Mensuration 2D & 3D Masterclass: Complete Formulas & Tricks
Crack SSC CGL, CHSL & Banking Exams with essential Mensuration Tricks for Rectangle, Circle, Cylinder,
Cone & Sphere.
Government exams jaise SSC CGL, CHSL, MTS, Banking (IBPS, SBI), aur Railways mein
Mensuration ek mandatory section hai. Geometry ke baad, Mensuration hi woh topic hai jisme students
sabse zyada confuse hote hain. Is topic ko do parts mein baata gaya hai: Mensuration 2D (Area &
Perimeter) aur Mensuration 3D (Volume & Surface Area).
Is exhaustive guide mein hum dono 2D aur 3D shapes ko detail mein cover karenge taaki aapka ek bhi mark
na kate.
Part 1: Mensuration 2D (Area & Perimeter)
1. Rectangle (Ayat)
Rectangle wo shape hai jiski aamne-saamne ki sides barabar aur parallel hoti hain, aur charo angles 90
degree ke hote hain.
- Length (l): Lambai
- Breadth (b): Chaudai
Area = l — b
Perimeter = 2(l + b)
Diagonal = v(l² + b²)
2. Square (Varg)
Square ek special rectangle hai jiski charo sides barabar hoti hain.
Area = a²
Perimeter = 4a
Diagonal = av2
3. Circle (Vrat)
Circle ek round shape hai jisme center se boundary tak ki doori (Radius) same hoti hai.
- Radius (r): Trijya
- Diameter (d): Vyas (d = 2r)
Area = pr²
Circumference (Perimeter) = 2pr
4. Triangle (Tribhuj)
Triangle ke area ka basic formula hai:
Area = — — Base — Height
Equilateral Triangle (Sambahu Tribhuj): Jiski teeno sides barabar hon.
Area = (v3/4)a²
Part 2: Mensuration 3D (Volume & Surface Area)
1. Cylinder (Belan)
Cylinder ek aisi 3D shape hai jiska base aur top dono circular (vratakar) hote hain aur unke beech ek
fixed height hoti hai.
- Radius (r): Base circle ka radius.
- Height (h): Top aur bottom ke beech ki vertical doori.
CSA = 2πrh
TSA = 2pr(r + h)
Volume = πr²h
Exam Pro-Tip: Jab bhi kisi wire (taar) ki baat ho, toh yaad rakhen ki wire hamesha ek
Cylinder hota hai. Wire ki length hi uski height (h) ban jati hai.
2. Cone (Shanku)
Cone ek aisi shape hai jiska base toh circular hota hai, lekin upar jate-jate woh ek point (Apex) par mil
jata hai. Example: Ice-cream cone ya Birthday cap.
- Slant Height (l): l = v(r² + h-)
CSA = πrl
TSA = pr(l + r)
Volume = (1/3)πr²h
3. Sphere and Hemisphere (Gola aur Ardh-Gola)
Sphere ek perfect symmetrical shape hai. Example: Football.
Sphere Volume = (4/3)πr²
Sphere Surface Area = 4pr —
Hemisphere (Ardh-Gola):
CSA = 2πr²
TSA = 3πr²
Volume = (2/3)pr —
Comparison Table for Quick Revision
| Shape |
Curved/Lateral Area |
Total Surface Area |
Volume |
| Cylinder |
2πrh |
2πr(r+h) |
πr²h |
| Cone |
πrl |
πr(l+r) |
(1/3)πr²h |
| Sphere |
4πr² |
4πr² |
(4/3)πr² |
| Hemisphere |
2πr² |
3πr² |
(2/3)πr² |
🔥 Advanced Strategy: The 2D vs 3D Transformation Trick
In Mains exams like SSC CGL Tier 2, questions often ask about cutting shapes or melting them. Instead of long calculations, use the Constant Volume Concept.
Number of smaller objects formed =
Volume of Bigger Object / Volume of One Smaller Object
Topper's Edge: Avoid Calculation Traps
Never multiply the values of Volume or Area until the absolute final step. Leave them as (22/7) × 14 × 14 × 20. When you divide by the smaller object's volume, most numbers (like π and radius squares) will cancel out instantly, saving you 2 minutes of rough work!
Most Important Exam Tricks (Shortcuts) 🚀
A. The π Trick (Divisibility by 11)
Mensuration ke adhiktat formulas mein π (22/7) hota hai. Iska matlab hai ki aapka answer hamesha
11 ka multiple hoga. Options check karein aur jo 11 se divide ho wahi answer choose
karein.
B. Melting and Recasting
Jab ek shape ko pighla kar dusri shape banayi jati hai, toh hamesha Volume Equal hota
hai.
Volume(Initial) = n — Volume(Final)
C. Ratio Method for Percentage Change
Agar radius 10% badh jaye, toh area ya volume par kya asar padega?
- For Area (2D): 10 + 10 + (10 × 10)/100 = 21% Increase
- For Volume (3D): 33.1% Increase (Successive Percentage)
Conclusion: Practice is Key
Mensuration 2D aur 3D dono hi scoring topics hain. Agar aapko formulas yaad hain aur calculation tricks
aati hain, toh aap seconds mein answer nikal sakte hain. Ikkish Prep ka maqsad aapko inhi complex topics
ko aasan bhasha mein samjhana hai.
Apni calculation speed badhane ke liye hamare Speed Math Tool ka use karein.
Expert
Analysis: Mensuration in IBPS PO & SSC CGL (2020-2024)
Mensuration is one of the highest-value topics for SSC CGL Tier-I and -II because it tests both
formula recall AND application under time pressure. Analysis of past 5 years of SSC CGL and
IBPS PO papers shows these recurring question archetypes:
- Type 1 — Shape Conversion: "A cube of side X is melted and reshaped into N
cylinders of radius r. Find height." (Volume conservation = key principle) — appears in 90% of SSC
papers.
- Type 2 — Percentage Change in Dimensions: "Radius increases by 20%, what % does the
volume increase?" (Use successive percentage: 3 dimensions for volume = complex, but there's a
shortcut below).
- Type 3 — Diagonal/Slant Relation: "Find the slant height of a cone whose radius is
5 and height is 12." (Pythagoras: l = v(r²+h-) = 13 — a 5-12-13 right triangle).
- Type 4 — Two Shapes Combined: "A toy is made of a hemisphere and a cone on the
base. Find total surface area." (TSA = Curved Surface of hemisphere + Curved Surface of cone, not
the flat bases).
Expert Tricks for Percentage Change in
Mensuration
This is where most aspirants lose marks. Here's the definitive guide:
If length = +a%, breadth = +b%: Change in Area = a + b + (ab)/100
If all 3 dimensions of a cube/cuboid change by r%: Change in Volume = 3r + 3r²/100 + r²/10000
Practical Application: If radius increases by 10%: Area change = 10+10+(10 × 10)/100 =
21%. Volume change (sphere) = 3(10) + 3(100)/100 + 1000/10000 = 30+3+0.1 = 33.1%.
The "No-Formula" Approach for Cylinder vs
Cone
One of the most elegant relationships in 3D Mensuration: A cone with the same base and height as
a cylinder contains exactly 1/3 the volume. Similarly, a sphere fitting inside a cube:
Volume of sphere / Volume of cube = p/6 × 0.524, meaning a sphere wastes about 47.6% of the cube's
space.
These ratio relationships help you answer "comparison" questions without any calculation: "Which has more
volume: a cylinder of radius R and height 2R, or a sphere of radius R?" Answer: Cylinder = pR-(2R) =
2pR-. Sphere = (4/3)pR-. Ratio = 2/(4/3) = 3/2. Cylinder wins by 50%.
Must-Know Value Pairs for Quick Calculation
| Shape |
Common Exam Values |
Quick Answer |
| Circle Area |
r = 7 |
pr² = 22/7 × 49 = 154 |
| Cylinder Volume |
r = 7, h = 10 |
p²49 × 10 = 1540 |
| Sphere Volume |
r = 3 |
(4/3)p²27 = 36p — 113.1 |
| Cone Volume |
r = 3, h = 4 |
(1/3)p²9 × 4 = 12p — 37.7 |
Practice these common value-pairs until the answers come instantly. Then start the Ikkish Prep Speed Math Tool to
build lightning calculation reflexes!
🔥 Part 4: Advanced Shortcuts — SSC CGL Tier 2 & IBPS Mains Level
📐 2D Shape Relationships You Must Know
In SSC CGL and Bank Mains, questions often ask for the largest shape inscribed inside another. These are the most important relationships:
- Largest circle in a square (side a): Radius = a/2. Area = πa²/4
- Largest square in a circle (radius r): Side = r√2. Area = 2r²
- Largest equilateral triangle in a circle (radius r): Side = r√3
- Largest circle in equilateral triangle (side a): Inradius = a/(2√3)
- Diagonal of a rectangle: d = √(l² + b²) — the classic Pythagoras!
🧊 3D Melting & Recasting — The Volume Conservation Law
When a shape is melted and recast into smaller shapes, remember: Volume never changes!
n × (Volume of small shape) = Volume of big shape
Solve for n to find number of small shapes formed
Mains-Level Example (SSC CGL 2023 Pattern)
Q: A solid metallic cylinder (r=3cm, h=14cm) is melted and recast into small cones (r=1.5cm, h=2cm). How many cones are formed?
Solution (Cancel π — never multiply first!):
Volume of cylinder = π × 3² × 14 = 126π cm³
Volume of one cone = (1/3) × π × 1.5² × 2 = (1/3) × 2.25π × 2 = 1.5π cm³
Number of cones = 126π / 1.5π = 84 cones ✅
💡 Key: π cancels completely. Always write in terms of π first!
⚡ The Slant Height Trap (Most Common Error)
Students use slant height l instead of vertical height h (or vice versa) in cone formulas. Remember:
l² = r² + h² → Slant height l = √(r² + h²)
CSA of Cone = πrl (uses slant height l)
Volume of Cone = (1/3)πr²h (uses vertical height h)
⚠️ Exam Trap: When a question gives r=3, h=4 — immediately find l = √(9+16) = √25 = 5. This is the 3-4-5 Pythagorean triplet! Memorize triplets: (3,4,5), (5,12,13), (8,15,17), (7,24,25). These appear in 70% of cone CSA questions.
Part 5: PYQ-Style Practice — Exam Ready Questions
Based on actual patterns from SSC CGL 2021–2023, IBPS PO Mains 2022, and RRB NTPC 2023:
Q1: (SSC CGL 2023 — Mensuration)
The radius of a sphere is increased by 50%. By what percent does its volume increase?
Solution: New radius = 1.5r. Volume ratio = (1.5)³ = 3.375
Increase = (3.375 − 1)/1 × 100 = 237.5% ✅
Q2: (IBPS PO Mains 2022)
A hollow cylinder has outer radius 5cm, inner radius 3cm and height 10cm. Find its volume.
Solution: Volume = π(R² − r²) × h = π(25 − 9) × 10 = 160π = 160 × 22/7 = 502.85 cm³ ✅
Q3: (RRB NTPC 2023)
A rectangle room is 15m × 10m. Its floor is to be covered with tiles of size 50cm × 25cm. How many tiles are needed?
Solution (Convert units first!):
Room area = 1500 × 1000 = 1,500,000 cm²
Tile area = 50 × 25 = 1,250 cm²
Tiles needed = 1,500,000 / 1,250 = 1,200 tiles ✅
Q4: (SSC CHSL 2022 — Perimeter)
The area of a square is 484 cm². Find its perimeter.
Solution: Side = √484 = 22 cm. Perimeter = 4 × 22 = 88 cm ✅
💡 √484: Note that 22² = 484. Memorize squares up to 30!
Target Your Exam — Practice Now
Mensuration appears in every SSC, Bank, and Railway exam. Start your targeted practice today:
