Mensuration — 2D & 3D — IBPS PO Quant
Mensuration is the most formula-dependent quant topic: get the right formula and the answer is one substitution away; reach for the wrong one and no cleverness saves you. IBPS asks 1–2 directly, but the same area/volume formulas appear inside DI and geometry questions, so the payoff is broader than the direct count. The whole chapter is a compact formula set plus the discipline of matching the question's words ("curved surface", "diagonal", "perimeter") to the exact formula. Memorise the set, and these are certain marks.
1. What IBPS actually asks
- 2D: area and perimeter of squares, rectangles, triangles, circles, and the "cost of fencing/flooring" word problems.
- 3D: volume and surface area of cubes, cuboids, cylinders, cones, spheres — and "how many small cubes fit", "water in a tank", "melting and recasting".
2. The 2D formula set
| Shape | Area | Perimeter |
|---|---|---|
| Square (side a) | a² | 4a; diagonal a√2 |
| Rectangle (l, b) | l·b | 2(l+b); diagonal √(l²+b²) |
| Triangle (base b, height h) | ½·b·h | sum of sides |
| Equilateral triangle (side a) | (√3/4)a² | 3a |
| Circle (radius r) | πr² | 2πr (circumference) |
| Parallelogram | base × height | 2(sum of adjacent sides) |
| Rhombus (diagonals d₁, d₂) | ½·d₁·d₂ | 4 × side |
| Trapezium (parallel sides a, b; height h) | ½(a+b)·h | sum of sides |
Heron's formula (triangle from 3 sides a,b,c): s = (a+b+c)/2, Area = √(s(s−a)(s−b)(s−c)).
3. The 3D formula set
| Solid | Volume | Total surface area |
|---|---|---|
| Cube (a) | a³ | 6a² |
| Cuboid (l,b,h) | l·b·h | 2(lb+bh+hl) |
| Cylinder (r,h) | πr²h | 2πr(r+h); curved = 2πrh |
| Cone (r,h,slant l) | ⅓πr²h | πr(r+l); curved = πrl |
| Sphere (r) | (4/3)πr³ | 4πr² |
| Hemisphere (r) | (2/3)πr³ | 3πr² (total); curved 2πr² |
Cone slant height: l = √(r² + h²).
4. Match the words to the formula
The trap is not the arithmetic — it's picking the right formula from the wording:
- "Curved / lateral surface area" ≠ total surface area (exclude the top/bottom).
- "Cost of fencing / painting a boundary" → perimeter/circumference. "Cost of flooring / turfing" → area.
- "Diagonal" of a rectangle = √(l²+b²); of a cube = a√3; of a cuboid = √(l²+b²+h²).
- "Melted and recast" → volume is conserved; set old volume = new volume.
Underline the key noun (area / perimeter / curved surface / volume) before choosing a formula. Most mensuration errors are a right calculation of the wrong quantity.
5. Handy shortcuts
- Volume conserved on recasting: a sphere melted into small spheres or a cylinder into a cone — equate volumes to find the count or the new dimension.
- Ratio scaling: if all linear dimensions scale by k, area scales by k² and volume by k³. ("Radius doubles ⇒ volume ×8".)
- π cancels in most ratio/comparison questions — don't carry it if it will cancel.
- Number of small cubes in a big cube = (big edge / small edge)³.
Solved examples
Q1 (2D). The area of a circle is 154 cm². Its circumference?
Show explanation
Solution. πr² = 154 ⇒ (22/7)r² = 154 ⇒ r² = 49 ⇒ r = 7. Circumference = 2πr = 2×(22/7)×7 = 44 cm.
Q2 (cost). A rectangular field 20 m × 15 m is fenced at ₹10/m. Cost?
Show explanation
Solution. Perimeter = 2(20+15) = 70 m ⇒ cost = 70 × 10 = ₹700.
Q3 (3D volume). A cylinder has radius 7 cm and height 10 cm. Volume?
Show explanation
Solution. πr²h = (22/7)×49×10 = 1540 cm³.
Q4 (recast). A sphere of radius 6 cm is melted into small spheres of radius 3 cm. How many?
Show explanation
Solution. Volumes: (4/3)π6³ ÷ (4/3)π3³ = 6³/3³ = 216/27 = 8.
Q5 (scaling). If a cube's edge is increased by 50%, its volume increases by?
Show explanation
Solution. Volume ×(1.5)³ = ×3.375 ⇒ 237.5% increase.
7. The protocol
- Underline the quantity asked — area, perimeter, curved vs total surface, volume.
- Pick the exact formula from the memorised set; use π = 22/7 when the radius is a multiple of 7.
- For recasting/melting, set old volume = new volume.
- For scaling ratios, remember area ∝ k², volume ∝ k³, and cancel π.
- Substitute and compute — the answer is one clean step once the formula is right.
