Math of Space: Surface Area and Volume: Class 9 Maths Chapter 14 Hinglish mein

Boxes, tins, cones, pyramids aur balls ko wrap karna aur bharna

Ganita Manjari Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

80 min
52 practice questions
21 MCQs

Is chapter mein aap kya seekhenge

Cuboid aur cube ka surface area aur volume derive karna aur use karna, volumes ke cube roots samet
Samjhana ki barabar volume wale solids ke surface areas alag kyun ho sakte hain, aur cube ko sabse kam kyun chahiye
Right aur oblique cylinders aur unke cross-sections describe karna, aur open aur closed cylinders ke curved aur total surface areas aur volume nikaalna
Cone ki slant height nikaalna, use sector mein unroll karke uska curved surface area derive karna, aur V = ⅓πr²h use karna
Kisi bhi pyramid ka volume base area times height ka one third maan kar nikaalna, aur faces jod kar uska surface area nikaalna
Spheres aur hemispheres ke surface area aur volume formulas use karna, aur unhe Archimedes ke cylinder aur thin cones se jodna
Asli objects ke liye guesstimates lagana, assumptions batana aur upper bounds pehchaanna

NCERT Mathematics Class 9, Chapter 14, pp. 123–148 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Wrap karna aur bharna

Ghar par koi bhi box utha lo: doodh ka carton, steel ka tiffin, chai ka packet. Asli zindagi mein iske baare mein do sawaal kaam ke hain. Ise banane ya dhakne mein kitna material lagta hai? Ismein kitna aa sakta hai? Pehla sawaal iske surface area (पृष्ठीय क्षेत्रफल) ke baare mein hai; doosra iske volume (आयतन) ke baare mein (Fig. 14.1).

2/2Andar ko bharna volume naapta hai, cubic units mein.

Fig. 14.1Surface area batata hai ki kisi solid ko wrap karne mein kitna lagta hai; volume batata hai ki use bharne mein kitna lagta hai.

Step 1 mein orange wrapping paper se dhaka ek box hai, saath mein note lapeto aur label surface area square centimetres mein; right mein ek halka sa doosra box hai. Step 2 mein wrapped box fade ho jaata hai aur right wala box aage aata hai, jo rows, columns aur layers mein chhote unit cubes se bhara hai, saath mein note bharo aur label volume cubic centimetres mein.

Tum pehle se bahut kuch jaante ho jo kaam aayega. Tumne nets ko fold karke cubes, cylinders aur cones banaye hain. Part 1 ke Chapter 6 mein tumne rectangles, triangles, circles aur sectors ke areas nikaale, aur Baudhāyana–Pythagoras theorem use kiya. Ab hum yeh sab solids ki paanch families (cuboids, cylinders, cones, pyramids aur spheres) par lagayenge, aur aakhir mein asli objects ke baare mein samajhdaari wale estimates lagayenge.

Definition

Surface area

Kisi solid ke bahar ke saare faces aur curved hisson ka total area. Ise $c m^{2}$ ya $m^{2}$ jaise square units mein naapa jaata hai.

ExampleEk band box banane ke liye lagne wala cardboard.

Definition

Volume

Koi solid kitni jagah gherta hai, ya koi container kitna rakh sakta hai, unit cubes mein gina hua: $c m^{3}$ , $m^{3}$ waghaira.

ExampleEk carton mein jitna doodh aa sakta hai.

14.1 Cuboids aur Cubes

Eent, maachis ki dibbi, steel ki almirah aur shoebox, sabki shape ek hi hai. Har ek cuboid (घनाभ) hai: ek box jiske saare faces rectangles hain, jo 12 edges aur 8 corners par right angles par milte hain. Teen measurements kisi bhi cuboid ko tay kar dete hain: uski length $l$ , width $w$ aur height $h$ . Rectangle ko sirf do chahiye; cuboid ek teesri direction jodta hai.

Iska surface area nikaalne ke liye dhyaan do ki faces pairs mein aate hain. Top aur bottom barabar hain, front aur back barabar hain, aur dono sides barabar hain (Fig. 14.2). Isliye humein sirf teen areas chahiye, har ek do baar gina hua. Agar tumhare paas textbook se cuboid ka printed net hai, to use fold karo aur check karo ki kaun se faces pair banate hain.

4/4Teenon areas jodo aur result ko double karo.

Fig. 14.2Chhah faces, teen barabar pairs: isiliye formula ka har term double hai.NCERT Fig. 14.2 · p. 123

Ek cuboid bana hai jiski length l front ke saath, width w peeche ki taraf aur height h side par upar ki taraf hai. Step 2 mein dikhne wale top, front aur side faces blue, orange aur green rang ke hain, aur ek note kehta hai ki har ek ka ek chhupa jodidaar hai. Step 3 mein chhahon faces right mein teen pairs ki tarah bichhe hain: do blue l by w rectangles, do orange l by h rectangles aur do green w by h rectangles. Step 4 mein pairs ke neeche formula TSA = 2(lw + lh + wh) aata hai.

Teenon pairs jodne se cuboid ka total surface area (TSA) milta hai. TSA ko $S$ likhkar:

Cuboid ka total surface area (TSA)

S = 2 (l w + l h + w h)

(14.1)

Volume naapne ke liye humein ek naapne wala block chahiye, jaise area ke liye humne unit squares use kiye the. Woh block hai unit cube, jiska har edge 1 unit lamba hai. Uske volume ko 1 cubic unit kehte hain: 1 cm wale cube ke liye $1 c m^{3}$ , 1 m wale cube ke liye $1 m^{3}$ . Bina gaps ke kitne unit cubes ek cuboid ko bharte hain, yeh ginne se uska volume milta hai.

Cuboid ko layer by layer bharo (Fig. 14.3). Neeche wali layer $l \times w$ cubes ka ek rectangle hai. Aisi $h$ layers hain, ek ke upar ek. Isliye cuboid mein $l \times w \times h$ unit cubes aate hain. Dhyaan do ki $l \times w$ bas base ka area hai, wahi product jo rectangle ka area deta hai.

3/4Teen barabar layers mein $12 \times 3 = 36$ cubes aate hain.

Fig. 14.3Volume unit cubes ginta hai:

l \times w

ki ek layer,

h

baar dohrayi gayi.NCERT Fig. 14.3 · p. 124

Left mein 1 cm edge wala ek akela unit cube hai. Step 2 mein 4 by 3 cubes ki ek layer aati hai, jis par l = 4 aur w = 3 likha hai, aur ginti 4 × 3 = 12 hai. Step 3 mein do aur layers khisak kar uske upar aa jaati hain, to stack 3 layers ooncha ho jaata hai, jis par h = 3 likha hai, aur ginti 12 × 3 = 36 hai. Step 4 mein stack ke bagal mein rule V = (l × w) × h likha hai.

Cuboid ka volume

\begin{aligned} V & = l w h \\ = base area \times h \end{aligned}

(14.2)

Teenon measurements barabar kar do aur cuboid ek cube (घन) ban jaata hai, jaise ek dice ya sugar cube. Har edge $a$ ke barabar ho to formula (14.1) chhota hokar $2 (a^{2} + a^{2} + a^{2}) = 6 a^{2}$ ban jaata hai aur formula (14.2) $a \times a \times a = a^{3}$ :

Cube ka total surface area

TSA = 6 a^{2}

(14.3)

Cube ka volume

V = a^{3}

(14.4)

Ab cube number naam ka matlab samajh aata hai: $5^{3} = 125$ theek utne unit cubes hain jitne edge 5 wala cube banate hain. Cubes ki table ko ulta padhne se cube roots milte hain, aur cube ka edge uske volume ka cube root hota hai. Jis cube mein $343 c m^{3}$ aata hai, uska edge hai $\sqrt[3]{343} = 7 c m .$

Example 14.1NCERT Example 1 · p. 125

Gur ke do blocks

Ek dukaan gur (jaggery) ko blocks mein dhaalti hai. Block A edge 4 cm wala ek cube hai. Block B 8 cm × 4 cm × 2 cm ka ek cuboid hai. Unke volumes aur surface areas compare karo. Kis block ko zyada wrapping paper chahiye, aur thanda hote samay kaun jaldi sakht hoga?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yeh example ek badi baat dikhata hai. Same volume wale boxes ke alag surface areas ho sakte hain, aur cube ko sabse kam material chahiye. Packaging designers card bachaane ke liye isko use karte hain. Engineers iska ulta use karte hain: car ka radiator ya motorbike engine ke fins metal ko patla phailaate hain taaki zyada surface khule aur heat jaldi nikle.

InteractiveBox lab

Do tabs. Build a box mein teen sliders length, width aur height ko 1 se 10 cm tak set karte hain; box ke faces par 1 cm grid bana hai aur teen cards uska volume, surface area aur dimensions dikhate hain, saath mein ek sentence layers ginta hai. Same volume mein buttons 24, 36 ya 64 cubic centimetres ka volume chunte hain aur us volume wale whole-number edges ke har box ki list dete hain, har ek ke saath uske surface area ka bar; box chunne par woh draw hota hai, aur status line batati hai ki kis box ko sabse kam card chahiye.

Fig. 14.4Unit cubes se box banao, ya same volume wale har whole-number box ko compare karo.

Quick check

3 mein se 1

Chaar band boxes mein se har ek mein theek

64 c m^{3}

aata hai. Kise banane mein sabse kam card lagega?

4 cm × 4 cm × 4 cm8 cm × 4 cm × 2 cm16 cm × 2 cm × 2 cm8 cm × 8 cm × 1 cm

Answer dekho

Answer (A) 4 cm × 4 cm × 4 cm

Cube

4 \times 4 \times 4

96 c m^{2}

chahiye. Baakiyon ko 112, 136 aur

160 c m^{2}

chahiye. Barabar volume ka matlab barabar surface nahi hai; jo shape cube ke sabse kareeb hai, use sabse kam chahiye.

Practice · 8 questions

Practice 14.1: answers ke saath

NumericalEasyNCERT Exercise Set 14.1, Q1
Cube ke shape wale ek water tank mein $512 m^{3}$ paani aata hai. Uske chhahon faces ke liye lagne wali steel sheet ka area nikaalo.
m²
Solution dikhao
Edge $a = \sqrt[3]{512} = 8 m$ hai, kyunki $8^{3} = 512$ .
TSA $= 6 a^{2} = 6 \times 64 = 384 m^{2} .$
NumericalEasyNCERT Exercise Set 14.1, Q2
1.5 m inner edge wale ek cubical crate mein, bina gaps ke, 25 cm edge wale kitne wooden cubes pack ho sakte hain?
cubes
Solution dikhao
1.5 m = 150 cm, aur $150 \div 25 = 6$ cubes har edge ke saath fit hote hain.
Isliye crate mein $6 \times 6 \times 6 = 216$ cubes aate hain. (Check: $150^{3} \div 25^{3} = 3 375 000 \div 15 625 = 216$ .)
NumericalMediumNCERT Exercise Set 14.1, Q3
Ek warehouse andar se 30 m lamba, 18 m chauda aur 8 m ooncha hai. Ise cartons se bharna hai, jinmein se har ek 1.5 m × 1.2 m × 0.8 m ka hai. Kitne cartons fit honge?
cartons
Solution dikhao
Length ke saath: $30 \div 1.5 = 20$ . Width ke saath: $18 \div 1.2 = 15$ . Height mein upar: $8 \div 0.8 = 10$ .
Cartons ki sankhya $= 20 \times 15 \times 10 = 3000$ .
Volumes se check karo: $(30 \times 18 \times 8) \div (1.5 \times 1.2 \times 0.8) = 4320 \div 1.44 = 3000.$ Ginti exactly divide hoti hai, isliye koi jagah waste nahi hoti.
NumericalMediumNCERT Exercise Set 14.1, Q4
Teen cubes, har ek ka volume $64 c m^{3}$ , ek seedhi row mein face se face chipkaaye gaye hain. Naye solid ka total surface area nikaalo.
cm²
Solution dikhao
Har cube ka edge $\sqrt[3]{64} = 4 c m$ hai, isliye row ek cuboid $12 c m \times 4 c m \times 4 c m$ hai.
TSA $= 2 (12 \times 4 + 12 \times 4 + 4 \times 4) = 2 (48 + 48 + 16) = 224 c m^{2} .$
Check: teen alag cubes ka $3 \times 96 = 288 c m^{2}$ hota hai; chipkaane se $16 c m^{2}$ ke 4 faces chhup jaate hain, aur $288 - 64 = 224$ .
NumericalMediumNCERT Exercise Set 14.1, Q5
Edge 3 cm wale ek cube ko edge 1 cm wale cubes mein kaata jaata hai. Bade cube ke surface area aur saare chhote cubes ke total surface area ka ratio nikaalo.
Solution dikhao
Answer1 : 3
Bada cube: $6 \times 3^{2} = 54 c m^{2}$ .
$3^{3} = 27$ chhote cubes hain, har ek ka $6 c m^{2}$ , isliye kul $162 c m^{2}$ .
Ratio $= 54 : 162 = 1 : 3$ . Volume nahi badalta, lekin kaatne se surface teen guna ho jaata hai. Isiliye ek chammach cheeni sugar cube se jaldi ghulti hai: zyada surface paani ko chhoota hai.
NumericalHardNCERT Exercise Set 14.1, Q6
Cuboid ke teen faces ek hi corner share karte hain. Unke areas $12 c m^{2}$ , $15 c m^{2}$ aur $20 c m^{2}$ hain. Cuboid mein kitna aata hai, aur uske edges kya hain?
cm³
Solution dikhao
Maan lo $l w = 12$ , $w h = 15$ aur $l h = 20$ . Multiply karne par: $(l w) (w h) (l h) = l^{2} w^{2} h^{2} = (l w h)^{2}$ .
Isliye $(l w h)^{2} = 12 \times 15 \times 20 = 3600$ , aur $V = l w h = 60 c m^{3} .$
Phir $h = 60 \div 12 = 5$ , $l = 60 \div 15 = 4$ aur $w = 60 \div 20 = 3$ (cm mein). Check: $4 \times 3 = 12$ , $3 \times 5 = 15$ , $4 \times 5 = 20$ .
7
Short answerHardNCERT Exercise Set 14.1, Q7
Edge 4 cm wale ek wooden cube ke chhahon faces laal rang se paint kiye jaate hain aur phir use $1 c m^{3}$ ke cubes mein kaata jaata hai. Kitne chhote cubes par paint hai
1. theek teen faces par?
  cubes
  Solution dikhao
  Sirf 8 corner cubes ke teen painted faces hain.
2. theek do faces par?
  cubes
  Solution dikhao
  12 edges mein se har ek par corners ke beech $4 - 2 = 2$ cubes hain: $12 \times 2 = 24$ .
3. theek ek face par?
  cubes
  Solution dikhao
  6 faces mein se har ek par edges se door cubes ka ek $2 \times 2$ block hai: $6 \times 4 = 24$ .
4. kisi bhi face par nahi?
  cubes
  Solution dikhao
  Bina paint wala core ek $2 \times 2 \times 2$ cube hai: 8 cubes. Check: $8 + 24 + 24 + 8 = 64 = 4^{3}$ .
8
ProofHardNCERT Exercise Set 14.1, Q8
Aise cuboids dhoondho jinke edges centimetres ke whole numbers hain aur jinka total surface area theek $64 c m^{2}$ hai.
1. Kya aisa ek se zyada cuboid hai?
  HaanNahi
  Solution dikhao
  Haan. Jaise $4 \times 4 \times 2$ se $2 (16 + 8 + 8) = 64$ milta hai aur $7 \times 2 \times 2$ se $2 (14 + 4 + 14) = 64$ .
2. Sabhi dhoondho.
  Solution dikhao
  Answer10 × 2 × 1, 7 × 2 × 2, 4 × 4 × 2
  Theek teen hain: $10 \times 2 \times 1$ , $7 \times 2 \times 2$ aur $4 \times 4 \times 2$ (cm mein).
3. Dikhao ki aur koi nahi hai.
  Solution dikhao
  Humein $h \leq w \leq l$ ke saath $l w + w h + h l = 32$ chahiye. Kyunki har product kam se kam $h^{2}$ hai, $3 h^{2} \leq 32$ , isliye $h \leq 3$ .
  Agar $h = 1$ : $l w + l + w = 32$ , isliye $(l + 1) (w + 1) = 33 = 3 \times 11$ . Isse $w = 2$ , $l = 10$ milta hai.
  Agar $h = 2$ : $l w + 2 l + 2 w = 32$ , isliye $(l + 2) (w + 2) = 36$ jahan $w + 2 \geq 4$ . Pairs hain $4 \times 9$ ( $w = 2, l = 7$ ) aur $6 \times 6$ ( $w = l = 4$ ).
  Agar $h = 3$ : $(l + 3) (w + 3) = 41$ , ek prime, jiska aisa koi factor pair nahi jismein dono factors kam se kam 6 hon. Isliye aur cuboids nahi hain.

14.2 Right Circular Cylinder

Paani ka pipe, gas cylinder, bina chheeli pencil aur ghee ka tin, sab cylinders hain. Cylinder ke do barabar, parallel circular ends hote hain jo ek curved surface se jude hote hain. Dono circles ke centres ko jodne wali line uska axis hai.

Right circular cylinder (लंब वृत्तीय बेलन) mein ends circles hote hain aur axis unke perpendicular hota hai, jaise sikkon ka ek saaf stack. Stack ko side se dhakelo to axis jhuk jaata hai: yeh oblique cylinder hai. Iske ends ab bhi circles hain, lekin axis unse right angle par nahi rehta. Is chapter mein hum sirf right circular cylinders padhenge, aur unhe bas cylinders kahenge.

Definition

Cross-section

Woh flat shape jo kisi solid ko ek bahut patli, seedhi slice se kaatne par dikhti hai.

ExampleKheere ko aar-paar kaatne se circles milte hain.

Cross-section (अनुप्रस्थ काट) is par depend karta hai ki tum kaise kaatte ho (Fig. 14.5). Cylinder ko uske base ke parallel kaato to har slice same circle hoti hai. Use axis ke saath kaato to rectangle milta hai. Jab hum 'cylinder ka cross-section' kehte hain, to aam taur par circle ki baat hoti hai. Salad ke liye kata kheera lagbhag circles deta hai, aur bread loaf ki har slice loaf ka ek cross-section hai. CT ya MRI scan bhi aise hi kaam karta hai: woh doctors ko body ek-ek patli slice karke dikhata hai.

1/3Right: axis base ke perpendicular hai. Oblique: axis jhukta hai.

Fig. 14.5Right cylinder ka axis uske base ke perpendicular hota hai. Aar-paar ki slices circles hain; axis ke saath ki slices rectangles hain.NCERT Cross-sections · p. 127

Step 1 mein do cylinders aamne-saamne hain: left mein ek right cylinder jiska dashed axis base se marked right angle par milta hai, aur right mein ek oblique cylinder jiska dashed axis jhukta hai. Step 2 mein oblique cylinder fade hota hai, right cylinder ke aar-paar ek horizontal slice mark hoti hai aur ek arrow flat bani slice tak le jaata hai: ek circle. Step 3 mein axis se hokar ek vertical slice mark hoti hai aur uske bagal mein flat slice ek rectangle hai.

Definition

Right circular cylinder

Ek solid jiske do barabar, parallel circular ends hain aur jiska axis ends ke perpendicular hai.

ExampleNayi pencil, bina cap ka gas cylinder.

Curved surface. Socho ki tum ek tin se uska paper label chheel rahe ho (Fig. 14.6). Seedha neeche tak kaato aur kholo: label ek rectangle ban jaata hai. Iski height tin ki height $h$ hai. Iski length tin ke ek poore chakkar jitni thi, isliye yeh circumference $2 π r$ ke barabar hai. Is rectangle ka area cylinder ka curved surface area (CSA) hai.

4/4Band tin ke do circular dhakkan bhi hote hain, har ek $π r^{2}$ .

Fig. 14.6Kholne par curved surface

2 π r

lamba aur

h

ooncha rectangle hai.NCERT Fig. 14.5A · p. 128

Step 1 mein radius r aur height h wala ek tin hai jiski curved side orange label se dhaki hai. Step 2 label ke front par neeche tak ek dashed cut line mark karta hai. Step 3 mein label right ki taraf khul kar height h aur length 2πr wala ek lamba rectangle ban jaata hai, jis par CSA = 2πrh likha hai. Step 4 mein do blue circles, yaani upar aur neeche ke dhakkan, rectangle ko chhoote hue aate hain, aur label badal kar TSA = 2πrh + 2πr² ho jaata hai.

Cylinder ka curved surface area

CSA = 2 π r h

(14.5)

Band cylinder, jaise dhakkan laga steel ka dabba, ke do circular ends bhi hote hain, har ek ka area $π r^{2}$ . Iska total surface area hai:

Dono ends se band cylinder

\begin{aligned} TSA & = 2 π r h + 2 π r^{2} \\ = 2 π r (h + r) \end{aligned}

(14.6)

Ek end se khule cylinder, jaise glass ya pencil stand, ka sirf ek circular end hota hai:

Ek end se khula cylinder

\begin{aligned} TSA & = 2 π r h + π r^{2} \\ = π r (2 h + r) \end{aligned}

(14.7)

Volume. Cylinder ki har aar-paar slice $π r^{2}$ area wala same circle hai. Cuboid ki tarah hi, volume cross-section ka area times height hai:

Cylinder ka volume

V = π r^{2} h

(14.8)

14.2.1 Upar wale formulas sahi kyun hain, ise samajhne ka ek aur tareeka

Ek jaise sikkon ko ek oonche tower mein jamao (Fig. 14.7). Har sikka ek bahut chhota cylinder hai: $π r^{2}$ area wala circle jiski motai bahut kam, $t$ , hai. Agar tower mein $n$ sikke hain, to uski height $h = n t$ hai aur volume $n \times π r^{2} t = π r^{2} (n t) = π r^{2} h$ hai. Ab edges ko dekho. Har sikke ka rim lagbhag $2 π r$ lambi aur $t$ chaudi ek patli strip hai. Saare rims milkar tower ki side dhakte hain, jiska area $n \times 2 π r t = 2 π r h$ hai. Dono formulas ek hi picture se nikal aate hain. Rotiyon ya papad ka stack bhi yahi kahaani batata hai.

3/3Faces se $V = π r^{2} h$ milta hai; rims se $2 π r h$ .

Fig. 14.7Cylinder patli discs ka stack hai: faces volume banate hain, rims curved surface banate hain.NCERT 14.2.1 · p. 128

Step 1 mein thoda upar se dekha gaya ek akela sikka hai, jis par radius r aur motai t mark hai; uske top face par area πr² likha hai. Step 2 mein right mein baarah sikkon ka tower aata hai jiske bagal mein uski height h mark hai. Step 3 mein saare sikkon ke rims ek saath highlight hote hain aur left mein do formulas aate hain: V = πr² × h aur rims = 2πr × h.

Example 14.2NCERT Example 2 · p. 129

Pencil stand ko dhakna

Ritu ek cylindrical pencil stand ke bahar ke curved hisse ko rangeen paper se dhakti hai. Stand ka radius 4.2 cm aur height 10 cm hai. Overlaps ko ignore karke use kitna paper chahiye? $π \approx \frac{22}{7}$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Chhat par paani ki tanki

Ek band cylindrical water tank ka radius 0.7 m aur height 2 m hai. (a) Ismein kitne litres aate hain? (b) Ise banane mein kitni steel sheet lagegi? $π \approx \frac{22}{7}$ aur $1 m^{3} = 1000$ litres use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 5 questions

Practice 14.2: answers ke saath

1
NumericalMediumNCERT Exercise Set 14.2, Q1
Cylinder Q ka radius cylinder P ka 3 guna hai lekin height sirf one third hai. Unke curved surface areas (P : Q) aur volumes (P : Q) ka ratio nikaalo.
1. P ka CSA : Q ka CSA
  Solution dikhao
  Answer1 : 1
  $2 π r h : 2 π (3 r) (\frac{h}{3}) = 2 π r h : 2 π r h = 1 : 1$ . Curved surfaces barabar hain.
2. P ka volume : Q ka volume
  Solution dikhao
  Answer1 : 3
  $π r^{2} h : π (3 r)^{2} (\frac{h}{3}) = π r^{2} h : 3 π r^{2} h = 1 : 3$ . Q mein teen guna aata hai.
2
NumericalMediumNCERT Exercise Set 14.2, Q2
Do cylinders ke radii ka ratio $3 : 4$ aur heights ka ratio $2 : 5$ hai. Ratios ke roop mein compare karo (a) unmein kitna aata hai aur (b) unke curved surfaces.
1. volumes ka ratio
  Solution dikhao
  Answer9 : 40
  $π (3 k)^{2} (2 m) : π (4 k)^{2} (5 m) = 18 : 80 = 9 : 40.$
2. curved surface areas ka ratio
  Solution dikhao
  Answer3 : 10
  $2 π (3 k) (2 m) : 2 π (4 k) (5 m) = 6 : 20 = 3 : 10.$
NumericalMediumNCERT Exercise Set 14.2, Q3
Edge 14 cm wale ek wooden cube mein se sabse bada possible right circular cylinder taraasha jaata hai. Cylinder ka volume nikaalo, aur cube ka kitna fraction use hua. $π \approx \frac{22}{7}$ use karo.
cm³
Solution dikhao
Cylinder ka diameter aur height dono edge ke barabar hain, isliye $r = 7 c m$ aur $h = 14 c m$ .
$V = \frac{22}{7} \times 7^{2} \times 14 = 22 \times 7 \times 14 = 2156 c m^{3} .$
Cube ka volume $14^{3} = 2744 c m^{3}$ hai, aur $2156 \div 2744 = \frac{11}{14} \approx 0.79$ . Aam taur par kisi bhi cube ke liye fraction $\frac{π}{4}$ hai, lagbhag 79 %.
NumericalHardNCERT Exercise Set 14.2, Q4
Ek cylinder ka radius 25 % badhaya jaata hai. Uski height kitne percent ghataani hogi taaki volume same rahe?
%
Solution dikhao
$π (1.25 r)^{2} (k h) = π r^{2} h$ se $1.5625 k = 1$ milta hai, isliye $k = 0.64$ .
Height pehle ki 64 % ho jaati hai, yaani 36 % ki kami.
5
NumericalMediumNCERT Exercise Set 14.2, Q5
Edge 14 cm wala ek solid copper cube pighlaaya jaata hai. Phir saara copper cylinders ke shape wali solid rods mein dhaala jaata hai, har rod 14 cm lambi aur radius 1 cm. $π \approx \frac{22}{7}$ use karo.
1. Cube ka volume nikaalo.
  cm³
  Solution dikhao
  $14^{3} = 2744 c m^{3}$ .
2. Ek rod ka volume nikaalo.
  cm³
  Solution dikhao
  $\frac{22}{7} \times 1^{2} \times 14 = 44 c m^{3}$ .
3. Kitni poori rods ban sakti hain?
  rods
  Solution dikhao
  $2744 \div 44 \approx 62.4$ , isliye 62 poori rods (thoda copper bach jaata hai).

14.3 Cones

Ice-cream cone, birthday cap, funnel aur khalihaan par anaaj ka dher, sab lagbhag cones (शंकु) hain. Ek right-angled triangle lo aur use uski do chhoti sides mein se ek ke around ek baar ghumao (Fig. 14.8). Doosri chhoti side ek circle banati hai, yaani base. Lambi side tirchhi surface banati hai. Jo solid milta hai woh right circular cone hai.

Teen lengths ise describe karti hain. Radius $r$ base ka hai. Height $h$ apex se seedha neeche base ke centre tak jaati hai. Slant height $l$ (तिर्यक ऊँचाई) surface ke saath apex se rim tak jaati hai. Yeh ghoomte triangle ki teen sides hain, isliye Baudhāyana–Pythagoras theorem inhe jodta hai:

Cone ki slant height

l^{2} = h^{2} + r^{2}

(14.9)

5/5Height $l$ ke patle triangles, jinke bases jud kar $2 π r$ bante hain.

Fig. 14.8Right triangle ko ghumao to cone milta hai; cone ko kholo to radius

l

aur arc

2 π r

wala sector milta hai.NCERT Fig. 14.8, 14.9 · p. 130

Step 1 mein ek right triangle hai jiski vertical side h, horizontal side r aur tirchhi side l hai, jo ek dashed axis par khada hai. Step 2 mein uski mirror image dashed dikhti hai aur base ke around ek curved arrow hai: triangle ghoom raha hai. Step 3 mein bana hua cone hai jis par h, r aur l likhe hain aur formula l² = h² + r² hai. Step 4 mein cone par neeche tak ek dashed seam bani hai aur right mein curved surface flat padi hai, radius l wale circle ke ek sector ki tarah jiska arc 2πr lamba hai. Step 5 mein sector bahut saare patle triangles mein kata hai, unmein se ek shaded hai, aur formula CSA = ½ × 2πr × l = πrl aata hai.

Definition

Slant height

Cone ke apex se uske base ke rim ke kisi bhi point tak ki doori, surface ke saath naapi gayi.

Example $r = 5$ aur $h = 12$ wale cone ka $l = 13$ hai.

Paper ki party cap ko ek slant line ke saath kaato aur dabaa kar flat karo: tumhe circle ka ek sector milta hai. Sector ka radius slant height $l$ hai. Iska curved edge base ka rim tha, isliye arc $2 π r$ lamba hai. Sector ko tab tak lapeto jab tak uske do seedhe edges mil na jaayein, aur cone wapas ban jaata hai. Textbook ke aakhir mein diye cut-out nets se tum yeh try kar sakte ho.

DerivationCone ka curved surface area

Pehla tareeka: patle triangles. Flat kiye sector ko pizza ki tarah $n$ patle wedges mein kaato (Fig. 14.8 ka aakhri step). Jab $n$ bahut bada ho, to har wedge lagbhag height $l$ ka ek lamba, patla triangle hai. Uska chhota base arc ka ek bahut chhota tukda hai: in tukdon ko $b_{1}, b_{2}, \dots, b_{n}$ kaho. Triangle areas jodne se curved surface area milta hai: $A = \frac{1}{2} b_{1} l + \frac{1}{2} b_{2} l + \dots + \frac{1}{2} b_{n} l .$ Common factor bahar nikaalne par, $A = \frac{1}{2} (b_{1} + b_{2} + \dots + b_{n}) l .$

Tukde $b_{1}, \dots, b_{n}$ milkar poora arc banate hain, jo $2 π r$ hai. Isliye:

\begin{aligned} A & = \frac{1}{2} \times 2 π r \times l \\ = π r l \end{aligned}

Doosra tareeka: sector angle. Maano sector ke centre par angle $θ$ hai. Part 1 se, radius $l$ wale sector ka area $π l^{2} \times \frac{θ}{360}$ aur arc length $2 π l \times \frac{θ}{360}$ hai. Arc base ke rim, $2 π r$ , ke barabar hona chahiye:

2 π l \times \frac{θ}{360} = 2 π r

isliye \frac{θ}{360} = \frac{r}{l}

Phir sector ka area $π l^{2} \times \frac{r}{l} = π r l$ hai, wahi answer. Circular base jodne se total surface area milta hai.

Cone ka curved surface area

CSA = π r l

(14.10)

Cone ka total surface area

\begin{aligned} TSA & = π r l + π r^{2} \\ = π r (l + r) \end{aligned}

(14.11)

Volume. Is section ka surprise yeh hai: cone mein theek one third aata hai us cylinder ka jo uske around fit hota hai (same base, same height).

Cone ka volume

\begin{aligned} V & = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times A \times h \end{aligned}

(14.12)

jahan

$A$: base ka area, $π r^{2}$

One third hi kyun? Poore proof ke liye ek limiting argument chahiye, jaisa tum Class 12 mein calculus mein dekhoge. Lekin neeche wali activity mein tum is rule ko apne haathon se check kar sakte ho, aur is chapter mein aage tum dekhoge ki har pyramid ke liye bhi wahi $\frac{1}{3}$ aata hai.

ActivityNCERT Experiment (Fig. 14.10) · p. 132

Teen cones se ek tin bharta hai

Tumhe chahiye

ek khula tin
patle card ki ek sheet
tape aur kainchi
sookhe chawal ya ret
ek ruler

Pehle guess karo

Tin ko bharne mein kitne cone bhar chawal lagenge?

234lagbhag 1.5

Upar se khula ek tin dhoondho (chhota powder ya coffee ka tin chalega). Uska radius aur height naapo.
Card ki sheet ko lapet kar ek aisa cone banao jo tin se match kare: uska rim tin ke munh jitna chauda ho, aur uski nok tin ki height jitni gehri ho. Tape lagao aur rim ko barabar kaato.
Cone ko sookhe chawal ya ret se oopar tak bharo. Ruler se top ko barabar karo.
Ise tin mein daalo. Cone bhar ginte hue dohraate raho, jab tak tin oopar tak bhar na jaaye. Apne results note karo.

Table 14.1

Kitne cone bhar daale	Tin kitna bhara hai?
1
2
3

Tumhari entries apne aap save hoti hain.

Doosra tareeka. Modelling clay ke ek gole ko solid cylinder mein roll karo aur uska radius aur height note karo. Use daba do, phir saari clay se aise cones banao jinmein se har ek cylinder ke radius aur height se match kare. Gino kitne cones bane.

3/4Do cone bhar ke baad woh two thirds bhara hai.

Fig. 14.9Teen cone bhar same radius aur height wale tin ko bhar dete hain.NCERT Fig. 14.10 · p. 132

Left mein peele chawal se bhara ek cone hai, nok neeche ki taraf, jis par radius r aur height h mark hai. Right mein same r aur h wala ek khaali tin hai jiski height ke one third aur two thirds par do dashed lines hain. Step 2 mein ek arrow cone ko daalte hue dikhata hai aur tin mein chawal pehli dashed line tak aa jaata hai, jis par tihaai bhara likha hai. Step 3 mein woh doosri line tak aata hai, do tihaai bhara. Step 4 mein teen cones ke baad tin oopar tak bhar jaata hai.

Example 14.3NCERT Example 3 · p. 133

Ek triangle, do cones

Ek right triangle ki sides 20 cm, 21 cm aur 29 cm hain. Use 21 cm wali side ke around 360° ghumaya jaata hai. Bane hue cone ka volume nikaalo. Agar ise 20 cm wali side ke around ghumaya jaaye to kya hoga? $π \approx \frac{22}{7}$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Gehun ka dher

Taaza kata gehun ek conical dher mein pada hai jo 4.2 m chauda aur 2.8 m ooncha hai. Nikaalo (a) iski slant height, (b) ise dhakne ke liye tarpaulin ka area, aur (c) gehun ka volume. $π \approx \frac{22}{7}$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 9 questions

Practice 14.3: answers ke saath

NumericalEasyNCERT Exercise Set 14.3, Q1
Ek conical tent ki slant height 25 m aur base diameter 14 m hai. Floor samet iska total surface area nikaalo. $π \approx \frac{22}{7}$ use karo.
m²
Solution dikhao
$r = 7 m$ , isliye $TSA = \frac{22}{7} \times 7 \times (25 + 7) = 22 \times 32 = 704 m^{2} .$
NumericalEasyNCERT Exercise Set 14.3, Q2
Base radius 3.5 cm aur slant height 12 cm wale cone ka curved surface area nikaalo. $π \approx \frac{22}{7}$ use karo.
cm²
Solution dikhao
$\frac{22}{7} \times 3.5 \times 12 = 11 \times 12 = 132 c m^{2} .$
3
NumericalMediumNCERT Exercise Set 14.3, Q3
Ek cone 15 cm ooncha hai aur uska base radius 8 cm hai. Uske curved aur total surface areas nikaalo. $π \approx 3.14$ use karo.
1. curved surface area nikaalo
  cm²
  Solution dikhao
  $l = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17 c m$ .
  CSA $= 3.14 \times 8 \times 17 = 427.04 c m^{2} .$
2. total surface area nikaalo
  cm²
  Solution dikhao
  TSA $= π r (l + r) = 3.14 \times 8 \times 25 = 628 c m^{2} .$
NumericalMediumNCERT Exercise Set 14.3, Q4
12 cm oonche cone ka volume $314 c m^{3}$ hai. Uske base ka radius nikaalo. $π \approx 3.14$ use karo.
cm
Solution dikhao
$r^{2} = \frac{3 V}{π h} = \frac{942}{3.14 \times 12} = \frac{942}{37.68} = 25$ , isliye $r = 5 c m .$
5
NumericalMediumNCERT Exercise Set 14.3, Q5
Ek cone ki slant height 25 cm hai, aur uski curved surface $550 c m^{2}$ dhakti hai. $π \approx \frac{22}{7}$ use karo.
1. Base ka radius nikaalo.
  cm
  Solution dikhao
  $\frac{22}{7} \times r \times 25 = 550$ se $r = \frac{550 \times 7}{550} = 7 c m .$ milta hai.
2. Total surface area nikaalo.
  cm²
  Solution dikhao
  TSA $= 550 + \frac{22}{7} \times 7^{2} = 550 + 154 = 704 c m^{2} .$
NumericalMediumNCERT Exercise Set 14.3, Q6
Ek party cap base radius 3.5 cm aur height 12 cm ka cone hai. Aisi 20 caps ke liye kitna card chahiye (sirf curved surface)? $π \approx \frac{22}{7}$ use karo.
cm²
Solution dikhao
$l = \sqrt{12^{2} + {3.5}^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 c m$ .
Ek cap: $\frac{22}{7} \times 3.5 \times 12.5 = 11 \times 12.5 = 137.5 c m^{2}$ .
Bees caps: $20 \times 137.5 = 2750 c m^{2} .$
NumericalHardNCERT Exercise Set 14.3, Q7
Ek conical tent 12 m ooncha hai aur uska base radius 5 m hai. Yeh 2.5 m chaude canvas se bana hai. Silaai aur kataai ke liye 0.36 m extra canvas maan kar, kitni lambai ka canvas chahiye? $π \approx 3.14$ use karo.
m
Solution dikhao
$l = \sqrt{12^{2} + 5^{2}} = 13 m$ , isliye CSA $= 3.14 \times 5 \times 13 = 204.1 m^{2}$ .
Canvas ki length $= 204.1 \div 2.5 = 81.64 m$ .
Extra ke saath: $81.64 + 0.36 = 82 m .$
8
NumericalMediumNCERT Exercise Set 14.3, Q8
Sides 9 cm, 12 cm aur 15 cm wale right triangle ko uski 12 cm wali side ke around 360° ghumaya jaata hai. Bana solid kitni jagah gherta hai, aur uski curved surface ka area kitna hai? $π \approx 3.14$ use karo.
1. volume nikaalo
  cm³
  Solution dikhao
  Solid ek cone hai jiska $h = 12$ , $r = 9$ , $l = 15$ (cm).
  $V = \frac{1}{3} \times 3.14 \times 81 \times 12 = 3.14 \times 324 = 1017.36 c m^{3} .$
2. curved surface area nikaalo
  cm²
  Solution dikhao
  CSA $= 3.14 \times 9 \times 15 = 423.9 c m^{2} .$
NumericalHardNCERT Exercise Set 14.3, Q9
Peene ka ek cup cone hai, nok neeche ki taraf. Usmein paani tab tak daala jaata hai jab tak woh cup ki gehraai ke two thirds tak na pahunch jaaye. Cup ka kitna fraction bhara hai?
$\frac{2}{3}$ $\frac{4}{9}$ $\frac{8}{27}$ $\frac{1}{3}$
Solution dikhao
Paani same shape ka chhota kiya hua cone hai: height $\frac{2}{3} h$ aur radius $\frac{2}{3} r$ (similar triangles).
Iska volume $\frac{1}{3} π (\frac{2}{3} r)^{2} (\frac{2}{3} h) = (\frac{2}{3})^{3} \times \frac{1}{3} π r^{2} h$ hai.
Isliye cup $\frac{8}{27}$ bhara hai, one third se bhi kam, jabki paani two thirds oopar tak hai. Cone ka zyaadatar volume uske chaude end ke paas hota hai.

14.4 Pyramid jaisi Shapes

Pyramid banane ke liye base ke roop mein ek flat shape se shuru karo: triangle, square, hexagon ya koi bhi aur band outline. Us base ke plane ke bahar ek point chuno, yaani apex, aur base se apex tak ki har seedhi line bhar do. Jab base ek polygon hota hai, to tirchhi sides triangles hoti hain jo apex par milti hain (Fig. 14.10).

Egypt ke bade pyramids ke bases square hain. Thanjavur ke Brihadeeswara temple mein garbhagriha ke upar ka ooncha tower (vimana) seedhiyon wale pyramid ki tarah uthta hai. Kuch tea bags triangular pyramids ki tarah mode jaate hain. Pahaad bhi yeh shape le sakte hain: jahan glaciers kisi chotti ko kai taraf se ghis dete hain, wahan woh ek nukeela pyramid jaisa horn chhod jaate hain, jaise Uttarakhand mein Shivling.

3/3Square pyramid ko kholo: ek square aur chaar triangles.

Fig. 14.10Koi bhi base, ek apex: volume base area aur height use karta hai; surface faces ka jod hai.NCERT Fig. 14.11 · p. 134

Step 1 mein perspective mein bana ek square base wala pyramid hai, jiska chhupa edge dashed hai aur jiski height h apex se seedha neeche base ke centre tak dashed line ki tarah bani hai. Step 2 mein right mein triangular base wala pyramid aata hai, uski height h bhi dashed hai. Step 3 mein triangular pyramid ki jagah square pyramid ka net aata hai: ek square jis par base likha hai aur uski chaaron sides par ek-ek triangle, jis par base plus 4 triangles likha hai.

Definition

Pyramid

Ek solid jo flat base (polygon ya koi aur band shape) ke har point ko base ke plane ke bahar ek apex se jod kar banta hai.

Har pyramid ka volume cone wale rule se hi milta hai, chahe uske base ki shape kuch bhi ho:

Pyramid ka volume

V = \frac{1}{3} \times A \times h

(14.13)

jahan

$A$: base ka area
$h$: perpendicular height

Yahan $h$ perpendicular height hai, apex se seedha neeche base ke plane tak, kisi dhalaan wale edge ki length nahi. (14.13) ko (14.12) se compare karo: cone bas ek pyramid hai jiska base circle hai. Surface area ke liye kuch naya nahi chahiye: faces ki list banao (base aur triangles), har ek ka area nikaalo, aur jodo.

Example

Pyramid wala tent

Ek tent square pyramid hai. Uska floor 10 m side ka square hai aur uska apex floor ke centre se 12 m upar hai. Nikaalo (a) andar ki hawa ka volume aur (b) chaar dhalaan wali sides ke liye canvas.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek triangular pyramid ka base area

30 c m^{2}

aur height 10 cm hai. Uska volume kya hai?

cm³

Answer dekho

Answer 100 cm³

V = \frac{1}{3} \times 30 \times 10 = 100 c m^{3} .

14.5 Spheres aur Hemispheres

Achhi tarah hawa bhari football ki skin ka har point uske centre se barabar doori par hai. Yahi ek property sphere (गोला) ko define karti hai: space ke un sab points ka set jo ek fixed point $O$ (centre) se fixed doori $r$ (radius) par hain. Flat plane mein yahi rule circle deta hai. Circle ko uske diameter ke around ghuma kar bhi sphere mil sakta hai, bilkul jaise right triangle ghumaane se cone mila tha. Kanche, laddoo aur pake santare, sab lagbhag spheres hain.

Definition

Sphere

Space ke un sab points ka set jo ek fixed point $O$ (centre) se same doori $r$ (radius) par hain.

ExampleFootball ki surface.

Surface area. Archimedes ne sphere ki surface ko dekhne ka ek sundar tareeka dhoondha. Sphere ko sabse tight possible cylinder mein khiskao, jiska radius $r$ aur height $2 r$ ho (Fig. 14.11). Unki khoj: ball ki surface aur cylinder ki curved side ke areas barabar hain. Sphere ka koi bhi patla horizontal band usi height par cylinder ke band se match karta hai. Sphere ka band axis ke zyada paas hai, isliye uska ghera chhota hai, lekin woh dhalaan par hai, isliye zyada chauda hai; dono effects bilkul cancel ho jaate hain. Cylinder ki curved surface $2 π r \times 2 r$ hai, isliye:

Sphere ka surface area

S = 4 π r^{2}

(14.14)

4/4Isliye sphere ka area $2 π r \times 2 r = 4 π r^{2}$ hai.

Fig. 14.11Archimedes: sphere ki surface us cylinder ki curved surface ke barabar hai jo use bas gher leta hai.NCERT Fig. 14.12 · p. 136

Step 1 mein radius r wala sphere hai, uska centre aur ek dashed equator. Step 2 mein uske around ek cylinder bana hai jo use bas chhoota hai, jiski height 2r mark hai; text kehta hai ki cylinder ball mein bilkul fit hota hai. Step 3 mein sphere ka ek patla band aur usi height par cylinder ka band orange shade hain, note ke saath ki dono bands ka area same hai. Step 4 mein shading hat jaati hai aur right mein result aata hai: 2πr × 2r = 4πr².

Volume. Archimedes ne sphere ka volume bhi nikaala:

Sphere ka volume

V = \frac{4}{3} π r^{3}

(14.15)

Aage ki classes mein tum ise calculus se theek se derive karoge. Lekin ek baat par dhyaan do: $\frac{4}{3} π r^{3} = \frac{1}{3} \times 4 π r^{2} \times r$ . Yeh (surface area) × (radius) ka one third hai, jo cone ke formula $\frac{1}{3} \times$ (base area) × (height) jaisa hi dikhta hai.

DerivationPatle cones se sphere ka volume

Ball ki skin ko hazaaron chhote patches mein baanto, jitne baareek chaho, taaki har ek lagbhag flat ho. Har patch ko seedhi lines se centre $O$ se jodo (Fig. 14.12). Har patch aur uski lines ek lamba, patla cone (asal mein, patla pyramid) banate hain jiska apex $O$ hai aur height radius $r$ hai.

Yeh cones milkar poori ball banate hain. Agar patch areas $a_{1}, a_{2}, \dots, a_{n}$ hain, to $V = \frac{1}{3} a_{1} r + \frac{1}{3} a_{2} r + \dots + \frac{1}{3} a_{n} r,$ jo $V = \frac{r}{3} (a_{1} + a_{2} + \dots + a_{n}) .$ hai.

Patches milkar poori surface $4 π r^{2}$ banate hain, isliye:

\begin{aligned} V & = \frac{r}{3} \times 4 π r^{2} \\ = \frac{4}{3} π r^{3} \end{aligned}

3/3Unka volume $\frac{1}{3} \times$ (total base area) $\times r$ hai.

Fig. 14.12Har patle cone ki height

r

hai, aur unke bases jud kar poori surface bante hain.NCERT Volume Formula ko Visualise karna · p. 138

Step 1 mein centre O wala sphere hai jiski surface par ek chhota patch orange shade hai, jo O se jud kar height r ka patla cone banata hai; text kehta hai chhota patch + thin cone to O. Step 2 mein poori ball baari-baari rang ke bahut saare patle wedges mein bant jaati hai, sab O par milte hain, text ke saath kai thin cones poori ball bharte. Step 3 mein right mein formula V = ⅓ × 4πr² × r = 4⁄3 πr³ aata hai.

Example

Football mein leather aur hawa

Ek football ka radius 11 cm hai. Leather ki motai ignore karke nikaalo (a) uski surface par leather ka area aur (b) andar ki hawa ka volume. $π \approx 3.14$ use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

14.5.1 Sphere ke surface area ka formula verify karne ki hands-on activity

ActivityNCERT 14.5.1 Activity · p. 138

Santara chheelo

Tumhe chahiye

ek gol santara (ya rubber ball aur mota dhaaga)
do rulers
paper aur compass

Pehle guess karo

Santare ka chhilka uske radius wale kitne circles bharega?

lagbhag 2lagbhag 3lagbhag 4lagbhag 6

Jitna ho sake utna gol santara chuno. Dono taraf seedhe khade do rulers se uska diameter naapo, aur use aadha karke $r$ nikaalo.
Paper ki sheet par radius $r$ ke kai circles banao.
Santare ko dhyaan se chhote tukdon mein chheelo. Tukdon ko circles ke andar flat bichhao, agla circle shuru karne se pehle ek circle poora bharo, aur jitna ho sake utne kam gaps chhodo.
Gino ki chhilka kitne circles bharta hai.

Table 14.2

Radius r (cm)	Chhilke se bhare circles

Tumhari entries apne aap save hoti hain.

Ek aur tareeka. Rubber ball ke top mein ek pin lagao. Pin se shuru karke ball ke chaaron taraf mota dhaaga lapet te jao jab tak woh dhak na jaaye, aur use pins se tikaaye rakho. Dhaaga kholo, aur use ball ke radius wale circles ke andar gol-gol rakho.

3/3Chhilka lagbhag chaar circles bharta hai: area $\approx 4 π r^{2}$ .

Fig. 14.13Santare ka chhilka santare ke radius wale lagbhag chaar circles bharta hai.NCERT Fig. 14.14 · p. 139

Step 1 mein ek santara hai jiska radius r mark hai. Step 2 mein same radius r ke chaar khaali circles two by two grid mein aate hain, jin par 4 circles, radius r likha hai. Step 3 mein chheela hua santara dashed outline ki tarah dikhta hai, aur chaaron circles santare ke chhilke ke flat tukdon se bhare hain.

14.5.2 Hemisphere

Ball ko bilkul uske centre se kaato to do ek jaise aadhe hisse milte hain, har ek ek hemisphere (अर्धगोला) hai. Solid aadhi ball ki do surfaces hoti hain (Fig. 14.14). Ek ball ki skin ka curved aadha hissa hai. Doosri woh flat circle hai jahan se chaaku guzra, jise iska base kehte hain, same radius $r$ ke saath. Katori, aadha nariyal ka khol, observatory ka dome aur Vijayapura mein Gol Gumbaz ka bada dome, sab lagbhag hemispheres hain. Geography bhi yeh shabd use karti hai: equator Earth ko Northern aur Southern Hemispheres mein baantta hai.

Definition

Hemisphere

Sphere ka aadha hissa, centre se guzarte plane se kata hua. Iski ek curved surface aur ek flat circular base hota hai.

3/3Curved hissa $2 π r^{2}$ , flat base $π r^{2}$ .

Fig. 14.14Hemisphere mein sphere ki aadhi curved surface plus ek flat circle hota hai.NCERT Fig. 14.15 · p. 139

Step 1 mein ek sphere hai jiske beech mein ek dashed circle hai jahan se use kaata jaayega. Step 2 mein upar wala aadha hissa hat jaata hai, aur neeche wale aadhe ka flat circular face dikhta hai, jo blue rang ka hai. Step 3 mein leader lines aur labels hisson ke naam batate hain: hemisphere ki curved surface 2πr² hai, flat base πr² hai, aur total surface area 3πr² hai.

Curved hissa sphere ki surface ka aadha hai, aur solid hemisphere ball ka aadha hai:

Hemisphere ki curved surface

CSA = 2 π r^{2}

(14.16)

Solid hemisphere

\begin{aligned} TSA & = 2 π r^{2} + π r^{2} \\ = 3 π r^{2} \end{aligned}

(14.17)

Hemisphere ka volume

V = \frac{2}{3} π r^{3}

(14.18)

Khokhle bowl ke saath dhyaan rakho: uska andar ya bahar bas curved surface $2 π r^{2}$ hai, kyunki bowl ka koi flat dhakkan nahi hota. $3 π r^{2}$ sirf solid hemisphere ke liye use karo.

Example 14.4NCERT Example 4 · p. 140

Dal ki katori

Ek steel ki katori inner radius 7 cm wala hemisphere hai. Oopar tak bharne par usmein kitni dal aa sakti hai? $π \approx \frac{22}{7}$ use karo aur answer millilitres mein do ( $1 c m^{3} = 1$ mL).

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek ball ka radius 7 cm hai. Uski surface kitna area gherti hai?

π \approx \frac{22}{7}

use karo.

cm²

Answer dekho

Answer 616 cm²

4 π r^{2} = 4 \times \frac{22}{7} \times 49 = 616 c m^{2} .

Practice · 8 questions

Practice 14.4: answers ke saath

NumericalEasyNCERT Exercise Set 14.4, Q1
Steel ke ek kanche ka radius 1.4 cm hai. Uska surface area nikaalo. $π \approx \frac{22}{7}$ use karo.
cm²
Solution dikhao
$4 \times \frac{22}{7} \times {1.4}^{2} = 4 \times \frac{22}{7} \times 1.96 = 4 \times 6.16 = 24.64 c m^{2} .$
NumericalHardNCERT Exercise Set 14.4, Q2
Do solid spheres ek hi brass se bane hain. Ek ka weight 200 g aur doosre ka 5400 g hai. Chhote wale ka diameter 4 cm hai. Bade wale ka radius nikaalo.
cm
Solution dikhao
Mass ratio $5400 \div 200 = 27$ hai, isliye volume ratio 27 hai.
Kyunki $V \propto r^{3}$ , radius ratio $\sqrt[3]{27} = 3$ hai.
Chhota radius 2 cm hai, isliye bada radius $3 \times 2 = 6 c m .$ hai.
NumericalMediumNCERT Exercise Set 14.4, Q3
Mars ka diameter Earth ke diameter ka lagbhag aadha hai. Dono ko spheres maan kar, Mars ke surface area aur Earth ke surface area ka ratio nikaalo.
Solution dikhao
Answer1 : 4
Agar Mars ka radius $r$ hai, to Earth ka radius $2 r$ hai.
Ratio $= 4 π r^{2} : 4 π (2 r)^{2} = 1 : 4$ . (Unke volumes ka ratio $1 : 8$ hai.)
4
NumericalEasyNCERT Exercise Set 14.4, Q4
Ek solid hemisphere ka radius 14 cm hai. $π \approx \frac{22}{7}$ use karo.
1. Iska curved surface area nikaalo.
  cm²
  Solution dikhao
  $2 \times \frac{22}{7} \times 196 = 1232 c m^{2} .$
2. Iska total surface area nikaalo.
  cm²
  Solution dikhao
  $3 \times \frac{22}{7} \times 196 = 1848 c m^{2} .$
NumericalMediumNCERT Exercise Set 14.4, Q5
3 cm radius wali chhah solid balls ek box mein ek layer mein pack ki gayi hain, jiska andar ka naap 18 cm × 12 cm × 6 cm hai. Phir box ko tel se bhar diya jaata hai. Tel ka volume nearest $c m^{3}$ tak nikaalo. $π \approx 3.14$ use karo.
cm³
Solution dikhao
Box: $18 \times 12 \times 6 = 1296 c m^{3}$ . (Balls $3 \times 2 \times 1$ ki tarah fit hoti hain, har ek 6 cm ke cube mein.)
Balls: $6 \times \frac{4}{3} \times 3.14 \times 27 = 678.24 c m^{3}$ .
Tel $= 1296 - 678.24 = 617.76 \approx 618 c m^{3} .$
6
NumericalMediumNCERT Exercise Set 14.4, Q6
Ek sphere ka radius 30 % badhta hai.
1. Uska surface area kitne percent badhta hai?
  %
  Solution dikhao
  ${1.3}^{2} = 1.69$ , yaani 69 % ki badhat.
2. Uska volume kitne percent badhta hai?
  %
  Solution dikhao
  ${1.3}^{3} = 2.197$ , yaani 119.7 % ki badhat: volume double se bhi zyada ho jaata hai.
NumericalHardNCERT Exercise Set 14.4, Q7
Ek spherical balloon se dheere-dheere hawa nikalti hai aur uska volume 48.8 % gir jaata hai. Maan lo woh sphere hi rehta hai, to uska radius kitne percent gira?
%
Solution dikhao
$(new r \div old r)^{3} = 1 - 0.488 = 0.512$ .
$\sqrt[3]{0.512} = 0.8$ , isliye radius pehle ka 80 % hai: woh 20 % gira hai.
NumericalHardNCERT Exercise Set 14.4, Q8
Ek hemispherical dome ke bahar paint karna hai. Uske base ki circumference 44 m hai, aur paint ka kharcha ₹5 per $100 c m^{2}$ hai. Kharcha nikaalo. $π \approx \frac{22}{7}$ use karo.
₹
Solution dikhao
$2 \times \frac{22}{7} \times r = 44$ , isliye $r = 7 m$ .
Curved surface $= 2 \times \frac{22}{7} \times 49 = 308 m^{2}$ .
$1 m^{2} = 10 000 c m^{2}$ , isliye rate ₹5 × 100 = ₹500 per $m^{2}$ hai.
Kharcha $= 308 \times 500 =$ ₹1,54,000.

14.6 Hamare Aas-paas Areas aur Volumes

Solids ki paanch families ko dhyaan mein rakh kar apne ghar aur school mein chaaron taraf dekho. Chhat par paani ki tanki cuboid ya cylinder ho sakti hai. LPG cylinder gol ends wala cylinder hai. Mitti ka matka lagbhag sphere hai, funnel cone hai, aur steel ki almirah cuboid hai. Asli objects shayad hi kabhi perfect hote hain, lekin achha model tumhe jaldi se kaam ka answer de deta hai.

ActivityNCERT Activity · p. 141

Ismein kitna aata hai?

Tumhe chahiye

ek measuring tape
ek ruler
ek notebook

Ghar par chaar containers chuno, jaise pressure cooker, balti, matka aur fridge ka andar ka hissa (ya kisi bade ki madad se overhead tank).
Tay karo ki har ek kis solid ke sabse kareeb hai, aur jo chahiye (radius, height, length waghaira) woh centimetres mein naapo.
Har volume $c m^{3}$ mein nikaalo, phir $1000 c m^{3} = 1$ litre use karke use litres mein badlo.
Agar label ya bill par capacity likhi hai, to use apne estimate se compare karo.

Table 14.3

Object	Model solid	Measurements (cm)	Volume (litres)
Pressure cooker
Balti
Matka
Fridge (andar)

Tumhari entries apne aap save hoti hain.

14.6.1 Guesstimates: kya tumne kabhi socha?

Tumhara school ek din mein kitne litres paani use karta hai? Ek kilo chawal mein kitne daane hote hain? Aise sawaalon ka koi exact data nahi hota, lekin tum phir bhi soch-samajh kar ek theek number tak pahunch sakte ho. Yeh ek guesstimate hai: simple models, reasonable assumptions aur round numbers se bana estimate. Do dhyaan se sochne wale log alag numbers par pahunch sakte hain, aur dono theek ho sakte hain. Guesstimate ko is baat se parkha jaata hai ki uske steps kitni saafi se check ho sakte hain, answer key se match karne se nahi.

Definition

Guesstimate

Kisi aisi cheez ki theek-thaak value jise hum exactly naap nahi sakte, jo simple model, batayi gayi assumptions aur round numbers se soch kar nikaali gayi ho.

Ek bharosemand routine ke paanch steps hain:

Koi bhi working se pehle guess karo, aur guess likh lo.
Model: asli object ki jagah aisi shapes rakho jinhe tum naap sako.
Assume: har woh size ya rate list karo jo tumhe khud maanna pada.
Round numbers ke saath calculate karo, units ko sahi rakhte hue.
Look back: apne guess se compare karo, aur poochho ki kaun si assumption answer ko sabse zyada badlegi.

Example 14.5NCERT Example 5 · p. 141

Kadhai se laddoo

Ek mithai ki dukaan mein laddoo mixture ki ek cylindrical kadhai hai jiska radius 20 cm hai, aur woh 30 cm gehraai tak bhari hai. Isse lagbhag kitne laddoo ban sakte hain? Pehle guess karo: dasiyon, saikdon ya hazaaron?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14.6NCERT Example 6 · p. 142

Storeroom mein footballs

School ka ek storeroom 4 m lamba, 3 m chauda aur 2.5 m ooncha hai. Usmein radius 11 cm wali lagbhag kitni footballs pack ho sakti hain? Pehle guess karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/2Rows ko gaddhon mein bithaane se usi box mein 48 balls aati hain.

Fig. 14.15Side se dekho: har row ko neeche ke gaddhon mein bithaane se usi box mein zyada balls aati hain.NCERT Fig. 14.17 · p. 143

Side se dekha gaya ek rectangular box. Step 1 mein usmein seedhi rows aur columns mein circles hain, 10 aade aur 4 oonche, jis par square rows: 40 balls likha hai, aur balls ke beech saaf gaps hain. Step 2 mein har doosri row aadhi ball khisak jaati hai taaki har ball neeche ki do balls ke beech ke gaddhe mein baithe; rows paas-paas aa jaati hain aur 5 rows fit hoti hain, jis par nestled rows: 48 balls likha hai.

Gaps se yeh bhi samajh aata hai ki hum volume cubes se kyun naapte hain, balls se nahi. Unit cubes bina koi jagah chhode aapas mein fit ho jaate hain, isliye unhe ginne se volume exactly milta hai. Spheres hamesha gaps chhodte hain, tum unhe kitni bhi chaturaai se jamao: zyada se zyada woh lagbhag 74 % jagah bharte hain.

Quick check

3 mein se 1

1 m^{3}

mein kitne litres hote hain?

litres

Answer dekho

Answer 1000 litres

1 m^{3} = 100 \times 100 \times 100 = 10^{6} c m^{3}

, aur

1000 c m^{3} = 1

litre, isliye

1 m^{3} = 1000

litres.

Chapter ke end ki exercises

Neeche har guesstimate ke liye calculate karne se pehle ek jaldi wala guess likho, aur apni assumptions list karo.

Table 14.4Is chapter ke saare formulasNCERT Chapter Summary · p. 147

Solid	Surface area	Volume
Cuboid	$2 (l w + l h + w h)$	$l w h$
Cube	$6 a^{2}$	$a^{3}$
Cylinder	CSA $2 π r h$ ; TSA $2 π r (h + r)$ ; ek end khula $π r (2 h + r)$	$π r^{2} h$
Cone	CSA $π r l$ ; TSA $π r (l + r)$	$\frac{1}{3} π r^{2} h$
Pyramid	faces ke areas ka jod	$\frac{1}{3} \times$ base area $\times h$
Sphere	$4 π r^{2}$	$\frac{4}{3} π r^{3}$
Hemisphere	CSA $2 π r^{2}$ ; TSA $3 π r^{2}$	$\frac{2}{3} π r^{3}$

Recap

Summary aur key points

Surface area solid ke bahar ko naapta hai (square units); volume woh jagah naapta hai jo solid gherta hai (cubic units); dekho Wrap karna aur bharna.
Cuboid: TSA $= 2 (l w + l h + w h)$ , $V = l w h$ ; cube: $6 a^{2}$ aur $a^{3}$ ; dekho Section 14.1.
Barabar volumes ke surface areas alag ho sakte hain; cube ko sabse kam chahiye; dekho Example 14.1.
Right circular cylinder ka axis uske circular ends ke perpendicular hota hai; aar-paar ki slices circles hain; dekho Section 14.2.
Cylinder: CSA $= 2 π r h$ , TSA $= 2 π r (h + r)$ , ek end khula $π r (2 h + r)$ , $V = π r^{2} h$ ; dekho Section 14.2.1.
Cone: $l^{2} = h^{2} + r^{2}$ , CSA $= π r l$ , TSA $= π r (l + r)$ , $V = \frac{1}{3} π r^{2} h$ ; dekho Section 14.3.
Koi bhi pyramid: $V = \frac{1}{3} \times$ base area $\times$ height; surface area = faces ka jod; dekho Section 14.4.
Sphere: $S = 4 π r^{2}$ (Archimedes ka cylinder) aur $V = \frac{4}{3} π r^{3} = \frac{1}{3} \times S \times r$ ; dekho Section 14.5.
Santare ka chhilka uske radius ke lagbhag chaar circles bharta hai; dekho Section 14.5.1.
Hemisphere: CSA $2 π r^{2}$ , solid TSA $3 π r^{2}$ , $V = \frac{2}{3} π r^{3}$ ; dekho Section 14.5.2.
Har length ko $k$ se scale karne par areas $k^{2}$ se aur volumes $k^{3}$ se multiply hote hain; dekho Practice 14.4.
Guesstimates: guess, model, assume, calculate, look back; volume ÷ object ka volume sirf upper bound deta hai; dekho Section 14.6.1.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise Set 14.1, Q1	Practice 14.1 · Q1
NCERT Exercise Set 14.1, Q2	Practice 14.1 · Q2
NCERT Exercise Set 14.1, Q3	Practice 14.1 · Q3
NCERT Exercise Set 14.1, Q4	Practice 14.1 · Q4
NCERT Exercise Set 14.1, Q5	Practice 14.1 · Q5
NCERT Exercise Set 14.1, Q6	Practice 14.1 · Q6
NCERT Exercise Set 14.1, Q7	Practice 14.1 · Q7
NCERT Exercise Set 14.1, Q8	Practice 14.1 · Q8
NCERT Exercise Set 14.2, Q1	Practice 14.2 · Q1
NCERT Exercise Set 14.2, Q2	Practice 14.2 · Q2
NCERT Exercise Set 14.2, Q3	Practice 14.2 · Q3
NCERT Exercise Set 14.2, Q4	Practice 14.2 · Q4
NCERT Exercise Set 14.2, Q5	Practice 14.2 · Q5
NCERT Exercise Set 14.3, Q1	Practice 14.3 · Q1
NCERT Exercise Set 14.3, Q2	Practice 14.3 · Q2
NCERT Exercise Set 14.3, Q3	Practice 14.3 · Q3
NCERT Exercise Set 14.3, Q4	Practice 14.3 · Q4
NCERT Exercise Set 14.3, Q5	Practice 14.3 · Q5
NCERT Exercise Set 14.3, Q6	Practice 14.3 · Q6
NCERT Exercise Set 14.3, Q7	Practice 14.3 · Q7
NCERT Exercise Set 14.3, Q8	Practice 14.3 · Q8
NCERT Exercise Set 14.3, Q9	Practice 14.3 · Q9
NCERT Exercise Set 14.4, Q1	Practice 14.4 · Q1
NCERT Exercise Set 14.4, Q2	Practice 14.4 · Q2
NCERT Exercise Set 14.4, Q3	Practice 14.4 · Q3
NCERT Exercise Set 14.4, Q4	Practice 14.4 · Q4
NCERT Exercise Set 14.4, Q5	Practice 14.4 · Q5
NCERT Exercise Set 14.4, Q6	Practice 14.4 · Q6
NCERT Exercise Set 14.4, Q7	Practice 14.4 · Q7
NCERT Exercise Set 14.4, Q8	Practice 14.4 · Q8
NCERT End-of-chapter Exercises Q1	Chapter ke end ki exercises · Q1
NCERT End-of-chapter Exercises Q2	Chapter ke end ki exercises · Q2
NCERT End-of-chapter Exercises Q3	Chapter ke end ki exercises · Q3
NCERT End-of-chapter Exercises Q4	Chapter ke end ki exercises · Q4
NCERT End-of-chapter Exercises Q5	Chapter ke end ki exercises · Q5
NCERT End-of-chapter Exercises Q6	Chapter ke end ki exercises · Q6
NCERT End-of-chapter Exercises Q7	Chapter ke end ki exercises · Q7
NCERT End-of-chapter Exercises Q8	Chapter ke end ki exercises · Q8
NCERT End-of-chapter Exercises Q9	Chapter ke end ki exercises · Q9
NCERT End-of-chapter Exercises Q10	Chapter ke end ki exercises · Q10
NCERT End-of-chapter Exercises Q11	Chapter ke end ki exercises · Q11
NCERT End-of-chapter Exercises Q12	Chapter ke end ki exercises · Q12
NCERT End-of-chapter Exercises Q13	Chapter ke end ki exercises · Q13
NCERT End-of-chapter Exercises Q14	Chapter ke end ki exercises · Q14
NCERT End-of-chapter Exercises Q15	Chapter ke end ki exercises · Q15
NCERT End-of-chapter Exercises Q16	Chapter ke end ki exercises · Q16
NCERT End-of-chapter Exercises Q17	Chapter ke end ki exercises · Q17
NCERT End-of-chapter Exercises Q18	Chapter ke end ki exercises · Q18
NCERT End-of-chapter Exercises Q19	Chapter ke end ki exercises · Q19
NCERT End-of-chapter Exercises Q20	Chapter ke end ki exercises · Q20
NCERT End-of-chapter Exercises Q21	Chapter ke end ki exercises · Q21
NCERT End-of-chapter Exercises Q22	Chapter ke end ki exercises · Q22

Quick check: MCQs

14.1 Cuboids aur Cubes3 questions
14.2.1 Upar wale formulas sahi kyun hain, ise samajhne ka ek aur tareeka4 questions
14.3 Cones4 questions
14.4 Pyramid jaisi Shapes3 questions
14.5.2 Hemisphere4 questions
14.6.1 Guesstimates: kya tumne kabhi socha?3 questions

Glossary · 14

Key terms (English + Hindi)

coneशंकुshanku: Ek solid jiska circular base ek apex se juda hai; right circular cone ka apex centre ke theek upar hota hai.Chapter mein pehli baar yahan aaya →
cross-sectionअनुप्रस्थ काटanuprasth kaat: Solid ke aar-paar ek patli, seedhi slice se bani flat shape.Chapter mein pehli baar yahan aaya →
cubeघनghan: Aisa cuboid jiski length, width aur height barabar hain.Chapter mein pehli baar yahan aaya →
cuboidघनाभghanaabh: Box ke shape ka solid jiske chhah rectangular faces right angles par milte hain.Chapter mein pehli baar yahan aaya →
curved surface areaवक्र पृष्ठीय क्षेत्रफलvakra prishthiya kshetrafal: Cylinder, cone ya sphere ke sirf curved hisse ka area.Chapter mein pehli baar yahan aaya →

guesstimateअनुमानanumaan: Exact data na hone par simple model aur batayi gayi assumptions se soch-samajh kar lagaya gaya estimate.Chapter mein pehli baar yahan aaya →

hemisphereअर्धगोलाardhgola: Sphere ka aadha hissa, centre se guzarte plane se kata hua.Chapter mein pehli baar yahan aaya →

pyramidपिरामिडpiraamid: Flat base wala solid jiska har point base ke bahar ek apex se juda hai.Chapter mein pehli baar yahan aaya →

right circular cylinderलंब वृत्तीय बेलनlamb vrittiya belan: Ek solid jiske do barabar parallel circular ends hain aur jiska axis unke perpendicular hai.Chapter mein pehli baar yahan aaya →

slant heightतिर्यक ऊँचाईtiryak oonchaai: Cone ke apex se uske rim tak surface ke saath doori: $l = \sqrt{h^{2} + r^{2}}$ .Chapter mein pehli baar yahan aaya →
sphereगोलाgola: Space ke woh sab points jo ek centre se fixed doori $r$ par hain.Chapter mein pehli baar yahan aaya →
surface areaपृष्ठीय क्षेत्रफलprishthiya kshetrafal: Solid ke bahar ka total area, square units mein.Chapter mein pehli baar yahan aaya →

total surface areaसंपूर्ण पृष्ठीय क्षेत्रफलsampoorn prishthiya kshetrafal: Solid ke har face ka area, flat aur curved, sab jod kar.Chapter mein pehli baar yahan aaya →

volumeआयतनaayatan: Kitni jagah koi solid gherta hai ya koi container rakhta hai, cubic units mein.Chapter mein pehli baar yahan aaya →

Math of Space: Surface Area and Volume: Class 9 Maths Chapter 14 Hinglish mein

Is chapter mein aap kya seekhenge

Wrap karna aur bharna

14.1 Cuboids aur Cubes

Practice 14.1: answers ke saath

14.2 Right Circular Cylinder

14.2.1 Upar wale formulas sahi kyun hain, ise samajhne ka ek aur tareeka

Practice 14.2: answers ke saath

14.3 Cones

Practice 14.3: answers ke saath

14.4 Pyramid jaisi Shapes

14.5 Spheres aur Hemispheres

14.5.1 Sphere ke surface area ka formula verify karne ki hands-on activity

14.5.2 Hemisphere

Practice 14.4: answers ke saath

14.6 Hamare Aas-paas Areas aur Volumes

14.6.1 Guesstimates: kya tumne kabhi socha?

Chapter ke end ki exercises

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye