🌍 3D Shapes — Duniya Hamare Aas Paas!
Hum jo duniya mein rehte hain woh 2D (flat) nahi balki 3D (teen-dimensional) hai. Ek football 3D hai, ek tin can 3D hai, ek diya 3D hai. In 3D shapes ke baare mein do important sawaal hote hain:
- Surface Area (Bahari Jagah): Shape ki poori bahari surface ka area — jaise ek dabba (box) paint karne ke liye kitna paint chahiye. Units: cm², m².
- Volume (Andar ki Jagah): Shape ke andar ki jagah — jaise ek tank mein kitna paani aayega. Units: cm³, m³.
Kumbhar (potter) ko pata hona chahiye ki matke mein kitna paani aayega (Volume). Painter ko pata hona chahiye ki kamre ki diwar paint karne ke liye kitna paint chahiye (Surface Area). Dono ek hi shape ke alag-alag sawaal hain!
Is chapter mein hum seekhenge: Cuboid → Cube → Cylinder → Cone → Pyramid → Sphere → Hemisphere → Combinations.
1️⃣ Cuboid (Aayatakaar Ghanda / Box)
Cuboid ek aisi 3D shape hai jiske 6 rectangular faces hote hain. Ek room, brick (eent), matchbox — sab cuboid hain. Teen dimensions hote hain: l (length), b (breadth), h (height).
🧠 TSA Formula Ki Derivation:
Cuboid ke 6 faces teen pairs mein hote hain:
- 2 faces: \( l \times b \) (Top + Bottom)
- 2 faces: \( b \times h \) (Front + Back)
- 2 faces: \( l \times h \) (Left + Right)
Total = \( 2lb + 2bh + 2lh = 2(lb + bh + hl) \) ✅
📐 Cuboid ke Formulas
Ek kamre ki length 5 m, breadth 4 m, height 3 m hai. 4 walls + ceiling paint karni hai. Kitna area?
Solution:
LSA (4 walls) = \( 2h(l+b) = 2 \times 3 \times (5+4) = 54 \text{ m}^2 \)
Ceiling = \( l \times b = 20 \text{ m}^2 \)
Total = \( 54 + 20 = \mathbf{74 \text{ m}^2} \) 🎨
2️⃣ Cube (Ghanda Samakaon / Perfect Box)
Cube ek special cuboid hai jisme l = b = h = a. Ice cube, Rubik's cube, sugar cube — sab examples hain!
Cuboid mein \( l = b = h = a \) rakh do:
TSA = \( 2(a \cdot a + a \cdot a + a \cdot a) = 6a^2 \)
Volume = \( a \times a \times a = a^3 \)
📐 Cube ke Formulas (side = a)
Ek cube ki side 5 cm hai. TSA aur Volume nikalo.
Solution:
TSA = \( 6 \times 5^2 = 6 \times 25 = \mathbf{150 \text{ cm}^2} \)
Volume = \( 5^3 = \mathbf{125 \text{ cm}^3} \)
3️⃣ Right Circular Cylinder (Balaen / Tin Can)
Cylinder ek aisi shape hai jo circle ko seedha upar uthane se banti hai. Road roller, drum, pencil — sab cylinders hain. Dimensions: r (base radius), h (height).
🧠 Curved SA Ka Logic — Unroll Karo!
Cylinder ko upar-niche ke circles hata kar side ko khol do → ek Rectangle banta hai!
- Rectangle ki height = h
- Rectangle ki length = Circle circumference = 2πr
- CSA = \( 2\pi r \times h = 2\pi rh \)
📐 Cylinder ke Formulas
Ek tin can: radius = 7 cm, height = 20 cm. TSA aur Volume nikalo. (\(\pi = 22/7\))
Solution:
TSA = \( 2\pi r(r+h) = 2 \times \frac{22}{7} \times 7 \times 27 = 44 \times 27 = \mathbf{1188 \text{ cm}^2} \)
Volume = \( \pi r^2 h = \frac{22}{7} \times 49 \times 20 = 22 \times 7 \times 20 = \mathbf{3080 \text{ cm}^3} \)
4️⃣ Right Circular Cone (Shankh / Ice Cream Cone)
Cone ek aisi shape hai jiska base circular hota hai aur sirf ek apex (top point) hota hai. Dimensions: r (base radius), h (height), l (slant height).
📏 Slant Height: Pythagoras se: \( l = \sqrt{r^2 + h^2} \)
(r, h aur l ek right triangle banate hain!)
🧠 Curved SA Ka Logic — Cone Unroll Karo!
Cone ko side se kaat kar flat karein → ek Sector of a circle milta hai jiska radius = slant height l aur arc = \(2\pi r\).
Sector Area = \( \frac{1}{2} \times \text{arc} \times \text{radius} = \frac{1}{2} \times 2\pi r \times l = \pi r l \)
📐 Cone ke Formulas
Cone ka volume exactly ⅓ hota hai us cylinder ke volume ka jiska same base radius aur same height ho! 3 cones fill karo aur cylinder mein daalo — bilkul bhar jaayega! Ghar mein try karo 🧪
Ice-cream cone: radius = 3.5 cm, height = 12 cm. CSA aur Volume nikalo.
Solution:
\( l = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm} \)
CSA = \( \pi r l = \frac{22}{7} \times 3.5 \times 12.5 = 22 \times 6.25 = \mathbf{137.5 \text{ cm}^2} \)
Volume = \( \frac{1}{3} \times \frac{22}{7} \times 3.5^2 \times 12 = \frac{1}{3} \times \frac{22}{7} \times 147 = \frac{1}{3} \times 462 = \mathbf{154 \text{ cm}^3} \)
5️⃣ Pyramid (Piramid)
Pyramid ek 3D shape hai jiska ek polygonal base aur triangular side faces hain jo apex par milte hain. Egypt ke pyramids square base ke hain! Hum Square Pyramid padhenge: base side a, height h, slant height l.
📏 Slant Height of Face: \( l = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} \)
📐 Square Pyramid ke Formulas
Egypt ka Great Pyramid of Giza (c. 2560 BCE) ek square pyramid hai! Base side ≈ 230 m, height ≈ 138 m. Volume ≈ 2.6 million m³! Ancient engineers ne ye sab bina calculators ke banaya. 🤯
Square pyramid: base side = 6 cm, height = 4 cm. TSA aur Volume nikalo.
Solution:
\( l = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ cm} \)
TSA = \( a^2 + 2al = 36 + 60 = \mathbf{96 \text{ cm}^2} \)
Volume = \( \frac{1}{3} \times 36 \times 4 = \mathbf{48 \text{ cm}^3} \)
6️⃣ Sphere (Gola — Football, Marble)
Sphere mein center se surface tak ki har jagah ki doori same (= r) hoti hai. Football, marble, Earth — sab approximate spheres hain. Ek hi dimension: r (radius).
🧠 SA Ka Logic — Archimedes Ki Discovery!
Mahan Greek mathematician Archimedes ne prove kiya ki sphere ka surface area bilkul equal hai us cylinder ke CSA ke jo sphere ko exactly fit kare (radius = r, height = 2r):
Cylinder CSA = \( 2\pi r \times 2r = 4\pi r^2 \) = Sphere SA ✅
📐 Sphere ke Formulas
Football ka diameter = 21 cm. SA aur Volume nikalo.
Solution: r = 10.5 cm
SA = \( 4 \times \frac{22}{7} \times 10.5^2 = \frac{88}{7} \times 110.25 = 88 \times 15.75 = \mathbf{1386 \text{ cm}^2} \)
Volume = \( \frac{4}{3} \times \frac{22}{7} \times 10.5^3 = \frac{4}{3} \times \frac{22}{7} \times 1157.625 \approx \mathbf{4851 \text{ cm}^3} \)
7️⃣ Hemisphere (Aadha Gola — Bowl, Igloo)
Hemisphere = sphere ka exactly aadha! Ek bowl, igloo, ya Earth ka koi ek half. Dimension: r (radius).
📐 Hemisphere ke Formulas
Ek bowl (hemisphere) ka radius = 10.5 cm. CSA aur Volume nikalo.
Solution:
CSA = \( 2 \times \frac{22}{7} \times 10.5^2 = \frac{44}{7} \times 110.25 = 44 \times 15.75 = \mathbf{693 \text{ cm}^2} \)
Volume = \( \frac{2}{3} \times \frac{22}{7} \times 10.5^3 = \frac{2}{3} \times \frac{22}{7} \times 1157.625 \approx \mathbf{2425.5 \text{ cm}^3} \)
8️⃣ Combinations of Shapes (Milaao aur Banao!)
Real life mein shapes akele nahi aati — combine hoti hain. Toy rocket = Cylinder + Cone. Oil barrel = Cylinder + 2 Hemispheres. Khilona temple = Cylinder + Hemisphere on top.
📌 Combined Shapes ke Rules:
- Volume: Hamesha add karo → Total V = V₁ + V₂ + ...
- Surface Area: Sirf dikhne wali (exposed) surfaces count karo! Jahan do shapes milti hain woh area add nahi hota.
Cylinder: r = 3 cm, h = 10 cm. Upar cone: same r, height = 4 cm. Total Volume nikalo.
Solution:
V(Cylinder) = \( \pi \times 9 \times 10 = 90\pi \)
V(Cone) = \( \frac{1}{3} \times \pi \times 9 \times 4 = 12\pi \)
Total = \( 102\pi \approx 102 \times 3.14 = \mathbf{320.28 \text{ cm}^3} \) 🚀
📋 Quick Reference Table
| Shape | CSA / LSA | Total SA | Volume |
|---|---|---|---|
| Cuboid (l,b,h) | 2h(l+b) | 2(lb+bh+hl) | lbh |
| Cube (a) | 4a² | 6a² | a³ |
| Cylinder (r,h) | 2πrh | 2πr(r+h) | πr²h |
| Cone (r,h,l) | πrl | πr(r+l) | ⅓πr²h |
| Sq. Pyramid (a,h,l) | 2al | a²+2al | ⅓a²h |
| Sphere (r) | 4πr² | ⁴⁄₃πr³ | |
| Hemisphere (r) | 2πr² | 3πr² | ⅔πr³ |
🎯 Self-Assessment: Practice Questions
❓ Q1: Cuboid
Ek cuboid: 8 cm × 6 cm × 4 cm. TSA nikalo.
Answer: TSA = \(2(48+24+32) = 2 \times 104 = \mathbf{208 \text{ cm}^2}\)
❓ Q2: Cylinder — Reverse
Cylinder ka Volume = 1540 cm³, height = 10 cm. Radius nikalo. (\(\pi=22/7\))
Answer: \( \pi r^2 h = 1540 \Rightarrow r^2 = \frac{1540 \times 7}{22 \times 10} = 49 \Rightarrow r = \mathbf{7 \text{ cm}}\)
❓ Q3: Sphere → Cylinder (Classic NCERT)
Ek metallic sphere (r = 6 cm) ko pighla kar cylinder (r = 4 cm) banaya. Height nikalo.
Answer: \(\frac{4}{3}\pi \times 216 = \pi \times 16 \times h \Rightarrow 288 = 16h \Rightarrow h = \mathbf{18 \text{ cm}} \) 🔥
❓ Q4 (Challenge): Cone TSA
Cone: diameter = 14 cm, slant height = 8 cm. TSA nikalo.
Answer: r = 7, l = 8
TSA = \(\pi r(r+l) = \frac{22}{7} \times 7 \times 15 = 22 \times 15 = \mathbf{330 \text{ cm}^2}\)