Class 10 Maths · Chapter 12
Short answer:
Is chapter me total 16 exercise questions cover kiye gaye hain — 9 questions surface area of combination of solids pe (cylinder+cone, hemisphere+cylinder, cube+hemisphere, capsule shape, tent shape, cylinder me conical cavity, cylinder ke dono end pe hemisphere scoop) aur 7 questions volume of combination of solids pe (cone+hemisphere solid, cylinder+cone rocket shape, gulab jamun/capsule volume with syrup, pen stand with conical depressions, lead shots dropped in cone vessel, iron pole made of two cylinders, aur cone+hemisphere dooba hua cylinder me). Rationalised NCERT print me "Conversion of Solid from One Shape to Another" aur "Frustum of a Cone" sections trim ho chuke hain — is spec me frustum ka koi exercise question nahi hai, sirf reference table me formula diya gaya hai completeness ke liye.
Jab do ya zyada basic solids (cone, cylinder, sphere, hemisphere, cube) ko jodkar ek naya solid banaya jaata hai, to uska surface area aur volume nikalna hi is chapter ka core skill hai. Real life me plastic ke khilone, capsule, tent, ice-cream cone with scoop — ye sab combination of solids hi hote hain. Trick simple hai: volume me sab solids ka volume seedha jud jaata hai, lekin surface area me jo common surface (jaise joining wala circle) chhup jaata hai uska hisaab rakhna padta hai.
Chapter 12 Summary — 5 Minute Revision
1. Combination of Solids kya hote hain
Jab do standard solids ek doosre se attach kiye jaate hain (ek ke upar doosra, ya ek ke andar cavity), to result ek naya "composite" solid hota hai. NCERT me common combinations: cone-on-cylinder (tent, rocket), hemisphere-on-cylinder (vessel, capsule), cone-on-hemisphere (toy), cube-with-hemisphere (mound ya depression), cylinder-with-conical-cavity.
2. Surface Area ka Golden Rule
- Jab do solids jode jaate hain (jaise cone upar, cylinder neeche), to total surface area me jud'ne wali common surface (base circle) count NAHI hoti — bas curved surfaces jodte hain.
- Formula: TSA of combination = sum of CURVED surfaces only (jab dono open ends touching hon), plus koi bhi FLAT face jo abhi bhi expose (uncut/unjoined) hai.
- Cavity/depression wale case me bhi same logic: jitna circle "cut" hua, uska flat area minus karo, aur jo curved surface expose hua (cone ya hemisphere ki lining) wo add karo.
3. Volume ka Golden Rule
- Volume me koi subtraction/cancellation nahi hota — Volume of combination = simple sum (ya difference, agar cavity hollow ki gayi hai) of individual volumes.
- Cavity/hollowed-out solid: Volume of remaining solid = Volume of original solid − Volume of removed part.
4. Slant height — cone ke saath hamesha check karo
Jab bhi cone ka CSA ya TSA nikalna ho, pehle check karo diya hua height "h" (vertical height) hai ya "l" (slant height). Agar h diya hai to pehle l = √(r² + h²) nikalo, tabhi πrl formula lagao. Volume me hamesha h (vertical height) use hota hai, l kabhi nahi.
5. Common Combinations — kaunsi surfaces count hoti hain
| Combination | Surface Area me kya jode | Volume me kya jode |
|---|---|---|
| Cone on Cylinder (tent) | CSA cone + CSA cylinder (+ base circle agar open bottom nahi hai) | Vol cone + Vol cylinder |
| Hemisphere on Cylinder (vessel) | CSA hemisphere + CSA cylinder | Vol hemisphere + Vol cylinder |
| Cone on Hemisphere (toy) | CSA cone + CSA hemisphere | Vol cone + Vol hemisphere |
| Cube + mounted hemisphere | TSA cube − base circle of hemisphere + CSA hemisphere | Vol cube + Vol hemisphere |
| Cube with hemispherical depression | TSA cube − circle removed + CSA hemisphere (inner) | Vol cube − Vol hemisphere |
| Cylinder with conical cavity (one end) | CSA cylinder + flat uncut base + CSA cone (cavity slant) | Vol cylinder − Vol cone |
| Capsule (cylinder + 2 hemispherical ends) | CSA cylinder + 4πr² (2 hemispheres = 1 full sphere CSA) | Vol cylinder + Vol sphere |
↔ Table ko side me swipe karein
6. Units — hamesha final answer ke units check karo
Surface area → cm², m², mm² (square units). Volume → cm³, m³, mm³ (cube units). Agar liquid capacity poochi jaaye (litre me), to 1000 cm³ = 1 litre use karke convert karo.
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q16)
Q1. Do cubes hain, har ek ka volume 64 cm³. Inhe end-to-end jod kar ek cuboid banaya gaya. Is resulting cuboid ka surface area nikalo.
Har cube ka side a: a³ = 64, to a = 4 cm.
End-to-end jodne se banega cuboid: length = 4+4 = 8 cm, breadth = 4 cm, height = 4 cm
TSA (direct formula) = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32+16+32) = 2×80 = 160 cm²
Check (jodne wala logic): 2 cubes ki total TSA = 2 × 6a² = 2×6×16 = 192 cm². Jodne par 2 faces (ek-ek har cube ki, area 16 cm² each) andar chhip jaati hain: 192 − 2×16 = 192−32 = 160 cm². Match ho gaya.
Answer: 160 cm²
Q2. Ek vessel hollow hemisphere ke upar hollow cylinder mount karke bana hai. Hemisphere ka diameter 14 cm hai aur vessel ki total height 13 cm hai. Vessel ka inner surface area nikalo.
r = 14/2 = 7 cm. Cylinder ki height h = total height − r = 13 − 7 = 6 cm.
Inner surface area = CSA cylinder + CSA hemisphere = 2πrh + 2πr² = 2πr(h + r)
= 2 × (22/7) × 7 × (6+7) = 2 × 22 × 13 = 572 cm²
Answer: 572 cm²
Q3. Ek khilona cone (radius 3.5 cm) ko hemisphere (same radius) ke upar mount karke bana hai. Khilone ki total height 15.5 cm hai. Total surface area nikalo.
r = 3.5 cm. Cone ki height h = 15.5 − 3.5 = 12 cm.
Slant height l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm
TSA = CSA cone + CSA hemisphere = πrl + 2πr² = πr(l + 2r)
= (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm²
Answer: 214.5 cm²
Q4. Ek cubical block ka side 7 cm hai, upar ek hemisphere mount kiya gaya hai jiska diameter block ke side ke barabar hai. Poore solid ka surface area nikalo.
Hemisphere ka greatest possible diameter = 7 cm (cube ke top face ke barabar), to r = 3.5 cm.
TSA solid = TSA cube − base circle of hemisphere + CSA hemisphere = 6a² − πr² + 2πr² = 6a² + πr²
= 6×7² + (22/7)×3.5² = 6×49 + (22/7)×12.25 = 294 + 38.5 = 332.5 cm²
Answer: 332.5 cm²
Q5. Ek cubical wooden block ka side 14 cm hai. Ek face me se ek hemispherical depression cut ki gayi hai jiska diameter block ke side ke barabar hai. Remaining solid ka surface area nikalo.
r = 14/2 = 7 cm.
TSA remaining = TSA cube − area of circle removed + CSA hemisphere (andar ki lining) = 6l² − πr² + 2πr² = 6l² + πr²
= 6×14² + (22/7)×7² = 6×196 + 22×7 = 1176 + 154 = 1330 cm²
Note: mound (Q4) aur depression (Q5) — dono me formula same aata hai (6l² + πr²), kyunki dono cases me ek circle ki flat area hemisphere ki curved surface se replace hoti hai.
Answer: 1330 cm²
Q6. Ek medicine capsule cylinder ke dono ends par hemisphere laga kar bana hai. Poore capsule ki length 14 mm hai aur diameter 5 mm hai. Capsule ka surface area nikalo.
r = 5/2 = 2.5 mm. Cylindrical part ki length h = total length − 2r = 14 − 5 = 9 mm.
Surface area = CSA cylinder + 2 × CSA hemisphere = 2πrh + 4πr² (do hemisphere = ek poore sphere ki CSA)
= 2×π×2.5×9 + 4×π×2.5² = 45π + 25π = 70π mm²
= 70 × 22/7 = 220 mm²
Answer: 220 mm²
Q7. Ek tent cylinder (height 2.1 m, diameter 4 m) ke upar cone (slant height 2.8 m) laga kar bana hai. Tent banane me lagne wale canvas ka area nikalo, aur Rs 500 per m² ki dar se canvas ka cost bhi batao.
r = 4/2 = 2 m, cylinder height h = 2.1 m, cone slant height l = 2.8 m.
CSA cylinder = 2πrh = 2 × (22/7) × 2 × 2.1 = 26.4 m²
CSA cone = πrl = (22/7) × 2 × 2.8 = 17.6 m²
Total canvas area = 26.4 + 17.6 = 44 m²
Cost = 44 × Rs 500 = Rs 22,000
Answer: Canvas area = 44 m², Cost = Rs 22,000
Q8. Ek solid cylinder ki height 2.4 cm aur diameter 1.4 cm hai. Isi height aur diameter ka ek conical cavity cylinder ke ek end se hollow kiya gaya hai (doosra end flat/uncut hai). Remaining solid ka total surface area nikalo.
r = 0.7 cm, h = 2.4 cm.
Slant height of cavity l = √(r² + h²) = √(0.49 + 5.76) = √6.25 = 2.5 cm
TSA remaining = CSA cylinder (lateral) + flat uncut base circle + CSA cone (cavity slant surface)
= 2πrh + πr² + πrl = 2π(0.7)(2.4) + π(0.7)² + π(0.7)(2.5)
= 3.36π + 0.49π + 1.75π = 5.6π = 5.6 × 22/7 = 17.6 cm²
Answer: 17.6 cm²
Q9. Ek wooden article ek solid cylinder (height 10 cm, radius 3.5 cm) se banaya gaya hai jisme se dono flat ends par ek-ek hemisphere scoop out kiya gaya hai (radius cylinder ke radius ke barabar). Article ka total surface area nikalo.
r = 3.5 cm, h = 10 cm.
TSA = CSA cylinder + 2 × CSA hemisphere (dono cavities) = 2πrh + 4πr² = 2πr(h + 2r)
= 2 × (22/7) × 3.5 × (10 + 7) = 22 × 17 = 374 cm²
Answer: 374 cm²
Q10. Ek solid cone hemisphere ke upar khada hai, dono ka radius 1 cm hai, aur cone ki height uske radius ke barabar hai. Solid ka volume π ke terms me nikalo.
r = 1 cm, cone height h = 1 cm.
Volume = Vol cone + Vol hemisphere = (1/3)πr²h + (2/3)πr³
= (1/3)π(1)²(1) + (2/3)π(1)³ = π/3 + 2π/3 = π cm³
Answer: π cm³
Q11. Ek solid ek cylinder (band niche wale end se, diameter 3 cm, height 8 cm) ke upar same radius ka cone (height 4 cm) laga kar bana hai — rocket jaisi shape. Poore solid ka volume nikalo.
r = 3/2 = 1.5 cm.
Vol cylinder = πr²h = π × 1.5² × 8 = 18π cm³
Vol cone = (1/3)πr²h = (1/3) × π × 1.5² × 4 = 3π cm³
Total volume = 18π + 3π = 21π = 21 × 22/7 = 66 cm³
Answer: 66 cm³
Q12. Ek gulab jamun cylinder shape ka hai jiske dono ends hemispherical hain — poori length 5 cm aur diameter 2.8 cm. Ismein sugar syrup uske volume ka lagbhag 30% hota hai. 45 gulab jamuns me kitna syrup hoga, approx nikalo (π = 22/7 lo).
r = 2.8/2 = 1.4 cm. Cylindrical part ki length h = 5 − 2×1.4 = 2.2 cm.
Vol cylinder = πr²h = (22/7) × 1.4² × 2.2 = (22/7) × 1.96 × 2.2 ≈ 13.55 cm³
Vol of 2 hemispheres (= 1 sphere) = (4/3)πr³ = (4/3) × (22/7) × 1.4³ ≈ 11.50 cm³
Volume of one gulab jamun ≈ 13.55 + 11.50 = 25.05 cm³
45 gulab jamuns ka total volume ≈ 45 × 25.05 = 1127.3 cm³
Syrup (30%) = 0.30 × 1127.3 ≈ 338 cm³
Answer: Approx 338 cm³ syrup
Q13. Ek wooden pen stand cuboid shape ka hai (15 cm × 10 cm × 3.5 cm) jisme 4 conical depressions hain, har ek ka radius 0.5 cm aur depth 1.4 cm, pens rakhne ke liye. Stand me lage wood ka volume nikalo.
Vol cuboid = 15 × 10 × 3.5 = 525 cm³
Vol of one conical depression = (1/3)πr²h = (1/3) × (22/7) × 0.5² × 1.4 = (1/3) × 1.1 ≈ 0.367 cm³
Vol of 4 depressions ≈ 4 × 0.367 = 1.467 cm³
Volume of wood = 525 − 1.467 ≈ 523.53 cm³
Answer: ≈ 523.53 cm³
Q14. Ek inverted-cone-shape vessel (height 8 cm, top radius 5 cm) paani se bhara hua hai. Isme radius 0.5 cm ke lead shots (spheres) dale jaate hain jab tak paani ka 1/4 bahar nahi bah jaata. Kitne lead shots dale gaye?
Overflow ka volume = dale gaye lead shots ka total volume (kyunki displace hua paani hi overflow hota hai).
Vol cone (vessel) = (1/3)πr²h = (1/3)π(5²)(8) = (200/3)π cm³
Overflow volume = (1/4) of vessel volume = (1/4) × (200/3)π = (50/3)π cm³
Vol of one lead shot = (4/3)πr³ = (4/3)π(0.5)³ = (1/6)π cm³
Number of shots n = (50/3)π ÷ (1/6)π = (50/3) × 6 = 100
Answer: 100 lead shots
Q15. Ek solid iron pole ek cylinder (height 220 cm, base diameter 24 cm) ke upar ek chhota cylinder (height 60 cm, radius 8 cm) laga kar bana hai. Agar 1 cm³ iron ka mass 8 g hai, to pole ka mass nikalo.
Cylinder 1: r₁ = 24/2 = 12 cm, h₁ = 220 cm. Cylinder 2: r₂ = 8 cm, h₂ = 60 cm.
Vol1 = πr₁²h₁ = π × 144 × 220 = 31680π cm³
Vol2 = πr₂²h₂ = π × 64 × 60 = 3840π cm³
Total volume = 35520π = 35520 × 22/7 ≈ 111634.29 cm³
Mass = 111634.29 × 8 g ≈ 893074.3 g ≈ 893.07 kg
Answer: ≈ 893.07 kg
Q16. Ek solid, cone hemisphere ke upar khada hua (dono ka radius 6 cm, cone ki height 12 cm), ek cylinder (radius 6 cm, height 18 cm) ke andar paani se poora bhara hua rakha gaya hai aur cylinder ki base ko touch karta hai. Cylinder me bacha hua paani ka volume nikalo.
Cylinder: r = 6 cm, h = 18 cm. Cone + hemisphere: r = 6 cm, cone height = 12 cm.
Vol cylinder = πr²h = π × 36 × 18 = 648π cm³
Vol cone = (1/3)πr²h = (1/3)π × 36 × 12 = 144π cm³
Vol hemisphere = (2/3)πr³ = (2/3)π × 216 = 144π cm³
Vol solid (cone+hemisphere) = 144π + 144π = 288π cm³
Water left = 648π − 288π = 360π = 360 × 22/7 ≈ 1131.43 cm³
Answer: ≈ 1131.43 cm³
Important Equations — Ek Nazar Me
| Solid | Curved/Lateral Surface Area | Total Surface Area | Volume |
|---|---|---|---|
| Cylinder (radius r, height h) | 2πrh | 2πrh + 2πr² = 2πr(h+r) | πr²h |
| Cone (radius r, height h, slant height l) | πrl, jahan l = √(r² + h²) | πrl + πr² = πr(l+r) | (1/3)πr²h |
| Sphere (radius r) | — | 4πr² | (4/3)πr³ |
| Hemisphere (radius r) | 2πr² (curved only) | 3πr² (curved + base circle) | (2/3)πr³ |
| Cube (side a) | — | 6a² | a³ |
| Cuboid (l, b, h) | — | 2(lb+bh+hl) | l×b×h |
| Frustum of a cone (radii r₁, r₂, slant height l, height h) — reference only, current book me exercise nahi | π(r₁+r₂)l | π(r₁+r₂)l + πr₁² + πr₂² | (1/3)πh(r₁²+r₂²+r₁r₂) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Dono circular bases add kar dena jab solids jode gaye hon. Jab cone/hemisphere kisi cylinder ke upar mount ho, joining wala circle andar chhup jaata hai — usko surface area me count mat karo, sirf curved surfaces jodo.
- Slant height nikalna bhool jaana. Cone ki CSA/TSA formula me l chahiye hota hai, h nahi. Height diya ho to pehle l = √(r²+h²) nikalo, tabhi πrl lagao — warna poora answer galat aayega.
- CSA aur TSA me confuse ho jaana. Question 'curved surface area' maange to sirf curved part do; 'total surface area' maange to flat faces bhi jodo. Dono formulas alag-alag yaad rakho, exam me galat formula lagana bahut common mistake hai.
- Unit conversion me galti — cm³ ko litre me convert karte waqt. 1000 cm³ = 1 litre yaad rakho. Bina convert kiye directly cm³ ka number litre bata dena bahut common galti hai.
- Diameter ko seedha radius ki jagah formula me daal dena. Question me diameter diya ho to pehle r = d/2 karo, phir formula lagao — nahi to answer 4× galat aayega.
- Uncut/flat face ka area add karna bhool jaana. Jab ek solid me sirf ek end par hemisphere/cone attach ya cavity ho aur doosra end flat rahe (jaise cylinder me ek taraf conical cavity), us flat circle ka area bhi total surface area me jodna padta hai — students isko chhod dete hain.
Board-Style Important Questions
- 1 mark: Ek hemisphere ki curved surface area ka formula likho.
- 2 marks: Ek cone ka radius 7 cm aur height 24 cm hai. Iski slant height nikalo.
- 3 marks: Ek khilona cone ko hemisphere ke upar mount karke banaya gaya hai, dono ka radius same hai. Iska total surface area nikalne ka tarika (steps) likho, formula ke saath.
- 3 marks: Ek tent cylinder ke upar cone laga kar banaya gaya hai. Diye gaye radius, cylinder height aur cone slant height se canvas ka area kaise nikaloge, poori working dikhao.
- 5 marks: Ek solid cylinder me se conical cavity same radius aur height ki hollow ki gayi hai. Remaining solid ka volume aur total surface area dono nikalne ka complete method likho.
Quick Quiz — Score Check Karein
Q1. Class 10 Maths Chapter 12 (Surface Areas and Volumes) me is chapter ka core skill kya hai?
Q2. Volume of combination of solids ka Golden Rule kya hai?
Q3. Surface Area ka Golden Rule kya bolta hai jab do solids jode jaate hain (jaise cone upar, cylinder neeche)?
Q4. Do cubes hain, har ek ka volume 64 cm³. Inhe end-to-end jod kar cuboid banaya gaya. Resulting cuboid ka surface area kya hai?
Q5. Ek vessel hollow hemisphere (diameter 14 cm) ke upar hollow cylinder mount karke bana hai, total height 13 cm. Vessel ka inner surface area kya hoga?
Q6. Cone-on-cone jaisa khilona (cone radius 3.5 cm, hemisphere same radius, total height 15.5 cm) ka slant height nikalne ke liye pehle kya nikalna padega?
Q7. Isi khilone (cone r=3.5 cm, total height 15.5 cm) ka total surface area kitna hai?
Q8. Cubical block (side 7 cm) ke upar hemisphere (diameter = block ka side) mount kiya gaya hai. Poore solid ka surface area formula kya banega?
Q9. Capsule (cylinder + do hemispherical ends) ke surface area me do hemispheres milkar kya banate hain?
Q10. Rationalised NCERT print me is chapter (Ch 12) se konsi sections trim ho chuki hain, jinke exercise questions is spec me nahi hain?
Aksar Poochhe Jaane Wale Sawaal
Combination of solids me common surface kyun subtract/ignore hoti hai?
Kyunki jab do solids jode jaate hain, unka touching surface (jaise ek circle) bahar nazar nahi aata — wo andar chhup jaata hai. Surface area sirf us hisse ka nikalte hain jo bahar se dikhta/expose hota hai, isliye common circle ko count nahi karte.
CSA aur TSA me exact farak kya hai jab solids combine ho?
CSA (curved surface area) me sirf curved/lateral parts aate hain, flat circular faces nahi. TSA (total surface area) me curved parts + jo bhi flat faces abhi bhi expose (uncovered) hain, dono aate hain. Combination solids me joining wali flat face hamesha excluded hoti hai, chahe CSA poocha ho ya TSA.
Slant height (l) kab chahiye hoti hai aur kaise nikalte hain?
Jab bhi cone ki curved ya total surface area nikalni ho, l chahiye. Formula: l = √(r² + h²), jahan r radius hai aur h vertical height. Volume nikalne me l ki zaroorat nahi hoti, sirf h use hota hai.
Kya is chapter me Frustum of a Cone padhna padega?
Rationalised NCERT print me 'Frustum of a Cone' aur 'Conversion of Solid from One Shape to Another' sections trim kar diye gaye hain. Apni school ki current textbook/syllabus check karo — agar wo sections nahi hain to sirf combination of solids (surface area + volume) practice karo. Frustum ka formula reference ke liye upar table me diya hai.
cm³ ko litre me kaise convert karte hain?
1000 cm³ = 1 litre. Toh answer cm³ me aane ke baad 1000 se divide karke litre me convert karo. Ye conversion capacity/paani wale questions me bahut common hai.
Exam me π ki value 22/7 use karein ya 3.14?
Jab tak question me specifically 3.14 use karne ko na kaha ho, aur radius/diameter 7 ke multiple me ho (jaise 7, 14, 3.5), to 22/7 use karo — calculation clean aati hai. Agar numbers 7 ke multiple nahi hain aur question 3.14 bole, to 3.14 use karo.
Class 10 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 10 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.