ULTIMATE MENSURATION (क्षेत्रमिति 2D & 3D)

Bihar STET Paper 1 - Complete Mensuration (क्षेत्रमिति 2D & 3D)

Deep Concept • Pure SVG Vector Diagrams • 11-Divisibility Option Hack • All Repeated PYQs (Pg 213–249)

Weightage: 10–12 Marks All 6 Repeated PYQ Types 100% Blogger Compatible

⚡ Master Weapon: 11 & 7 Divisibility Option Elimination (5-Second Hack)

Exam me 3D figures jaise Cylinder, Cone, Sphere, Hemisphere, Circle, Frustum ka sawal ho jisme π (π = 22 / 7) multiply hota hai:

Trick 1 (11 ka Rule): Kyunki 22 = 2 × 11 hota hai, isliye answer 11 se poora cut-na chahiye.
Test: (Sum of Odd digits) - (Sum of Even digits) = 0 ya 11 ka multiple.
Trick 2 (7 ka Rule): Agar radius (r) ya height (h) 7 ka multiple na ho, toh answer 7 se divide ho sakta hai.
Caution: Agar sawal me π = 3.14 explicitly likha ho ya do options 11 se cut rahe hon, tabhi full solve karo, warna 90% sawal direct option dekh kar 5 second me tick hote hain!

Type 1: Circle Sector, Arc Length & Circular Ring (त्रिज्यखंड एवं वलय)

2D Repeated

Area = (θ/360°) × πr²  |  l = (θ/360°) × 2πr

Ring Area = π(R² - r²) = π(R + r)(R - r)

Concept & Shortcuts:
• Area of Sector: A = (1 / 2) × l × r (Agar arc length l di ho toh direct nikalta hai)
• Revolutions of a Wheel (पहिए के चक्कर): Number of Revolutions = Total Distance / (2πr)

[PYQ Example 1] Radius of a sector is 6 cm, central angle θ = 60°. Find its area.

(a) 12π cm²    (b) 6π cm²    (c) 18π cm²    (d) 3π cm²

Area = (θ / 360°) × π × r² = (60° / 360°) × π × 6²
= (1 / 6) × 36π = 6π cm²
✔ Correct Answer: (b) 6π cm²

[PYQ Example 2] Ek circular circular ground ka radius 21 m hai. Uske bahar 7 m chauda rasta hai. Raste ka area find karein.

(a) 1078 m²    (b) 1056 m²    (c) 1120 m²    (d) 980 m²

Inner radius r = 21 m, Outer radius R = 21 + 7 = 28 m
Area = π(R + r)(R - r) = (22 / 7) × (28 + 21) × (28 - 21)
= (22 / 7) × 49 × 7 = 22 × 49 = 1078 m²
⚡ Option Check (11 Divisibility): 1078 ➔ (1+7) - (0+8) = 8 - 8 = 0 (Direct divisible!)
✔ Correct Answer: (a) 1078 m²

Type 2: Right Circular Cylinder (लंब वृत्तीय बेलन)

Cylinder Type

Volume = πr²h  |  Curved Surface Area = 2πrh  |  TSA = 2πr(h + r)

⚡ Wire Drawing Exam Hack: Kisi metallic solid ko kheenchte hue cylindrical wire banaya jaye, toh Volume of Solid = π × (wire radius)² × (wire length L) hota hai!

[PYQ Example] Ratio of radii of two cylinders is 2 : 3 and heights ratio is 5 : 3. Find ratio of their volumes.

(a) 10 : 9    (b) 20 : 27    (c) 4 : 9    (d) 5 : 3

V1 / V2 = (π·r1²·h1) / (π·r2²·h2) = (r1 / r2)² × (h1 / h2)
= (2 / 3)² × (5 / 3) = (4 / 9) × (5 / 3) = 20 / 27
✔ Correct Answer: (b) 20 : 27

Type 3: Right Circular Cone (शंकु एवं तंबू का कैनवास)

Slant Height Triplet

Volume = (1/3)πr²h  |  Curved Surface Area = πrl  |  TSA = πr(l + r)

⚡ Pythagorean Triplet Trick for Cone:
Radius r, vertical height h, aur slant height l milkar ek Right Triangle banate hain: l² = r² + h².
Triplets directly use karo: (r=3, h=4 ➔ l=5) | (r=5, h=12 ➔ l=13) | (r=7, h=24 ➔ l=25) | (r=8, h=15 ➔ l=17).

[PYQ Example] A conical tent is 12 m high and base radius is 5 m. Find the canvas area required to make the tent (use π = 22/7).

(a) 204.28 m²    (b) 220 m²    (c) 195.5 m²    (d) 215 m²

Radius r = 5 m, Height h = 12 m
Triplet: (5, 12, 13) ➔ Slant height l = 13 m
Canvas area = Curved Surface Area = π × r × l
= (22 / 7) × 5 × 13 = 1430 / 7 = 204.28 m²
✔ Correct Answer: (a) 204.28 m²

Type 4: Sphere & Hemisphere (गोला एवं अर्धगोला Exam Trap)

Beware of 3πr²

Volume = (4/3)πr³  |  Surface Area = 4πr²

Volume = (2/3)πr³  |  CSA = 2πr²  |  TSA = 3πr²

⚠️ Biggest Exam Trap: Solid Hemisphere ka Total Surface Area 3πr² hota hai! Kyunki curved part 2πr² hota hai aur upar ka flat circular base πr² hota hai (2πr² + πr² = 3πr²).

[PYQ Example] Find the Total Surface Area (TSA) of a solid hemisphere of radius 7 cm (take π = 22/7).

(a) 308 cm²    (b) 462 cm²    (c) 616 cm²    (d) 154 cm²

Solid Hemisphere TSA = 3 × π × r²
= 3 × (22 / 7) × 7 × 7 = 3 × 22 × 7 = 462 cm²
(⚡ 11 Divisibility check: 462 ➔ (4+2) - 6 = 6 - 6 = 0!)
✔ Correct Answer: (b) 462 cm²

Type 5: Melting, Recasting & Number of Lead Shots (रूपांतरण)

Conservation Rule

Volume of Big Solid = n × Volume of 1 Small Solid

⚡ 5-Second Sphere-to-Sphere Hack:
Agar ek bade sphere (Radius R) ko pighla kar chhote spheres (Radius r) banaye jayein, toh:
Number of Spheres (n) = (R / r)³

[PYQ Example] A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones formed.

(a) 126    (b) 136    (c) 112    (d) 140

n = (Volume of Sphere) / (Volume of 1 Cone)
n = [(4/3)π × R³] / [(1/3)π × r² × h] = (4 × R³) / (r² × h)
n = [4 × (10.5)³] / [(3.5)² × 3]
10.5 / 3.5 = 3 hota hai, toh: (10.5)³ / (3.5)² = 10.5 × 3² = 10.5 × 9 = 94.5
n = (4 × 94.5) / 3 = 378 / 3 = 126 cones
✔ Correct Answer: (a) 126

Type 6: Frustum of a Cone (शंकु का छिन्नक - बाल्टी/ग्लास)

100% Fixed PYQ

Slant Height l = √[h² + (R - r)²]

Master Frustum Formulas:
• Volume (आयतन) = (1/3) × π × h × [R² + r² + (R × r)]
• Curved Surface Area (वक्र पृष्ठ) = π × l × (R + r)
• Total Surface Area (कुल पृष्ठ) = π × l × (R + r) + πR² + πr²

[PYQ Example - Bihar STET / Class 10th Core] A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity (volume) of the glass.

(a) 102.67 cm³    (b) 105.5 cm³    (c) 110.2 cm³    (d) 98.4 cm³

Top Radius R = 4 / 2 = 2 cm, Bottom Radius r = 2 / 2 = 1 cm, Height h = 14 cm
Volume = (1/3) × (22 / 7) × 14 × [2² + 1² + (2 × 1)]
= (1/3) × 22 × 2 × [4 + 1 + 2]
= (1/3) × 44 × 7 = 308 / 3 = 102.67 cm³ (ya 102 ⅔ cm³)
✔ Correct Answer: (a) 102.67 cm³