Measuring Space: Perimeter and Area (Jagah ko Naapna: Perimeter aur Area): Class 9 Maths Chapter 6 Hinglish mein

Running tracks, π ki khoj, aur kaat kar dobara jodne ki kala

Ganita Manjari Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

130 min
1 simulation
60 practice questions
43 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki kisi shape ki scaled copies ke liye perimeter ÷ size fixed kyun rehta hai, aur C = πd = 2πr ka use karo
Batao ki itihaas mein π ko polygons ke beech kaise trap kiya gaya aur kaise approximate kiya gaya, aur samjhao ki π irrational kyun hai aur 22/7 ke barabar kyun nahi hai
Arc lengths nikalo, aur arcs aur seedhe tukdon se bani shapes ke perimeters nikalo, running track par lanes ke stagger bhi
Kaat kar aur dobara jod kar parallelograms aur triangles ke areas derive karo, aur prove karo ki median triangle ke area ko aadha-aadha baant deta hai
Heron's formula aur Brahmagupta's formula ka use karo, aur special cases se unhe check karo
Kisi rectangle jitne hi area wala square construct karo aur samjhao ki yeh construction kyun kaam karta hai
Archimedes ke polygons aur Nīlakaṇṭha ki slices se A = πr² samjhao, aur sectors aur segments ke areas nikalo

NCERT Mathematics Class 9, Chapter 6, pp. 118–154 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Runners alag-alag jagah se start kyun karte hain?

Sports day ki ek 400 m race socho jismein har runner poore lap mein apni hi lane mein rehti hai. Sabse bahar wali lane, lane 1 se kahin zyada chaude curve par ghoomti hai, isliye agar sab ek hi line se start karein, to bahar wali runner ko kai metre zyada daudna padega. Isi wajah se starting marks pehle bend ke around phaile hote hain, jabki saare athletes ek hi common line par finish karte hain.

Paas-paas wali lanes ke starting marks ke beech ke farak ko stagger kehte hain. Yeh kitna lamba hona chahiye taaki na koi aage nikle, na koi peeche rahe? Kya yeh track ke size par depend karta hai, ya is par ki hum kaun si lanes compare kar rahe hain? Fig. 6.1 track ko tukdon mein khol kar iska jawab dhoondhta hai.

3/4Bends par bahar wali lane ek bade curve par ghoomti hai.

Fig. 6.1Straights sabke liye barabar hain; farak sirf bends mein hai, aur ek lane ke dono bends milkar ek poora circle banate hain.NCERT Fig. 6.1 · p. 118

Upar se dekha gaya stadium jaisi shape ka running track, jismein ek green infield ke chaaron taraf teen lanes hain. Step 1 mein neeche wale straight ke end par finish line hai, lane 1 ki runner us par khadi hai, aur lane 2 aur lane 3 ki runners right wale bend par aur aage khadi hain, jinke paas staggered starts likha hai. Step 2 mein lanes 1 aur 2 ke dono straight hisse blue aur orange mein highlight hain, aur likha hai ki straights har lane ke liye same hain. Step 3 mein lane 1 ke dono bends blue mein aur lane 2 ke orange mein highlight hain; orange bends saaf taur par lambe dikhte hain. Step 4 mein left wale bends khiska kar right wale bends se jod diye jaate hain, to har lane ke bends ek poora circle bana lete hain; blue circle ka radius r hai, orange ka radius r + w, aur labels par 2πr, 2π(r + w) aur extra = 2πw likha hai.

Bends circles ke tukde hain, isliye pehle hamein circle ke chaaron taraf ki distance naapne ka tareeka chahiye. Phir chapter lengths se areas ki taraf badhta hai: rectangles, parallelograms, triangles, chaar sides wale figures, aur aakhir mein circles aur unki slices.

6.1 Shape ka Perimeter

Kisi flat shape ka perimeter (परिमाप) uski boundary ki length hota hai. Ek dhaaga kisi patte ya sikke ke edge ke bilkul saath-saath rakho, phir dhaage ko seedha kheench kar naapo: wahi length perimeter hai. Seedhe edges wali shape ke liye dhaage ki zaroorat nahi, bas side lengths ko jod do.

Definition

Perimeter

Kisi closed shape ki boundary ki total length, jo ek baar poora chakkar laga kar naapi jaati hai.

Example3 cm × 5 cm wale card ka perimeter $3 + 5 + 3 + 5 = 16 c m$ hai.

Kai shapes ke liye ek chhota formula hamare liye yeh jodne ka kaam kar deta hai. Length $a$ aur breadth $b$ wale rectangle mein har length ki do sides hoti hain, isliye uska perimeter $2 (a + b)$ hai. Regular polygon ki $n$ barabar sides hoti hain; agar har side $a$ lambi ho, to uska perimeter $n a$ hai. $n = 3$ par equilateral triangle ke liye yeh $3 a$ deta hai, $n = 6$ par regular hexagon ke liye $6 a$ , aur $n = 4$ par square ke liye $4 a$ .

Square dono families mein aata hai. Yeh $n = 4$ wala regular polygon hai, aur yeh woh rectangle bhi hai jiski breadth uski length ke barabar hai. $2 (a + b)$ mein $b = a$ rakho to tumhe phir se $2 (2 a) = 4 a$ milta hai, jaisa milna chahiye. Kisi zyada general formula mein $b = a$ jaisi condition lagakar naya formula paana special case lena kehlata hai; yeh idea tum Cyclic 4-gon ke liye Brahmagupta's formula mein phir dekhoge.

Ab running track ke curved end ko dekho, ya kisi choodi ko. Circle ki koi sides nahi hoti jinhe joda jaaye, to hum uska perimeter kaise nikalein? Pehla clue yeh hai. Alag-alag size ke regular hexagons banao, jaise Fig. 6.2 mein. Jab side double hoti hai, to har side double hoti hai, isliye perimeter bhi double ho jaata hai. Jab side teen guna hoti hai, to perimeter bhi teen guna ho jaata hai. Hexagon chahe kitna bhi bada ya chhota ho, perimeter ko side se divide karne par hamesha 6 aata hai.

3/4Side teen guna karo: 18 cm. Perimeter ÷ side har baar 6 hai.

Fig. 6.2Shape ko bada ya chhota karo to uska perimeter bhi utne hi factor se badalta hai, isliye perimeter ÷ side fixed rehta hai.NCERT Fig. 6.4 · p. 119

Teen regular hexagons ka centre ek hi hai. Step 1 mein sabse chhota hexagon hai, side 1 cm, aur uske paas perimeter 6 cm likha hai. Step 2 mein side 2 cm aur perimeter 12 cm wala hexagon judta hai. Step 3 mein side 3 cm aur perimeter 18 cm wala hexagon judta hai, aur likha hai ki har size ke liye perimeter ÷ side = 6. Step 4 mein hexagons halke ho jaate hain aur har hexagon ke corners se hokar ek dashed circle banta hai, aur sawaal poocha jaata hai ki kya circumference ko diameter se divide karne par bhi har circle ke liye same number aata hai.

Yahi baat squares ke liye bhi sach hai (ratio hamesha $4 : 1$ hai) aur equilateral triangles ke liye bhi ( $3 : 1$ ). Har circle bhi baaki har circle ki bas ek scaled copy hai. Circle ke perimeter ko uska circumference (परिधि) kehte hain. Isliye hum ummeed kar sakte hain ki circumference ko diameter se divide karne par saare circles ke liye ek hi number aayega. Agla section pata lagata hai ki woh number kya hai.

Quick check

3 mein se 1

Ek photo frame 18 cm lamba aur 12.5 cm chauda rectangle hai. Uska perimeter kitna hai, cm mein?

Answer dekho

Answer 61 cm

Perimeter

= 2 (18 + 12.5) = 2 \times 30.5 = 61 c m

6.2 Circle ka Perimeter — C/D Ratio

Ek sikka, ek steel ki plate aur ek cycle ka pahiya lo. Har ek ke circumference $C$ ko uske diameter $D$ se divide karo, to teeno baar lagbhag bilkul same number milta hai. Kai purani sabhyataon ke logon ne yeh notice kiya tha. Hum is number ko C/D ratio kahenge, aur is section ka baaki hissa iski value ko pakka karne ke baare mein hai.

Iske do raaste hain. Tum asli circles ko dhyaan se naap sakte ho. Ya geometry se reasoning karke is number ko aisi do values ke beech daba sakte ho jinhe exactly calculate kiya ja sake. Pehla raasta khud try karo.

Chalo naapeinNCERT Home measurement · p. 120

Kuch gol cheezein naapo

Tumhe chahiye

lagbhag 1 cm chaudi kaagaz ki ek lambi patti, ya darzi ka tape
ek ruler
do moti kitaabein
teen ya chaar gol cheezein: steel ka glass (tumbler), choodi, jar ka dhakkan, cycle ka pahiya

Pehle guess karo

Naapne se pehle: circle ke chaaron taraf lagbhag kitne diameters fit honge?

Lagbhag 2Lagbhag 33 se thoda zyadaLagbhag 4

Patti ko cheez ke chaaron taraf ek baar lapeto aur jahan woh overlap kare wahan nishaan lagao. Use khol kar $C$ naapo.
Dono kitaabon ko cheez ke dono taraf seedha khada karo taaki woh use bas chhoo lein. Kitaabon ke beech ka gap diameter $D$ hai.
$C \div D$ ko do decimal places tak nikalo.
Pahiye ke liye tum tyre par ek nishaan laga sakte ho, pahiye ko farsh par ek poora chakkar ghuma sakte ho aur jitni distance chali use naap sakte ho.

Table 6.1Tumhare measurements

Cheez	C (cm)	D (cm)	C ÷ D
Tumbler
Choodi
Dhakkan
Tumhari pasand

Tumhari entries apne aap save hoti hain.

Measurement kabhi bhi bilkul perfect exact nahi ho sakta. Is number ko pakka karne ke liye mathematicians ne pure reasoning ka sahara liya. Aaj C/D ratio ko $π$ (पाई) likhte hain, yeh Greek letter pi hai, jo sky ke saath rhyme karta hai. Definition se,

Circle ka circumference

C = π D = 2 π r

jahan

$C$: circumference
$D$: diameter
$r$: radius, isliye $D = 2 r$

π ki lambi khoj

Sabse purani reasoning sabse simple bhi hai. Radius 1 wale circle ke andar ek regular hexagon banao, jiske corners circle par hon. Yeh chhah equilateral triangles mein bant jaata hai, isliye har side radius ke barabar, yaani 1 hai, aur perimeter 6 hai. Circle har side ke bahar ubhra hua hai, isliye woh 6 se lamba hona chahiye. Kyunki circumference $2 π$ hai, hamein $2 π > 6$ milta hai, yaani $π > 3$ . Mesopotamia ke mathematicians lagbhag 1900 BCE mein yeh jaante the aur $3 \frac{1}{8} = 3.125$ use karte the.

3/3Dono milakar: $3 < π < 3.47$ .

Fig. 6.3Andar wala hexagon circle se chhota hai aur bahar wala hexagon lamba, isliye

π

un dono ke beech trap ho jaata hai.NCERT Fig. 6.7 · p. 121

Left mein radius 1 wala circle hai, jiske andar ek regular hexagon hai jo chhah equilateral triangles mein bata hai, aur baad mein uske bahar ek regular hexagon aata hai jo har side ke beech mein circle ko chhoota hai. Right mein picture ka ek chhatha hissa bada karke dikhaya gaya hai: ek 60 degree ka wedge jismein 1 length ki andar wali side, π/3 length ka circular arc, aur 1.155 length ki bahar wali side hai. Step 1 andar wala hexagon dikhata hai aur nateeja nikalta hai 6 < 2π, isliye π > 3. Step 2 bahar wala hexagon dikhata hai, jiska perimeter 6 × 1.155 = 6.93 hai, aur nateeja nikalta hai π < 3.47. Step 3 dono ko saath dikhata hai: 3 < π < 3.47.

PropertyHexagons π ko trap karte hain

$3 < π < 2 \sqrt{3} \approx 3.46$ .

Lower bound. Radius 1 wale circle mein, andar wala hexagon side 1 ke chhah equilateral triangles se bana hai, isliye uska perimeter 6 hai. Har side circle ke do points ke beech ka seedha raasta hai aur us arc se chhoti hai jise woh kaatti hai. Sab jodne par $6 < 2 π$ , isliye $π > 3$ .

Upper bound. Bahar wale hexagon ki har side circle ko apne midpoint par chhooti hai, isliye centre se bane chhah triangles equilateral hain aur unki height 1 hai. Agar side $s$ ho, to Baudhāyana–Pythagoras theorem se $s^{2} = 1^{2} + {(\frac{s}{2})}^{2}$ milta hai, isliye $\frac{3}{4} s^{2} = 1$ aur $s = \frac{2}{\sqrt{3}} \approx 1.155$ .

Circle bahar wale hexagon ke andar hai, isliye woh uske perimeter se chhota hai: $2 π < 6 \times \frac{2}{\sqrt{3}} = 4 \sqrt{3}$ . Isliye $π < 2 \sqrt{3} \approx 3.464$ .

Lagbhag 250 BCE mein, Syracuse ke Archimedes ne is trap ko aur tight kiya. Woh sides ki sankhya double karte gaye: 6, 12, 24, 48 aur aakhir mein 96. Har doubling dono polygons ko circle ke aur paas le aati hai. 96 sides aur dhairya se haath se kiye calculation ke saath woh bounds $3 \frac{10}{71} < π < 3 \frac{1}{7}$ tak pahunche, yaani $π$ lagbhag 3.1408 aur 3.1429 ke beech hai. Unki doubling khud Fig. 6.4 mein try karo.

Interactiveπ ko trap karna

Do tabs. Polygon trap mein, ek circle andar ke blue regular polygon aur bahar ke orange regular polygon ke beech hai. Ek slider aur Halve ya Double buttons sides ki sankhya 6, 12, 24, 48, 96, 192, 384 ya 768 set karte hain, aur do cards har polygon ke perimeter ko diameter se divide karke dikhate hain, jo π ko trap karte hain; ek zoom button ek side ko bada karta hai taaki gap dikhta rahe. Mādhava's series mein, buttons 1 − 1/3 + 1/5 − … ke 1, 10 ya 100 terms jodte hain; ek graph running total ka chaar guna dikhata hai jo π par bani dashed line ke upar-neeche zig-zag karta hai, cards estimate aur π se uski doori dikhate hain, aur ek toggle Mādhava ka end correction jodta hai.

Fig. 6.4Sides ki sankhya double karo aur dekho kaise andar aur bahar ke polygons circle ke paas aate jaate hain; doosra tab Mādhava ki series ko ek-ek term karke jodta hai.

Is khoj mein kai aur log bhi jude. Timeline kuch khaas padaav dikhati hai. Dhyaan do ki behtar value ke liye ya to zyada chalaak method chahiye tha, ya bahut zyada sides.

Timeline

π ki kahani ke khaas padaav

lagbhag 1900 BCE
Mesopotamia
Circle apne andar wale hexagon se thoda lamba hai, isliye π 3 se zyada hai; woh 3⅛ use karte the.
lagbhag 250 BCE
Archimedes
96 sides wale polygons se 3 10⁄71 < π < 3 1⁄7 milta hai.
lagbhag 150 CE
Ptolemy
Apni astronomical tables mein 377⁄120 ≈ 3.14167 use karte hain.
263 CE
Liu Hui
China mein polygons se “circle-cutting” method banate hain.
lagbhag 480 CE
Zu Chongzhi
24,576 sides wale polygons ke saath kaam karte hain; 22⁄7 aur kamaal ka 355⁄113 ≈ 3.1415929 dete hain.
499 CE
Āryabhaṭa
62832⁄20000 = 3.1416 dete hain aur ise āsanna, yaani “approximate”, kehte hain.
628 CE
Brahmagupta
√10 ≈ 3.162 use karte hain, ek aisi value jiske saath calculations karna aasaan hai.
lagbhag 1400 CE
Saṅgamagrāma ke Mādhava
π ke liye ek exact infinite series dhoondhte hain aur use 11 decimal places tak compute karte hain.
1706
William Jones
C/D ratio ke liye pehli baar Greek letter π use karte hain.
1761
Lambert
Prove karte hain ki π irrational hai.
Aaj
Computers
Ramanujan aur Chudnovsky brothers ke kaam se nikli series trillions digits deti hain.

Polygons hamesha sirf bounds hi de sakte hain. Lagbhag 1400 CE mein, Kerala ke Saṅgamagrāma ke Mādhava ne ek alag tarah ki cheez dhoondhi: ek exact formula. Odd numbers ke reciprocals ko baari-baari jodo aur ghatao, hamesha ke liye, aur total bilkul $π$ ka ek chauthai hota hai:

Mādhava's series

\frac{π}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \dots

Kitne bhi terms ke baad ruko, tumhe $π$ nahi milega; aage badhte raho to running total bina ruke $\frac{π}{4}$ ke paas aata jaata hai. Is tarah step-by-step pahunchi gayi value ko limit kehte hain, aur limits calculus ka central idea hain, jo tum aage ki classes mein padhoge. Akele yeh series bahut dheere chalti hai, isliye Mādhava ne ek correction bhi nikala jo wahan joda jaata hai jahan tum rukte ho. Iske saath unhone $π$ ko 11 decimal places tak sahi nikala. Baad ke mathematicians jaise Nīlakaṇṭha, Machin aur Ramanujan ne aur bhi tez raaste dhoondhe. Fig. 6.4 ke doosre tab mein dono versions compare karo.

Quick check

4 mein se 1

Ek regular hexagon ke chhah corners radius 5 cm wale circle par hain. Uska perimeter kitna hai, cm mein?

Answer dekho

Answer 30 cm

Centre ko corners se jodne par chhah equilateral triangles bante hain, isliye har side radius ke barabar, yaani 5 cm hai. Perimeter

6 \times 5 = 30 c m

hai, jo circumference

10 π \approx 31.4 c m

se thoda kam hai.

6.3 π Irrational Hai

Kya koi chalaak fraction, shayad bahut bade numerator aur denominator ke saath, bilkul $π$ ke barabar ho sakta hai? Decimals se is par sochne ka ek tareeka milta hai. 3 ko 7 se long division se divide karo: har step par remainder 0, 1, 2, …, 6 mein se koi ek hota hai. Sirf saat possible remainders hone se, zyada se zyada aathve step tak inmein se koi ek wapas aana hi hai, aur wahan se wahi digits phir se aane lagte hain: $\frac{3}{7} = 0.428571 428571 \dots$ Yahi reasoning har fraction par kaam karti hai, isliye uska decimal ya to ruk jaata hai, jaise $\frac{3}{8} = 0.375$ , ya ek repeating block mein ban jaata hai, jaise $\frac{5}{11} = 0.4545 \dots$

Ab $π = 3.14159 26535 89793 23846 \dots$ ko dekho. Computers ne iske trillions digits nikale hain, aur woh kabhi kisi repeating block mein nahi bandhte. Isliye koi bhi fraction $π$ ke barabar nahi ho sakta. Aisa number, jise integers $a$ aur $b$ ke saath $\frac{a}{b}$ ke roop mein nahi likha ja sakta, irrational (अपरिमेय) kehlata hai. Fig. 6.5 dono tarah ke decimals ko saath-saath rakhta hai.

Fig. 6.5Fractions repeating blocks mein bandh jaate hain;

π

ke digits kabhi nahi.

Digits ki teen rows. Pehli row 5/11 = 0.45 45 45 45 vagairah dikhati hai, jismein har repeating block 45 blue box mein hai. Doosri row 3/7 = 0.428571 428571 vagairah dikhati hai, jismein chhah digits ka har block green box mein hai. Teesri row π = 3.14159 26535 89793 23846 vagairah dikhati hai, jismein koi box nahi hai aur note likha hai ki koi block kabhi repeat nahi hota.

Definition

Irrational number

Aisa number jise $\frac{a}{b}$ ke roop mein nahi likha ja sakta, jahan $a$ aur $b$ integers hain aur $b \neq 0$ . Iska decimal na kabhi khatam hota hai, na kabhi repeat hota hai.

Example $\sqrt{2}$ aur $π$ irrational hain; $- \frac{5}{8}$ , $2 \frac{3}{4}$ aur $0.6 = \frac{3}{5}$ rational hain.

Bas digits dekhna proof nahi hai: ho sakta hai koi pattern trillionth digit ke baad shuru ho. Is baat ka asli proof ki koi fraction $π$ ke barabar nahi, Johann Lambert ne 1761 mein diya, aise tools se jo is kitaab se bahut aage ke hain. Usse bahut pehle, Āryabhaṭa ne apni value 3.1416 ko āsanna, yaani “lagbhag”, kaha tha, jaise unhe pehle se andaaza ho ki koi bhi fraction exact nahi hoga; lagta hai Zu Chongzhi ko bhi aisa hi laga tha. Ek aur irrational number, $\sqrt{2}$ , tum pehle hi dekh chuke ho.

Ab table ka aakhri column dekho. $\frac{22}{7}$ , $π$ se lagbhag ek hazaarve hisse se chookta hai, $\frac{333}{106}$ das-hazaarve hisse se bhi kam se, aur $\frac{355}{113}$ sirf kuch das-millionth hisson se. Kya koi sabse paas wala fraction hai? Nahi. Tum jo bhi fraction chuno, woh $π$ ke barabar nahi hai, isliye dono ke beech ek chhota sa gap hai. $π$ ke decimal ko kaafi digits ke baad kaat do to tumhe ek fraction milta hai, jaise $\frac{3141592}{1000000}$ , jo usi gap ke andar aata hai. Isliye $π$ ke liye har fraction sirf ek approximation hai. Hum $π \approx \frac{22}{7}$ (“ke kareeb hai”) likhte hain, aur $π \neq \frac{22}{7}$ bhi sach hai.

Table 6.2Kuch fractions π ≈ 3.1415927 ke kitne paas aate hain?NCERT p. 124

Fraction	Decimal	Error (fraction − π)
$3$	3	$- 0.1416$
$\frac{22}{7}$	3.1428571…	$+ 0.0012645$
$\frac{333}{106}$	3.1415094…	$- 0.0000832$
$\frac{355}{113}$	3.1415929…	$+ 0.0000003$

Roz ke kaam ke liye $\frac{22}{7}$ ya $3.14$ kaafi hai, aur is chapter mein hum $π \approx \frac{22}{7}$ use karte hain jab tak sawaal kuch aur na kahe. Jin engineers ko zyada accuracy chahiye, woh zyada digits use karte hain.

6.4 Circle ke Arc ki Length

Circle ke do points ke beech ke tukde ko arc (चाप) kehte hain. Jhoola, gaadi ka wiper aur ghadi ki sui ka sira, sab arcs par chalte hain. Arc kitna lamba hota hai? Aasaan tukdon se shuru karte hain.

Iski chaabi hai circle ka bilkul perfect gol hona. Circle ko uske centre ke around kisi bhi angle se ghumao, woh bilkul apne aap par hi aa baithta hai. Isliye agar do arcs centre par same angle banate hain, to ek ghumaav ek ko doosre par le aata hai, aur unki length same hoti hai. Centre par barabar angles se barabar arcs milte hain.

Ab poore chakkar ko baanto. Centre ke 360° ko 90° ke 4 barabar angles mein baanto: circle 4 barabar arcs mein bant jaata hai, isliye har ek $2 π r$ ka $\frac{1}{4}$ hai, yaani $\frac{1}{2} π r$ . Ise 180° ke 2 angles mein baanto to har half-circle $π r$ hai; 60° ke 6 angles mein baanto to har arc $\frac{1}{3} π r$ hai.

Har case mein arc ko circumference ka utna hi hissa milta hai jitna uske angle ko 360° ka milta hai. Hum kehte hain ki arc centre par woh angle subtend karta hai. Fig. 6.6 yeh pattern dikhata hai, jo kisi bhi angle $θ$ ke liye sach hai, sirf chakkar ke saaf-suthre fractions ke liye nahi.

4/4Aam taur par arc, $2 π r$ ka $\frac{θ}{360}$ hota hai.

Fig. 6.6Jo arc centre par

θ^{\circ}

subtend karta hai, woh poore circle ka

\frac{θ}{360}

hai.NCERT Fig. 6.10 · p. 126

Centre O wala ek circle, jiska ek fixed radius r right ki taraf point karta hai. Ek doosra radius ghoom kar centre par ek angle banata hai, aur dono radii ke beech ka arc mota banaya gaya hai. Step 1: angle 180 degrees hai, arc aadha circle hai, aur uski length 2πr × 180/360 = πr hai. Step 2: angle 90 degrees hai, circle ka ek chauthai, length πr ka aadha. Step 3: angle 60 degrees hai, circle ka chhatha hissa, length πr ka ek tihaai. Step 4: ek general angle θ, rule l = 2πr × θ/360 ke saath.

Definition

Arc

Circle ka woh hissa jo us par ke do points ke beech hai. Yeh centre par ek angle subtend karta hai, jo iske dono ends tak jaane wale radii se banta hai.

ExampleGhadi ki minute wali sui har 15 minute mein 90° ka arc ghoom jaati hai.

Arc ki length

l = 2 π r \times \frac{θ}{360}

jahan

$l$: arc ki length
$r$: radius
$θ$: centre par subtend hua angle, degrees mein

Example

Jhoole ki seat

Garden ke jhoole ki seat bar se 2.1 m neeche latakti hai. Ek jhonke mein rassiyan 60° ghoomti hain. Seat kitni door chalti hai? $π \approx \frac{22}{7}$ use karo.

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

Example

Ulta chal kar angle tak

Radius 7 cm wale circle ka ek arc 11 cm lamba hai. Yeh centre par kitna angle subtend karta hai?

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

Running track ko aur dhyaan se dekhna

Ab hum runners par wapas aa sakte hain. Ek standard track mein do seedhe hisse hote hain jo do semicircular bends se jude hote hain, aur kisi bhi lane ke dono bends milkar ek poora circle banate hain, jaise Fig. 6.1 mein. Isliye ek lane mein ek lap ki length do straights jitni hai, plus us lane ke circle ka circumference. Rules kehte hain ki distance lane ke andar wale edge se thoda sa bahar ki ek line par naapi jaati hai, lagbhag wahan jahan runner ke pair padte hain.

Example

200 m ka school track banana

Ek school ek track banata hai jiske seedhe hisse 34 m ke hain. Lane 1 mein, har bend par measuring line radius 21 m ka semicircle hai. Lanes 1.2 m chaudi hain. (a) Check karo ki lane 1 ka ek lap 200 m hai. (b) Lane 2 ka ek lap kitna zyada lamba hai?

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

Aakhri step ko numbers ki jagah letters ke saath dekho. Agar lane 1 ke bends ka radius $r$ ho aur lanes $w$ chaudi hon, to circles ke circumferences $2 π r$ aur $2 π (r + w)$ hain. Farak hai

Agli bahar wali lane mein ek lap ki extra distance

2 π (r + w) - 2 π r = 2 π w

Radius $r$ cancel ho gaya! Extra distance sirf lane ki chaudai par aur race mein kitna ghoomna hai us par depend karti hai, track ke size par nahi. Isliye lanes 2 aur 3 ke beech ka stagger utna hi hai jitna lanes 1 aur 2 ke beech, aur aise hi aage bahar tak. 1.22 m chaudi lanes wale full-size track par, lanes mein ek lap ke liye har lane ka stagger $2 π \times 1.22 \approx 7.67 m$ chahiye.Asli rules mein lane 1 ko kerb se 30 cm par naapte hain, lekin baaki lanes ko unki andar wali line se 20 cm par, isliye pehla stagger thoda chhota, lagbhag 7.04 m, hota hai. Stagger koi tohfa nahi hai: yeh lambe bends ko bilkul barabar kar deta hai.

Quick check

4 mein se 1

Gaadi ke wiper ka sira uske pivot se 14 cm door hai, aur wiper 45° ghoomta hai. Sira kitni door chalta hai?

π \approx \frac{22}{7}

use karo.

Answer dekho

Answer 11 cm

l = 2 \times \frac{22}{7} \times 14 \times \frac{45}{360} = 88 \times \frac{1}{8} = 11 c m

6.5 Perimeter par Problems, Puzzles aur Paradoxes

Arc lengths kuch hairaan karne wali puzzles tak le jaati hain. Har problem ka solution kholne se pehle use khud try karo.

Example 1NCERT Example 1 · p. 127

Curved triangle ka perimeter

Side $r$ wala ek equilateral triangle $A B C$ banao. Compass ki nok $A$ par rakh kar radius $r$ se $B$ se $C$ tak arc banao. Yahi $B$ se (arc $C A$ ) aur $C$ se (arc $A B$ ) karo. Teeno arcs ek gol-si shape gher lete hain jise Reuleaux triangle kehte hain. Iska perimeter nikalo. $r = 6 c m$ hone par yeh kitna hai?

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

3/4Teen aise arcs: perimeter $π r$ .

Fig. 6.7Teen 60° arcs ek Reuleaux triangle banate hain jiska perimeter,

π r

, diameter

r

wale circle ke perimeter ke barabar hai.NCERT Fig. 6.12 · p. 128

Step 1 mein equilateral triangle ABC hai, jiski har side par r likha hai aur har angle 60 degrees mark hai. Step 2 mein A par 60 degree ka sector shade hota hai aur A centre wala B se C tak ka arc banta hai, saath mein working 60/360 × 2πr = πr/3. Step 3 mein teeno arcs bante hain, jo bahar ki taraf ubhar kar ek gol-sa triangle banate hain, saath mein 3 × πr/3 = πr. Step 4 mein iske paas diameter r wala ek circle do parallel dashed lines ke beech rakha hai jo r doori par hain: dono shapes ki chaudai har direction mein r hai aur dono ka perimeter πr hai.

Example 2NCERT Example 2 · p. 128

Kaun si sadak lambi hai?

Do gaon $P$ aur $Q$ 12 km door hain. Road 1 ek akela semicircle hai jiska diameter $P Q$ hai. Road 2, $P Q$ ke dono halves par bane do semicircles se bani hai, ek north ki taraf ubhra aur ek south ki taraf. Road 3 chaar barabar semicircles use karti hai, jo baari-baari north aur south ki taraf ubhare hain. Kaun si road sabse chhoti hai?

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

Fig. 6.8 aage badhta rehta hai: 8, 16, 32 semicircles. Lehraati sadak seedhe segment $P Q$ se aur zyada chipakti jaati hai, isliye tum soch sakte ho ki uski length 12 ke paas pahunchegi. Aisa nahi hota. Semicircles kitne bhi hon, yeh $6 π$ , lagbhag 18.9, hi rehti hai. Lagbhag same dikhna aur lagbhag same length hona do alag baatein hain.

4/4Aath chhote-chhote: raasta lagbhag $P Q$ par hi pada hai, phir bhi woh $6 π \approx 18.9 c m$ hai.

Fig. 6.8Hum kitne bhi barabar semicircles use karein, unke diameters jud kar

P Q

hote hain, isliye raasta hamesha

6 π

lamba hota hai.NCERT Fig. 6.13 · p. 128

Ek dashed segment PQ, 12 cm lamba. Step 1 mein PQ ke upar ek bada semicircle hai, calculation ke saath: π ka aadha guna 12 barabar 6π. Step 2 mein diameter 6 wale do semicircles hain, ek segment ke upar aur ek neeche, total phir 6π. Step 3 mein chaar chhote semicircles baari-baari upar aur neeche hain. Step 4 mein aath chhote-chhote semicircles hain jo lagbhag PQ par pade hain, aur note likha hai ki raasta phir bhi 6π, lagbhag 18.9 cm lamba hai.

Quick check

3 mein se 1

Side 14 cm wale equilateral triangle par ek Reuleaux triangle bana hai. Uska perimeter kitna hai, cm mein?

π \approx \frac{22}{7}

use karo.

Answer dekho

Answer 44 cm

Perimeter

π r = \frac{22}{7} \times 14 = 44 c m

hai.

Practice · 9 questions

Practice 6.1: answers ke saath

NumericalEasyNCERT Exercise Set 6.1, Q1
Ek rangoli ke circular border mein bilkul 132 cm ribbon lagta hai. Circle ka radius kitna hai? (Is practice set mein $π \approx \frac{22}{7}$ use karo.)
cm
Answer dekho
Answer 21 cm
$2 \times \frac{22}{7} \times r = 132$ se $\frac{44}{7} r = 132$ milta hai, isliye $r = 132 \times \frac{7}{44} = 21 c m$ .
2
NumericalEasyNCERT Exercise Set 6.1, Q2
Teen gol dabbon ke dhakkanon ka radius 3.5 cm, 9 cm aur 15 cm hai. Har dhakkan ke kinare ko ek baar seal karne ke liye tape ki kitni lambi patti chahiye? Har answer 3 significant figures tak sahi do.
1. 3.5 cm wala dhakkan
  cm
  Answer dekho
  Answer 22 cm
  $2 \times \frac{22}{7} \times 3.5 = 22.0 c m$ .
2. 9 cm wala dhakkan
  cm
  Answer dekho
  Answer 56.6 cm
  $2 \times \frac{22}{7} \times 9 = 56.57 \dots \approx 56.6 c m$ .
3. 15 cm wala dhakkan
  cm
  Answer dekho
  Answer 94.3 cm
  $2 \times \frac{22}{7} \times 15 = 94.28 \dots \approx 94.3 c m$ .
3
NumericalEasyNCERT Exercise Set 6.1, Q3
Har curved raasta kitna lamba hai?
1. Deewar ghadi ke pendulum ka bob uske pivot se 21 cm neeche latakta hai aur ek taraf se doosri taraf 30° jhoolta hai. Ek jhonke mein bob kitni door chalta hai?
  cm
  Answer dekho
  Answer 11 cm
  $2 \times \frac{22}{7} \times 21 = 132$ aur $132 \times \frac{30}{360} = 11 c m$ .
2. Garden ki ek curved deewar radius 4.2 m wale circle ke arc par bani hai, aur arc centre par 150° ka angle banata hai. Deewar kitni lambi hai?
  m
  Answer dekho
  Answer 11 m
  $2 \times \frac{22}{7} \times 4.2 = 26.4$ aur $26.4 \times \frac{150}{360} = 11 m$ .
NumericalMediumNCERT Exercise Set 6.1, Q4
Radius 7 cm wale gol cake se ek slice kaati jaati hai jiska centre par angle 135° hai. Slice ke upar wale hisse ka perimeter nikalo: curved edge plus dono seedhe cuts.
Solution dikhao
Answer $30.5 c m$
Curved edge $2 \times \frac{22}{7} \times 7 \times \frac{135}{360} = 44 \times \frac{3}{8} = 16.5 c m$ hai.
Dono seedhe cuts radii hain: $7 + 7 = 14 c m$ . Perimeter $16.5 + 14 = 30.5 c m$ hai.
5
DiagramMediumNCERT Exercise Set 6.1, Q5
Har shape ka perimeter nikalo. Har curved tukda circle ka aadha, chauthai ya teen-chauthai hai.
Fig. 6.9
Chaar shapes. (a) Arch wali khidki: 1.4 m chauda rectangle jiski seedhi sides 1.8 m oonchi hain aur upar ek semicircle hai. (b) Side 14 cm ka square jiski right side ki jagah bahar ubhra semicircle hai aur left side ki jagah andar ki taraf muda semicircle. (c) Radius 7 cm wale circle ka teen-chauthai hissa, ek right-angled gap ke saath. (d) 20 cm × 10 cm ka rectangle jiske chaaron corners radius 3.5 cm ke quarter circles mein gol kiye gaye hain.
1. arch wali khidki
  m
  Answer dekho
  Answer 7.2 m
  Neeche 1.4 m, sides $2 \times 1.8 = 3.6 m$ , upar semicircle $\frac{22}{7} \times 0.7 = 2.2 m$ ; total $7.2 m$ .
2. ubhaar aur khaanche wala square
  cm
  Answer dekho
  Answer 72 cm
  Upar aur neeche: $14 + 14 = 28$ . Har semicircle ki length $\frac{22}{7} \times 7 = 22$ hai, aur khaancha bhi ubhaar ki tarah hi ginta hai: $28 + 22 + 22 = 72 c m$ .
3. teen-chauthai circle
  cm
  Answer dekho
  Answer 47 cm
  Arc: $\frac{3}{4} \times 44 = 33$ ; do radii: 14; total $47 c m$ .
4. gol corners wala rectangle: 20 cm × 10 cm, jiska har corner radius 3.5 cm ka quarter circle hai
  cm
  Answer dekho
  Answer 54 cm
  Seedhe hisse: $2 (20 - 7) + 2 (10 - 7) = 26 + 6 = 32$ . Chaaron quarter circles milkar radius 3.5 ka ek poora circle banate hain: $2 \times \frac{22}{7} \times 3.5 = 22$ . Total $54 c m$ .
6
NumericalEasyNCERT Exercise Set 6.1, Q6
Ek surveyor trundle wheel se sadkein naapta hai: handle par laga diameter 42 cm ka ek pahiya, aur ek counter jo har chakkar par ek baar click karta hai.
1. Ek click kitni distance batata hai, cm mein?
  cm
  Answer dekho
  Answer 132 cm
  $π d = \frac{22}{7} \times 42 = 132 c m$ .
2. 3.3 km lambi sadak par counter kitne clicks dikhayega?
  Answer dekho
  Answer 2500
  $3.3 k m = 330 000 c m$ , aur $330 000 \div 132 = 2500$ clicks.
7
DiagramMediumNCERT Exercise Set 6.1, Q7
Har design ki outline ki length nikalo.
Fig. 6.10
Do designs. (a) Side 14 cm ka ek dashed square. Radius 14 cm ke do quarter circles, jinke centre bottom-left aur top-right corners par hain (dots se mark), baaki do corners par milte hain aur ek green patte jaisi shape gher lete hain. (b) Ek blue khidki: 21 cm chauda rectangle jiski sides 30 cm oonchi hain, aur upar radius 21 cm ke do arcs jinke centre dono upar wale corners par hain (dots se mark), jo beech ke upar ek nok par milte hain. Ek dashed line rectangle ka upar wala hissa mark karti hai.
1. patte jaisa badge: side 14 cm wale square ke andar, do aamne-saamne ke corners par centre wale do quarter circles ke beech ka saajha hissa
  cm
  Answer dekho
  Answer 44 cm
  Patte ka har edge radius 14 ka quarter circle hai: $\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22 c m$ . Do edges milkar $44 c m$ .
2. nukeele arch wali khidki: 21 cm chauda rectangle jiski sides 30 cm oonchi hain, aur upar radius 21 cm ke do arcs, jinka centre saamne wali side ke upar wale corner par hai
  cm
  Answer dekho
  Answer 125 cm
  Arc ka centre, uska neeche wala end (doosra upar wala corner) aur arch ki nok, teeno ek doosre se 21 cm door hain, isliye woh equilateral triangle banate hain aur har arc 60° subtend karta hai: $\frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22 c m$ . Outline $= 21 + 30 + 30 + 22 + 22 = 125 c m$ .
8
Short answerMediumNCERT Exercise Set 6.1, Q8
Do gol talaabon ke circumferences $7 : 3$ ke ratio mein hain.
1. Unke radii ka ratio kya hai?
  Solution dikhao
  Answer $7 : 3$
  $\frac{C_{1}}{C_{2}} = \frac{2 π r_{1}}{2 π r_{2}} = \frac{r_{1}}{r_{2}}$ , isliye radii bhi $7 : 3$ ke ratio mein hain (aur diameters bhi).
2. Agar chhote talaab ka radius 6 m hai, to bade talaab ka radius kitna hai?
  m
  Answer dekho
  Answer 14 m
  $r_{1} = 6 \times \frac{7}{3} = 14 m$ .
Quick answerEasy
Garden ka ek pipe (hose) radius 35 cm ke drum par lipta hai. Drum ko 40 baar ghumane par kitne metre pipe khulta hai? (Pipe ki motaai ko ignore karo.)
m
Answer dekho
Answer 88 m
Ek chakkar: $2 \times \frac{22}{7} \times 35 = 220 c m$ . 40 chakkar: $8800 c m = 88 m$ .

6.6 Rectangle ka Area

Perimeter batata hai ki kisi khet ko kitni fencing chahiye; area (क्षेत्रफल) batata hai ki use kitne beej chahiye. Area kisi boundary ke andar ki flat surface ki maatra hai, aur hum ise side 1 unit wale squares gin kar naapte hain. Aise har square ka area 1 square unit (1 unit²) hota hai.

4 units lamba aur 3 units chauda rectangle 4 unit squares ki 3 rows rakhta hai, isliye uska area $4 \times 3 = 12$ square units hai. Isi tarah sides $a$ aur $b$ wale rectangle ka area $a b$ hai, aur side $a$ wale square ka, jo special case $b = a$ hai, area $a^{2}$ hai. Sides whole numbers na hon tab bhi rule kaam karta hai. 2.5 by 1.5 ka rectangle 2 poore squares, 3 aadhe squares aur ek chauthai square rakhta hai: $2 + 1.5 + 0.25 = 3.75 = 2.5 \times 1.5$ .

4/4Corner ek chauthai square hai: $2 + 1.5 + 0.25 = 3.75 = 2.5 \times 1.5$ .

Fig. 6.11Unit squares ginna, kate hue squares bhi, length × breadth deta hai.

Step 1: left mein 4 units × 3 units ka rectangle 12 unit squares se bhara hai. Step 2: right mein 2.5 units × 1.5 units ke rectangle ki outline aati hai aur uske andar ke 2 poore unit squares blue shade hain. Step 3: teen aadhe squares, right mein ek lambi patti aur upar do chapti pattiyan, green shade hain. Step 4: aakhri corner ka tukda, ek chauthai square, orange shade hai, aur total 2 + 3 × 1/2 + 1/4 = 3.75 = 2.5 × 1.5 dikhaya gaya hai.

Quick check

3 mein se 1

Ek classroom ka farsh 7.5 m lamba aur 6 m chauda hai. Uska area kitna hai, m² mein?

m²

Answer dekho

Answer 45 m²

Area

= 7.5 \times 6 = 45 m^{2}

. Jo side whole number nahi hai, woh bhi bilkul 2.5 by 1.5 rectangle ki tarah kaam karti hai.

6.7 Parallelogram ka Area

Parallelogram mein parallel sides ke do pairs hote hain. Ek side ko base $b$ chuno. Height $h$ base aur uske saamne wali side ke beech ki perpendicular distance hai. Yeh tirchhi side ki length nahi hai.

Fig. 6.12 dikhata hai ki area $b h$ kyun hai. Upar ke ek corner se height ke saath kaato. Ek right-angled triangle alag ho jaata hai. Use khiska kar doosre end par le jao, woh gap mein bilkul fit ho jaata hai, aur parallelogram sides $b$ aur $h$ wala rectangle ban jaata hai. Na koi kaagaz joda gaya, na phenka gaya, isliye areas barabar hain.

3/6Use khiska do: ek $b \times h$ rectangle.

Fig. 6.12Kaat kar khiskao, ya gher kar ghatao: dono tarah se parallelogram ka area base × height hai.NCERT Fig. 6.17 · p. 131

Steps 1 se 3 mein base b aur height h wala, left ki taraf jhuka parallelogram hai. Step 1 use poora dikhata hai. Step 2 top right corner se dashed height ke saath kaatta hai, jisse ek orange right-angled triangle banta hai. Step 3 triangle ko khiska kar left end par le jaata hai, jahan woh gap bhar deta hai; purani outline halki ho jaati hai aur nayi shape ke chaaron taraf b × h ka solid rectangle banta hai, saath mein A = b × h. Steps 4 se 6 mein base b wala ek patla parallelogram hai. Step 4 dikhata hai ki uske upar ke corner se height base ke bahar girti hai. Step 5 use width b + x aur height h wale dashed rectangle mein gherta hai. Step 6 corner ke do right triangles ko shade karta hai, jinke legs x aur h hain, aur (b + x)h − xh = bh dikhata hai.

Kaat-kar-khiskao wali trick ke liye zaroori hai ki upar ke corner se height base par hi gire. Bahut jhuke hue, patle parallelogram mein aisa nahi hota, isliye hamein ek aur argument chahiye. Parallelogram ko height $h$ aur width $b + x$ wale rectangle mein ghero, jahan $x$ batata hai ki upar wali side kitni bahar nikli hai. Parallelogram ke bahar jo bachta hai, woh legs $x$ aur $h$ wale do right triangles hain. Isliye area $(b + x) h - 2 \times \frac{1}{2} x h = b h$ hai. Yeh argument har parallelogram ke liye kaam karta hai, patla ho ya nahi.

TheoremParallelogram ka area

Base $b$ aur uski corresponding height $h$ wale parallelogram ka area $b h$ hota hai.

Parallelogram ko us rectangle mein ghero jiski neeche wali side mein base hai aur upar wali side mein saamne wali side. Maan lo upar wali side base se $x$ bahar nikli hai, to rectangle $(b + x)$ chauda aur $h$ ooncha hai.

Rectangle ke jo do tukde parallelogram ke bahar hain, woh legs $x$ aur $h$ wale right triangles hain; woh congruent hain (RHS), aur har ek ka area $\frac{1}{2} x h$ hai.

Parallelogram ka area $= (b + x) h - x h = b h$ .

InteractiveSame sides, alag areas

Teen tabs, har ek mein scale par bani drawing, ek slider aur area ka ek chhota graph. Rectangle, perimeter 36: ek slider width ko 0.5 se 17.5 tak set karta hai, length 18 minus width hai, aur cards dimensions, perimeter 36 aur area dikhate hain; graph width 9 par sabse upar hai. Hinged frame: sides 6 aur 4 wala parallelogram jiska angle 15 se 165 degrees tak jaata hai; cards height aur area dikhate hain, jo 90 degrees par sabse bada hai. Four rods: rods 5, 5, 5, 5 ya 10, 5, 4, 5 ya 5, 12, 12, 5 chuno aur A par angle se shape ko modo; cards angle A plus angle C, Brahmagupta ki value aur area dikhate hain, A, B aur D se hokar jaata ek dashed circle tab solid ho jaata hai jab shape cyclic ban jaati hai, aur ek button seedha us position par le jaata hai.

Fig. 6.13Perimeter ya sides fixed rakho aur area ko badalte dekho: perimeter 36 wale rectangles, ek hinged 6 by 4 frame, aur chaar rods jinhe moda ja sakta hai.

Example

Sirf sides kaafi nahi hain

Farsh ki ek tile parallelogram hai jiski sides 12 cm aur 8 cm hain. Dono 12 cm wali sides ke beech ki perpendicular distance 6 cm hai. Tile ka area nikalo, aur dono 8 cm wali sides ke beech ki distance bhi.

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

Quick check

3 mein se 1

Ek parallelogram ka base 15 cm aur corresponding height 8 cm hai. Uska area kitna hai, cm² mein?

cm²

Answer dekho

Answer 120 cm²

A = b h = 15 \times 8 = 120 c m^{2}

6.8 Triangle ka Area

Tum saalon se triangle ka area = ½ × base × height use kar rahe ho; yahan dekho ki yeh sach kyun hai. Ek right-angled triangle aur uski copy milkar rectangle banate hain, isliye woh us rectangle ka aadha hai. Baaki triangles ke liye upar wale vertex se height girao. Agar woh base ke andar girti hai, to triangle ko bases $b_{1}$ aur $b_{2}$ wale do right triangles mein kaat deti hai, jahan $b_{1} + b_{2} = b$ , aur total $\frac{1}{2} b_{1} h + \frac{1}{2} b_{2} h = \frac{1}{2} b h$ hai.

Us argument ne chupchaap maan liya tha ki height base ke andar girti hai. Jis triangle mein base ke ek end par obtuse angle ho, usmein aisa nahi hota: height bahar, badhayi gayi base line par girti hai. Tab hum jodne ki jagah ghatate hain. Neeche ke figure mein, triangle $P Q R$ bada right triangle $P S R$ hai jismein se chhota right triangle $P S Q$ hata diya gaya hai.

Fig. 6.14Jab height bahar girti hai, to ghatao:

\frac{1}{2} (b + x) h - \frac{1}{2} x h = \frac{1}{2} b h

Triangle PQR mein Q par obtuse angle hai. P se height, line RQ ko badhane par S par milti hai, jo base ke bahar hai. Bade right triangle PSR ka base b + x aur height h hai; chhote right triangle PSQ ka, jo orange shade hai, base x aur height h hai. Working mein likha hai: (b + x)h ka aadha minus xh ka aadha barabar bh ka aadha.

Ek aur smart raasta bhi hai jismein koi cases nahi chahiye. Triangle ki ek bilkul same copy banao aur use ek side ke midpoint ke around aadha chakkar ghumao. Copy original ke saath fit hokar ek parallelogram bana leti hai, jaise Fig. 6.15 dikhata hai.

4/4 $A B C D$ area $b h$ wala parallelogram hai, isliye triangle ka area $\frac{1}{2} b h$ hai.

Fig. 6.15Kisi bhi triangle ki do copies ek parallelogram banati hain, isliye triangle ka area uska aadha hai.NCERT Fig. 6.21 · p. 132

Blue triangle ABC, jismein side AC ka midpoint M mark hai. Step 1 mein sirf triangle hai. Step 2 mein uski ek orange copy thodi upar uthti hai. Step 3 mein copy M ke around aadha chakkar ghoomti hai, to uske corners C, A aur ek naye point D par aate hain; ab dono triangles ki side AC common hai. Step 4 mein parallelogram ABCD par label hai, BC ke saath base b aur A se ek dashed height h, aur parallelogram ka area bh aur triangle ka area bh ka aadha diya gaya hai.

TheoremTriangle ka area

Base $b$ aur corresponding height $h$ wale triangle ka area $\frac{1}{2} b h$ hota hai. Koi bhi side base ban sakti hai.

Triangle $A B C$ ki copy ko $A C$ ke midpoint $M$ ke around aadha chakkar ghumao. Tab $A$ aur $C$ apni jagah badal lete hain aur $B$ ek point $D$ par jaata hai, aur $M$ , $B D$ ka bhi midpoint hai.

Aadha chakkar har line ko ek parallel line par bhejta hai. Isliye $C D$ , jo $A B$ ki image hai, $A B$ ke parallel hai, aur $A D$ , jo $C B$ ki image hai, $C B$ ke parallel hai. (Angles ki bhasha mein, $∠ D C A = ∠ B A C$ alternate angles hain.) Isliye $A B C D$ ek parallelogram hai.

Iska base $B C = b$ aur height $h$ hai, isliye iska area $b h$ hai. Yeh do congruent triangles se bana hai, isliye har ek ka area $\frac{1}{2} b h$ hai.

Ek challenge hai. Koi bhi triangle banao aur ek corner se uske saamne wali side tak ek seedha cut lagao. Cut us side se kahan mile taaki dono tukdon ke areas barabar hon? Neeche ka theorem padhne se pehle tay karo. Corner se saamne wali side ke beech wale point tak ke segment ko median (माध्यिका) kehte hain; har triangle ke teen medians hote hain.

TheoremMedian area ko aadha karta hai

Tum kisi bhi vertex se shuru karo, median triangle ko same area wale do tukdon mein kaatta hai.

Fig. 6.16Barabar bases aur ek common height se barabar areas milte hain, bhale hi dono tukdon ki shapes alag hon.NCERT Fig. 6.22 · p. 133

Triangle ABC jiska base BC lamba hai aur D, BC ka beech wala point hai; tick marks BD = DC dikhate hain. Purple mein bana median AD triangle ko ek blue triangle ABD mein, jo patla aur seedha khada hai, aur ek orange triangle ADC mein, jo chauda aur jhuka hua hai, baant deta hai. A se BC tak ek dashed height h girti hai jo dono triangles ke kaam aati hai. Note mein likha hai: barabar areas, alag shapes, BD ka aadha guna h barabar DC ka aadha guna h.

Fig. 6.16 ke do tukde dikhate hain ki proof ko kitni kam cheezon ki zaroorat hai. Ek tukda patla aur seedha hai, doosra chauda aur jhuka hua, aur koi bhi ghumana ya palatna ek ko doosre par fit nahi karta: barabar area ke liye matching shapes zaroori nahi. Proof ne sirf barabar bases aur common height use ki. Isliye ek hi base par bane aise saare triangles, jinke teesre corners us base ke parallel ek line par hon, unke areas bhi barabar hote hain. Fig. 6.17 mein tum ise test kar sakte ho.

SimulationFixed base par triangles

Ek coordinate grid jismein fixed points A (0, 0) par aur B (14, 0) par hain, aur ek draggable point C hai jo (5, 12) se shuru hota hai. Triangle ABC ki teeno sides bani hain aur un par unki lengths likhi hain, jo shuru mein 14, 15 aur 13 hain, aur ek panel points ke coordinates aur distances AB, BC aur CA dikhata hai. Kyunki AB, x-axis par hai, C ka y-coordinate triangle ki height hai.

Fig. 6.17Base

A B

x

-axis par hai, isliye triangle ki height

C

y

-coordinate hai.

C

ko drag karo aur teeno side lengths padho.NCERT p. 133

Try karo

$C$ ko line $y = 12$ ke saath $(9, 12)$ tak aur phir $(14, 12)$ tak drag karo. Sides badalti hain, lekin base 14 aur height 12 nahi badalte. Har baar area kitna hai?
$C$ ko neeche $(5, 6)$ par le jao. Area ka kya hota hai, aur kyun?
$C$ ko wapas $(5, 12)$ par rakho aur sides 13, 14, 15 note karo. Is triangle ko tum Section 6.8.1 mein ek naye formula se check karoge.

Try karke dekhoNCERT Try this · p. 134

36 m fencing se bade (pens)

Ek kisaan ke paas bakriyon ke rectangular bade ke liye bilkul 36 m fencing hai. Woh koi bhi length aur breadth chun sakti hai, aisi lengths bhi jo poore metres mein nahi hain.

Paanch alag-alag bade likho jo woh bana sakti hai. Kya alag-alag bado ki sankhya ki koi limit hai?
Kaun sa bada bakriyon ko sabse zyada jagah deta hai? Kya uski shape tumhe hairaan karti hai?
Kya woh jitna chaho utne chhote area wala bada bana sakti hai? Kya sabse chhote area wala koi bada hai?

Pehle predict karo, phir apne answers Fig. 6.13 ke Rectangle, perimeter 36 tab mein test karo.

6.8.1 Heron's formula

Maan lo ek kisaan ne tape se ek triangular khet ki teeno sides naap li hain, lekin uski height kisi ko nahi pata. Phir bhi area sirf sides se nikala ja sakta hai, ek formula se jo Alexandria ke Heron ne likha tha, jo pehli sadi CE ke ek Greek engineer aur teacher the. Pehla step hai perimeter ko aadha karna; jo milta hai use semi-perimeter $s$ kehte hain.

Heron's formula

s = \frac{a + b + c}{2}, area = \sqrt{s (s - a) (s - b) (s - c)}

jahan

$a, b, c$: triangle ki teeno sides
$s$: semi-perimeter

Definition

Semi-perimeter

Kisi shape ke perimeter ka aadha. Sides $a$ , $b$ , $c$ wale triangle ke liye yeh $s = \frac{1}{2} (a + b + c)$ hai.

Chaar numbers ke product ka square root ek area kyun hona chahiye? Proof ki fikr karne se pehle, chalo saboot ikattha karein: formula ko aise triangles par lagao jinke areas hum kisi aur tareeke se nikal sakte hain, aur dekho ki answers milte hain ya nahi.

Example 3NCERT Example 3 · p. 134

Equilateral road sign

Ek triangular warning sign equilateral triangle hai jiski sides 90 cm hain. Heron's formula se iska area nikalo, aur ½ × base × height se check karo. $\sqrt{3} \approx 1.732$ lo.

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

90 ki jagah side $a$ lekar working dohrao, to equilateral triangle ka general result milta hai: area $= \frac{\sqrt{3}}{4} a^{2}$ .Mathematicians aksar “therefore” (isliye) ke liye $∴$ likhte hain, jaise $h^{2} = 3 \times 45^{2} ∴ h = 45 \sqrt{3}$ . Ise yaad rakhna; hum ise Section 6.10 mein phir use karenge.

Example 4NCERT Example 4 · p. 135

Shed ki saamne wali deewar

Garden shed ki triangular saamne wali deewar ke do barabar dhalaan wale edges 5 m ke hain aur base 8 m ka hai. Kitni deewar paint karni hai?

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

Example 5NCERT Example 5 · p. 136

Triangular sabzi ki kyaari

Sabzi ki ek kyaari triangle hai jiski sides 8 m, 15 m aur 17 m hain. Iska area do tareekon se nikalo.

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

Teeno tests pass ho gaye, lekin har triangle special tha: equilateral, isosceles ya right-angled. Sides 13, 14 aur 15 wala triangle inmein se koi nahi hai. Fig. 6.18 yeh solve karke uski height nikalta hai ki height base se kahan milti hai.

4/4 $\frac{1}{2} \times 14 \times 12 = 84$ , aur Heron bhi 84 deta hai.

Fig. 6.18Do right triangles se height 12 milti hai, aur dono methods 84 dete hain.NCERT p. 136

Triangle ABC jismein AB = 13, AC = 15 aur base BC = 14. Step 1 triangle dikhata hai aur note karta hai ki usmein koi right angle nahi hai. Step 2 A se BC par ek dashed height h girata hai, jo base ko x aur 14 − x mein baantti hai, aur h square ko do tarah likhta hai: 13 square minus x square, aur 15 square minus (14 − x) square. Step 3 dono ko barabar rakhta hai, 28x = 140 paata hai, isliye x = 5, aur phir h = 12; base ke hisse 5 aur 9 ho jaate hain aur dono right triangles green aur blue shade hote hain. Step 4 area ½ × 14 × 12 = 84 dikhata hai, aur s = 21 ke saath Heron's formula, jo 21 × 8 × 7 × 6 ka square root = 84 deta hai.

Example

Bina right angle wala triangle

Sides 13 cm, 14 cm aur 15 cm wale triangle ka area nikalo, pehle Heron's formula se aur phir height nikal kar.

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

Har triangle ke saath do circles hote hain. Incircle woh sabse bada circle hai jo triangle ke andar fit hota hai; yeh har side ko ek point par chhoota hai, aur iske radius ko hum $r$ kehte hain. Circumcircle woh circle hai jo teeno corners se hokar jaata hai, jo tumne Chapter 5 mein dekha tha; iske radius ko $R$ kaho. Har radius sides $a$ , $b$ , $c$ se area nikalne ka ek aur tareeka deta hai:

Area ke do aur formulas

area = \frac{a b c}{4 R} aur area = r \cdot s = \frac{r (a + b + c)}{2}

PropertyIncircle se area

Agar kisi triangle ki sides $a$ , $b$ , $c$ , semi-perimeter $s$ aur incircle ka radius $r$ ho, to uska area $r s$ hai.

Sides 6, 8 aur 10 wale right triangle mein hypotenuse circumcircle ka diameter hai, isliye $R = 5$ aur $\frac{a b c}{4 R} = \frac{480}{20} = 24$ , jo sach mein $\frac{1}{2} \times 6 \times 8$ hai. Formula $a b c / 4 R$ ke liye aage ki class ke results chahiye, isliye hum yahan sirf use batate hain.

Quick check

3 mein se 1

Sides 10 cm, 17 cm aur 21 cm wale triangle ka area Heron's formula se nikalo, cm² mein.

cm²

Answer dekho

Answer 84 cm²

s = \frac{1}{2} (10 + 17 + 21) = 24

, isliye area

= \sqrt{24 \times 14 \times 7 \times 3} = \sqrt{7056} = 84 c m^{2}

Cyclic 4-gon ke liye Brahmagupta's formula

Kya chaar side lengths kisi chaar sides wale figure (4-gon, yaani quadrilateral) ka area tay kar sakti hain? Nahi. Hinged frame mein tumne dekha ki kyun: ends par pin se judi chaar rods ko dhakel kar kai shapes mein badla ja sakta hai, aur frame ke mudne par area badalta hai. Ek hi answer paane ke liye hamein shape ke baare mein kuch aur jaanna hoga.

Brahmagupta ne frame ki woh ek khaas position dhoondhi jismein chaaron corners ek hi circle par hon. Is property wale 4-gon ko cyclic kehte hain. 628 CE mein, apne Brāhmasphuṭasiddhānta mein, unhone ek rule diya jo aise 4-gon ka area sirf uski sides se nikalta hai, aur yeh hairaani ki had tak Heron ke formula jaisa dikhta hai.

Definition

Cyclic 4-gon

Chaar sides wala figure (quadrilateral) jiske chaaron vertices ek hi circle par hon. Iske saamne wale angles jud kar 180° hote hain.

Brahmagupta's formula (sirf cyclic 4-gons ke liye)

s = \frac{a + b + c + d}{2}, area = \sqrt{(s - a) (s - b) (s - c) (s - d)}

Example 6NCERT Example 6 · p. 138

Rectangle se check karna

Har rectangle cyclic hota hai, kyunki uske diagonals barabar hote hain aur ek doosre ko aadha-aadha kaatte hain, isliye unka milne ka point chaaron corners se barabar door hota hai. 12 m × 5 m ke rectangle ke liye Brahmagupta's formula check karo.

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

Example 7NCERT Example 7 · p. 138

Isosceles trapezium se check karna

Khidki ka ek sheesha isosceles trapezium hai: parallel sides 10 cm aur 4 cm, aur dono tirchhi sides 5 cm ki. Isosceles trapezia cyclic hote hain. Brahmagupta's formula se iska area nikalo aur seedhe tareeke se check karo.

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

Key ideaNCERT Special Cases and Generalisation in Mathematics · p. 139

Special cases aur generalisation

Kisi general result mein ek extra condition daalo to tumhe special case milta hai; kisi special case se ek condition hata do to tum generalisation (व्यापकीकरण) ki taraf badhte ho. Special cases check karna kisi bhi naye formula ka jaldi wala test hai, aur generalise karke hi kai naye formulas dhoondhe gaye hain.

Parallel sides $a$ aur $b$ wala trapezium formula $\frac{1}{2} (a + b) h$ : $b = a$ rakhne par parallelogram, $a h$ , milta hai; $b = 0$ rakhne par chhoti side dab kar point ban jaati hai aur triangle, $\frac{1}{2} a h$ , milta hai.
Arc formula $2 π r \times \frac{θ}{360}$ mein $θ = 360$ rakhne par poora circumference, $2 π r$ , milta hai.
Heron's formula mein $a = b = c$ rakhne par equilateral triangle ka area, $\frac{\sqrt{3}}{4} a^{2}$ , milta hai (Example 3 dekho).

Brahmagupta's formula Heron's formula ka generalisation hai. Ek cyclic 4-gon $A B C D$ socho aur corner $D$ ko circle ke saath $A$ ki taraf khiskao, jaise Fig. 6.19 mein. Side $d = D A$ chhoti hoti jaati hai, aur jab $D$ , $A$ tak pahunchta hai to 4-gon triangle $A B C$ ban jaata hai. Har triangle ka ek circumcircle hota hai, isliye use $d = 0$ wala cyclic 4-gon maanne mein kuch nahi khota.

4/4 $d = 0$ rakho: Brahmagupta's formula Heron's formula ban jaata hai.

Fig. 6.19Heron's formula, Brahmagupta's formula ka special case

d = 0

hai.NCERT p. 140

Ek dashed circle par ek green chaar sides wala figure ABCD hai jiske corners circle par hain; sides par a = AB, b = BC, c = CD aur d = DA likha hai, aur side d vermillion rang mein hai. Step 1 4-gon ko Brahmagupta's formula ke saath dikhata hai. Step 2 D ko circle ke saath A ki taraf khiskata hai, isliye side d chhoti ho jaati hai. Step 3 mein D, A par baitha hai: shape triangle ABC hai aur d = 0. Step 4 formula mein d = 0 rakhta hai: s, a + b + c ka aadha ban jaata hai aur square root, s(s − a)(s − b)(s − c) ka square root ban jaata hai, jo Heron's formula hai.

DerivationBrahmagupta se Heron tak

$d = 0$ ke saath semi-perimeter $s = \frac{1}{2} (a + b + c + 0) = \frac{1}{2} (a + b + c)$ hai, jo triangle ka semi-perimeter hai. Aakhri bracket $s - d = s - 0 = s$ hai. Isliye

\sqrt{(s - a) (s - b) (s - c) (s - d)} = \sqrt{(s - a) (s - b) (s - c) s} = \sqrt{s (s - a) (s - b) (s - c)}

Yeh bilkul Heron's formula hai. Cyclic 4-gons ka ek formula triangle wale formula ko special case ki tarah apne andar rakhta hai.

Quick check

3 mein se 1

Ek cyclic 4-gon ki sides 7, 15, 20 aur 24 units hain. Brahmagupta's formula se uska area nikalo.

Answer dekho

Answer 234

s = \frac{1}{2} (7 + 15 + 20 + 24) = 33

, isliye area

= \sqrt{26 \times 18 \times 13 \times 9} = \sqrt{54756} = 234

square units.

6.9 Rectangle ko Square Banana

Maan lo ek pujari ko ek square yagya vedi (fire altar) chahiye jiska area bilkul ek pehle se bani rectangular vedi jitna ho. Vedic vediyan banane walon ke saamne aisi problems aati thi, aur Śulbasūtras, lagbhag 800 BCE ki unki construction manuals, inhe rassiyon aur khoonton se solve karti hain. Kisi diye gaye figure jitne hi area wala square banane ko us figure ko square karna (squaring) kehte hain. Neeche ka method Baudhāyana ke idea par chalta hai, do stages mein.

Stage 1: rectangle ko L-shape banao. Lambi side $a$ aur chhoti side $b$ wala rectangle lo. Ek $b \times b$ square kaat lo. Bachi hui patti ko beech se do tukdon mein kaato, har ek $\frac{a - b}{2}$ by $b$ ka, aur ek tukde ko square ke upar rakh do. Jo L-shape banti hai, woh side $\frac{a + b}{2}$ wale ek bade square mein fit hoti hai jiske corner se side $\frac{a - b}{2}$ ka ek chhota square gayab hai. Isliye

Do squares ke difference ke roop mein rectangle

a b = {(\frac{a + b}{2})}^{2} - {(\frac{a - b}{2})}^{2}

Stage 2: do squares ke difference ko ek square banao. Ek right triangle banao jiska hypotenuse $\frac{a + b}{2}$ ho aur ek leg $\frac{a - b}{2}$ . Zameen par, chhote leg ke ek end par $\frac{a + b}{2}$ lambi rassi baandho aur use ghumao jab tak woh right angle par kheenchi line se na mil jaaye. Baudhāyana–Pythagoras theorem se, doosra leg $x$ , $x^{2} = {(\frac{a + b}{2})}^{2} - {(\frac{a - b}{2})}^{2} = a b$ ko satisfy karta hai. $x$ par bana square hi woh square hai jo hamein chahiye. Fig. 6.20 yeh sab 9 by 4 rectangle ke liye karta hai.

5/7L, 6.5 ka square minus 2.5 ka square hai: ${6.5}^{2} - {2.5}^{2} = 36$ .

Fig. 6.20Rectangle → L-shape → do squares ka difference → ek square, aur aakhri step Baudhāyana–Pythagoras theorem karta hai.NCERT Fig. 6.30 · p. 141

Step 1: 9 by 4 ka rectangle, area 36. Step 2: 4 by 4 ka square kaata jaata hai, aur 5 by 4 ki patti bachti hai. Step 3: patti aadhi karke 2.5 by 4 ki do pattiyan banti hain. Step 4: ek patti ghoom kar square ke upar aa jaati hai, aur L-shape banti hai. Step 5: L side 6.5 ke dashed square mein fit hoti hai jiska 2.5 by 2.5 corner gayab hai, isliye 6.5 square minus 2.5 square barabar 36. Step 6: hypotenuse 6.5 aur ek leg 2.5 wale right triangle ka doosra leg x hai, jahan x square barabar 36. Step 7: side x = 6 ka green square aata hai, jiska area 6 × 6 = 36 hai, rectangle jitna hi.

TheoremConstruction kaam kyun karta hai

Sides $a > b$ wale kisi bhi rectangle ke liye, hypotenuse $\frac{a + b}{2}$ aur leg $\frac{a - b}{2}$ wale right triangle ka doosra leg $x$ aisa hota hai ki $x^{2} = a b$ .

Example

Aangan ko square banana

Ek aangan 9 m × 4 m ka rectangle hai. Construction se utne hi area wale square aangan ki side nikalo.

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

Quick check

3 mein se 1

Ek rectangle 16 cm × 9 cm ka hai. Utne hi area wale square ki side kitni hai, cm mein?

Answer dekho

Answer 12 cm

\frac{16 + 9}{2} = 12.5

aur

\frac{16 - 9}{2} = 3.5

, isliye

x^{2} = 156.25 - 12.25 = 144

aur

x = 12 c m

. Check:

16 \times 9 = 144

Practice · 11 questions

Practice 6.2: answers ke saath

1
NumericalEasyNCERT Exercise Set 6.2, Q1
$P Q R S$ ek trapezium hai jismein $P Q ∥ S R$ , $P Q = 12 c m$ aur $S R = 8 c m$ . In parallel sides ke beech ki distance $7 c m$ hai, aur $T$ , $S R$ par koi bhi point hai.
1. Triangle $P Q T$ ka area nikalo.
  cm²
  Answer dekho
  Answer 42 cm²
  $P Q = 12$ ko base lo. $S R$ ka har point $P Q$ se 7 cm door hai, isliye $T$ kahin bhi ho, height 7 hai: $\frac{1}{2} \times 12 \times 7 = 42 c m^{2}$ .
2. Triangles $P S T$ aur $Q R T$ ka total area nikalo.
  cm²
  Answer dekho
  Answer 28 cm²
  Trapezium ka area $\frac{1}{2} (12 + 8) \times 7 = 70 c m^{2}$ hai aur teeno triangles use bhar dete hain, isliye baaki do ka area $70 - 42 = 28 c m^{2}$ hai. Check: unke bases $S T + T R = 8$ ki height 7 common hai, jisse $\frac{1}{2} \times 8 \times 7 = 28$ milta hai.
NumericalMediumNCERT Exercise Set 6.2, Q2
Phoolon ki ek kyaari trapezium jaisi hai. Uske parallel edges 25 m aur 11 m lambe hain, aur tirchhe edges 13 m aur 15 m. Yeh kitni zameen gherti hai?
Solution dikhao
Answer $216 m^{2}$
Line sides 11 m aur 15 m wala ek parallelogram kaat deti hai aur sides 13 m, 15 m aur $25 - 11 = 14 m$ wala triangle chhodti hai.
Triangle ke liye Heron: $s = 21$ , area $= \sqrt{21 \times 8 \times 7 \times 6} = 84 m^{2}$ . 14 m wali side par iski height $\frac{2 \times 84}{14} = 12 m$ hai, aur yahi parallel edges ke beech ki distance bhi hai.
Area $= \frac{1}{2} (25 + 11) \times 12 = 18 \times 12 = 216 m^{2}$ . Check: parallelogram $11 \times 12 = 132$ plus triangle $84$ se bhi $216$ milta hai.
NumericalMediumNCERT Exercise Set 6.2, Q3
Ek triangular garden ko kul 36 m wire se ghera gaya hai. Uski do sides 9 m aur 10 m lambi hain. Uska area kitna hai?
m²
Answer dekho
Answer 36 m²
Teesri side $= 36 - 9 - 10 = 17 m$ . $s = 18$ , isliye area $= \sqrt{18 \times 9 \times 8 \times 1} = \sqrt{1296} = 36 m^{2}$ .
NumericalMediumNCERT Exercise Set 6.2, Q4
Ek jogger ek triangular park ka ek chakkar lagata hai aur 210 m daudta hai. Map par park ki sides $13 : 14 : 15$ ke ratio mein hain. Park ka area kitna hai?
m²
Answer dekho
Answer 2100 m²
$42 k = 210$ se $k = 5$ : sides 65, 70, 75. $s = 105$ , isliye area $= \sqrt{105 \times 40 \times 35 \times 30} = \sqrt{4 410 000} = 2100 m^{2}$ .
NumericalMediumNCERT Exercise Set 6.2, Q5
Ek patang rhombus jaisi hai aur uske diagonals ke saath lagi do cross-dandiyan use shape mein rakhti hain. Lambi dandi chhoti dandi se teen guna lambi hai, aur patang ka area 150 cm² hai. Har dandi kitni lambi hai?
Solution dikhao
Answer10 cm aur 30 cm
Maan lo diagonals $d$ aur $3 d$ hain. Rhombus ka area $\frac{1}{2} \times d \times 3 d = \frac{3}{2} d^{2}$ hai.
$\frac{3}{2} d^{2} = 150$ se $d^{2} = 100$ , isliye $d = 10 c m$ aur doosra diagonal $30 c m$ hai.
ProofMediumNCERT Exercise Set 6.2, Q6
Do seedhi nahrein ek doosre ke parallel behti hain. Dakshini kinare ka ek hissa $X Y$ do triangular plots ka base hai. Ek plot ka teesra corner uttari kinare par ek mandir $T$ par hai; doosre ka teesra corner usi kinare par aage ek kiosk $K$ par hai. Dono plots ke areas aapas mein kaise compare hote hain?
Solution dikhao
AnswerPlots ke areas barabar hain: ratio $1 : 1$ hai.
Dono plots same base $X Y$ wale triangles hain.
Kinare parallel hain, isliye uttari kinare ka har point, $T$ aur $K$ bhi, line $X Y$ se same distance par hai. Dono triangles ki height same hai.
Barabar bases aur barabar heights se barabar areas milte hain, isliye area $X Y T$ : area $X Y K = 1 : 1$ , chahe $T$ aur $K$ uttari kinare par kahin bhi hon.
ProofHardNCERT Exercise Set 6.2, Q7
$A B C D$ ek parallelogram hai. $O$ diagonal $B D$ ko $D$ se aage badhane par us par ek point hai. Prove karo ki triangles $A B O$ aur $C B O$ ke areas barabar hain.
Solution dikhao
Dono triangles ka base $B O$ hai, jo line $B D$ par hai. Isliye hamein dikhana hai ki $A$ aur $C$ is line se barabar door hain.
Parallelogram ke diagonals ek doosre ko aadha-aadha kaatte hain, isliye $B D$ , $A C$ ke midpoint $M$ se hokar jaata hai.
Line $B D$ par perpendiculars $A E$ aur $C F$ girao. Triangles $A E M$ aur $C F M$ congruent hain (AAS: right angles, $M$ par vertically opposite angles, aur $A M = C M$ ). Isliye $A E = C F$ .
Barabar base, barabar heights: area $A B O$ = area $C B O$ .
ProofHardNCERT Exercise Set 6.2, Q8
Ek kisaan ka chaar sides wala khet $A B C D$ hai. Woh chaaron sides ke midpoints mark karti hai aur chaaron corner triangles ko fence se alag kar deti hai, jinmein se har ek ek corner aur wahan milne wali do sides ke midpoints se banta hai. Prove karo ki chaaron corner triangles milkar khet ke aadhe area ke barabar hain.
Solution dikhao
Maan lo $P$ , $Q$ , $R$ , $S$ kramshah $A B$ , $B C$ , $C D$ , $D A$ ke midpoints hain. Diagonal $B D$ kheencho.
Triangle $A B D$ mein $D P$ median hai, isliye area $A P D = \frac{1}{2}$ area $A B D$ . Triangle $A P D$ mein $P S$ median hai, isliye area $A P S = \frac{1}{2}$ area $A P D = \frac{1}{4}$ area $A B D$ .
Isi tarah, $C$ wala corner triangle $\frac{1}{4}$ area $C B D$ hai. $A$ aur $C$ wale corners milkar poore khet ka $\frac{1}{4}$ banate hain.
Iski jagah diagonal $A C$ use karo, to $B$ aur $D$ wale corners bhi khet ka $\frac{1}{4}$ banate hain. Isliye chaaron corners milkar $\frac{1}{2}$ banate hain, aur beech ka tukda $P Q R S$ baaki aadha hai. ( $P Q R S$ hamesha parallelogram kyun hota hai, yeh quadrilaterals wale chapter mein dikhaya gaya hai.)
ProofMediumNCERT Exercise Set 6.2, Q9
Triangle $P Q R$ mein $S$ , $Q R$ ka midpoint hai. Median $P S$ ko $S$ se aage ek point $T$ tak badhaya gaya hai. Prove karo ki triangles $P Q T$ aur $P R T$ ke areas barabar hain.
Solution dikhao
$P S$ triangle $P Q R$ ka median hai, isliye area $P Q S$ = area $P R S$ .
Triangle $Q T R$ mein $T S$ median hai (kyunki $S$ , $Q R$ ka midpoint hai), isliye area $Q S T$ = area $R S T$ .
Jodne par, area $P Q T$ = area $P Q S$ + area $Q S T$ = area $P R S$ + area $R S T$ = area $P R T$ .
ProofMediumNCERT Exercise Set 6.2, Q10
Ek garden $W X Y Z$ parallelogram jaisa hai, aur uske andar kahin ek fountain $F$ hai. Seedhe raaste $F$ ko chaaron corners se jodte hain aur garden ko chaar triangular kyaariyon mein baant dete hain. Dikhao ki kyaariyan $F W X$ aur $F Y Z$ milkar bilkul aadha garden banati hain, fountain chahe kahin bhi ho. Isse baaki do kyaariyon ke baare mein kya pata chalta hai?
Solution dikhao
Maan lo $W X = Y Z = b$ , aur $h$ parallel sides $W X$ aur $Y Z$ ke beech ki distance hai. $F$ se $W X$ tak ki distance $h_{1}$ aur $F$ se $Y Z$ tak ki distance $h_{2}$ jud kar $h$ hoti hain.
Area $F W X$ + area $F Y Z = \frac{1}{2} b h_{1} + \frac{1}{2} b h_{2} = \frac{1}{2} b (h_{1} + h_{2}) = \frac{1}{2} b h$ , jo garden ka aadha hai.
Kyaariyan $F X Y$ aur $F Z W$ baaki aadha banati hain, isliye aamne-saamne ki kyaariyon ke dono pairs ka total area hamesha barabar hota hai: ratio $1 : 1$ hai.
ProofHardNCERT Exercise Set 6.2, Q11
Ek triangular plot $A B C$ mein side $B C$ par ek point $P$ par kuaan hai, jo $B$ ki tulna mein $C$ ke zyada paas hai. Maalik $P$ se ek seedhi fence chahta hai jo plot ko barabar area ke do hisson mein baant de. Method: $B C$ ka midpoint $D$ mark karo, $A P$ jodo, aur $D$ se $A P$ ke parallel line kheencho; maan lo woh $A B$ ko $Q$ par milti hai. Prove karo ki fence $P Q$ triangle $B P Q$ ko kaat kar alag karti hai jiska area bilkul $A B C$ ka aadha hai.
Solution dikhao
Median $A D$ se area $A B D = \frac{1}{2}$ area $A B C$ milta hai.
Triangles $D Q A$ aur $D Q P$ ka base same $D Q$ hai, aur $A$ aur $P$ line $A P$ par hain, jo $D Q$ ke parallel hai. Isliye unki height same hai aur areas barabar hain.
Isliye area $B P Q$ = area $B D Q$ + area $D Q P$ = area $B D Q$ + area $D Q A$ = area $A B D$ = $\frac{1}{2}$ area $A B C$ .

6.10 Circle ka Area

Kisi gaon ya shehar mein chaaron taraf dekho, circles har jagah hain: kuen, gol jhopdiyan, khalihaan (threshing floors), choodiyan, pahiye aur stupas ke gumbad.

Yeh ek sawaal hai jiska jawab tum circle ke area ke kisi formula ke bina de sakte ho. Ek gol khalihaan ka kinara doosre ke kinare se do guna lamba hai. Use kitne guna zyada paving chahiye? Do guna lamba kinara matlab do guna lamba radius, isliye bada khalihaan chhote wale ko har direction mein 2 ke factor se bada kiya hua roop hai. Isliye paving $2^{2} = 4$ ke factor se badhti hai, 2 se nahi. Purane zamane mein gol farsh aur khet naapne wale log yeh pattern jaante the: circle ka area uske circumference ke square ke saath chalta hai.

Yahi reasoning har shape ke liye kaam karti hai. Use $k$ ke factor se bada karo: har length, aur isliye perimeter $P$ bhi, $k$ se multiply hoti hai, jabki area $A$ , $k^{2}$ se multiply hota hai. Tab $P^{2}$ aur $A$ dono $k^{2}$ se multiply hote hain, aur ratio $\frac{P^{2}}{A}$ same rehta hai. Yeh shape par depend karta hai lekin uske size par nahi, jaisa table dikhati hai.

Table 6.3Perimeter ka square, area se divide karkeNCERT p. 144

Shape	Perimeter $P$	Area $A$	$P^{2} \div A$
Square, side $a$	$4 a$	$a^{2}$	16
Equilateral triangle, jiski side $a$ hai	$3 a$	$\frac{\sqrt{3}}{4} a^{2}$	$12 \sqrt{3} \approx 20.78$
Regular hexagon, jiski side $a$ hai	$6 a$	$\frac{3 \sqrt{3}}{2} a^{2}$	$8 \sqrt{3} \approx 13.86$
Circle, radius $r$	$2 π r$	?	? (ek fixed number)

Isliye circles ke liye bhi $C^{2} \div A$ ek fixed number hona chahiye, aur purane rules ne iska alag-alag andaaza lagaya. 1500 BCE se pehle ka ek Babylonian rule ise 12 maanta hai, isliye $A \approx \frac{C^{2}}{12}$ . Lagbhag usi samay ke Egyptian scribes ne circle ko ek aise square se milaya jiski side diameter ka $\frac{8}{9}$ ho: $A \approx {(\frac{8}{9} d)}^{2}$ . Baudhāyana ka Śulbasūtra bhi rassi-aur-khoonte wale constructions deta hai jo circle ko lagbhag barabar area ke square mein badalte hain. Kaun sa rule behtar hai? Agla example unhe compare karta hai.

Example

Ek talaab ke liye teen purane rules

Ek gol talaab ka radius 7 m hai. Iska area nikalo (a) $A = π r^{2}$ aur $π \approx \frac{22}{7}$ se, jise hum neeche prove karenge, (b) Babylonian rule se, (c) Egyptian rule se. Purane rules kitne door hain?

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

Archimedes aur Nīlakaṇṭha: area πr² kyun hai

Lagbhag 250 BCE mein Archimedes ne yeh maamla suljha diya, yeh prove karke ki circle ka area bilkul legs $r$ aur $C$ wale right triangle ke area jitna hai. Us triangle ka area $\frac{1}{2} \times 2 π r \times r = π r^{2}$ hai, isliye fixed number $\frac{C^{2}}{A}$ , $\frac{4 π^{2} r^{2}}{π r^{2}} = 4 π \approx 12.57$ hai, jo Babylonian 12 se zyada door nahi.

Unka argument regular polygons se shuru hota hai. Regular polygon ke centre ko uske corners se jodo. Jo triangles bante hain, sabki height same hai: centre se side tak ki distance $r$ . Triangles ko ek line ke saath khol kar bichha do, jaise Fig. 6.21 mein. Unke bases jud kar perimeter $P$ banate hain, aur same height wale triangles apne upar ke corners ko area badle bina ek saath khiska sakte hain. Isliye polygon ka area $\frac{1}{2} \times P \times r$ hai. Ab sides ki sankhya badhne do: polygon circle se chipak jaata hai, $P$ circumference $C$ ke paas pahunchta hai, aur area $\frac{1}{2} C r$ ke paas pahunchta hai.

3/4Same bases aur heights: area $= \frac{1}{2} P r$ .

Fig. 6.21Regular polygon ka area ½ × perimeter × r hai; circle, jo aise polygons ki limit hai, uska area ½ × C × r = πr² hai.NCERT Fig. 6.36 · p. 145

Step 1 mein ek regular hexagon centre se chhah triangles mein kata hai, jinke rang baari-baari badalte hain, aur ek triangle ki height r mark hai. Step 2 chhahon triangles ko ek line ke saath ek ke baad ek rakhta hai; unke bases jud kar perimeter P banate hain aur sabki height r hai. Step 3 row ke upar base P aur height r wala ek bada triangle banata hai, area ½ P r dikhate hue. Step 4 mein radius r ka circle aur ek right triangle hai jiske legs r aur circumference C hain, saath mein ½ C r = πr square.

Lagbhag 1500 CE mein, Kerala ke Nīlakaṇṭha Somayājī ne yahi result ek aisi picture se samjhaya jo koi bhi bana sakta hai. Āryabhaṭīya par apni commentary mein unhone circle ko bahut si patli slices mein kaata aur unhe ulta-seedha ek ke baad ek rakha. Aadhi slices upar point karti hain aur aadhi neeche, aur milkar ek aisi shape banati hain jo parallelogram ke bahut kareeb hai. Iski lambi side aadhe circumference, $π r$ , ke barabar hai, kyunki aadhe arcs uske saath hain; iski height radius $r$ hai. Ise Fig. 6.22 mein try karo.

InteractiveCircle ko kaato aur dobara jamao

Radius 5 cm ka ek circle barabar slices mein kata hai, upar wale aadhe mein blue aur neeche wale aadhe mein orange. Ek slider slices ki sankhya 4, 6, 8, 12, 16, 24, 36, 48 ya 64 set karta hai. Ek button slices ko ek row mein jamata hai, upar wale aadhe ki slices neeche point karti hain aur neeche wale aadhe ki upar, aur unhe wapas bhi rakh deta hai. Cards dikhate hain ki neeche ka edge lagbhag aadha circumference, 15.71 cm, hai, height radius, 5 cm, hai, aur edge ke saath ke ubhaar kitne oonche hain. Ek aur button comparison ke liye row ke upar πr by r ka dashed rectangle banata hai.

Fig. 6.22Circle ko zyada se zyada slices mein kaato aur unhe ek row mein rakho: row lagbhag

π r

lambe aur

r

oonche rectangle mein badal jaati hai.

TheoremCircle ka area

Radius $r$ wale circle ka area $A = π r^{2}$ hota hai.

Circle ko $n$ barabar slices mein kaato aur unhe ek row mein rakho jaise Fig. 6.22 mein. $n$ kuch bhi ho, row ka area bilkul circle jitna hai, kyunki woh unhi tukdon se bani hai.

Aadhe arcs upar ke edge ke saath hain aur aadhe neeche ke, isliye har edge ki length $\frac{1}{2} \times 2 π r = π r$ hai. Jaise-jaise $n$ badhta hai, arcs chapte hote jaate hain, edges seedhe hote jaate hain aur tirchhe ends seedhe khade ho jaate hain. Row $π r$ lambe aur $r$ oonche rectangle ke jitna chaho utna paas aa jaati hai.

Jo area fixed rehta hai aur saath hi $π r \times r$ ke jitna chaho utna paas aa jaata hai, woh $π r^{2}$ ke barabar hi hona chahiye.

Example

Talaab ke chaaron taraf raasta

Radius 7 m wale gol talaab ke chaaron taraf 3.5 m chauda gravel ka raasta hai. Raaste ka area nikalo.

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

Quick check

4 mein se 1

Radius 14 cm wale circle ka area nikalo, cm² mein.

π \approx \frac{22}{7}

use karo.

cm²

Answer dekho

Answer 616 cm²

\frac{22}{7} \times 14 \times 14 = 22 \times 28 = 616 c m^{2}

6.10.1 Circle ke Sector ka Area

Disc ka woh tukda jo do radii aur unke beech ke arc se ghira ho, sector (त्रिज्यखंड) kehlata hai. Iska area bhi arc wala hi rule maanta hai. Centre ke around ghumaav kisi sector ko same angle wale kisi bhi doosre sector par le aata hai, isliye barabar angles se barabar areas milte hain. Disc ko 360 barabar ek-degree wale sectors mein kaato: har ek ke paas area $π r^{2}$ ka $\frac{1}{360}$ hai, aur $θ^{\circ}$ ka sector unmein se $θ$ se bana hai. (Yeh rule un angles ke liye bhi sach hai jo whole numbers nahi hain.) Jaise, aadhi disc $\frac{180}{360} π r^{2} = \frac{1}{2} π r^{2}$ hai.

Sector ka area

sector ka area = π r^{2} \times \frac{θ}{360}

jahan

$r$: radius
$θ$: centre par sector ka angle, degrees mein

Definition

Sector

Disc ka woh hissa jo do radii aur unke ends ko jodne wale arc ke beech hai.

ExampleGol pizza ki ek slice.

Iski jagah disc ko ek seedhe cut se kaato, to cut (ek chord) use do tukdon mein baant deta hai jinhe segments (वृत्तखंड) kehte hain: ek chhota minor segment aur ek bada major segment. Jaisa Fig. 6.23 dikhata hai, minor segment ek sector hai jismein se ek triangle hata diya gaya hai.

Definition

Segment

Disc ka woh hissa jo ek chord aur us chord se kate arc ke beech hai. Chhota wala minor segment hai aur bada wala major segment.

3/4Sector minus triangle se chord $A B$ aur arc ke beech ka segment bachta hai.

Fig. 6.23Minor segment yaani sector − triangle; major segment yaani circle − minor segment.NCERT Fig. 6.38 · p. 147

Centre O wala ek circle jis par points A aur B hain, aur angle AOB right angle hai. Step 1 sector AOB ko blue mein shade karta hai, working 90/360 × πr² = πr² ka ek chauthai ke saath. Step 2 triangle AOB ko orange mein shade karta hai, area r guna r ka aadha. Step 3 sirf chord AB aur arc ke beech ke pink hisse ko shade karta hai, yaani segment, jiska area πr² ka ek chauthai minus r² ka aadha hai. Step 4 disc ke baaki hisse ko green mein major segment ke roop mein shade karta hai, πr² minus minor segment.

Example

Haath ke pankhe ka kaagaz

Ek folding haath ka pankha khul kar 150° ka sector banata hai. Uski teeliyan 21 cm lambi hain, lekin kaagaz pivot se 7 cm door se shuru hota hai. Kaagaz ka area nikalo.

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

Example

Raaste se bata ek lawn

Centre $O$ wale ek gol lawn ka radius 14 m hai. Ek seedha raasta $A B$ uske aar-paar jaata hai, aur $∠ A O B = 90^{\circ}$ . Raaste ko ek line maan kar lawn ke dono hisson ke areas nikalo.

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

Quick check

3 mein se 1

Radius 21 cm ka ek gol pizza nau barabar slices mein kaata jaata hai. Har slice 40° angle wala sector hai. Ek slice ka area kitna hai, cm² mein?

π \approx \frac{22}{7}

use karo.

cm²

Answer dekho

Answer 154 cm²

\frac{40}{360} \times \frac{22}{7} \times 441 = \frac{1}{9} \times 1386 = 154 c m^{2}

Practice · 10 questions

Practice 6.3: answers ke saath

NumericalEasyNCERT Exercise Set 6.3, Q1
Radius 14 m wali ek gol phoolon ki kyaari ko centre se nikalne wale seedhe raaston se 8 barabar sectors mein baanta gaya hai, har tarah ke phool ke liye ek. Har tarah ke phool ko kitna area milta hai? ( $π \approx \frac{22}{7}$ use karo jab tak kuch aur na kaha jaaye.)
m²
Answer dekho
Answer 77 m²
$\frac{45}{360} \times \frac{22}{7} \times 196 = \frac{1}{8} \times 616 = 77 m^{2}$ .
NumericalEasyNCERT Exercise Set 6.3, Q2
Gol table top ka kinara 88 cm lamba hai. Ek kapda top ke bilkul ek chauthai hisse ko, centre se kinare tak, dhakta hai. Kapda kitna area dhakta hai?
cm²
Answer dekho
Answer 154 cm²
$2 \times \frac{22}{7} \times r = 88$ se $r = 14$ . Quadrant $= \frac{1}{4} \times \frac{22}{7} \times 196 = 154 c m^{2}$ .
NumericalMediumNCERT Exercise Set 6.3, Q3
Ek gol radar screen par ghoomne wali line 21 cm lambi hai aur har 12 seconds mein ek poora chakkar lagati hai. 4 seconds mein woh screen ka kitna area cover karti hai?
cm²
Answer dekho
Answer 462 cm²
4 seconds ek chakkar ka $\frac{4}{12}$ hai, yaani 120°. Area $= \frac{120}{360} \times \frac{22}{7} \times 21^{2} = \frac{1}{3} \times 1386 = 462 c m^{2}$ .
4
NumericalMediumNCERT Exercise Set 6.3, Q4
Radius 12 m wale ek gol lawn ko do seedhe raaste kaatte hain jo centre se kinare tak jaate hain aur aapas mein 120° ka angle banate hain. Chhote hisse mein phool lagaye jaate hain aur bada hissa ghaas hi rehta hai. $π \approx 3.14$ use karke har hisse ka area nikalo.
1. phoolon ki kyaari (minor sector)
  m²
  Answer dekho
  Answer 150.72 m²
  $\frac{120}{360} \times 3.14 \times 144 = 150.72 m^{2}$ .
2. ghaas (major sector)
  m²
  Answer dekho
  Answer 301.44 m²
  $\frac{240}{360} \times 3.14 \times 144 = 301.44 m^{2}$ (ya $452.16 - 150.72$ ).
NumericalMediumNCERT Exercise Set 6.3, Q5
Radius 12 cm wali ek gol khidki ke sheeshe ke aar-paar lakdi ki ek seedhi patti lagi hai. Patti ke dono ends tak ke radii centre par 60° ka angle banate hain. $π \approx 3.14$ aur $\sqrt{3} \approx 1.73$ use karke patti ke dono taraf ke sheeshe ka area nikalo.
Solution dikhao
AnswerChhota hissa (minor segment) $13.08 c m^{2}$ ; bada hissa (major segment) $439.08 c m^{2}$ .
Sector: $\frac{60}{360} \times 3.14 \times 144 = 75.36 c m^{2}$ .
Triangle $A O B$ mein $O A = O B$ aur unke beech 60° ka angle hai, isliye yeh side 12 wala equilateral triangle hai. Iska area $\frac{\sqrt{3}}{4} \times 144 = 36 \sqrt{3} \approx 36 \times 1.73 = 62.28 c m^{2}$ hai.
Minor segment $= 75.36 - 62.28 = 13.08 c m^{2}$ .
Circle $= 3.14 \times 144 = 452.16 c m^{2}$ , isliye major segment $= 452.16 - 13.08 = 439.08 c m^{2}$ .
NumericalEasyNCERT Exercise Set 6.3, Q6
Garden ke do sprinklers mein se har ek 7 m tak paani phenkta hai aur 150° ghoomta hai. Jin areas ko woh seenchte hain, woh overlap nahi karte. Woh kul kitna area seenchte hain?
m²
Answer dekho
Answer 128.33 m²
Har ek $\frac{150}{360} \times \frac{22}{7} \times 49 = \frac{5}{12} \times 154 \approx 64.17 m^{2}$ seenchta hai; dono milkar lagbhag $128.33 m^{2}$ .
ProofHardNCERT Exercise Set 6.3, Q7
Circle ke do radii centre par 120° par milte hain, aur ek chord unke bahar wale ends ko jodti hai. Agar radius $r$ hai, to dikhao ki is chord se kate minor segment ka area $r^{2} (\frac{π}{3} - \frac{\sqrt{3}}{4})$ hai.
Solution dikhao
Sector ka area $\frac{120}{360} π r^{2} = \frac{π r^{2}}{3}$ hai.
Maan lo $M$ , $O$ se chord $A B$ par gire perpendicular ka foot hai. Yeh angle ko aadha karta hai, isliye $∠ A O M = 60 °$ aur $∠ O A M = 30 °$ . Triangle $O M A$ ki do copies ko $A M$ ke saath peeth se peeth milakar rakhne par side $r$ wala equilateral triangle banta hai, isliye $O M = \frac{r}{2}$ .
Tab $A M = \sqrt{r^{2} - \frac{r^{2}}{4}} = \frac{\sqrt{3}}{2} r$ , isliye $A B = \sqrt{3} r$ aur area $A O B = \frac{1}{2} \times \sqrt{3} r \times \frac{r}{2} = \frac{\sqrt{3}}{4} r^{2}$ .
Minor segment $= \frac{π r^{2}}{3} - \frac{\sqrt{3}}{4} r^{2} = r^{2} (\frac{π}{3} - \frac{\sqrt{3}}{4})$ .
ProofMediumNCERT Exercise Set 6.3, Q9
Radius $r$ ka ek circle ek square ke andar aise banaya gaya hai ki woh chaaron sides ko chhoota hai. Dikhao ki circle square ka $\frac{π}{4} \approx 0.785$ hissa dhakta hai.
Solution dikhao
Circle aamne-saamne ki sides ko chhoota hai, isliye square ki side diameter $2 r$ hai aur uska area $4 r^{2}$ hai.
Ratio $= \frac{π r^{2}}{4 r^{2}} = \frac{π}{4} \approx 0.785$ .
ProofHardNCERT Exercise Set 6.3, Q8
Radius $r$ ka ek circle ek equilateral triangle ke andar aise banaya gaya hai ki woh teeno sides ko chhoota hai. Dikhao ki circle triangle ka $\frac{π}{3 \sqrt{3}} \approx 0.605$ hissa dhakta hai.
Solution dikhao
Maan lo side $a$ hai. Tab $s = \frac{3 a}{2}$ aur area $= r s = \frac{3 a r}{2}$ . Saath hi area $= \frac{\sqrt{3}}{4} a^{2}$ .
Isliye $\frac{\sqrt{3}}{4} a^{2} = \frac{3 a r}{2}$ , jisse $a = \frac{6 r}{\sqrt{3}} = 2 \sqrt{3} r$ aur area $= \frac{\sqrt{3}}{4} \times 12 r^{2} = 3 \sqrt{3} r^{2}$ milta hai.
Ratio $= \frac{π r^{2}}{3 \sqrt{3} r^{2}} = \frac{π}{3 \sqrt{3}} \approx 0.605$ .
ProofHardNCERT Exercise Set 6.3, Q10
Radius $r$ ka ek circle ek regular hexagon ke andar aise banaya gaya hai ki woh chhahon sides ko chhoota hai. Dikhao ki circle hexagon ka $\frac{π}{2 \sqrt{3}} \approx 0.907$ hissa dhakta hai. Questions 8 aur 9 se compare karo: sides ki sankhya badhne ke saath tumhe kya pattern dikhta hai?
Solution dikhao
Centre se bane 6 triangles mein se har ek height $r$ wala equilateral triangle hai. $\frac{\sqrt{3}}{2} a = r$ se side $a = \frac{2 r}{\sqrt{3}}$ hai.
Hexagon ka area $= 6 \times \frac{1}{2} \times a \times r = 3 a r = \frac{6 r^{2}}{\sqrt{3}} = 2 \sqrt{3} r^{2}$ .
Ratio $= \frac{π r^{2}}{2 \sqrt{3} r^{2}} = \frac{π}{2 \sqrt{3}} \approx 0.907$ .
Triangle 0.605, square 0.785, hexagon 0.907: polygon ki jitni zyada sides hon, woh circle se utna hi kas kar chipakta hai aur ratio 1 ke utna hi paas pahunchta hai.

Recap

Summary aur key points

Perimeter kisi shape ke chaaron taraf ki poori length hai. Similar shapes ke liye perimeter ÷ size fixed hota hai (Section 6.1).
Har circle ke liye circumference ÷ diameter same number $π \approx \frac{22}{7} \approx 3.14$ hai, isliye $C = π d = 2 π r$ (Section 6.2).
Andar aur bahar ke polygons $π$ ko trap karte hain: hexagons se $3 < π < 2 \sqrt{3}$ milta hai, aur Archimedes ke 96-gons se $3 \frac{10}{71} < π < 3 \frac{1}{7}$ (Property).
Behtar values Zu Chongzhi ( $\frac{355}{113}$ ), Āryabhaṭa (3.1416) aur Mādhava ki exact series $\frac{π}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \dots$ se aayin (π ki lambi khoj).
$π$ irrational hai: iska decimal na khatam hota hai na repeat hota hai, aur $π \neq \frac{22}{7}$ (Section 6.3).
Jo arc centre par $θ^{\circ}$ subtend karta hai, uski length $2 π r \times \frac{θ}{360}$ hai; $w$ zyada chaudi lane har lap mein $2 π w$ jodti hai (Section 6.4).
Parallelogram ka area base × height hai, side × side nahi (Theorem).
Triangle ka area $\frac{1}{2}$ × base × height hai, aur median use do barabar areas mein baant deta hai (Theorem).
Heron's formula: area $= \sqrt{s (s - a) (s - b) (s - c)}$ , jahan $s = \frac{1}{2} (a + b + c)$ (Section 6.8.1).
Brahmagupta's formula $\sqrt{(s - a) (s - b) (s - c) (s - d)}$ cyclic 4-gon ka area deta hai aur $d = 0$ par Heron's formula ban jaata hai (Cyclic 4-gon ke liye Brahmagupta's formula).
Baudhāyana ne $a b = {(\frac{a + b}{2})}^{2} - {(\frac{a - b}{2})}^{2}$ use karke rectangle ko square banaya (Section 6.9).
Circle ka area $π r^{2} = \frac{1}{2} \times C \times r$ hai, jo Archimedes ke polygons aur Nīlakaṇṭha ki slices se dikhaya gaya (Theorem).
Sector ka area $π r^{2} \times \frac{θ}{360}$ hai; minor segment ek sector minus ek triangle hai (Section 6.10.1).

Flash cards · 10

Formula cards

1 / 10

Chapter ke end ki exercises

$π \approx \frac{22}{7}$ use karo jab tak sawaal kuch aur na kahe. Questions 16 se 27 challenge problems hain: pehle ek saaf figure banao.

Practice · 30 questions

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

1
DiagramMediumNCERT End-of-chapter Exercises Q1
Identity $(a + b) (c + d) = a c + a d + b c + b d$ ko ek area ki tarah dekha ja sakta hai: sides $a + b$ aur $c + d$ wala rectangle chaar chhote rectangles mein bant jaata hai. Isi tarah har identity ke liye area wali picture banao aur use samjhao.
1. $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
  Solution dikhao
  Side $a$ ka square banao. Upar ke saath $a \times b$ ki ek patti aur right side ke saath $a \times b$ ki ek patti hatao. Corner ka $b \times b$ square do baar hat gaya hai, isliye use ek baar wapas jodo.
  Jo bachta hai woh side $a - b$ ka square hai: $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ .
2. $(x + a) (x + b) = x^{2} + (a + b) x + a b$
  Solution dikhao
  Sides $x + a$ aur $x + b$ wala rectangle banao. Har side ke saath ek corner se distance $x$ par lines kheench kar use kaato.
  Chaar tukde hain: ek $x \times x$ square, ek $x \times b$ patti, ek $a \times x$ patti aur ek $a \times b$ rectangle, jo jud kar $x^{2} + (a + b) x + a b$ banate hain.
NumericalEasyNCERT End-of-chapter Exercises Q2
Patang banane wala ek triangular sail kaatta hai jiske do barabar edges 17 cm-17 cm ke hain. Poore sail ke kinare par lagne wala tape 50 cm lamba hai. Sail mein kitna kapda hai?
cm²
Answer dekho
Answer 120 cm²
Base $= 50 - 34 = 16$ . Height base ko aadha karti hai: $h = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15$ . Area $= \frac{1}{2} \times 16 \times 15 = 120 c m^{2}$ . (Heron: $s = 25$ , $\sqrt{25 \times 8 \times 8 \times 9} = 120$ .)
NumericalMediumNCERT End-of-chapter Exercises Q3
Ek triangular jhandi ka chhota edge 24 cm hai, aur uske do lambe edges barabar hain. Ismein 108 cm² kapda lagta hai. Dono lambe edges ke liye kul kitni lambi piping chahiye?
cm
Answer dekho
Answer 30 cm
$\frac{1}{2} \times 24 \times h = 108$ se $h = 9$ . Height chhote edge ko aadha karti hai, isliye har lamba edge $\sqrt{12^{2} + 9^{2}} = \sqrt{225} = 15 c m$ hai, aur piping $2 \times 15 = 30 c m$ hai.
NumericalMediumNCERT End-of-chapter Exercises Q4
Plastic ka set square ek right-angled triangle hai. Uske right angle par ek edge 7 cm lamba hai, aur uske face ka area 84 cm² hai. Uske chaaron taraf kinara lagane ke liye kitni lambi patti chahiye?
cm
Answer dekho
Answer 56 cm
$\frac{1}{2} \times 7 \times b = 84$ se $b = 24$ . Hypotenuse $\sqrt{7^{2} + 24^{2}} = 25$ hai. Perimeter $= 7 + 24 + 25 = 56 c m$ .
NumericalMediumNCERT End-of-chapter Exercises Q5
Ek surveyor ek triangular plot ka ek chakkar lagati hai aur 60 m chalti hai. Uske sketch mein sides $4 : 5 : 6$ ke ratio mein hain. Plot ka area ek decimal place tak sahi nikalo.
m²
Answer dekho
Answer 158.7 m²
$15 k = 60$ se $k = 4$ : sides 16 m, 20 m aur 24 m, aur $s = 30$ . Area $= \sqrt{30 \times 14 \times 10 \times 6} = \sqrt{25 200} = 60 \sqrt{7} \approx 158.7 m^{2}$ .
NumericalMediumNCERT End-of-chapter Exercises Q6
Kya sides 20 cm, 21 cm aur 29 cm wala triangle right-angled hai? Apne answer se uska area $\frac{1}{2} \times$ base $\times$ height ke roop mein nikalo, aur phir Heron's formula se area confirm karo.
Solution dikhao
Answer $210 c m^{2}$
$20^{2} + 21^{2} = 400 + 441 = 841 = 29^{2}$ , isliye Baudhāyana–Pythagoras theorem ke converse se 20 cm aur 21 cm wali sides ke beech ka angle right angle hai. Area $= \frac{1}{2} \times 20 \times 21 = 210 c m^{2}$ .
Heron: $s = 35$ , area $= \sqrt{35 \times 15 \times 14 \times 6} = \sqrt{44 100} = 210 c m^{2}$ . Dono methods ka answer same hai.
NumericalEasyNCERT End-of-chapter Exercises Q7
Road roller ka drum 1.4 m chauda (diameter) hai. Nayi sadak ke 440 m lambe hisse ki poori length par chalne ke liye drum ko kitne chakkar lagane honge?
Answer dekho
Answer 100
Ek chakkar roller ko $\frac{22}{7} \times 1.4 = 4.4 m$ aage le jaata hai, isliye drum $440 \div 4.4 = 100$ chakkar lagata hai.
NumericalMediumNCERT End-of-chapter Exercises Q8
Ek semicircular phoolon ki kyaari ka curved edge 33 m lamba hai. Kyaari ka area nikalo.
m²
Answer dekho
Answer 173.25 m²
$\frac{22}{7} r = 33$ se $r = 10.5$ . Area $= \frac{1}{2} \times \frac{22}{7} \times {10.5}^{2} = 173.25 m^{2}$ .
9
NumericalEasyNCERT End-of-chapter Exercises Q9
Gaon ke kuen ki rassi radius 17.5 cm wale lakdi ke drum par lipatti hai. (Rassi ki motaai ko ignore karo.)
1. Ek chakkar mein kitni rassi lipatti hai, cm mein?
  cm
  Answer dekho
  Answer 110 cm
  $2 \times \frac{22}{7} \times 17.5 = 110 c m$ .
2. Paani drum se 11 m neeche hai. Balti ko upar tak laane mein kitne chakkar lagenge?
  Answer dekho
  Answer 10
  $11 m = 1100 c m$ ; $1100 \div 110 = 10$ chakkar.
10
ProofHardNCERT End-of-chapter Exercises Q10
Same area aur same perimeter: kya isse shape bhi same honi hi chahiye?
1. Do rectangles dono ka area 24 m² aur perimeter 20 m hai. Dikhao ki woh congruent hone hi chahiye. Kya yeh kisi bhi common area aur perimeter ke liye kaam karta hai?
  Solution dikhao
  $l + w = 10$ aur $l w = 24$ . Tab $(l - w)^{2} = (l + w)^{2} - 4 l w = 100 - 96 = 4$ , isliye $l - w = 2$ ( $l \geq w$ lekar). Isliye dono rectangles ke liye $l = 6$ aur $w = 4$ : woh congruent hain.
  Aam taur par $l + w$ aur $l w$ se $(l - w)^{2} = (l + w)^{2} - 4 l w$ fix hota hai, isliye $l - w$ bhi, aur isliye dono sides bhi. Isliye barabar area aur barabar perimeter wale do rectangles hamesha congruent hote hain.
2. Check karo ki sides 20, 21, 29 wale aur sides 17, 25, 28 wale triangles ka perimeter aur area same hai. Kya woh congruent hain?
  Solution dikhao
  Dono perimeters 70 hain, isliye dono ke liye $s = 35$ .
  Pehla: $\sqrt{35 \times 15 \times 14 \times 6} = 210$ . Doosra: $\sqrt{35 \times 18 \times 10 \times 7} = \sqrt{44 100} = 210$ .
  Same perimeter, same area, lekin sides alag, isliye woh congruent nahi hain. Triangles ke liye perimeter aur area milkar bhi shape fix nahi karte.
ProofMediumNCERT End-of-chapter Exercises Q11
Ek trapezium ki parallel sides $a$ aur $b$ ( $a > b$ ) hain aur height $h$ hai. Chhoti side ke ek end se saamne wali tirchhi side ke parallel ek line kheencho. Tukdon se dikhao ki area $\frac{1}{2} (a + b) h$ hai.
Solution dikhao
Line trapezium ko base $b$ aur height $h$ wale parallelogram mein, aur base $a - b$ aur height $h$ wale triangle mein kaat deti hai.
Area $= b h + \frac{1}{2} (a - b) h = \frac{1}{2} (2 b + a - b) h = \frac{1}{2} (a + b) h$ .
ProofMediumNCERT End-of-chapter Exercises Q12
Ek diagonal kheench kar, jo trapezium ko do triangles mein baant deta hai, wahi trapezium formula prove karo.
Solution dikhao
Diagonal ek triangle side $a$ par aur ek side $b$ par banata hai. Har ek ka teesra vertex doosri parallel side par hai, isliye har ek ki height $h$ hai.
Area $= \frac{1}{2} a h + \frac{1}{2} b h = \frac{1}{2} (a + b) h$ .
ProofMediumNCERT End-of-chapter Exercises Q13
Kaagaz ke do bilkul same trapezium kaato. Ek ko ulta karo aur doosre ke paas aise rakho ki dono milkar ek parallelogram banayein. Is parallelogram ka base aur height kya hain? Inse trapezium formula teesri baar prove karo.
Solution dikhao
Copy ko aadha chakkar ghumao aur ek tirchhi side ke saath jod do. Ek copy ki lambi side $a$ , doosri ki chhoti side $b$ ke saath ek line mein aa jaati hai, upar aur neeche dono taraf.
Result ek parallelogram hai jiska base $a + b$ aur height $h$ hai, area $(a + b) h$ . Ek trapezium iska aadha hai: $\frac{1}{2} (a + b) h$ .
14
ProofHardNCERT End-of-chapter Exercises Q14
Ek 4-gon ke do diagonals ki lengths $d_{1}$ aur $d_{2}$ hain aur woh right angle par kaatte hain, jaise har kite aur har rhombus mein. Do tareekon se prove karo ki iska area $\frac{1}{2} d_{1} d_{2}$ hai:
1. ise ek diagonal ke saath do triangles mein kaat kar
  Solution dikhao
  Diagonal $d_{1}$ ke saath base $d_{1}$ wale do triangles mein baanto. Unki heights woh do hisse $p$ aur $q$ hain jinme $d_{1}$ doosre diagonal ko kaatta hai, isliye $p + q = d_{2}$ .
  Area $= \frac{1}{2} d_{1} p + \frac{1}{2} d_{1} q = \frac{1}{2} d_{1} (p + q) = \frac{1}{2} d_{1} d_{2}$ .
2. iske chaaron corners se hokar jaane wala, diagonals ke parallel sides wala rectangle bana kar
  Solution dikhao
  Woh rectangle banao jiski sides chaaron vertices se hokar diagonals ke parallel jaati hain. Yeh $d_{1}$ by $d_{2}$ ka hai.
  Diagonals is rectangle ko chaar chhote rectangles mein kaatte hain, aur har ek ko 4-gon ki ek side aadha-aadha kaat deti hai. Isliye 4-gon bade rectangle ka bilkul aadha hissa dhakta hai: $\frac{1}{2} d_{1} d_{2}$ .
15
Short answerMediumNCERT End-of-chapter Exercises Q15
Bada karna aur copies fit karna.
1. Sides $a$ aur $b$ wale rectangle ko bada karke sides $3 a$ aur $3 b$ kar diya jaata hai. Area kitne guna hai? Kya chhote rectangle ki utni copies use bhar sakti hain?
  Solution dikhao
  $3 a \times 3 b = 9 a b$ : 9 guna area. Haan: 9 copies 3-3 ki 3 rows mein use bhar deti hain.
2. Sides $a$ , $b$ , $c$ wale triangle ko bada karke $4 a$ , $4 b$ , $4 c$ kiya jaata hai. Heron's formula se nikalo ki area kitne guna badhta hai. Dikhao ki chhote triangle ki utni copies bade triangle ko kaise bharti hain.
  Solution dikhao
  $s (s - a) (s - b) (s - c)$ ka har factor 4 guna ho jaata hai, isliye product $4^{4} = 256$ guna aur uska square root 16 guna badhta hai.
  Bade triangle ki har side ko 4 barabar hisson mein baanto aur division points se sides ke parallel lines kheencho. Isse chhote triangles ki rows banti hain, har ek ki sides $a$ , $b$ , $c$ : kul $1 + 3 + 5 + 7 = 16$ , kuch seedhe aur kuch ulte. Isliye 16 copies use bilkul bhar deti hain.
3. Circle ka radius double kiya jaata hai. Dikhao ki uska area 4 guna ho jaata hai. Kya chhote circle ki 4 copies use bhar sakti hain?
  Solution dikhao
  $π (2 r)^{2} = 4 π r^{2}$ . Lekin circles tile nahi kar sakte: tum unhe kaise bhi rakho, gaps bachte hain. Asal mein bade circle ke andar chhote circles mein se sirf do hi bina overlap ke fit hote hain; sikkon ke saath try karo. Copies fit na hon tab bhi area 4 guna badh sakta hai.
16
DiagramHardNCERT End-of-chapter Exercises Q16
Har figure ka kitna hissa shaded hai?
Fig. 6.24
Figure (a): ek triangle jiski har side tick marks se teen barabar hisson mein bati hai; har corner ke paas wale division points ko jodne se teen corner triangles kat jaate hain, aur ek shaded hexagon bachta hai. Figure (b): ek square ABCD jismein E, AB ka midpoint hai aur F, BC ka midpoint; triangle DEF shaded hai.
1. Woh hexagon jo har side ko teen barabar hisson mein baantne wale points par corners kaat kar banta hai.
  Answer dekho
  Answer 0.6667
  $A$ ke paas wale division points ko $P$ ( $A B$ par) aur $Q$ ( $A C$ par) kaho. Triangle $A P Q$ ka base $A Q = \frac{1}{3} A C$ hai aur $P$ se height wahi hai jo triangle $A P C$ ki, isliye yeh $A P C$ ka $\frac{1}{3}$ hai. Triangle $A P C$ ka base $A P = \frac{1}{3} A B$ hai aur $C$ se height wahi hai jo $A B C$ ki, isliye yeh $A B C$ ka $\frac{1}{3}$ hai. Har corner $\frac{1}{9}$ hai, teen corners $\frac{1}{3}$ hain, aur hexagon $\frac{2}{3}$ hai.
2. Square $A B C D$ mein triangle $D E F$ , jahan $E$ aur $F$ , $A B$ aur $B C$ ke midpoints hain.
  Answer dekho
  Answer 0.375
  Side 2 lo. Teen safed triangles ke areas $\frac{1}{2} \times 1 \times 2 = 1$ , $\frac{1}{2} \times 1 \times 1 = 0.5$ aur $1$ hain. Shaded triangle $4 - 2.5 = 1.5$ hai, jo square ka $\frac{1.5}{4} = \frac{3}{8}$ hai.
17
NumericalMediumNCERT End-of-chapter Exercises Q17
Barabar size ke circles ek square ke andar square grid mein bhare gaye hain, har circle apne padosiyon aur square ki sides ko chhoota hai. Circles square ka kitna hissa dhakte hain jab:
1. 1 circle ho
  Answer dekho
  Answer 0.785
  Side $2 r$ wale square mein radius $r$ ka ek circle: $\frac{π r^{2}}{4 r^{2}} = \frac{π}{4} \approx 0.785$ .
2. 4 circles hon, 2-2 ki 2 rows mein
  Answer dekho
  Answer 0.785
  Har circle square ke apne chauthai hisse mein hai aur uska $\frac{π}{4}$ dhakta hai, isliye phir se $\frac{π}{4} \approx 0.785$ .
3. 9 circles hon, 3-3 ki 3 rows mein
  Answer dekho
  Answer 0.785
  Nau chhote squares, har ek ka $\frac{π}{4}$ dhaka hua: phir bhi $\frac{π}{4}$ .
ProofMediumNCERT End-of-chapter Exercises Q18
Is tarah bhare circles ke $n \times n$ grid se square ka kitna hissa dhakta hai, predict karo. Apna prediction 100 circles (10-10 ki 10 rows) aur 2500 circles ke liye check karo, aur phir samjhao ki yeh har $n$ ke liye sach kyun hai.
Solution dikhao
Conjecture: fraction hamesha $\frac{π}{4} \approx 78.5 %$ hai, circles chahe kitne bhi hon.
Proof: agar har circle ka radius $r$ hai, to square ki side $2 n r$ aur area $4 n^{2} r^{2}$ hai. Circles $n^{2} π r^{2}$ dhakte hain. Fraction $\frac{n^{2} π r^{2}}{4 n^{2} r^{2}} = \frac{π}{4}$ hai; $n^{2}$ cancel ho jaata hai. Isliye 100 circles ( $n = 10$ ) aur 2500 circles ( $n = 50$ ) dono $\frac{π}{4}$ dhakte hain.
(Rows ko khiska kar, honeycomb ki tarah rakhne se aur behtar hota hai, lagbhag 90.7%: Practice 6.3 ke question 10 se compare karo.)
19
DiagramHardNCERT End-of-chapter Exercises Q19
Ek carpenter paanch bilkul same lakdi ki tiles se ek rectangular panel banata hai, jaisa dikhaya gaya hai: upar ki row mein teen tiles seedhi khadi hain aur neeche ki row mein do leti hui hain. Panel 270 cm² dhakta hai.
Fig. 6.25
Paanch bilkul same rectangular tiles se bana ek rectangular panel. Upar ki row mein teen tiles ek ke paas ek seedhi khadi hain; neeche ki row mein do tiles ek ke aage ek leti hain. Dono rows ki kul chaudai same hai. Label par area 270 cm² likha hai.
1. Ek tile ke chaaron taraf ek baar kitni lambi edging lagti hai?
  cm
  Answer dekho
  Answer 30 cm
  Teen khadi tiles ki chaudai do leti tiles ki length ke barabar hai: $3 w = 2 l$ , isliye $l = 1.5 w$ . Paanch tiles $5 \times 1.5 w^{2} = 7.5 w^{2} = 270$ dhakti hain, isliye $w^{2} = 36$ , $w = 6$ aur $l = 9$ . Ek tile: $2 (9 + 6) = 30 c m$ .
2. Poore panel ke chaaron taraf ek baar kitni lambi edging lagti hai?
  cm
  Answer dekho
  Answer 66 cm
  Panel $3 w = 18 c m$ chauda aur $l + w = 15 c m$ ooncha hai (check: $18 \times 15 = 270$ ). Perimeter $= 2 (18 + 15) = 66 c m$ .
ProofHardNCERT End-of-chapter Exercises Q20
Chaar behnein ek triangular khet $A B C$ baant leti hain. Corner $A$ ke gate se teen seedhi fences saamne wali side $B C$ tak jaati hain aur use $P$ , $Q$ aur $R$ par milti hain, aise ki $B P = P Q = Q R = R C$ . Dikhao ki har behan ko same area milta hai. Phir samjhao ki ek behan ke plot ko tukdon mein kaat kar kaise jamaya ja sakta hai ki woh doosri behan ke plot ko bilkul dhak le.
Solution dikhao
Chaar plots triangles $A B P$ , $A P Q$ , $A Q R$ aur $A R C$ hain. $B C$ par unke bases barabar hain (har ek $B C$ ka chauthai) aur unki height same hai, $A$ se $B C$ tak ki distance. Isliye unke areas barabar hain.
Dobara jamane ke liye: plot ko uski do tirchhi sides ke midpoints ko jodne wali line ke saath kaato. Upar ke chhote tukde ko unmein se ek midpoint ke around aadha chakkar ghumao; woh neeche wale tukde ke saath fit ho jaata hai, aur dono milkar same base aur aadhi height wala parallelogram banate hain.
Doosre plot ke saath bhi yahi karo. Dono parallelograms ka base aur height same hai, isliye ek right triangle kaat kar use khiskane se, jaise Fig. 6.12 mein (agar koi bahut jhuka ho to dohra kar), ek doosre mein badal jaata hai.
ProofHardNCERT End-of-chapter Exercises Q21
Radius $r$ wale circle ke andar ek square aise banaya gaya hai ki uske chaaron corners circle par hain. Square ki har side par bahar ki taraf ek semicircle banaya gaya hai. Jo chaar aadhe-chaand jaise hisse (lunes) semicircles ke andar lekin circle ke bahar hain, woh shaded hain. Dikhao ki chaaron lunes ka kul area bilkul square ke area ke barabar hai. (Chios ke Hippocrates ne lagbhag 440 BCE mein aise results dhoondhe the: curved edges wale aise hisse jinke areas polygons ke areas ke barabar hain.)
Fig. 6.26
Centre O wala ek circle, aur ek orange square ke ek corner tak ek dashed radius r; square ke chaaron corners circle par hain. Square ki har side par ek semicircle circle ke bahar tak ubhra hua hai. Semicircles aur circle ke beech ke chaar green aadhe-chaand jaise hisse lunes hain; ek par lune likha hai.
Solution dikhao
Square ka diagonal ek diameter, $2 r$ , hai. Agar side $s$ hai, to $s^{2} + s^{2} = (2 r)^{2}$ , isliye $s^{2} = 2 r^{2}$ : square ka area $2 r^{2}$ hai.
Har semicircle ka radius $\frac{s}{2}$ hai, isliye uska area $\frac{1}{2} π \cdot \frac{s^{2}}{4} = \frac{π \cdot 2 r^{2}}{8} = \frac{π r^{2}}{4}$ hai. Chaaron semicircles ka kul area $π r^{2}$ hai.
Square ke bahar ki disc chaar tukdon se bani hai jinka kul area $π r^{2} - 2 r^{2}$ hai. Har semicircle ek lune plus in tukdon mein se ek hai.
Isliye chaar lunes $=$ chaar semicircles $-$ chaar tukde $= π r^{2} - (π r^{2} - 2 r^{2}) = 2 r^{2}$ , bilkul square ka area. $π$ cancel ho jaate hain!
22
NumericalHardNCERT End-of-chapter Exercises Q22
Ek square farsh ki tile ki side 10 cm hai. Har corner par, usi corner par centre wala radius 5 cm ka ek quarter circle paint kiya gaya hai. Tile ke beech mein bina paint wale, chaar nokon wale hisse ka area aur perimeter nikalo.
1. area ( $π \approx 3.14$ use karo)
  cm²
  Answer dekho
  Answer 21.5 cm²
  Chaaron quarter discs milkar area $π \times 5^{2} = 25 π \approx 78.5 c m^{2}$ ki ek disc banate hain. Beech ka hissa $100 - 78.5 = 21.5 c m^{2}$ hai.
2. perimeter ( $π \approx 3.14$ use karo)
  cm
  Answer dekho
  Answer 31.4 cm
  Iski boundary chaar quarter arcs hain, jo milkar radius 5 ka ek poora circle banate hain: $2 π \times 5 = 10 π \approx 31.4 c m$ .
NumericalHardNCERT End-of-chapter Exercises Q23
Metal ka washer same centre wale do circles ke beech ki flat ring hai, aur iska area $154 m m^{2}$ hai. Bahar wale circle ki ek seedhi chord aise kheenchi jaati hai ki woh andar wale circle ko bas chhoo le. Chord kitni lambi hai? ( $π \approx \frac{22}{7}$ use karo. Tumhe koi bhi radius nahi chahiye, aur tum use nikal bhi nahi sakte.)
mm
Answer dekho
Answer 14 mm
Agar radii $R$ aur $r$ hain, to chhoone wale point tak ka radius chord par perpendicular hai aur use aadha karta hai, isliye length $l$ wali chord ke liye $R^{2} - r^{2} = {(\frac{l}{2})}^{2}$ . Area se $\frac{22}{7} (R^{2} - r^{2}) = 154$ milta hai, isliye $R^{2} - r^{2} = 49$ , $\frac{l}{2} = 7$ aur $l = 14 m m$ .
ProofMediumNCERT End-of-chapter Exercises Q24
Legs $a$ , $b$ aur hypotenuse $c$ wale right-angled triangle ki har side par bahar ki taraf ek equilateral triangle banaya gaya hai. Dikhao ki hypotenuse wale triangle ka area baaki do ke sum ke barabar hai.
Solution dikhao
Teeno areas $\frac{\sqrt{3}}{4} a^{2}$ , $\frac{\sqrt{3}}{4} b^{2}$ aur $\frac{\sqrt{3}}{4} c^{2}$ hain.
Kyunki $a^{2} + b^{2} = c^{2}$ , $\frac{\sqrt{3}}{4}$ se multiply karne par result milta hai. Teeno sides par bani ek hi tarah ki kisi bhi shapes ke liye yahi kaam karta hai, jaise semicircles.
ProofHardNCERT End-of-chapter Exercises Q25
Radius $r$ wale do circles ke centres $r \sqrt{2}$ door hain. Dono circles ke common hisse ka area nikalo.
Solution dikhao
Answer $r^{2} (\frac{π}{2} - 1)$
Maan lo centres $K$ aur $L$ hain aur kaatne wale points $P$ aur $Q$ . Tab $K P = P L = L Q = Q K = r$ aur $K L = r \sqrt{2}$ .
Kyunki $r^{2} + r^{2} = (r \sqrt{2})^{2}$ , $P$ par angle right angle hai, aur right angle wala rhombus square hota hai. Isliye $∠ P K Q = 90 °$ .
Common hissa do segments hai, har ek chord $P Q$ se 90° ke sector mein se kata hua: har ek $\frac{π r^{2}}{4} - \frac{r^{2}}{2}$ hai.
Total $= 2 (\frac{π r^{2}}{4} - \frac{r^{2}}{2}) = r^{2} (\frac{π}{2} - 1) \approx 0.571 r^{2}$ .
26
NumericalHardNCERT End-of-chapter Exercises Q26
$P$ rectangle $A B C D$ ke andar ek point hai. Triangles $P A B$ , $P B C$ aur $P C D$ ke areas 8 cm², 10 cm² aur 12 cm² hain. Triangle $P D A$ ka aur rectangle ka area nikalo.
1. rectangle, cm² mein
  cm²
  Answer dekho
  Answer 40 cm²
  Area $P A B$ + area $P C D$ rectangle ka aadha hai: $8 + 12 = 20$ , isliye rectangle $40 c m^{2}$ hai.
2. triangle $P D A$ , cm² mein
  cm²
  Answer dekho
  Answer 10 cm²
  Area $P B C$ + area $P D A$ bhi 20 hai, isliye area $P D A = 20 - 10 = 10 c m^{2}$ .
ProofHardNCERT End-of-chapter Exercises Q27
$B$ segment $A C$ par ek point hai. $A B$ , $B C$ aur $A C$ par semicircles banaye gaye hain, sab ek hi taraf. Jo hissa bade semicircle ke andar lekin dono chhote semicircles ke bahar hai, use arbelos (mochi ka chaaku) kehte hain. $B$ par $A C$ ka perpendicular bade semicircle ko $D$ par milta hai. Dikhao ki arbelos ka area diameter $B D$ wale circle ke area ke barabar hai.
Fig. 6.27
Ek segment AC jis par ek point B hai. AC par ek bada semicircle aur AB aur BC par do chhote semicircles ek hi taraf bane hain; unke beech ka shaded hissa arbelos hai. B se ek vertical segment bade semicircle ko D par milta hai, aur ek dashed circle ka diameter BD hai.
Solution dikhao
$A B = 2 a$ aur $B C = 2 b$ ke saath radii $a$ , $b$ aur $a + b$ hain. Arbelos $= \frac{π}{2} [(a + b)^{2} - a^{2} - b^{2}] = π a b$ .
Bade semicircle ka centre $O$ , $A C$ ka midpoint hai, isliye $O B = ∣ a - b ∣$ aur $O D = a + b$ . Right triangle $O B D$ mein $B D^{2} = (a + b)^{2} - (a - b)^{2} = 4 a b$ .
Diameter $B D$ wale circle ka area $π {(\frac{B D}{2})}^{2} = π \cdot \frac{4 a b}{4} = π a b$ hai, arbelos jitna hi. Dhyaan do ki Section 6.9 wali identity phir se kaam kar rahi hai.
Quick answerEasy
$π$ ke baare mein kaun sa statement sahi hai?
$π = \frac{22}{7}$ bilkul exact $π$ irrational hai, aur $\frac{22}{7}$ sirf ek approximation hai $π$ circle ke size par depend karta hai $π = 3.14$ bilkul exact
Answer dekho
Answer (B) $π$ irrational hai, aur $\frac{22}{7}$ sirf ek approximation hai
C/D ratio har circle ke liye same hai, aur yeh irrational hai, isliye koi bhi fraction ya khatam hone wala decimal iske barabar nahi hai.
Quick answerEasy
Ek sector ka radius 6 cm aur angle 60° hai. Iska area $k π c m^{2}$ hai. $k$ nikalo.
Answer dekho
Answer 6
$\frac{60}{360} \times π \times 36 = 6 π$ , isliye $k = 6$ .
30
Quick answerMedium
Har region ko uske area formula se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
base $b$ aur height $h$ wala triangle : $\frac{1}{2} b h$
base $b$ aur height $h$ wala parallelogram : $b h$
radius $r$ ka circle : $π r^{2}$
radius $r$ , angle $θ^{\circ}$ ka sector : $π r^{2} \times \frac{θ}{360}$
sides $a$ , $b$ , $c$ wala triangle : $\sqrt{s (s - a) (s - b) (s - c)}$
Yeh chapter ke paanch area formulas hain. Heron's formula ko sirf teen sides chahiye.

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 6.1, Q1	Practice 6.1 · Q1
NCERT Exercise Set 6.1, Q2	Practice 6.1 · Q2
NCERT Exercise Set 6.1, Q3	Practice 6.1 · Q3
NCERT Exercise Set 6.1, Q4	Practice 6.1 · Q4
NCERT Exercise Set 6.1, Q5	Practice 6.1 · Q5
NCERT Exercise Set 6.1, Q6	Practice 6.1 · Q6
NCERT Exercise Set 6.1, Q7	Practice 6.1 · Q7
NCERT Exercise Set 6.1, Q8	Practice 6.1 · Q8
NCERT Exercise Set 6.2, Q1	Practice 6.2 · Q1
NCERT Exercise Set 6.2, Q2	Practice 6.2 · Q2
NCERT Exercise Set 6.2, Q3	Practice 6.2 · Q3
NCERT Exercise Set 6.2, Q4	Practice 6.2 · Q4
NCERT Exercise Set 6.2, Q5	Practice 6.2 · Q5
NCERT Exercise Set 6.2, Q6	Practice 6.2 · Q6
NCERT Exercise Set 6.2, Q7	Practice 6.2 · Q7
NCERT Exercise Set 6.2, Q8	Practice 6.2 · Q8
NCERT Exercise Set 6.2, Q9	Practice 6.2 · Q9
NCERT Exercise Set 6.2, Q10	Practice 6.2 · Q10
NCERT Exercise Set 6.2, Q11	Practice 6.2 · Q11
NCERT Exercise Set 6.3, Q1	Practice 6.3 · Q1
NCERT Exercise Set 6.3, Q2	Practice 6.3 · Q2
NCERT Exercise Set 6.3, Q3	Practice 6.3 · Q3
NCERT Exercise Set 6.3, Q4	Practice 6.3 · Q4
NCERT Exercise Set 6.3, Q5	Practice 6.3 · Q5
NCERT Exercise Set 6.3, Q6	Practice 6.3 · Q6
NCERT Exercise Set 6.3, Q7	Practice 6.3 · Q7
NCERT Exercise Set 6.3, Q9	Practice 6.3 · Q8
NCERT Exercise Set 6.3, Q8	Practice 6.3 · Q9
NCERT Exercise Set 6.3, Q10	Practice 6.3 · Q10
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
NCERT End-of-chapter Exercises Q23	Chapter ke end ki exercises · Q23
NCERT End-of-chapter Exercises Q24	Chapter ke end ki exercises · Q24
NCERT End-of-chapter Exercises Q25	Chapter ke end ki exercises · Q25
NCERT End-of-chapter Exercises Q26	Chapter ke end ki exercises · Q26
NCERT End-of-chapter Exercises Q27	Chapter ke end ki exercises · Q27

Quick check: MCQs

6.1 Shape ka Perimeter3 questions
π ki lambi khoj4 questions
6.3 π Irrational Hai3 questions
Running track ko aur dhyaan se dekhna4 questions
6.5 Perimeter par Problems, Puzzles aur Paradoxes3 questions
6.6 Rectangle ka Area3 questions
6.7 Parallelogram ka Area3 questions
6.8 Triangle ka Area4 questions
6.8.1 Heron's formula3 questions
Cyclic 4-gon ke liye Brahmagupta's formula3 questions
6.9 Rectangle ko Square Banana3 questions
Archimedes aur Nīlakaṇṭha: area πr² kyun hai4 questions
6.10.1 Circle ke Sector ka Area3 questions

Glossary · 14

Key terms (English + Hindi)

arcचापchaap: Circle ke do points ke beech ka tukda. $θ^{\circ}$ subtend karne wale arc ki length $2 π r \times \frac{θ}{360}$ hai.Chapter mein pehli baar yahan aaya →
areaक्षेत्रफलkshetraphal: Kisi shape se dhaki flat surface ki maatra, square units mein naapi jaati hai.Chapter mein pehli baar yahan aaya →

circumferenceपरिधिparidhi: Circle ka perimeter: $C = π d = 2 π r$ .Chapter mein pehli baar yahan aaya →
cyclic 4-gonचक्रीय चतुर्भुजchakriya chaturbhuj: Aisa quadrilateral jiske chaaron vertices ek hi circle par hon.Chapter mein pehli baar yahan aaya →

generalisationव्यापकीकरणvyapakikaran: Aisa result jo doosre ko special case ki tarah apne andar rakhta hai, jaise Brahmagupta's formula mein Heron's formula aata hai.Chapter mein pehli baar yahan aaya →

irrational numberअपरिमेय संख्याaparimey sankhya: Aisa number jise do integers ke fraction ki tarah nahi likha ja sakta; iska decimal na khatam hota hai na repeat hota hai.Chapter mein pehli baar yahan aaya →

medianमाध्यिकाmadhyika: Triangle ke ek corner ko saamne wali side ke beech wale point se jodne wala segment. Yeh area ko aadha karta hai.Chapter mein pehli baar yahan aaya →

perimeterपरिमापparimap: Kisi closed shape ki boundary ki total length.Chapter mein pehli baar yahan aaya →

sectorत्रिज्यखंडtrijyakhand: Disc ka do radii aur unke arc ke beech ka hissa; area $π r^{2} \times \frac{θ}{360}$ .Chapter mein pehli baar yahan aaya →
segmentवृत्तखंडvrittakhand: Disc ka ek chord aur uske arc ke beech ka hissa.Chapter mein pehli baar yahan aaya →
semi-perimeterअर्धपरिमापardhparimap: Perimeter ka aadha; triangle ke liye $s = \frac{1}{2} (a + b + c)$ .Chapter mein pehli baar yahan aaya →
special caseविशेष स्थितिvishesh sthiti: Kisi zyada general result mein ek extra condition, jaise $b = a$ , jod kar mila result.Chapter mein pehli baar yahan aaya →
stagger: Running track par paas-paas wali lanes ke starting marks ke beech ki distance; ek lap ke liye, $w$ chaudi lanes mein yeh $2 π w$ hai.Chapter mein pehli baar yahan aaya →

π (pi)पाईpai: Har circle ka fixed ratio circumference ÷ diameter, lagbhag 3.14159. Yeh irrational hai.Chapter mein pehli baar yahan aaya →

Runners alag-alag jagah se start kyun karte hain?

6.1 Shape ka Perimeter

6.2 Circle ka Perimeter — C/D Ratio

π ki lambi khoj

6.3 π Irrational Hai

6.4 Circle ke Arc ki Length

Running track ko aur dhyaan se dekhna

6.5 Perimeter par Problems, Puzzles aur Paradoxes

6.6 Rectangle ka Area

6.7 Parallelogram ka Area

6.8 Triangle ka Area

6.8.1 Heron's formula

Cyclic 4-gon ke liye Brahmagupta's formula

6.9 Rectangle ko Square Banana

6.10 Circle ka Area

Archimedes aur Nīlakaṇṭha: area πr² kyun hai

6.10.1 Circle ke Sector ka Area

Chapter ke end ki exercises

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