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Numerical Ability · AFCAT

NA09 — Mensuration (3D)

📦 Numerical Ability – NA09 AFCAT Level ☆ Low Priority
📌 AFCAT Focus 2022–2026: 3D mensuration is a low-weightage chapter (0–1 question). Focus only on cube, cuboid, cylinder, and sphere volume/surface area. Cone and frustum occasionally appear. The key trick: when a solid is melted/recast, volume stays constant.

1. 3D Shapes — Surface Area & Volume

Fig. 1.1 — 3D Solids: All Formulas at a Glance
3D SOLIDS — Surface Area & Volume Formulas SOLID LATERAL / CURVED SURFACE & TOTAL SURFACE AREA VOLUME CUBE side = a LSA = 4a²  |  TSA = 6a² Lateral = 4 faces · Total = 6 faces a³ CUBOID l, b, h LSA = 2h(l+b)  |  TSA = 2(lb+bh+hl) Diagonal = √(l²+b²+h²) l × b × h CYLINDER r, h CSA = 2πrh  |  TSA = 2πr(h + r) + 2 circular ends for TSA πr²h CONE r, h, l=slant CSA = πrl  |  TSA = πr(l + r) Slant height l = √(r² + h²) ⅓ πr²h SPHERE radius r SA = 4πr²  (entire surface) No separate LSA — it's a full curved surface ⁴⁄₃ πr³ HEMISPHERE CSA = 2πr²  |  TSA = 3πr²  (curved + flat base) ⅔ πr³
💡 Conversion of Solids (Melting & Recasting): When one solid is melted and recast into another, Volume is conserved.
Volume of original solid = Total volume of new solids formed.
Example: A sphere of radius 3 cm is melted into small spheres of radius 0.5 cm. Number = Volume(large) / Volume(small) = (4/3)π×27 / (4/3)π×0.125 = 27/0.125 = 216.
✎ Worked Example — Cylinder Volume
A cylindrical tank has radius 7 m and height 10 m. Find its capacity in litres. (π = 22/7)
Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 7 × 10 = 1,540 m³
1 m³ = 1,000 litres → Capacity = 15,40,000 litres
✔ Capacity = 15,40,000 litres
Topic BFrustum of a ConeAFCAT Occasionally Tested
What is it
A frustum is the portion of a cone that remains after cutting off the top with a plane parallel to the base. It has two circular faces (radii R at base and r at top) and a slant surface.
Slant Height
l = √[h² + (R−r)²] where h = vertical height, R = base radius, r = top radius.
CSA
Curved Surface Area = π(R + r)l
TSA
Total Surface Area = π(R + r)l + πR² + πr²   (CSA + both circular faces)
Volume
V = (πh/3)(R² + r² + Rr)   When r = 0, this becomes cone: ⅓πR²h ✓
Topic CCylinder & Cone — Detailed PropertiesAFCAT Direct
Cylinder CSA
2πrh — only the curved lateral surface (like the label of a can). Does not include top and bottom circular faces.
Cylinder TSA
2πr(h + r) = 2πrh + 2πr² — curved surface + 2 circular ends. Use TSA when the question asks for total material needed.
Cylinder Volume
πr²h — area of circular base × height. Most frequently tested formula in AFCAT 3D questions.
Cone Slant Height
l = √(r² + h²) — slant height is the distance from the apex to any point on the base circumference. Always calculate slant height first in cone problems.
Cone CSA
πrl — curved surface only. Think of unrolling the cone into a sector of a circle with radius = slant height.
Cone TSA
πr(l + r) = πrl + πr² — CSA + one circular base. Cone has only 1 base (unlike cylinder which has 2).
Volume Ratio
Cone : Hemisphere : Cylinder = 1 : 2 : 3 (same radius and height). Classic AFCAT comparison question.

📐 Formula Sheet — NA09

Cube & Cuboid
Cube: LSA=4a², TSA=6a², V=a³
Cuboid: TSA=2(lb+bh+hl)
V = l×b×h
Diag of cuboid = √(l²+b²+h²)
Cylinder & Cone
Cylinder: CSA=2πrh, TSA=2πr(r+h), V=πr²h
Cone: CSA=πrl, TSA=πr(l+r), V=⅓πr²h
Slant height l = √(r²+h²)
Sphere & Hemisphere
Sphere: SA=4πr², V=⁴⁄₃πr³
Hemisphere: CSA=2πr², TSA=3πr², V=⅔πr³
Ratio of cone:hemi:sphere volumes = 1:2:4 (same r=h)
Conversion Trick
Volume stays constant on melting
n small spheres from 1 large: n = R³/r³
Wire from cylinder: πR²H = πr²L
Cone from cylinder: V = ⅓ of cylinder

📝 Topic-Wise PYQs — NA09

Q1. The volume of a cube is 343 cm³. Find its surface area. AFCAT PYQ
(a) 196 cm²(b) 294 cm²(c) 343 cm²(d) 216 cm²
✔ Answer: (b) 294 cm²
V = a³ = 343 → a = 7 cm. TSA = 6a² = 6×49 = 294 cm².
Q2. A sphere of radius 6 cm is melted into small spheres of radius 1 cm. How many are formed? ⚡ Tricky
(a) 108(b) 216(c) 36(d) 72
✔ Answer: (b) 216
Volume constant. n = R³/r³ = 6³/1³ = 216/1 = 216. (Ratio of radii cubed gives ratio of volumes.)
Q3. CSA of a cylinder with radius 7 cm and height 10 cm is: (π = 22/7) AFCAT PYQ
(a) 440 cm²(b) 660 cm²(c) 880 cm²(d) 770 cm²
✔ Answer: (a) 440 cm²
CSA = 2πrh = 2×(22/7)×7×10 = 2×22×10 = 440 cm².

🧠 Quick Memory Chart — NA09

📦 Cube / Cuboid
  • Cube TSA: 6a²
  • Cube V: a³
  • Cuboid V: lbh
  • Diag: √(l²+b²+h²)
🔵 Cylinder / Cone
  • Cyl V: πr²h
  • Cyl CSA: 2πrh
  • Cone V: ⅓πr²h
  • Slant l: √(r²+h²)
⚪ Sphere
  • SA: 4πr²
  • V: ⁴⁄₃πr³
  • Hemi TSA: 3πr²
  • Melt: V constant
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Q4. A cone has radius 5 cm and height 12 cm. Find its slant height and CSA. (π=22/7) AFCAT PYQ
(a) l=13, CSA=204.2(b) l=13, CSA=204(c) l=17, CSA=267(d) l=13, CSA=220
✔ Answer: (a) l=13 cm, CSA ≈ 204.28 cm²
Slant height l = √(r²+h²) = √(25+144) = √169 = 13 cm. CSA = πrl = (22/7)×5×13 = 1430/7 ≈ 204.28 cm².
Q5. The volume ratio of cone:hemisphere:cylinder (same radius, same height as radius) is: ⚡ Tricky
(a) 1:2:3(b) 1:3:2(c) 2:3:1(d) 1:2:4
✔ Answer: (a) 1 : 2 : 3
For radius=height=r: Cone = ⅓πr³, Hemisphere = ⅔πr³, Cylinder = πr³. Ratio = ⅓ : ⅔ : 1 = 1 : 2 : 3. Classic AFCAT question — memorise this ratio.