Mensuration is the branch of mathematics that deals with measuring geometric shapes — their lengths, areas, and volumes. In 2D mensuration, you are working entirely on a flat plane: finding the perimeter (boundary length) and area (surface enclosed) of shapes like triangles, quadrilaterals, circles, and polygons.
Look — the reason mensuration is feared in SSC CGL is not because the formulas are hard. It is because questions disguise simple shapes behind word problems, or combine two shapes, or ask about ratios of areas when dimensions change. Once you can pattern-match the underlying shape, the formula is just a lookup.
Here is a useful way to think about it. Every 2D mensuration problem is doing one of three things:
The analogy that helps: think of a plot of land. The perimeter is the fencing you need to buy. The area is the tiles you need to lay. Different costs, different calculations. Many real-world SSC problems literally use plot-of-land language, so this mental model maps directly.
One more thing before the formulas — in SSC CGL, the 3:4:5 triangle and the Pythagorean relationship appear far more than any exotic polygon. Recognize right-triangle side ratios instantly. If sides are in ratio 3:4:5, or 5:12:13, or 8:15:17, you are dealing with a right triangle, and the area formula simplifies to (1/2) × leg₁ × leg₂ — no Heron's formula needed.
Triangle
| Quantity | Formula |
|---|---|
| Area (base-height known) | (1/2) × base × height |
| Area (all sides known) | Heron's: √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2 |
| Area (equilateral, side a) | (√3/4) × a² |
| Perimeter | a + b + c |
For a right triangle with legs p and q: Area = (1/2)pq. You don't need Heron's formula when you recognize it is a right triangle.
Rectangle and Square
| Shape | Area | Perimeter |
|---|---|---|
| Rectangle (l × b) | l × b | 2(l + b) |
| Square (side a) | a² | 4a |
| Square (diagonal d) | d²/2 | 2√2 × d |
Circle and Related
| Quantity | Formula |
|---|---|
| Area | πr² |
| Circumference | 2πr |
| Area of sector (angle θ°) | (θ/360) × πr² |
| Arc length (angle θ°) | (θ/360) × 2πr |
| Area of ring (radii R, r) | π(R² − r²) |
Parallelogram and Trapezium
| Shape | Area |
|---|---|
| Parallelogram | base × height |
| Rhombus | (1/2) × d₁ × d₂ (diagonals) |
| Trapezium | (1/2) × (sum of parallel sides) × height |
Regular Polygon (n sides, side a)
Area = (na²/4) × cot(π/n). You won't need this in the exam hall. What you need: hexagon area = (3√3/2) × a², and regular hexagon made of 6 equilateral triangles.
When two similar figures have sides in ratio k : 1, their:
k : 1k² : 1This single rule eliminates most of the calculation in ratio-type questions. If a circle's radius doubles, its area quadruples. If a square's side becomes 3 times, its area becomes 9 times.
Generalization for circles: A₁/A₂ = (r₁/r₂)². If r₁ = 1 and r₂ = √2, then A₂/A₁ = (√2/1)² = 2. This directly solves tyre-area type questions without computing πr² at all.
Heron's formula is slow. Use it only when:
Avoid it when:
(√3/4)a²)For the PYQ triangle with sides 15, 20, 25 (ratio 3:4:5), applying Heron's formula would take 60+ seconds. Spotting the right triangle takes 5 seconds: Area = (1/2) × 15 × 20 = 150 m².
A shockingly common error is computing area when perimeter is asked, and vice versa. SSC setters know this and sometimes put both area and perimeter as answer choices. Read the question word. Mark it before you start computing.
SSC CGL frequently gives you a figure that is a combination: rectangle + semicircle, square + triangle, circle − rhombus, etc. The method is always:
Draw the decomposition on paper. Do not try to hold it in your head.
If sides are given as a ratio, check immediately: does a² + b² = c²? For 3:4:5, check 9+16=25. Yes. For 5:12:13, check 25+144=169. Yes. Once confirmed right triangle, Area = (1/2)×leg×leg. No Heron's needed.
Micro-example: sides 15, 20, 25. Check: 15²+20²=225+400=625=25². Right triangle. Area=(1/2)×15×20=150. Time: 8 seconds. Heron's formula on the same: s=30, compute √(30×15×10×5)=√22500=150. Time: 45 seconds. You save 37 seconds per question.
When two circles (or similar figures) have radii r₁ and r₂, their areas are in ratio r₁²:r₂². Never compute πr² for both — just square the radii and take the ratio.
Micro-example: radii 1 and √2. Area ratio = 1²:(√2)² = 1:2. If smaller area is 100 cm², larger = 200 cm². Standard method (computing both πr² values): 30 seconds. This pattern: 6 seconds. 5-step reduction.
When a solid is melted and recast, total volume is conserved. For spheres: (4/3)πR³ = sum of (4/3)πrᵢ³, which simplifies to R³ = r₁³ + r₂³ + r₃³. Plug in cube values directly.
Micro-example: radii 3, 4, 5. R³ = 27+64+125 = 216. R = 6. You never write out (4/3)π at all. Standard method: 50 seconds. This substitution: 12 seconds. 4-step reduction.
If one sphere of radius R is melted into n identical spheres of radius r, then R³ = n × r³, so r = R/n^(1/3). Equivalently, n = (R/r)³.
Micro-example: hemisphere radius 12 melted into hemispheres of radius 3. n = (12/3)³ = 4³ = 64. Never write the (2/3)π factor — it cancels. Standard method: 40 seconds. This pattern: 8 seconds. You save 32 seconds.
When a sphere fits exactly inside a cylinder (sphere diameter = cylinder diameter = cylinder height), volume of cylinder = π r² × 2r = 2πr³. Volume of sphere = (4/3)πr³. Vacant space = 2πr³ − (4/3)πr³ = (2/3)πr³ = (1/3) × sphere volume. So vacant space is always exactly half the sphere's volume.
Micro-example: sphere volume = 36π. Vacant space = 36π/2 = 18π. You do zero extra computation once you internalize the ratio. Standard method (compute cylinder volume separately): 35 seconds. This shortcut: 5 seconds.
When a mensuration question hits, run this decision tree before picking up your pen:
Step 1 — Identify the shape(s). Is it a single standard shape or a compound? Write the shape name.
Step 2 — What is asked? Area, perimeter, volume, ratio, or a missing dimension? Mark it.
Step 3 — Do you have enough data? For area of triangle: do you have base+height, or all three sides? If all three sides, check for right-triangle ratio before reaching for Heron's.
Step 4 — Is this a ratio/scaling question? If so, use the ratio rule: areas scale as square of linear dimensions, volumes as cube. Avoid computing actual values when a ratio suffices.
Step 5 — Is this a melting/recasting question? Set volumes equal. Cancel the common (4/3)π factor. Work with cubes of radii directly.
Step 6 — Compute. Use the simplest applicable formula. Approximate π ≈ 22/7 or 3.14 as needed. Cross-check units at the end.
Total decision time: 10-15 seconds. You should be computing by second 15.
Why this question: Tests whether you recognize volume-conservation in melting problems and whether you can add cubes quickly.
Solving path: Write R³ = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216. Then R = ∛216 = 6. The (4/3)π cancels from both sides — never write it. Answer: 6 cm.
Why this question: Classic disguised right-triangle. The entire difficulty is in recognizing 3:4:5, not in any formula.
Solving path: Sides are 3k:4k:5k with 12k = 60, so k = 5. Sides = 15, 20, 25. Check: 15² + 20² = 225 + 400 = 625 = 25². Right triangle. Area = (1/2) × 15 × 20 = 150 m². Done.
Why this question: Tests the cylinder-sphere relationship. Many candidates compute cylinder volume from scratch; knowing the (1/3) × sphere volume shortcut makes this a 5-second question.
Solving path: (4/3)πr³ = 36π → r³ = 27 → r = 3. Cylinder: radius = 3, height = 6. Volume of cylinder = π × 9 × 6 = 54π. Vacant = 54π − 36π = 18π cm³. Or use the shortcut: vacant = (1/2) × sphere = 18π instantly.
Why this question: Equating two volume formulas. A direct algebraic manipulation that SSC repeats in varied forms.
Solving path: Set (2/3)πR³ = (1/3)πR²H. Cancel (1/3)πR² from both sides: 2R = H. So H = 2R.
Why this question: The ratio-of-areas shortcut. If you compute πr² for both circles, you lose 30 seconds. If you use the square-of-radius ratio, you spend 6 seconds.
Solving path: Area ratio = (√2)²/(1)² = 2/1. Larger area = 2 × 100 = 200 cm².
Why this question: Surface area of original vs. total surface area after breaking into n equal pieces. Different from volume-ratio. Many candidates confuse volume ratio with surface area ratio here.
Solving path: Volume of one small sphere = (1/8) of original. So r³ = (10³)/8 = 125, r = 5. Surface area of original = 4π × 100 = 400π. Total SA of 8 small = 8 × 4π × 25 = 800π. Ratio = 400π : 800π = 1 : 2.
Why this question: The n = (R/r)³ pattern for melting hemispheres. If you set up volume equations with (2/3)π, it works but takes longer. The cube-ratio shortcut is clean.
Solving path: n = (R/r)³ = (12/3)³ = 4³ = 64.
Why this question: Pyramid volume is a formula-recall question. Straightforward but often missed because candidates confuse (1/3) factor with (2/3) from hemisphere.
Solving path: Volume = (1/3) × base area × height = (1/3) × 60 × 9 = 180 cm³.
Using Heron's formula on a 3:4:5 triangle. This wastes 30-40 seconds. Always check for Pythagorean triplet ratios before reaching for Heron's. If sides are multiples of 3, 4, 5 (or 5, 12, 13 or 8, 15, 17), stop and use the legs directly.
Forgetting the (1/3) factor in pyramid and cone volume. Pyramid = (1/3) × base × height. Cone = (1/3) × πr²h. The (1/3) is structural — these are pointed solids. Hemisphere uses (2/3)πr³. These three factors get swapped under exam pressure.
Confusing diameter with radius. When a sphere fits inside a cylinder, the cylinder's height equals the sphere's diameter (= 2r), not the radius. Plugging r instead of 2r into cylinder height gives a volume off by factor of 2.
Applying the area-ratio rule to perimeters. If areas scale as k², perimeters scale as k (not k²). This is a trap in questions like: "If the area of a square doubles, by what factor does the perimeter change?" Answer: √2, not 2.
Not cancelling π or common factors before computing. In ratio questions involving circles or spheres, π always cancels. In melting questions, (4/3)π cancels. Writing these out and then cancelling wastes time and introduces arithmetic errors.
Adding volumes of different-shape solids incorrectly. When a composite solid has a hemisphere on top of a cone, the total volume is the sum of both individual volumes. A common error is using the wrong formula for one part (e.g., using full sphere formula instead of hemisphere).