COMPLETE MENSURATION (क्षेत्रमिति 2D & 3D)

Bihar STET Paper 1 - Mensuration (क्षेत्रमिति Visual Super Notes)

2D Plane Figures, 3D Solids, Melting & Recasting, Frustum • Solved PYQ Cheatsheet (Pg 213–249)

Weightage: 10–12 Marks 11 ka Multiple Elimination Hack 100% Blogger Ready

⚡ 5-Second Super Hack: "Divisibility by 11" Option Elimination

Jisme bhi π (π = 22 / 7) aata hai (Cylinder, Cone, Sphere, Circle, Frustum):
Kyunki 22 = 2 × 11 hota hai, isliye Volume ya Curved Surface Area ka correct answer lagbhag hamesha 11 se divisible hota hai!
Rule of 11: (Sum of Odd place digits) - (Sum of Even place digits) = 0 ya 11 ka multiple.
Agar calculation lamba ho ya time kam bacha ho, toh options me 11 ka divisibility test karke 3 options turant eliminate kar sakte ho!

Type 1: 2D Figures (Sector, Ring, Equilateral & Trapezium)

2D Formulas

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

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

• Equilateral Triangle: Area = (√3 / 4) × a²  |  Height = (√3 / 2) × a
• Trapezium (समलम्ब): Area = (1 / 2) × (Sum of parallel sides) × Height
• Rhombus (समचतुर्भुज): Area = (1 / 2) × d1 × d2  |  Side² = (d1/2)² + (d2/2)²

[PYQ Example] Radius of sector is 6 cm, angle is 60°. Find area:

Area = (60° / 360°) × π × 6² = (1 / 6) × π × 36 = 6π cm² (ya 18.85 cm²)
✔ Answer: 6π cm²

Type 2: 3D Standard Solids (बेलन, शंकु, गोला)

High Yield

Solid Shape	Curved Surface Area (CSA)	Total Surface Area (TSA)	Volume (आयतन)
Cylinder (बेलन)	2πrh	2πr(h + r)	πr²h
Cone (शंकु) [l = √(r²+h²)]	πrl	πr(l + r)	(1/3)πr²h
Sphere (गोला)	4πr²	4πr²	(4/3)πr³
Hemisphere (अर्धगोला)	2πr²	3πr²	(2/3)πr³

⚠️ Common Exam Trap: Hemisphere ka Total Surface Area 3πr² hota hai (kisi bhi condition me 2πr² mat lagana, upar ka circular base πr² bhi judta hai!).

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

Volume Ratio = (r1/r2)² × (h1/h2)
= (2/3)² × (3/2) = (4/9) × (3/2) = 2 / 3 = 2 : 3
✔ Answer: 2 : 3

Type 3: Melting, Recasting & Conservation of Volume

Conversion Rule

Golden Conversion Rule:
Jab kisi solid ko pighla kar (melt karke) koi doosra solid banaya jaye, toh Volume hamesha same rehta hai!
• Number of Smaller Objects (n) = (Volume of Bigger Solid) / (Volume of 1 Smaller Solid)
• Ek bade sphere (R) se chhote spheres (r) banane ho toh: n = (R / r)³

[PYQ Example] A metallic sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder:

Volume of Cylinder = Volume of Sphere
π × r_cyl² × h = (4/3) × π × R_sph³
π se π cancel:
4² × h = (4/3) × 6³ ⇒ 16 × h = (4/3) × 216 = 288
h = 288 / 16 = 18 cm
✔ Answer: 18 cm

Type 4: Frustum of a Cone (शंकु का छिन्नक - बाल्टी/ग्लास)

STET Repeated

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] A bucket has radii 28 cm and 7 cm, height 45 cm. Find its volume:

V = (1/3) × (22/7) × 45 × [28² + 7² + (28 × 7)]
= (1/3) × (22/7) × 45 × [784 + 49 + 196] = 15 × (22/7) × 1029
= 15 × 22 × 147 = 48,510 cm³
✔ Answer: 48,510 cm³ (⚡ Option Hack: 48510 is directly divisible by 11: (4+5+0) - (8+1) = 9 - 9 = 0!)