Prishthiya Kshetrafal Ayatan Class 10 Notes: Is Chapter Se Dar Kyun Lagta Hai
Mensuration ka naam sunte hi kai students ke dimag me formula ki lambi list ghoomne lagti hai. Cuboid ka formula alag, cylinder ka alag, cone ka alag, aur phir frustum ka to sabse lamba. Lekin sach ye hai ki ye chapter formula ratne ka nahi, samajhne ka hai. Ek baar ye samajh aa jaye ki prishthiya kshetrafal ka matlab kya hai aur ayatan ka matlab kya hai, to aadhe formula apne aap yaad ho jaate hain. Prishthiya kshetrafal ayatan class 10 notes ka ye guide isi samajh par bana hai.
Prishthiya kshetrafal ka matlab hai kisi thos vastu ki baahari satah ka kshetrafal. Sochiye ki aapko ek dibbe par kagaz chipkana hai, to kitna kagaz lagega, wahi uska prishthiya kshetrafal hai. Ayatan ka matlab hai us vastu ke andar kitni jagah hai, ya usme kitna saaman aa sakta hai. Dibbe me kitna paani aayega, wahi uska ayatan hai. Dono me farq saaf hai, aur board exam me sabse zyada galti isi farq ko bhoolne se hoti hai.
Is chapter me aap cuboid, cube, cylinder, cone, sphere, hemisphere aur frustum ke baare me padhenge. Saath me combination wale thos aur ek aakar se doosre aakar me badalne wale sawaal bhi aayenge. Har formula ke saath ek chhota example diya gaya hai taaki formula sirf kitaab me na rahe, balki haath me bhi aa jaye.
| Thos vastu | CSA (vakr prishth) | TSA (kul prishth) | Ayatan |
|---|---|---|---|
| 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) | (1/3)πr²h |
| Sphere (r) | 4πr² | 4πr² | (4/3)πr³ |
| Hemisphere (r) | 2πr² | 3πr² | (2/3)πr³ |
| Frustum (R, r, h, l) | πl(R + r) | πl(R + r) + πR² + πr² | (1/3)πh(R² + Rr + r²) |
Prishthiya Kshetrafal Ayatan Class 10 Notes Me CSA Aur TSA Ka Farq
CSA ka matlab hai vakr prishth ka kshetrafal, matlab sirf gol ya side wali satah. TSA ka matlab hai kul prishth ka kshetrafal, matlab saari satahen milakar. Cylinder ka example lein. Uska CSA 2πrh hai, sirf side ka gol hissa. Uska TSA 2πr(r + h) hai, jisme upar aur neeche ke dono vritt bhi jud jaate hain. Agar dibba khula hai, matlab upar ka dhakkan nahi hai, to TSA alag hoga. Isliye sawaal padhte samay ye zaroor dekhein ki vastu band hai ya khuli hai.
Sphere me CSA aur TSA ek hi hote hain, kyunki sphere ki ek hi satah hoti hai, upar-neeche ka farq nahi hai. Hemisphere me farq hai. Uska CSA 2πr² hai, sirf gol hissa. Uska TSA 3πr² hai, jisme neeche ka vritt bhi jud jaata hai. Prishthiya kshetrafal ayatan class 10 notes me ye farq sabse zyada poochha jaata hai, isliye hemisphere ke sawaal me hamesha ye check karein ki sawaal CSA maang raha hai ya TSA.
Combination wale thos me bhi yahi logic kaam aata hai. Jab do thos judte hain, to judne wali satah andar chali jaati hai aur bahar nahi dikhti. Isliye us satah ka kshetrafal nahi joda jaata. Jaise ek cylinder par ek hemisphere judta hai to us jagah ka vritt dono ke andar chhup jaata hai. Uska kshetrafal baad me minus karna hota hai. Ye baat samajh lene par combination ke sawaal aasan ho jaate hain.
Cuboid Aur Cube Ke Formula Samajh Ke Saath
Prishthiya kshetrafal ayatan class 10 notes ki shuruaat cuboid se hoti hai. Cuboid ek aayatakar dibba hai, jiski lambai l, chaudai b aur unchai h hoti hai. Iska ayatan lbh hota hai, matlab teeno ko guna kar dein. Iska TSA 2(lb + bh + hl) hota hai, matlab teeno alag-alag jode ke kshetrafal ka doguna. Iska CSA 2h(l + b) hota hai, matlab sirf chaaron side ki deewarein, upar-neeche ke dhakkan nahi. Cube cuboid ka hi khaas roop hai jisme l = b = h = a hota hai. Isliye cube ka ayatan a³, TSA 6a² aur CSA 4a² ho jaata hai.
Example: ek cuboid ki lambai 10 cm, chaudai 6 cm aur unchai 4 cm hai. Uska ayatan 10 × 6 × 4 = 240 cm³ hoga. Uska TSA 2(10×6 + 6×4 + 4×10) = 2(60 + 24 + 40) = 248 cm² hoga. Cube ka example: agar side 5 cm hai to ayatan 125 cm³ aur TSA 6 × 25 = 150 cm² hoga. Unit ka dhyan rakhein: kshetrafal hamesha square unit me aur ayatan cubic unit me aata hai. Unit galat likhne par board me number kat sakte hain.
Cylinder Aur Cone Ke Formula Aur Unka Rishta
Prishthiya kshetrafal ayatan class 10 notes me cylinder aur cone ka rishta sabse zyada poochha jaata hai. Cylinder ek belan jaisa aakar hai, jiski trijya r aur unchai h hoti hai. Iska CSA 2πrh hai aur ayatan πr²h hai. Cone ek shanku jaisa aakar hai, jiski trijya r, unchai h aur tiryak unchai l hoti hai. Tiryak unchai ka formula l = √(r² + h²) hai. Cone ka CSA πrl hai aur ayatan (1/3)πr²h hai. Dhyan dein ki cone ka ayatan usi trijya aur unchai wale cylinder ke ayatan ka ek-tihaai hota hai. Ye rishta board me aksar poochha jaata hai.
Example: ek cylinder ki trijya 7 cm aur unchai 10 cm hai. π = 22/7 lene par uska ayatan (22/7) × 49 × 10 = 1540 cm³ hoga. Usi trijya aur unchai wale cone ka ayatan iska ek-tihaai, matlab 1540/3 = 513.33 cm³ hoga. Cone ka CSA nikalne ke liye pehle l nikalna hoga. Agar r = 3 cm aur h = 4 cm hai to l = √(9 + 16) = 5 cm hoga. Phir CSA = πrl = (22/7) × 3 × 5 = 47.14 cm² hoga. Tiryak unchai aur seedhi unchai me farq ko kabhi confuse mat karein, ye is chapter ki doosri sabse badi galti hai.
Sphere Aur Hemisphere Ke Formula
Prishthiya kshetrafal ayatan class 10 notes me sphere aur hemisphere ke formula ek saath yaad rakhein. Sphere ek gend jaisa aakar hai, jiski sirf ek trijya r hoti hai. Iska prishthiya kshetrafal 4πr² hai aur ayatan (4/3)πr³ hai. Hemisphere aadha sphere hai, jaise ek katori. Iska CSA 2πr² hai, TSA 3πr² hai aur ayatan (2/3)πr³ hai. Hemisphere ke sawaal me sabse pehle ye tay karein ki sawaal kis kshetrafal ki maang kar raha hai. Katori ki andar ki satah poochhi jaye to CSA, aur poori katori ka baahari kshetrafal poochha jaye to TSA.
Example: ek sphere ki trijya 7 cm hai. Uska prishthiya kshetrafal 4 × (22/7) × 49 = 616 cm² hoga. Uska ayatan (4/3) × (22/7) × 343 = 1437.33 cm³ hoga. Usi trijya wale hemisphere ka ayatan iska aadha, matlab 718.67 cm³ hoga. Aur uska TSA 3 × (22/7) × 49 = 462 cm² hoga. Aadha karne ka ye tareeka tabhi kaam karta hai jab aakar bilkul aadha ho. Agar hemisphere ke saath cylinder juda ho to phir combination ka niyam lagega, seedha aadha nahi.
Prishthiya Kshetrafal Ayatan Class 10 Notes Me Combination Wale Thos
Combination ka matlab hai do ya zyada thos ka judna. Jaise ek cylinder ke upar ek cone, ya ek cuboid ke upar ek hemisphere. Aise sawaal me pehle har hisse ka alag-alag kshetrafal ya ayatan nikalte hain, phir judne wali satah ka hissa hata dete hain. Ayatan me koi hissa minus nahi hota, kyunki dono hisse ki jagah jud jaati hai. Lekin prishthiya kshetrafal me judne wali satah ko minus karna hota hai, kyunki wo bahar nahi dikhti.
Example: ek khilauna ek cone ke upar ek hemisphere se bana hai, dono ki trijya 7 cm hai. Cone ki tiryak unchai 25 cm hai. Hemisphere ka CSA 2 × (22/7) × 49 = 308 cm² hai. Cone ka CSA (22/7) × 7 × 25 = 550 cm² hai. Kul prishthiya kshetrafal 308 + 550 = 858 cm² hoga. Yahan koi minus nahi hua kyunki dono ke CSA me judne wala vritt pehle se nahi juda hai. Lekin agar poore thos ka TSA poochha jata to hisaab alag hota. Isliye sawaal ke shabdon par dhyan dein: “surface area” ka matlab aksar bahar dikhne wala poora kshetrafal hota hai.
Frustum: Balti Wala Aakar
Prishthiya kshetrafal ayatan class 10 notes me frustum ko balti wale aakar se yaad rakhein. Frustum ek cone ka kata hua hissa hai, jaise ek balti. Jab ek cone ko upar se kaat diya jata hai to neeche ka hissa frustum kehlata hai. Iski upar ki trijya r, neeche ki trijya R, unchai h aur tiryak unchai l hoti hai. Tiryak unchai ka formula l = √[h² + (R − r)²] hai. Frustum ka CSA πl(R + r) hai, TSA πl(R + r) + πR² + πr² hai, aur ayatan (1/3)πh(R² + Rr + r²) hai.
Example: ek balti ki upar ki trijya 14 cm, neeche ki trijya 7 cm aur unchai 24 cm hai. Pehle l nikalte hain: l = √[24² + (14 − 7)²] = √[576 + 49] = √625 = 25 cm. Phir CSA = (22/7) × 25 × (14 + 7) = (22/7) × 25 × 21 = 1650 cm². Ayatan = (1/3) × (22/7) × 24 × (196 + 98 + 49) = (1/3) × (22/7) × 24 × 343 = 8624 cm³. Frustum ke sawaal me R aur r ko ulat dena sabse aam galti hai. Badi trijya hamesha R aur chhoti r hoti hai, chahe balti ulti hi kyun na rakhi ho.
Ek Aakar Se Doosre Aakar Me Badalna
Prishthiya kshetrafal ayatan class 10 notes me conversion wale sawaal isi niyam par bante hain. Jab ek thos ko pighla kar doosra aakar banaya jata hai to ayatan wahi rehta hai. Ye is chapter ka sabse sundar niyam hai. Jaise ek gend ko pighla kar taar banaya jaye to gend ka ayatan taar ke ayatan ke barabar hoga. Is niyam se kitne chhote thos banenge, ye nikala ja sakta hai. Formula seedha hai: bade thos ka ayatan = chhote thos ke ayatan × sankhya.
Example: ek dhatu ka gend jiski trijya 6 cm hai, use pighla kar 2 mm trijya wali taar banayi jaati hai. Pehle unit ek karein: taar ki trijya 0.2 cm hogi. Gend ka ayatan (4/3)π × 216 = 288π cm³ hai. Taar ek lamba cylinder hai, uska ayatan π × 0.04 × L hai. Dono ko barabar rakhne par L = 288π / 0.04π = 7200 cm = 72 metre hoga. Unit badalna is tarah ke sawaal me sabse zaroori hai. mm ko cm me badle bina sawaal galat ho jayega.
Prishthiya Kshetrafal Ayatan Class 10 Notes Me Board Exam Tips
Prishthiya kshetrafal ayatan class 10 notes ke hisaab se board exam me is chapter se aam taur par 3-4 marks ke sawaal aate hain. Pehle formula likhna, phir value rakhna aur phir unit ke saath jawab dena, ye teen steps har sawaal me follow karein. π ki value sawaal me di gayi ho to wahi istemal karein. Agar sawaal me π = 22/7 likha hai to 3.14 mat istemal karein, warna jawab thoda alag aa jayega. Agar koi value nahi di gayi hai to 22/7 lena safe hai.
Sabse aam galtiyan ye hain: trijya ki jagah vyaas istemal karna, CSA ki jagah TSA nikal dena, tiryak unchai ki jagah seedhi unchai lena, aur unit bhool jaana. Har sawaal ke baad ek baar check karein ki aapne trijya hi li hai ya vyaas. Agar vyaas diya gaya hai to pehle use aadha karein. Aakhri me jawab ke saath sahi unit likhna mat bhoolein: kshetrafal ke liye cm² aur ayatan ke liye cm³.
Practice Questions With Answers
Sawaal 1: Ek cube ki side 8 cm hai. Uska TSA aur ayatan nikaliye. Jawab: TSA = 6 × 64 = 384 cm², ayatan = 512 cm³. Sawaal 2: Ek cylinder ki trijya 7 cm aur unchai 12 cm hai. Uska CSA nikaliye. Jawab: CSA = 2 × (22/7) × 7 × 12 = 528 cm². Sawaal 3: Ek cone ki trijya 6 cm aur unchai 8 cm hai. Uska ayatan nikaliye. Jawab: ayatan = (1/3) × (22/7) × 36 × 8 = 301.71 cm³.
Sawaal 4: Ek hemisphere ki trijya 7 cm hai. Uska TSA nikaliye. Jawab: TSA = 3 × (22/7) × 49 = 462 cm². Sawaal 5: Ek frustum ki trijyaen 10 cm aur 5 cm hain aur unchai 12 cm hai. Uska ayatan nikaliye. Jawab: ayatan = (1/3) × (22/7) × 12 × (100 + 50 + 25) = 2200 cm³. Prishthiya kshetrafal ayatan class 10 notes ke in practice questions ko pehle khud solve karein, phir jawab se milayein. Jo sawaal galat ho, uska formula dobara samajh kar phir se karein.
Aur bhi padhein: Vritt Class 10 Notes Chapter 10 aur Tribhuj Class 10 Notes Chapter 6.
Official Links
- NCERT Class 10 Maths Textbook (Chapter 12): https://ncert.nic.in
- NCERT Solutions aur syllabus: https://ncert.nic.in
- Is chapter se jude doosre notes: [Vritt Class 10 Notes](/vritt-class-10-notes-chapter-10/), [Tribhuj Class 10 Notes](/tribhuj-class-10-notes-chapter-6/)
Aksar Puche Jaane Wale Sawaal (FAQs)
CSA aur TSA me kya farq hai?
CSA sirf vakr ya side wali satah ka kshetrafal hai, jabki TSA saari satahon ka kul kshetrafal hai. Cylinder me CSA 2πrh aur TSA 2πr(r + h) hota hai. Hemisphere me CSA 2πr² aur TSA 3πr² hota hai. Sphere me dono barabar hote hain.
Cone ka ayatan cylinder ka ek-tihaai kyun hota hai?
Kyunki ek hi trijya aur unchai wale teen cone milkar ek cylinder bharte hain. Isliye cone ka ayatan (1/3)πr²h hota hai jabki cylinder ka πr²h hota hai. Ye rishta board me aksar poochha jaata hai.
Frustum ka l kaise nikalte hain?
Frustum ki tiryak unchai l = √[h² + (R − r)²] hoti hai. Isme R badi trijya aur r chhoti trijya hai. R aur r ko ulat dene par bhi fark nahi padta kyunki antar ka square liya jaata hai, lekin formula me antar ka square hi istemal karein.
Pighlane wale sawaal me kya barabar rehta hai?
Ayatan barabar rehta hai. Bade thos ka ayatan chhote thos ke ayatan ke barabar hota hai. Is niyam se taar ki lambai ya chhote thos ki sankhya nikali jaati hai. Unit ko ek jaisa karna sabse zaroori hai.
π ki kaun si value istemal karein?
Jo value sawaal me di gayi hai, wahi istemal karein. Agar 22/7 di gayi hai to 22/7 lein, agar 3.14 di gayi hai to 3.14 lein. Agar koi value nahi di gayi hai to 22/7 lena safe hai.
Aakhri Baat
Ye chapter tab aasan lagta hai jab formula ke peechhe ki samajh aa jaye. Prishthiya kshetrafal ka matlab baahari satah aur ayatan ka matlab andar ki jagah, bas itna yaad rakhein to aadha chapter khud samajh aa jayega. Prishthiya kshetrafal ayatan class 10 notes ka ye guide baar-baar padhein, formula ko example ke saath yaad karein, aur practice questions khud solve karein. Board me is chapter se number lena sabse aasan hai, bas unit aur trijya-vyaas ki galti se bachein.