(10.91) = \frac{1}{3}\pi (25)(10.91) \approx 28.33\pi \approx 89.0\text{ cm}^{3} ]
-
Volume of the smaller cone
It is ¼ of the original: [ V_{\text{small}} = \frac{1}{4} V_{\text{cone}} \approx 22.25\text{ cm}^{3} ] -
Find the height of the cylinder
The cylinder is formed by the remaining part after cutting the cone.
The volume of the cylinder equals the volume of the original cone minus the smaller cone: [ V_{\text{cyl}} = V_{\text{cone}} - V_{\text{small}} = \frac{3}{4} V_{\text{cone}} \approx 66.75\text{ cm}^{3} ]The radius of the cylinder is the same as the cone’s base radius, (r = 5) cm.
Volume of a cylinder: [ V_{\text{cyl}} = \pi r^{2} h_{\text{cyl}} \Rightarrow h_{\text{cyl}} = \frac{V_{\text{cyl}}}{\pi r^{2}} = \frac{66.75}{\pi \times 25} \approx \frac{66.75}{78.54} \approx 0.85\text{ cm} ]Answer: Height