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Ganita ManjariCh 12 Quadrilaterals

Quadrilaterals: Class 9 Maths Chapter 12 Hinglish mein

Parallelograms, midpoints, medians aur tilings

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

Published · Updated

  • 70 min
  • 1 simulation
  • 45 practice questions
  • 19 MCQs

Is chapter mein aap kya seekhenge

  • Quadrilateral ki precise definition dena aur convex, non-convex, self-intersecting aur non-planar 4-gons mein farq pehchanna
  • Parallelogram ki properties aur unke converses prove karna, aur parallelogram ke paanch tests use karna
  • Midpoint Theorem aur uska converse prove karna aur apply karna
  • Prove karna ki triangle ke medians centroid par milte hain, jo har median par vertex se do-tihaai doori par hota hai
  • Dikhana ki kisi bhi quadrilateral ki sides ke midpoints ek parallelogram banate hain, aur diagonals se uski sides nikaalna
  • Angle sum aur half-turns ki madad se samjhana ki kisi bhi quadrilateral ki copies plane ko tile kyun karti hain

NCERT Mathematics Class 9, Chapter 12, pp. 51–84 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Floors, tiles aur 4-gons

Railway platforms, school corridors aur kitchen floors par aksar chaar sides wali tiles edge se edge mila kar lagi hoti hain. Chaar seedhi sides wali shape ko quadrilateral kehte hain, ise 4-gon bhi kehte hain. Jab ek hi shape ki copies floor ko bina gaps aur bina overlaps ke dhak dein, to woh ek tiling banati hain; socho ki floor hamesha ke liye aage chalta jaata hai, to tum plane ko tile kar rahe ho.

Ek case aasaan hai: parallelogram. Iski opposite sides equal aur parallel hoti hain, isliye ek side ke saath slide ki gayi copy pehli tile se bilkul chipak kar fit ho jaati hai. Aisi copies ki ek line ek row banati hai jiske upar aur neeche ke edges seedhe hote hain (Fig. 12.1). Phir rows bhi isi tarah rows ke saath fit ho jaati hain. Isliye kisi bhi parallelogram ki copies plane ko tile karti hain.

3/3Rows, rows ke saath fit hoti hain, isliye parallelogram ki copies plane ko bina gaps ke dhak leti hain.

Fig. 12.1Ek parallelogram ki rows plane ko dhak leti hain.NCERT Fig. 12.1 · p. 52

Step 1 mein ek akeli parallelogram tile hai, jis par tick marks aur arrow marks dikhate hain ki uski opposite sides equal aur parallel hain. Step 2 mein chaar copies bottom side ki direction mein slide karke ek row banayi gayi hai, jiske upar aur neeche ke edges seedhi lines hain. Step 3 mein aisi chaar rows ek ke upar ek rakhi hain, har row tile ke slant jitni side mein khiski hui hai, aur poora region bina gaps ke dhak gaya hai.

Raaste mein hum parallelograms ko dhyaan se padhenge: kuch measurements se ek parallelogram kaise pehchanein, aur kaise yeh triangles aur quadrilaterals ke baare mein kuch hairaan karne wale facts kholte hain. Phir hum tile factory par wapas aayenge.

12.1 Quadrilateral Asal Mein Kya Hai?

Quadrilateral ko sab dekhte hi pehchan lete hain. Lekin ek aisa rule likh kar dekho jo computer ke liye kaafi precise ho: chaar points aur chaar segments diye hain, to figure quadrilateral hai ya nahi? Fig. 12.2 ke saat figures dekho. Har ek ki chaar seedhi sides hain. Kya tum har ek ko quadrilateral kahoge?

5/5Sirf normal shape aur dent wala arrowhead bachte hain. Dono quadrilaterals hain.

Fig. 12.2Teen rules tay karte hain ki kaun si chaar sides wali figures quadrilateral maani jaati hain.NCERT Fig. 12.2 · p. 52

Saat chhoti figures, har ek chaar seedhe segments se bani: ek normal chaar sides wali shape, ek triangle jiske base par ek extra corner hai, ek shape jiski sides ek doosre ke upar wapas mud jaati hain, ek crossed bow-tie, ek dent wala arrowhead, ek khula zigzag jo band nahi hota, aur ek card jo diagonal ke saath moda gaya hai taaki uske dono half alag planes mein hon. Step 2 mein bow-tie aur khule zigzag ko cross kiya gaya hai. Step 3 mein un do figures ko bhi cross kiya gaya hai jinke teen corners ek line mein hain. Step 4 mein mude hue card ko bhi cross kiya gaya hai. Step 5 mein bache hue do figures, normal shape aur arrowhead, highlight kiye gaye hain, jo dono quadrilaterals hain.

Triangles aasaan hain: koi bhi teen points jo ek line par nahi hain, ek triangle dete hain (ek line par teen points sirf ek segment dete hain). Chaar points A, B, C, D ko A→B→C→D→A order mein jodne par figure teen alag tareekon se fail ho sakti hai.

  1. Loop khula hai, ya do sides cross karti hain. Humein ek band loop chahiye jismein do sides sirf apne shared corner par chhuein. Bow-tie, jiski sides beech mein cross karti hain, self-intersecting 4-gon kehlata hai; hum use chhod dete hain.
  2. Teen corners ek line par hain. Tab do padosi sides usi line par padti hain, isliye figure asal mein ek extra point wala triangle hai, ya do sides ek doosre ke upar wapas mud jaati hain. Hum ise allow nahi karte: chaar points mein se koi bhi teen collinear nahi hone chahiye.
  3. Corners ek plane mein nahi hain. Ek square card ko diagonal ke saath fold karo aur dono halves ko alag jhukao: ab chaaron corners do alag planes mein hain. Aisi figure non-planar 4-gon hai. Mude hue 4-gons sirf chapter ke end ki ek-do exercises mein aate hain.

Teeno conditions ko saath rakhne par ek precise definition milti hai.

Definition
Quadrilateral

Ek plane mein chaar alag points A, B, C, D lo, jinmein se koi teen ek line par na hon. Chaar segments AB, BC, CD aur DA quadrilateral ABCD banate hain jab in segments mein se kisi do ka shared corner ke alawa koi bhi point common na ho.

ExampleFig. 12.2 mein dent wala arrowhead quadrilateral hai; bow-tie nahi hai, kyunki uski do sides cross karti hain.

Parts ke kuch naam. Quadrilateral ABCD mein points A, B, C, D uske vertices (corners) hain, aur uski chaar sides, jinhe edges bhi kehte hain, AB, BC, CD aur DA hain. Segments AC aur BD, jo opposite corners ko jodte hain, uske diagonals hain. Ek side ke do ends, jaise A aur B, adjacent vertices hain. Common end wali do sides, jaise AB aur BC, adjacent sides hain. Har vertex par wahan milne wali do sides ek internal angle banati hain, isliye chaar internal angles hote hain; adjacent vertices wale angles adjacent angles hain. Opposite sides aur opposite angles ko tum kaise describe karoge? Practice 12.1 ka Q1 try karo.

Ek quadrilateral ke kai naam hote hain. Uski boundary par ek baar chakkar lagao, chaaron corners mein se kisi se bhi shuru karke aur kisi bhi direction mein, aur letters padhte jao. Clockwise aur anticlockwise mein tumhe ABCD, BCDA, CDAB, DABC, ADCB, DCBA, CBAD aur BADC milte hain: yaani ek shape ke 4×2=8 naam. ACBD jaisa naam allowed nahi hai, kyunki woh diagonal par koodta hai, aur A, C, B, D ko is order mein jodne se ek alag figure banti hai.

Ab se quadrilateral ka matlab ek flat quadrilateral hai jiski sides cross nahi karti, jab tak hum kuch aur na kahein.

Convex quadrilaterals

Fig. 12.2 ke dono bache hue figures quadrilaterals hain, lekin arrowhead ek baat mein alag hai: uska ek corner andar ki taraf dhakela hua hai, jisse ek dent banta hai. Dent par shape ka andar wala hissa corner ke around lipat jaata hai, isliye wahan internal angle 180∘ se zyada hota hai (ek reflex angle). Quadrilateral convex hota hai jab uske chaaron internal angles 180∘ se kam hon, aur non-convex hota hai jab unmein se ek 180∘ se zyada ho.

2/3PQRS ka har angle 180∘ se kam hai. WXYZ mein Z par angle 254∘ hai. Dono sets ka total 360∘ hai.

Fig. 12.3Convexity ke do tests: angles, aur diagonals.NCERT Figs. 12.4–12.5 · p. 54

Do quadrilaterals aamne-saamne: left mein ek convex PQRS aur right mein ek arrowhead WXYZ, jismein Z par dent hai. Step 2 mein har internal angle mark kiya gaya hai: PQRS ke angles 100, 73, 99 aur 88 degrees hain, sab 180 se kam; WXYZ ke angles 27, 53 aur 26 degrees hain aur Z par 254 degrees ka reflex angle hai. Dono sets ka total 360 degrees hai. Step 3 mein diagonals khiche gaye hain: PQRS mein woh shape ke andar ek point par cross karte hain; WXYZ mein diagonal WY, Z ke neeche shape ke bahar se jaata hai aur dono diagonals kabhi nahi milte.

Doosre test ke liye protractor nahi chahiye: dono diagonals draw karo. Convex quadrilateral mein diagonals shape ke andar ek doosre ko cross karte hain. Non-convex mein ek diagonal shape ke bahar hota hai aur dono diagonals bilkul nahi milte. Koi bhi test bata deta hai ki tumhare paas kaun sa type hai. Neeche wale lab mein dono try karo.

InteractiveQuadrilateral lab

Ek grid jis par quadrilateral ABCD hai, jiske chaaron corners drag kiye ja sakte hain ya arrow keys se move kiye ja sakte hain. Neeche ek message shape ko convex, non-convex (reflex angle wale corner ka naam batakar), self-intersecting, ya teen corners ek line mein hon to quadrilateral nahi, classify karta hai. Ek card chaaron sides, dono diagonals, chaaron internal angles aur unka sum list karta hai; doosra card batata hai ki opposite sides parallel ya equal hain ya nahi aur diagonals ek doosre ko bisect karte hain ya nahi. Ek checkbox sides ke midpoints P, Q, R aur S ko jodta hai, aur buttons ek parallelogram, ek trapezium, ek dart aur ek crossed 4-gon load karte hain.

Fig. 12.4ABCD ke corners ko drag karo. Lab batata hai ki woh convex, non-convex ya crossed hai, uske angles list karta hai aur Section 12.2 ke parallelogram tests chalata hai.
Table 12.14-gons ki family
4-gon ka typeCorners ek plane mein?Sides cross karti hain?Koi angle 180° se zyada?Kya yahan quadrilateral hai?
Non-planarNahi——Nahi
Self-intersectingHaanHaan—Nahi
Non-convexHaanNahiHaan, ekHaan
ConvexHaanNahiNahiHaan
Quick check
4 mein se 1
Ek quadrilateral K ko L se, L ko M se, M ko N se aur N ko K se jod kar banaya gaya hai. Iska ek aur sahi naam kaun sa hai?
Answer dekho

Answer (B) NMLK

NMLK usi boundary par ulta chalta hai: NM, ML, LK, KN. Baaki har option do opposite corners (K aur M, ya L aur N) ko jodta hai, jo diagonal hai, side nahi.
Practice · 5 questions

Practice 12.1: answers ke saath

  1. Short answerEasyNCERT Exercise Set 12.1, Q1

    Quadrilateral PQRS mein:

    1. Side QR ki adjacent sides kaun si hain? Uski opposite side kaun si hai?

      Solution dikhao

      AnswerAdjacent: PQ aur RS. Opposite: SP.

      PQ, QR ke saath vertex Q share karti hai, aur RS vertex R share karti hai. Sirf SP hi aisi side hai jiska QR ke saath koi vertex common nahi hai.

    2. ∠S ke adjacent angles kaun se hain? Uska opposite angle kaun sa hai?

      Solution dikhao

      AnswerAdjacent: ∠R aur ∠P. Opposite: ∠Q.

      S sides se R aur P se juda hai, isliye ∠R aur ∠P, ∠S ke adjacent hain. Q kisi side se S se nahi juda, isliye ∠Q opposite hai.

    3. Aise shabdon mein batao jo har quadrilateral par kaam karein: do sides kab opposite hoti hain aur do angles kab opposite hote hain.

      Solution dikhao

      Opposite sides: aisi do sides jinka koi vertex common nahi hai. Opposite angles: aise do vertices ke angles jo kisi ek side ke do ends nahi hain.

  2. Long answerHardNCERT Exercise Set 12.1, Q2

    Quadrilateral ABCD ke vertex B par rays BA aur BC do angles banati hain: ek 180∘ se kam aur ek zyada, jinka total 360∘ hai. Baaki vertices ka use karke ek rule likho jo bataye ki in dono mein se B par internal angle kaun sa hai. Tumhara rule dent wale quadrilaterals par bhi kaam karna chahiye.

    Solution dikhao

    Baaki teen vertices se bane triangle ACD ko dekho.

    Rule: agar B, △ACD ke andar hai, to B par internal angle reflex angle 360∘−∠ABC hai; warna woh chhota angle ∠ABC hai.

    Yeh kyun kaam karta hai: B par internal angle tabhi 180∘ se zyada hota hai jab quadrilateral mein B par dent ho, aur aisa tabhi hota hai jab B baaki teen corners se bane triangle ke andar ho (jaise Fig. 12.3 mein Z, △WXY ke andar hai). Convex quadrilateral mein koi bhi vertex baaki teen ke triangle ke andar nahi hota, isliye har internal angle chhota wala hota hai.

  3. Short answerMediumNCERT Exercise Set 12.1, Q3

    Quadrilateral PQRS par har condition ke liye batao: kya PQRS phir bhi non-convex (dent wala) ho sakta hai?

    1. PQ∥SR

      Solution dikhao

      Nahi. Har vertex do parallel lines mein se kisi ek par hai, aur poora quadrilateral unke beech ki strip mein hai. Har vertex par shape us vertex se guzarne wali line ki ek hi side par hai, isliye koi bhi internal angle 180∘ se zyada nahi ho sakta.

    2. PS=QR

      Solution dikhao

      Haan. P(0,0), Q(3,4), R(10,0), S(8,1) lo. Tab PS=64+1=65 aur QR=49+16=65, aur S, △PQR ke andar hai, isliye PQRS mein S par dent hai.

    3. ∠P=∠R

      Solution dikhao

      Haan. Arrowhead P(0,0), Q(4,6), R(8,0), S(4,2) line x=4 ke about symmetric hai, isliye ∠P=∠R, aur ismein S par dent hai.

    4. ∠P+∠Q=180∘

      Solution dikhao

      Nahi. Co-interior angles ka total 180∘ hone se PS∥QR ho jaata hai, aur phir sides ke us pair par part (a) lagta hai.

  4. DiagramMediumNCERT Exercise Set 12.1, Q4

    Teen points K, L, M lo jo ek line par nahi hain, aur poori lines KL, LM aur MK draw karo; yeh plane ko 7 regions mein baant deti hain. Ek chautha point N, jo in lines par nahi hai, sides KL, LM, MN, NK wala 4-gon KLMN banata hai. KLMN convex hai, non-convex hai ya self-intersecting, jab N:

    1. △KLM ke andar ho?

      Solution dikhao

      AnswerNon-convex, N par dent.

      N baaki teen vertices ke triangle ke andar hai, isliye N par angle reflex hai.

    2. KM ke us taraf ho jahan L nahi hai, lines LK aur LM ke beech?

      Solution dikhao

      AnswerConvex.

      Koi side doosri side ko cross nahi karti, aur koi vertex baaki ke triangle ke andar nahi hai. Diagonals KM aur LN cross karte hain, isliye KLMN convex hai.

    3. KL ke us taraf ho jahan M nahi hai, lines MK aur ML ke beech?

      Solution dikhao

      AnswerSelf-intersecting.

      N tak pahunchne ke liye side MN ko line KL cross karni padti hai, aur woh use K aur L ke beech cross karti hai. Isliye sides MN aur KL cross karti hain. LM ke us taraf bhi yahi hota hai jahan K nahi hai; wahan NK, LM ko cross karti hai.

    4. L ke aage wale corner region mein ho, KL aur ML ke L se aage badhaaye gaye hisson ke beech?

      Solution dikhao

      AnswerNon-convex, L par dent.

      Ab L, △KMN ke andar hai, isliye L par angle reflex hai. K ke aage aur M ke aage wale corner regions bhi isi tarah behave karte hain.

      Kul milakar: 1 region convex 4-gon deta hai, 2 regions self-intersecting dete hain, aur 4 (andar wala aur teen corner regions) non-convex dete hain.

  5. Short answerMediumNCERT Exercise Set 12.1, Q5

    Ek wire ko chaar seedhi sides wale band loop mein moda gaya hai, aur do sides cross karti hain. Kya uske chaar corners ek plane mein hone se chook sakte hain? Samjhao.

    Solution dikhao

    AnswerNahi.

    Sirf opposite sides cross kar sakti hain, maan lo AB aur CD, jo E par milti hain. Milne wali do lines ek plane mein hoti hain, aur us plane mein dono lines hain, isliye usmein chaaron corners A, B, C aur D hain. Isliye self-intersecting 4-gon hamesha flat hota hai: woh saath mein non-planar nahi ho sakta.

12.2 Parallelograms

Window frame banane wala carpenter lengths, angles aur diagonals aasaani se naap sakta hai, lekin aasaani se check nahi kar sakta ki do lambi sides bilkul parallel hain. Phir bhi frame ko parallelogram hona chahiye, yaani aisa quadrilateral jiski opposite sides parallel hon (rectangles, rhombuses aur squares iske special cases hain). Kaun se measurements iski guarantee dete hain? Un facts se shuru karo jo tumne pichhle saal prove kiye the, yahan ek theorem mein ikatthe kiye gaye hain.

3/4Diagonals O par ek doosre ko aadha-aadha kaatte hain.

Fig. 12.5Ek parallelogram, chaar tarah ke saboot.NCERT Fig. 12.6 · p. 57

Parallelogram ABCD, jismein AB neeche aur DC upar hai. Step 1 mein diagonal AC khicha gaya hai, uske bane do triangles shade kiye gaye hain, aur tick marks se AB ko DC ke equal aur AD ko BC ke equal dikhaya gaya hai, aur AC par common likha hai. Step 2 mein angles mark kiye gaye hain: A aur C par 54 degrees, aur B aur D par 126 degrees. Step 3 mein dono diagonals O par milte hain, aur tick marks dikhate hain ki OA, OC ke equal hai aur OB, OD ke equal hai. Step 4 mein sirf AB aur DC highlight kiye gaye hain, parallel aur equal mark karke.

Theorem 1Parallelogram ki properties

Parallelogram ABCD mein: (a) opposite sides equal hoti hain; (b) opposite angles equal hote hain; (c) diagonals ek doosre ko bisect karte hain.

Theorem 1 parallelogram se measurements tak jaata hai. Carpenter ko ulti direction chahiye, measurements se parallelogram tak. Isliye har part ko agar … to … ke roop mein likho aur dono hisson ko badal kar uska converse banao, jaise tumne Chapter 9 mein kiya tha.

Table 12.2Theorem 1 aur uske converses
PartAgar quadrilateral parallelogram hai, to …Converse: quadrilateral parallelogram hai agar …
(a)uski opposite sides equal hainopposite sides ke dono pairs equal hain
(b)uske opposite angles equal hainopposite angles ke dono pairs equal hain
(c)uske diagonals ek doosre ko bisect karte hainuske diagonals ek doosre ko bisect karte hain

Ek sach statement ka converse galat ho sakta hai, isliye in mein se har ek ka apna proof chahiye. (Pehle lab mein inhe test karo: teeno sahi nikalte hain.) Ek natural plan hai purane proof ko ulta chalana, lekin har ulte step ko check karna zaroori hai.

Theorem 2Barabar opposite sides

Agar quadrilateral ki opposite sides ke dono pairs equal hain, to woh parallelogram hai.

SimulationChautha corner dhoondho

Ek coordinate grid jis par fixed points A(-4, -1), B(-2, 3) aur C(4, -1) hain aur ek draggable point D hai jo whole-number positions par snap hota hai. Chaaron sides AB, BC, CD aur DA unki lengths ke saath khichi hain, taaki learner D ko tab tak move kare jab tak CD, AB ke barabar aur DA, BC ke barabar na ho jaaye, aur un do positions ko compare kare jahan aisa hota hai.

Fig. 12.6A, B aur C fixed hain. D ko drag karo aur chaaron side lengths dekho.
Try karo
  • D ko tab tak drag karo jab tak CD, AB ke barabar aur DA, BC ke barabar na ho jaaye. Aise do points hain: dono dhoondho.
  • Unmein se ek par ABCD parallelogram hai. Doosre par kya gadbad hoti hai?
  • Parallelogram wali position par C se D tak ke step ko B se A tak ke step se compare karo.
Theorem 3Barabar opposite angles

Agar quadrilateral ke opposite angles ke dono pairs equal hain, to woh parallelogram hai.

Proof isse bhi zyada dikhata hai. Quadrilateral mein opposite angles ke dono pairs equal hain aur har do adjacent angles ka total 180∘ hai ek hi baat kehte hain. Pehla upar ki tarah doosra deta hai; aur agar ∠A+∠B=180∘=∠B+∠C, to ∠A=∠C. Isliye Theorem 3 ko adjacent angles ke terms mein bhi likha ja sakta hai (neeche wala quick check dekho).

Theorem 4Ek doosre ko bisect karne wale diagonals

Jis quadrilateral ke diagonals ek doosre ko aadha-aadha kaatte hain, woh parallelogram hai.

Theorem 5Ek pair equal aur parallel

Agar quadrilateral ki opposite sides ka ek pair equal bhi hai aur parallel bhi, to woh parallelogram hai.

Example 12.1

Angles jo parallelogram bana dete hain

Quadrilateral PQRS mein ∠P=∠R=(2x+10)∘ aur ∠Q=∠S=(3x−20)∘. x aur chaaron angles nikaalo. Kya PQRS parallelogram hai?

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

Example 12.2

Parallelogram ke andar parallelogram

ABCD ek parallelogram hai. Point P side AB par hai aur point Q side CD par hai, aur AP=CQ. Prove karo ki PBQD parallelogram hai.

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

Quick check
4 mein se 1
Ek carpenter 60 cm, 40 cm, 60 cm aur 40 cm lambi chaar patti, isi order mein, jod kar ek frame banata hai. Kaun sa result guarantee karta hai ki frame parallelogram hai, chahe use kaise bhi dhakela jaaye?
Answer dekho

Answer (A) Theorem 2: opposite sides equal hain

Sirf side lengths fixed hain, aur opposite sides ke dono pairs equal hain, isliye Theorem 2 lagta hai. Frame ko dhakelne par angles aur diagonals badalte hain, isliye diye gaye facts se baaki tests check nahi kiye ja sakte.
Practice · 4 questions

Practice 12.2: answers ke saath

  1. Short answerEasyNCERT Exercise Set 12.2, Q1

    True ya false?

    1. Do equal adjacent sides wala parallelogram rhombus hota hai.

      Answer dekho

      Answer Sahi

      Parallelogram ki opposite sides equal hoti hain, isliye agar do adjacent sides equal hain, to chaaron equal hain.

    2. Jis parallelogram ke diagonals ek doosre ko bisect karte hain, woh rectangle hai.

      Answer dekho

      Answer Galat

      Har parallelogram ke diagonals ek doosre ko bisect karte hain (Theorem 1), isliye isse kuch naya pata nahi chalta.

    3. Equal diagonals wala rhombus square hota hai.

      Answer dekho

      Answer Sahi

      Rhombus ek parallelogram hai. Parallelogram ABCD mein AC=BD ho to: △ABC≅△BAD (SSS: AB common, BC=AD, AC=BD), isliye ∠ABC=∠BAD. In co-interior angles ka total 180∘ hai, isliye har ek 90∘ hai: equal diagonals wala parallelogram rectangle hota hai. Right angles wala rhombus square hota hai.

    4. Perpendicular diagonals wala rectangle square hota hai.

      Answer dekho

      Answer Sahi

      Diagonals ek doosre ko right angles par bisect karte hain, isliye unke milne ke point ke around chaar triangles congruent hain (SAS) aur chaaron sides equal hain.
  2. ProofMediumNCERT Exercise Set 12.2, Q2

    Parallelogram KLMN mein ∠LKM=∠NKM=35∘. ∠KML aur ∠KMN nikaalo, dikhao ki KLMN rhombus hai, aur uske chaaron angles nikaalo.

    Solution dikhao

    Answer∠KML=∠KMN=35∘. Angles 70∘,110∘,70∘,110∘ hain

    Kyunki KN∥LM, ∠KML=∠NKM=35∘ (alternate angles). Kyunki KL∥NM, ∠KMN=∠LKM=35∘. Isliye diagonal KM, ∠K ke saath-saath ∠M ko bhi bisect karta hai.

    △KLM mein K aur M par dono angles 35∘ hain, isliye unke opposite sides equal hain: LM=KL. Parallelogram ki do adjacent sides equal hain, isliye chaaron equal hain, aur KLMN rhombus hai.

    ∠K=∠M=35∘+35∘=70∘, aur ∠L=∠N=180∘−70∘=110∘.

  3. Short answerMediumNCERT Exercise Set 12.2, Q3

    Haan ya Nahi mein jawab do. Agar Nahi, to ek condition jodo jisse jawab Haan ho jaaye.

    1. Quadrilateral WXYZ ka diagonal WY, ∠W aur ∠Y dono ko bisect karta hai. Kya WXYZ ka rhombus hona zaroori hai?

      Solution dikhao

      AnswerNahi. Jodo: diagonal XZ bhi ∠X aur ∠Z ko bisect karta hai.

      △WXY aur △WZY mein WY common hai, ∠XWY=∠ZWY aur ∠XYW=∠ZYW, isliye triangles congruent hain (ASA): WX=WZ aur YX=YZ. Yeh ek kite hai, aur kite ki chaaron sides equal hona zaroori nahi.

      Agar diagonal XZ bhi ∠X aur ∠Z ko bisect kare, to wahi argument XW=XY aur ZW=ZY deta hai. Tab chaaron sides equal hain: rhombus.

    2. Ek quadrilateral ke diagonals right angles par cross karte hain. Kya uska rhombus hona zaroori hai?

      Solution dikhao

      AnswerNahi. Jodo: diagonals ek doosre ko bisect bhi karte hain.

      Kite ke diagonals perpendicular hote hain, lekin uski sides do alag pairs mein hoti hain. Agar diagonals ek doosre ko bisect bhi karein, to quadrilateral parallelogram hai (Theorem 4), aur centre ke around chaar right triangles congruent hain (SAS), isliye saari sides equal hain: rhombus.

    3. Ek mistri kamre ki outline dono diagonals naap kar check karta hai, aur woh equal hain. Kya outline ka rectangle hona zaroori hai?

      Solution dikhao

      AnswerNahi. Jodo: opposite sides ke dono pairs bhi equal hain.

      Isosceles trapezium ke diagonals bhi equal hote hain. Agar opposite sides bhi equal hon, to outline parallelogram hai (Theorem 2), aur equal diagonals wala parallelogram rectangle hota hai (Q1(c)). Isi wajah se mistri diagonals ke saath opposite sides bhi check karte hain.

  4. ProofHardNCERT Exercise Set 12.2, Q4

    Ek park PQRS parallelogram hai jismein PQ=40 m aur QR=25 m. Raaste uske chaaron corner angles ke bisectors ke saath chalte hain aur jodi mein chaar points par cross karte hain. Dikhao ki yeh chaar points ek rectangle banate hain. Agar park rhombus hota to kya hota?

    Solution dikhao

    ∠P aur ∠Q jaise adjacent angles ka total 180∘ hai, isliye unke aadhon ka total 90∘ hai. Side PQ aur do bisectors se bane triangle mein teesra angle isliye 180∘−90∘=90∘ hai: P aur Q ke bisectors right angles par milte hain.

    Adjacent corners ke har pair ke liye yahi hota hai, isliye chaar milne wale points se bane quadrilateral ke chaaron angles right angles hain: woh rectangle hai. Uske chaar corners alag-alag points hain kyunki PQ≠QR.

    Rhombus mein har angle bisector ek diagonal hota hai. P aur R ke bisectors ek hi line par hain, aur Q aur S ke bhi, isliye chaaron ek hi point par milte hain jahan diagonals cross karte hain: koi rectangle nahi banta.

12.3 Parallelograms ke Applications

Hinge wale parallelogram ko dhakel kar uski shape badlo, phir bhi uski opposite sides parallel rehti hain; engineers is property ko lamps aur drawing tools mein kaam mein lete hain (box dekho). Hamare liye bada inaam geometry ke andar hai: parallelograms ki madad se hum triangles ke baare mein aise facts prove kar sakte hain jin tak seedhe pahunchna mushkil hai. (Physics tumhe ek aur use dikhayegi: parallelogram complete karke do forces ko combine karna.)

12.3.1 Triangles ki Ek Important Property

2/4Midpoints jodo: chaar triangles, 1 se 4 tak numbered.

Fig. 12.7Midpoints ko jodne se koi bhi triangle chaar ek jaise tukdon mein kat jaata hai.NCERT Fig. 12.12 · p. 61

Ek scalene triangle. Step 1 mein har side ka midpoint mark kiya gaya hai, tick marks equal halves dikhate hain. Step 2 mein midpoints jode gaye hain, jisse triangle 1 se 4 numbered chaar tukdon mein bant jaata hai, aur tukda 4 beech mein ulta hai. Step 3 mein tukdon ko alag kiya gaya hai: tukde 1, 2 aur 3 ek hi taraf point karte hain. Step 4 mein tukde 4 ko half-turn dekar baaki ke paas rakha gaya hai, aur chaaron tukde ek jaise dikhte hain.

Squares ginna ek triangle ke liye kaam kar gaya; proof ko sab ke liye kaam karna chahiye. Step counts raasta dikhate hain. △ABC lo jismein M, AB ka midpoint hai aur N, AC ka midpoint hai, aur akele join MN ke baare mein fact prove karo: woh BC ke parallel hai aur usse aadha lamba hai. Teeno joins par lagane se yahi fact chaar congruent triangles dega (Practice 12.3 ka Q1).

4/4MBCK mein CK, BM ke equal aur parallel hai, isliye woh parallelogram hai: MK∥BC aur MK=BC.

Fig. 12.8MN ko double karne se problem parallelogram mein badal jaati hai.NCERT Fig. 12.14 · p. 62

Triangle ABC jismein M, AB ka midpoint hai aur N, AC ka midpoint hai, aur dono ek orange segment MN se jude hain. Step 2 mein MN ko N ke aage uski apni length jitna ek point K tak badhaya gaya hai, tick marks dikhate hain ki MN, NK ke equal hai, aur C ko K se joda gaya hai. Step 3 mein triangles AMN aur CKN shade kiye gaye hain, tick marks dikhate hain ki AN, NC ke equal hai aur AM, CK ke equal hai. Step 4 mein parallelogram MBCK shade kiya gaya hai, arrow marks dikhate hain ki MN, BC ke parallel hai aur BM, CK ke parallel hai.

Theorem 6Midpoint Theorem

Kisi bhi triangle mein do sides ke midpoints ke beech ka segment baaki side ke parallel hota hai aur usse bilkul aadha lamba hota hai.

Example 12.3

Gap ka aadha

Ek step-ladder level zameen par letter A ki tarah khadi hai. Uski do taangon ke midpoints se bandhi ek rassi unhe failne se rokti hai. Taangon ke neeche ke sire 1.3 m door hain. Rassi kitni lambi hai, aur kis direction mein jaati hai?

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

Ab roles ulat do. Sirf AB ke midpoint M se shuru karo, aur M se BC ke parallel line draw karo. Yeh AC ko kahan cross karti hai? Activity wale triangle mein (1,5) se guzarne wali parallel line har 1 upar ke liye 6 aar-paar jaati hai, isliye woh (7,6) tak pahunchti hai, jo bilkul AC ka midpoint hai. (Midpoint Theorem ko ulatne ke aur tareeke chapter ke end ki exercises ke Q22 mein hain.)

3/3M se BC ke parallel sirf ek line hai, isliye dono lines ek hi hain aur Q=N.

Fig. 12.9M se BC ke parallel sirf ek line jaati hai.NCERT Fig. 12.16 · p. 64

Triangle ABC jismein M, AB ka midpoint hai. Step 1 mein M se BC ke parallel ek dashed line khichi gayi hai, jo AC ko ek point Q par milti hai, jis par question mark laga hai. Step 2 mein AC ka midpoint N mark karke MN joda gaya hai, aur AB aur AC ke equal halves par tick marks hain. Step 3 dikhata hai ki dashed line aur MN ek hi line hain, isliye Q aur N ek hi point hain.

Theorem 7Midpoint Theorem ka Converse

Triangle mein ek side ka midpoint lo aur usse doosri side ke parallel line draw karo. Yeh line teesri side ko uske midpoint par milti hai, aur iska triangle ke andar wala hissa us side ka aadha hota hai jiske yeh parallel hai.

Quick check
4 mein se 1
△PQR mein PQ=9 cm, QR=12 cm aur RP=15 cm. Iski sides ke midpoints ko jodne se bane triangle ka perimeter cm mein nikaalo.
cm
Answer dekho

Answer 18 cm

Midpoint triangle ki har side △PQR ki ek side ki aadhi hai, isliye uska perimeter 12(9+12+15)=18 cm hai.

12.3.2 Medians ki Ek Hairaan Karne Wali Property

Cardboard ke triangle ko kahan pakdo ki woh ek ungli ki tip par flat balance ho jaaye? Jawab uske medians se milta hai: woh teen segments jo har vertex ko opposite side ke midpoint se jodte hain. △ABC mein, jahan BC, CA aur AB ke midpoints D, E, F hain, medians AD, BE aur CF hain. Jab teen ya zyada lines ek common point se guzarti hain, to unhe concurrent kehte hain, aur neeche ki activity batati hai ki medians hamesha concurrent hote hain.

Claim yeh hai ki medians concurrent hain aur unka common point har median ko 2:1 ratio mein do hisson mein kaatta hai. Aadhon wala theorem tihaai kaise de sakta hai? Do lambe hisson ke midpoints mark karo; agar woh midpoints aur milne wala point har median ko teen equal tukdon mein kaatein, to kaam ho gaya. Sirf do medians se shuru karo.

4/4Teesra median AD bhi G se guzarta hai, aur AG=2 GD.

Fig. 12.10Triangle ke andar chhupa ek parallelogram har median ko teen barabar hisson mein kaatta hai.NCERT Fig. 12.18 · p. 65

Triangle ABC jismein BC, CA aur AB ke midpoints D, E aur F hain. Step 1 mein medians BE aur CF khiche gaye hain jo G par milte hain. Step 2 mein X aur Y, yaani BG aur CG ke midpoints, mark kiye gaye hain, aur parallelogram FEYX shade kiya gaya hai, jismein FE aur XY dono BC ke parallel hain. Step 3 mein har median teen equal hisson mein kata dikhaya gaya hai: BX, XG aur GE equal, aur CY, YG aur GF equal. Step 4 mein teesra median AD joda gaya hai jo G se guzarta hai, saath mein note AG = 2 GD.

Theorem 8Centroid Theorem

Triangle ke teeno medians ek point se guzarte hain, jo har median ko 2:1 ratio mein baantta hai, aur lamba hissa vertex ki taraf hota hai. Yeh point triangle ka centroid hai.

Example 12.4

Centroid par ek pump

Gaon ke ek triangular talaab ke centroid G par ek fountain pump hai. Corner A par nahane ki seedhiyon se opposite kinare ke midpoint D tak ka median 27 m lamba hai. Pump A se kitni door hai, aur D se? Corner B se median ka hissa BG=14 m hai; poora median BE kitna lamba hai?

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

Quadrilateral ke midpoints

Midpoints ek aur surprise rakhte hain. Lab mein Join the midpoints par tick karo aur A, B, C, D ko kahin bhi drag karo: shape ko lamba aur patla banao, ya dent wala. Andar wali shape PQRS, jo AB, BC, CD, DA ke midpoints P, Q, R, S ko jodti hai, poore samay opposite sides ke dono pairs ko parallel rakhti hai.

4/4PQRS ki opposite sides ke dono pairs parallel hain: yeh parallelogram hai.

Fig. 12.11Do diagonals, Midpoint Theorem ke chaar uses.NCERT Fig. 12.21 · p. 67

Ek general quadrilateral ABCD jiski sides ke midpoints P, Q, R aur S hain. Step 2 mein diagonal AC dashed khicha gaya hai aur PQ aur SR highlight kiye gaye hain, dono AC ke parallel. Step 3 mein diagonal BD dashed khicha gaya hai aur QR aur PS highlight kiye gaye hain, dono BD ke parallel. Step 4 mein PQRS shade kiya gaya hai, jo ek parallelogram hai.

Theorem 9Quadrilaterals ke liye Midpoint Theorem

Kisi bhi quadrilateral ki sides ke midpoints ko order mein jodo: result hamesha ek parallelogram hota hai, jise uska Varignon parallelogram kehte hain.

Example 12.5

Diagonals se perimeter

Quadrilateral ABCD ke diagonals AC=10 cm aur BD=14 cm hain. Iske Varignon parallelogram PQRS ka perimeter nikaalo. Phir batao ki PQRS kaun sa special parallelogram hai jab quadrilateral ke diagonals (a) perpendicular hon, (b) equal hon.

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

Quick check
4 mein se 1
G, △PQR ka centroid hai, aur median PM 15 cm lamba hai. PG cm mein nikaalo.
cm
Answer dekho

Answer 10 cm

PG=23×15=10 cm, aur GM=5 cm.
Practice · 5 questions

Practice 12.3: answers ke saath

  1. ProofMediumNCERT Exercise Set 12.3, Q1

    △XYZ mein points L, M, N, YZ, ZX aur XY ke midpoints hain.

    1. Dikhao ki △LMN, △XNM ke aur do aur triangles ke congruent hai. Unke naam batao.

      Solution dikhao

      Answer△XNM, △NYL aur △MLZ.

      Midpoint Theorem se MN=12YZ, LM=12XY aur LN=12XZ.

      △XNM mein: XN=12XY=LM, XM=12XZ=LN aur NM common hai, isliye △XNM≅△LMN (SSS). Isi tarah △NYL aur △MLZ ki sides 12XY, 12YZ aur 12XZ hain, isliye woh bhi △LMN ke congruent hain. Isse woh prove hota hai jo tracing ne dikhaya tha.

    2. △XYZ ki ek purani drawing feeki pad gayi hai, aur sirf teen joins LM, MN, NL dikh rahe hain. △XYZ ko bilkul waisa hi dobara kaise draw karoge?

      Solution dikhao

      △LMN ke har vertex se opposite side ke parallel line draw karo. Yeh teen lines jodi mein X, Y aur Z par milti hain.

      Yeh isliye kaam karta hai kyunki Midpoint Theorem se side YZ, L se guzarti hai aur MN ke parallel hai, aur L se MN ke parallel sirf ek line hai. Baaki do sides ke liye bhi yahi sach hai.

  2. ProofMediumNCERT Exercise Set 12.3, Q2

    Ek triangular field PQR mein side PQ ke midpoint X se side PR ke midpoint Y tak ek rassi taani gayi hai. Dikhao ki P se door wali side QR ke kisi point T tak ka har seedha raasta rassi se aadha kat jaata hai.

    Solution dikhao

    Maan lo XY, PT ko U par milti hai. Midpoint Theorem se XY∥QR, isliye XU∥QT.

    △PQT mein line XU, PQ ke midpoint X se guzarti hai aur QT ke parallel hai. Midpoint Theorem ke converse se yeh PT ko bisect karti hai: U, PT ka midpoint hai.

  3. ProofHardNCERT Exercise Set 12.3, Q3

    Trapezium ABCD mein AB∥DC, AB=14 cm aur DC=8 cm. G aur H, diagonals AC aur BD ke midpoints hain.

    1. Prove karo ki GH∥AB.

      Solution dikhao

      Maan lo M, AD ka midpoint hai. △ADC mein Theorem 6 se MG∥DC aur MG=12DC; △DAB mein MH∥AB aur MH=12AB.

      Kyunki DC∥AB, MG aur MH dono M se AB ke parallel lines hain, isliye woh ek hi line hain. Isliye M, G, H, AB ke parallel ek line par hain, aur GH∥AB.

    2. GH cm mein nikaalo.

      cm
      Answer dekho

      Answer 3 cm

      GH=MH−MG=12(14)−12(8)=7−4=3 cm.

    3. Is question ka matlab banne ke liye AB≠DC kyun hona chahiye?

      Solution dikhao

      Agar AB=DC, to ABCD parallelogram hai (Theorem 5), uske diagonals ek doosre ko bisect karte hain, aur G aur H ek hi point hain. Tab koi segment GH hota hi nahi.

  4. ProofMediumNCERT Exercise Set 12.3, Q4

    P, Q, R, S, quadrilateral ABCD ki sides AB, BC, CD, DA ke midpoints hain.

    1. Prove karo ki PR aur QS ek doosre ko aadha-aadha kaatte hain.

      Solution dikhao

      PR aur QS parallelogram PQRS ke diagonals hain (Theorem 9), aur parallelogram ke diagonals ek doosre ko bisect karte hain.

    2. Agar AC=BD=12 cm, to PQRS ki har side ki length nikaalo.

      cm
      Answer dekho

      Answer 6 cm

      Har side ek diagonal ki aadhi hai: 12(12)=6 cm. Chaaron sides equal hain, isliye PQRS rhombus hai, aur uske diagonals PR aur QS perpendicular hain.

    3. Ulta, agar PR⊥QS, to kya AC=BD hona zaroori hai?

      Solution dikhao

      AnswerHaan.

      PQRS ek aisa parallelogram hai jiske diagonals ek doosre ko right angles par bisect karte hain, isliye woh rhombus hai (Practice 12.2, Q3(b)). Tab PQ=QR, yaani 12AC=12BD, isliye AC=BD.

  5. ProofHardNCERT Exercise Set 12.3, Q5

    PQRS quadrilateral ABCD ka Varignon parallelogram hai (P, AB par, Q, BC par, R, CD par, S, DA par).

    1. Tumhe sirf PQRS aur vertex A ki position di gayi hai. Dikhao ki B, C aur D ko dobara kaise banaoge.

      Solution dikhao

      AP ko P ke aage uski apni length jitna badhao to B milta hai (P ke about A ka half-turn). Isi tarah C, Q ke about B ka half-turn hai, aur D, R ke about C ka half-turn hai.

    2. Samjhao ki dobara bani figure mein S, DA ka midpoint kyun nikalta hai.

      Solution dikhao

      Maan lo S′, dobara bane quadrilateral mein DA ka midpoint hai. Theorem 9 se PQRS′ parallelogram hai. Lekin PQRS bhi parallelogram hai, aur P, Q, R parallelogram ka chautha vertex uniquely tay kar dete hain. Isliye S′=S.

    3. Kya akela PQRS, ABCD ki shape tay kar deta hai?

      Solution dikhao

      AnswerNahi.

      Alag point A se shuru karne par same Varignon parallelogram wala alag quadrilateral milta hai. Tumhe ek vertex bhi chahiye. (Yeh tiling ke Method 2 mein maayne rakhta hai, jahan tile ko ek grid cell par rakha jaata hai.)

    4. Agar PQRS square hai, to dikhao ki AC aur BD equal aur perpendicular hain. Kya converse sach hai?

      Solution dikhao

      AnswerHaan, converse sach hai.

      PQ=12AC aur QR=12BD, isliye PQ=QR se AC=BD milta hai. Saath hi PQ∥AC aur QR∥BD, isliye PQ⊥QR se AC⊥BD milta hai.

      Ulta, agar AC=BD aur AC⊥BD, to parallelogram PQRS ki do equal adjacent sides aur ek right angle hai, isliye woh square hai.

12.4 Kisi Bhi 4-gon se Plane ko Tile Karna

Tilings dhoondho to woh har jagah milti hain: madhumakhi ke chhatte ke cells, Mughal jaali screen ki patthar ki jaali, charpai ke bune hue squares, rangoli ke neeche ka grid. Kuch ek shape use karti hain aur kuch kai shapes. Tile factory ka sawaal ek shape ke baare mein hai: kya kisi bhi quadrilateral ki copies plane ko tile kar sakti hain?

Chaaron corners fit kyun hone chahiye? Jis bhi point par tiles milti hain, uske around angles ko 360∘ ka poora turn bharna hota hai, aur quadrilateral ke chaaron angles ka total exactly itna hi hota hai. Angle sum pehla clue hai.

Method 1: half-turns. Pin ne bilkul yahi kiya. Edge ke midpoint ke about 180∘ ka turn (ek half-turn) edge ke dono ends ko aapas mein badal deta hai. Isliye ghumayi gayi copy us edge ki doosri taraf baithti hai aur poora edge share karti hai (Fig. 12.12). Har nayi copy ke har edge par half-turns lagane se copies poore plane par phail jaati hain.

4/4Har corner par angles 1, 2, 3 aur 4 ek-ek baar milte hain aur 360∘ bharte hain.

Fig. 12.12Sides ke midpoints ke about half-turns har nayi copy ko jagah par rakhte hain.NCERT Figs. 12.29–12.30 · p. 71

Blue rang ka ek irregular convex 4-gon jiske corners par 1 se 4 number hain. Step 1 mein ek copy uske upar rakhi hai aur uski right-hand side ke midpoint par ek dot hai. Step 2 mein copy ko dot ke about 180 degrees ghumaya gaya hai, isliye woh tile ke paas aa jaati hai, woh side share karti hai, aur uske numbers ulte ho jaate hain. Step 3 mein baaki teen sides ke midpoints ke about ghumayi gayi copies jodi gayi hain, jo tile ko gher leti hain. Step 4 mein ek aur blue copy jodi gayi hai aur ek corner par gola banaya gaya hai jahan angles 1, 2, 3 aur 4 ek-ek baar milte hain.

Yeh method kabhi atakta ya overlap kyun nahi karta? Ek tile ko uske half-turned padosi ke saath jodo: dono milkar chhe sides wali shape banate hain jiski opposite sides equal aur parallel hain. Aise hexagons rows mein fit ho jaate hain, madhumakhi ke chhatte ke cells ki tarah. Isliye har blue copy pehli tile ki slide hai, aur har orange copy uske ghumaye gaye partner ki slide hai. Chapter ke end ki exercises ka Q23 details poori karta hai.

InteractiveKisi bhi 4-gon se tile karo

Screen par ek 4-gon ki copies se bani tiling: blue copies moti tile ki slides hain aur orange copies ek side ke midpoint ke about uske half-turns hain. Moti tile ke chaaron corners drag kiye ja sakte hain, ya arrow keys se move kiye ja sakte hain, aur tiling turant dobara ban jaati hai. Har copy mein 1 se 4 tak ke numbers har vertex par milne wale chaar angles dikhate hain. Ek checkbox sides ke midpoints se guzarne wali lines draw karta hai, jo ek jaise parallelograms ka grid banati hain, aur buttons ek general 4-gon, ek dart aur ek kite load karte hain. Ek message batata hai ki tile convex hai, dent wali hai ya crossed, aur warning deta hai ki crossed 4-gon tiling ki jagah overlaps deta hai.

Fig. 12.13Moti tile ke corners drag karo aur poori tiling saath chalti hai. Ek dart try karo, phir sides ko cross karane ki koshish karo.

Method 2: grid se shuru karo. Tile ke Varignon parallelogram ki copies se bana ek grid draw karo, chapter ki shuruaat wali parallelograms ki rows ki tarah. Fig. 12.14 ke blue cells jaisi doori par cells chuno, har row aur har column mein ek cell chhod kar. Har chune gaye cell par ek tile aise rakho ki uski sides ke midpoints cell ke corners par hon. Rakhi gayi tiles sirf corners par chhooti hain, aur unke beech ka har gap tile ki ek half-turned copy hai. Yeh kaam kyun karna chahiye? Tiler mein Midpoint grid par tick karo: half-turn tiling ke Varignon parallelograms bilkul aise hi grid mein line up ho jaate hain.

3/3Har gap tile ki ek half-turned copy (orange) hai, aur tiling poori ho jaati hai.

Fig. 12.14Tiles ke Varignon parallelograms ek grid banate hain.NCERT Figs. 12.31–12.32 · p. 72

Dashed lines ka ek grid jo ek jaise parallelograms banata hai. Step 1 mein chune gaye cells blue shade kiye gaye hain, dono directions mein ek cell ki doori par. Step 2 mein har chune gaye cell par ek blue 4-gon rakha gaya hai, jiski sides ke midpoints cell ke corners par hain; padosi tiles sirf corners par chhooti hain. Step 3 mein har gap ek orange half-turned copy se bhara gaya hai, aur tiling poori ho jaati hai.

Kisi bhi method ke liye tile ka convex hona zaroori nahi: dent wala 4-gon bhi bilkul isi tarah tile karta hai (Practice 12.4 ka Q2). Crossed 4-gon alag hai, kyunki uski copies overlap karti hain. Isliye hamare shuruaati sawaal ka jawab haan hai: kisi bhi quadrilateral ki copies plane ko tile kar sakti hain. Lekin har shape tile nahi karti; Practice 12.4 mein tum ek aisi shape se miloge jo nahi kar sakti.

Quick check
3 mein se 1
Ek quadrilateral ki chaar copies ek point par milti hain, aur angles 1, 2, 3 aur 4 ek-ek baar use hote hain. Woh bilkul fit kyun hoti hain?
Answer dekho

Answer (C) Kyunki quadrilateral ke chaaron angles ka total 360∘ hai

Ek point ke around poora turn 360∘ hota hai, aur yeh exactly kisi bhi quadrilateral ke chaaron angles ka total hai. Baaki options sirf special shapes ke liye sach hain.
Practice · 2 questions

Practice 12.4: answers ke saath

  1. Short answerMediumNCERT Exercise Set 12.4, Q1

    Kya regular octagon ki copies akele plane ko tile kar sakti hain? Angles se samjhao. Kaun se regular polygons kar sakte hain?

    Solution dikhao

    AnswerNahi. Sirf equilateral triangle, square aur regular hexagon kar sakte hain.

    Regular octagon ke around chalte hue tum har corner par 360∘÷8=45∘ ghoomte ho, isliye har internal angle 180∘−45∘=135∘ hai.

    Jahan tiles ek point par milti hain, unke angles ka total 360∘ hona chahiye; agar point kisi tile ke edge par hai, to baaki tiles ko 180∘ bharna hoga. Lekin 360÷135 aur 180÷135 whole numbers nahi hain, isliye 135∘ ke corners kabhi bhi ek point ke around ki jagah nahi bhar sakte. Octagons wale floors mein hamesha chhote squares bhi lagte hain.

    Regular polygon akele tabhi tile karta hai jab uska angle 360∘ ko divide kare: equilateral triangle (60∘), square (90∘) aur regular hexagon (120∘). Regular pentagon (108∘) fail hota hai, aur 7 ya zyada sides wale har polygon ka angle 120∘ aur 180∘ ke beech hota hai, isliye woh bhi fail hota hai.

  2. ActivityMediumNCERT Exercise Set 12.4, Q2

    Ek step-by-step recipe likho jise follow karke tumhara dost half-turns ki madad se dent wale 4-gon WXYZ (dent Z par hai) ki copies se floor tile kar sake. Phir batao ki grid method kaise shuru hoga, aur dono mein se tum kaun sa use karna pasand karoge.

    Solution dikhao

    Method 1: ek copy ko har side ke midpoint ke about half-turn do, aur har naye edge par yahi dohrao. Reflex corner Z padosi copies ke corners se bhar jaata hai; har point ke around chaaron angles ka total phir bhi 360∘ hai.

    Method 2: sides ke midpoints jod kar WXYZ ka Varignon parallelogram banao, use repeat karke grid banao, aur har direction mein ek-ek cell chhod kar har cell par ek copy rakho, jiski sides ke midpoints cell ke corners par hon. Gaps half-turned copies hain.

    Dono theek hain: Method 1 ke liye sirf midpoints chahiye, jabki Method 2 ke liye pehle grid chahiye. Apni tiling ko tiler mein dart ke saath check karo.

Recap

Summary aur key points

  1. Ek plane mein chaar points, jinmein se koi teen ek line par na hon, quadrilateral ABCD banate hain jab sides AB, BC, CD, DA sirf shared corners par milein (Section 12.1).
  2. Quadrilateral convex hota hai jab har internal angle 180∘ se kam ho; yaani, uske diagonals cross karte hain (Convex quadrilaterals).
  3. Parallelogram ki opposite sides equal, opposite angles equal hote hain aur diagonals ek doosre ko bisect karte hain (Theorem 1).
  4. In mein se har converse sach hai: equal opposite sides, equal opposite angles, ya bisect karne wale diagonals quadrilateral ko parallelogram bana dete hain (Theorem 2, Theorem 3, Theorem 4).
  5. Opposite sides ka ek pair jo equal bhi ho aur parallel bhi, parallelogram banane ke liye kaafi hai (Theorem 5).
  6. Midpoint Theorem: triangle mein do sides ke midpoints ke beech ka segment baaki side ke parallel hota hai aur uski length ka aadha hota hai (Theorem 6).
  7. Converse: ek side ke midpoint se doosri side ke parallel line draw karo, to woh teesri side ko aadha kaatti hai (Theorem 7).
  8. Teeno medians centroid par milte hain, jo har median par vertex se do-tihaai doori par hota hai (Theorem 8).
  9. Kisi bhi quadrilateral ki sides ke midpoints ek parallelogram banate hain jiski sides diagonals ki aadhi hoti hain (Theorem 9).
  10. Kisi bhi quadrilateral ki copies plane ko tile karti hain: sides ke midpoints ke about half-turns copies ko jagah par rakhte hain, aur har corner par milne wale chaar angles ka total 360∘ hota hai (Section 12.4).

Chapter ke end ki exercises

Practice · 29 questions

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

  1. Short answerEasyNCERT End-of-chapter Exercises, Q1

    Dikhao ki kisi bhi triangle ki copies plane ko tile kar sakti hain.

    Solution dikhao

    △ABC ki ek copy ko BC ke midpoint M ke about half-turn do. Tab B aur C aapas mein badal jaate hain, aur A ek point A′ par jaata hai jahan M, AA′ ka midpoint hai.

    Quadrilateral ABA′C mein diagonals AA′ aur BC ek doosre ko M par bisect karte hain, isliye woh parallelogram hai (Theorem 4). Parallelograms rows mein plane ko tile karte hain (Fig. 12.1), aur har ek triangle ki do copies se bana hai. Isliye triangle plane ko tile karta hai.

  2. Short answerMediumNCERT End-of-chapter Exercises, Q2

    Notebook ki ruled lines barabar doori par hain. Reena line 2 se line 8 tak ek tircha segment PQ draw karti hai. PQ ka midpoint kahan hai, aur kyun?

    Solution dikhao

    AnswerJahan PQ line 5 ko cross karta hai.

    Q se lines ke perpendicular segment draw karo, jo line 2 ko T par mile. Line 2 se line 8 tak 6 equal gaps hain, isliye line 5, jo har end se teen gaps door hai, QT ko uske midpoint par cross karti hai.

    Line 5, line 2 ke parallel hai, jis par TP hai. △QTP mein line 5, QT ke midpoint se guzarti hai aur TP ke parallel hai, isliye woh QP ko bisect karti hai (Theorem 7). PQ ka midpoint wahan hai jahan PQ line 5 ko cross karta hai.

  3. ProofHardNCERT End-of-chapter Exercises, Q3

    4-gon ABCD mein sides AB aur CD, E par cross karti hain (bow-tie). Har corner par angle wahan milne wali do sides ke beech ka angle hai.

    1. Dikhao ki ∠A+∠B+∠C+∠D=360∘−2∠AED, isliye sum 360∘ se kam hai.

      Solution dikhao

      E, AB par aur CD par hai, isliye 4-gon ke A aur D wale angles △ADE ke angles hain, aur B aur C wale △BCE ke angles hain.

      Isliye ∠A+∠D=180∘−∠AED aur ∠B+∠C=180∘−∠BEC. Kyunki ∠BEC=∠AED (vertically opposite), jodne par 360∘−2∠AED milta hai, jo 360∘ se kam hai.

    2. Agar ∠AED=70∘, to angle sum degrees mein nikaalo.

      °
      Answer dekho

      Answer 220 °

      360∘−2×70∘=220∘.

    3. Kya angle sum 10∘ jitna chhota ho sakta hai? Agar haan, to kaise?

      Solution dikhao

      AnswerHaan, ∠AED=175∘ ke saath.

      Humein 360∘−2∠AED=10∘ chahiye, isliye ∠AED=175∘. Ek bahut chapta bow-tie draw karo jiski crossing sides 175∘ par milein: tab dono triangles lambe aur patle hain, aur unke baaki angles bahut chhote hain.

  4. ProofMediumNCERT End-of-chapter Exercises, Q4

    ABCD ek parallelogram (rhombus nahi) hai jiske diagonals O par milte hain. A aur C se diagonal BD par daale gaye perpendiculars use P aur Q par milte hain. Dikhao ki APCQ parallelogram hai.

    Solution dikhao

    ABCD ke diagonals ek doosre ko bisect karte hain, isliye OA=OC.

    △AOP aur △COQ mein: OA=OC, ∠AOP=∠COQ (vertically opposite) aur ∠APO=∠CQO=90∘. Isliye triangles congruent hain (AAS), aur OP=OQ.

    APCQ ke diagonals AC aur PQ ek doosre ko O par bisect karte hain, isliye APCQ parallelogram hai (Theorem 4).

  5. ProofMediumNCERT End-of-chapter Exercises, Q5

    Paper ka set-square ek right triangle PQR hai jiska right angle Q par hai. Tum use aise fold karte ho ki R bilkul Q par aa jaaye, aur crease banate ho. Dikhao ki crease QR ke midpoint se aur PR ke midpoint se bhi guzarti hai.

    Solution dikhao

    R ko Q par fold karne se crease QR ka perpendicular bisector ban jaati hai. Isliye woh QR ke midpoint M se, QR ke right angles par, guzarti hai.

    PQ bhi QR ke perpendicular hai, isliye crease PQ ke parallel hai. △RQP mein crease RQ ke midpoint M se guzarti hai aur QP ke parallel hai, isliye woh RP ko bisect karti hai (Theorem 7).

  6. ActivityHardNCERT End-of-chapter Exercises, Q6

    Triangles ko kaatna aur dobara banana. △ABC mein medians AD, BE, CF centroid G par milte hain.

    1. Teeno medians ke saath kaat kar chhe chhote triangles banao. △GDB ko D ke about half-turn do, taaki B, C par aa jaaye. Dikhao ki yeh △GDC ke saath fit hokar ek triangle banata hai, aur us triangle ki sides medians ke terms mein nikaalo.

      Solution dikhao

      Half-turn B ko C par aur G ko ray GD par ek point G′ par bhejta hai jahan DG′=DG. Kyunki ∠GDB+∠GDC=180∘, ghumaya gaya tukda DC ke saath △GDC se fit ho jaata hai, aur G, D, G′ ek line par hain. Dono milkar △GG′C banate hain.

      Iski sides GG′=2 GD=AG=23AD, GC=23CF aur G′C=GB=23BE hain.

    2. Baaki chaar tukdon ko bhi isi tarah jodo. Dikhao ki tumhe milne wale teeno triangles congruent hain.

      Solution dikhao

      Har jodi ek aisa triangle deti hai jiski sides teeno medians ki do-tihaai hain, isliye teeno triangles congruent hain (SSS). Yeh mazedaar trick sirf 2014 mein puzzle maker Lee Sallows ne notice ki thi.

    3. Ab △ABC ko AD ke saath kaato, aur △ADB ko DF ke saath kaato. Dikhao ki teeno tukde dobara jud kar do congruent triangles banate hain.

      Solution dikhao

      △DFB ko F ke about half-turn do: B, A par aata hai, aur D, D′ par aata hai jahan F, DD′ ka midpoint hai. (a) ki tarah yeh △AFD ke saath fit hokar △ADD′ banata hai.

      Iski sides AD, AD′=BD=DC aur DD′=2 DF=AC hain (△BAC mein Midpoint Theorem). Isliye △ADD′≅△DAC (SSS), matching A→D, D→A, D′→C.

    4. (b) ke teen triangles mein se ek par (a) aur (b) dohrao. Uske medians 12BC, 12CA aur 12AB nikalte hain. Aakhri nau triangles ki sides kya hain?

      Solution dikhao

      Answer13BC, 13CA aur 13AB.

      Har side ek median ki do-tihaai hai: 23×12BC=13BC, aur baaki ke liye bhi isi tarah. In medians mein se ek tum seedha dekh sakte ho: △GG′C mein D, GG′ ka midpoint hai, isliye CD=12BC ek median hai.

  7. ProofMediumNCERT End-of-chapter Exercises, Q7

    Trapezium ABCD (AB∥DC) mein tirchi side AD ka midpoint E mark karo, aur E se doosri tirchi side BC par ek point F tak segment draw karo.

    1. Agar EF∥AB, to prove karo ki F, BC ka midpoint hai aur EF=12(AB+DC).

      Solution dikhao

      Maan lo EF, AC ko G par milti hai. △ADC mein E, AD ka midpoint hai aur EG∥DC, isliye G, AC ka midpoint hai aur EG=12DC (Theorem 7).

      △CAB mein G, CA ka midpoint hai aur GF∥AB, isliye F, CB ka midpoint hai aur GF=12AB. Jodne par EF=12(AB+DC).

    2. Ek khet ka plot trapezium hai jiski parallel sides 30 m aur 18 m hain. Ek fence uski do tirchi sides ke midpoints ko jodti hai. Fence kitni lambi hai, metres mein?

      m
      Answer dekho

      Answer 24 m

      12(30+18)=24 m.

    3. Ulta, agar F, BC ka midpoint hai, to prove karo ki EF∥AB.

      Solution dikhao

      E se AB ke parallel line draw karo. (a) se yeh BC ko uske midpoint par milti hai, jo F hai. E aur F se sirf ek line guzarti hai, isliye EF wahi parallel line hai: EF∥AB.

  8. ProofMediumNCERT End-of-chapter Exercises, Q8

    Card ka ek tukda parallelogram ABCD hai, aur uske diagonals O par cross karte hain. O se guzarta ek seedha knife cut AB par P se CD par Q tak jaata hai.

    1. Prove karo ki PQ, O par bisect hota hai.

      Solution dikhao

      △OAP aur △OCQ mein: OA=OC (Theorem 1), ∠OAP=∠OCQ (alternate angles, AB∥DC) aur ∠AOP=∠COQ (vertically opposite). Isliye triangles congruent hain (ASA), aur OP=OQ.

    2. Dikhao ki cut parallelogram ko equal area wale do tukdon mein baantta hai.

      Solution dikhao

      O, AC, BD aur PQ ka midpoint hai, isliye O ke about half-turn P→Q, B→D, C→A aur Q→P bhejta hai. Yeh tukde PBCQ ko bilkul tukde QDAP par le jaata hai, isliye dono tukde congruent hain aur unke areas equal hain.

  9. NumericalHardNCERT End-of-chapter Exercises, Q9

    Trapezium PQRS ki parallel sides PS=4 cm aur QR=10 cm hain. K aur L tirchi sides PQ aur SR ke midpoints hain. Segment KL trapezium ko do chhote trapeziums mein baantta hai. PKLS aur KQRL ke areas ka ratio nikaalo.

    Solution dikhao

    Answer11:17

    Q7 se KL∥PS aur KL=12(4+10)=7 cm.

    Maan lo height h hai. P se QR par perpendicular daalo; Theorem 7 se (P, Q aur is perpendicular ke foot se bane triangle mein) line KL use aadha kaatti hai. Isliye har chhote trapezium ki height h2 hai.

    [PKLS][KQRL]=12(4+7)⋅h212(7+10)⋅h2=1117
  10. Short answerMediumNCERT End-of-chapter Exercises, Q10

    Chaar points, jinmein se koi teen ek line par nahi, kai tareekon se 4-gons mein jode ja sakte hain (crossed ya dent wale bhi allowed).

    1. Corners (0,0), (4,0), (4,3), (0,3) wale kitne alag 4-gons hain?

      Answer dekho

      Answer 3

      Pehle (0,0) fix karo. Baaki teen 3×2×1=6 orders mein aa sakte hain, lekin har loop har direction mein ek baar gina jaata hai, isliye 6÷2=3 hain. Yahan: rectangle (convex) aur do crossed bow-ties.

    2. (0,0), (6,0), (3,6) aur (3,2) kis tarah ke 4-gons dete hain?

      Solution dikhao

      AnswerTeen 4-gons, sab non-convex.

      (3,2) baaki teen points se bane triangle ke andar hai. Teeno possible 4-gons mein (3,2) par dent hai: koi convex nahi aur koi crossed nahi.

    3. General case mein kitne 4-gons convex hain aur kitne self-intersecting?

      Solution dikhao

      Agar koi point baaki teen ke triangle ke andar nahi hai: 1 convex aur 2 self-intersecting. Agar ek point andar hai: 0 convex, 0 self-intersecting aur 3 non-convex.

  11. NumericalHardNCERT End-of-chapter Exercises, Q11

    △ABC mein P, AB ka ek point hai, aur P se BC ke parallel line AC ko Q par kaatti hai. AP=2 cm aur AQ=3 cm lo.

    1. Agar PB=4 cm, to QC cm mein nikaalo.

      cm
      Solution dikhao

      PB par P2 mark karo jahan PP2=2, taaki AP=PP2=P2B=2. P2 se BC ke parallel line draw karo, jo AC ko Q2 par mile.

      △AP2Q2 mein Theorem 7 se Q, AQ2 ka midpoint hai, isliye QQ2=3. Trapezium PBCQ mein Q7(a) se Q2, QC ka midpoint hai, isliye Q2C=3. Isliye QC=6 cm.

    2. Agar PB=1 cm, to QC cm mein nikaalo.

      cm
      Solution dikhao

      AP ka midpoint P1 mark karo, taaki AP1=P1P=PB=1. P1 aur P se BC ke parallel lines AC ko teen equal hisson mein kaatti hain, (a) ki tarah. AQ=3 unmein se do hisse hain, isliye har ek 1.5 hai aur QC=1.5 cm.

    3. Agar PB=43 cm, to QC cm mein nikaalo.

      cm
      Solution dikhao

      23 cm ke tukde use karo: AP 3 tukde hai aur PB 2. Cut points se parallel lines AC ko 5 equal tukdon mein baantti hain, jinmein se 3 milkar AQ=3 banate hain. Har tukda 1 cm hai, isliye QC=2 cm.

    4. Dikhao ki jab bhi APPB whole numbers ke fraction mn ke barabar ho, APPB=AQQC.

      Solution dikhao

      AB ko m+n equal tukdon mein kaato, jinmein se m milkar AP banate hain. Cut points se BC ke parallel lines AB ke saath barabar doori par hain, isliye ((a) ki tarah) woh AC ko m+n equal tukdon mein kaatti hain, jinmein se m milkar AQ banate hain. Isliye AQ:QC=m:n. Class 10 mein tum ise yeh maane bina prove karoge ki ratio ek fraction hai.

  12. ProofHardNCERT End-of-chapter Exercises, Q12

    Parallelogram PQRS mein K, side PQ ka midpoint hai aur L, side RS ka midpoint hai.

    1. Dikhao ki SK aur QL diagonal PR ko teen equal hisson mein kaatte hain.

      Solution dikhao

      KQ∥SL aur KQ=SL (equal opposite sides ke aadhe), isliye KQLS parallelogram hai (Theorem 5) aur SK∥QL.

      Maan lo SK, PR ko X par milti hai aur QL, PR ko Y par. △PQY mein K, PQ ka midpoint hai aur KX∥QY, isliye X, PY ka midpoint hai (Theorem 7). △RSX mein L, RS ka midpoint hai aur LY∥SX, isliye Y, RX ka midpoint hai. Isliye PX=XY=YR.

    2. (a) ka use karke diye gaye segment UV ko trisect karo.

      Solution dikhao

      Line UV ke bahar koi bhi point S chuno, aur Q ko UV ke midpoint ke about uska half-turn maano; tab UQVS ek parallelogram hai jiska diagonal UV hai. S ko UQ ke midpoint se jodo, aur Q ko VS ke midpoint se. Yeh do lines UV ko teen equal hisson mein kaatti hain.

    3. Centroid Theorem use karke UV ko trisect karne ka doosra tareeka batao.

      Solution dikhao

      UV ko median banao: line ke bahar koi bhi point B chuno, aur C ko V ke about uska half-turn maano, taaki V, BC ka midpoint ho. △UBC ka centroid G, UV par hai aur UG=23UV, aur UG ka midpoint doosra trisection point hai. (Q11 teesra tareeka deta hai: barabar doori wali parallel lines.)

  13. Short answerHardNCERT End-of-chapter Exercises, Q13

    Kya Quadrilaterals ke liye Midpoint Theorem (Theorem 9) ajeeb 4-gons ke liye bhi sach rehta hai?

    1. Samjhao ki jab ABCD dent wala, ya self-intersecting bhi ho, tab bhi yeh kyun sach hai.

      Solution dikhao

      Proof ne sirf △ABC, △ADC, △BCD aur △BAD mein Midpoint Theorem use kiya, jo ABCD ki koi bhi shape ho, exist karte hain. Isliye PQ∥AC∥SR aur QR∥BD∥PS phir bhi sach hain.

    2. Crossed 4-gon A(0,0), B(0,4), C(6,0), D(6,4) lo. AB, BC, CD, DA ke midpoints P, Q, R, S nikaalo. Kya gadbad hoti hai, aur general case mein aisa kab hota hai?

      Solution dikhao

      P(0,2), Q(3,2), R(6,2) aur S(3,2). Chaaron line y=2 par hain, aur Q=S: parallelogram sikud kar ek segment ban gaya hai.

      Aisa tabhi hota hai jab diagonals AC aur BD parallel hon (yahan y=0 aur y=4), kyunki tab PQRS ki sides, jo AC aur BD ke parallel hain, sab ek hi line par padti hain.

    3. Kya reasoning tab kaam karti hai jab ABCD non-planar ho?

      Solution dikhao

      AnswerHaan.

      Har triangle ABC, ADC, BCD, BAD apne plane mein hai, aur har ek mein Midpoint Theorem sach hai. PQ aur SR dono AC ke parallel hain, isliye ek doosre ke parallel hain; do parallel lines ek plane mein hoti hain, isliye P, Q, R, S ek plane mein hain aur PQRS parallelogram hai.

  14. ProofMediumNCERT End-of-chapter Exercises, Q14

    ABCD 6 cm side wala square hai. AB, BC, CD, DA par points P, Q, R, S aise hain ki AP=BQ=CR=DS=2 cm. Dikhao ki PQRS square hai, jabki P, Q, R, S midpoints nahi hain. Isse Theorem 9 ke converse ke baare mein kya pata chalta hai?

    Solution dikhao

    Har corner triangle, jaise △APS, mein 2 cm aur 4 cm ke legs ke beech right angle hai (AS=6−2=4). Isliye chaaron corner triangles congruent hain (SAS), aur PQ=QR=RS=SP.

    Saath hi ∠BPQ=∠ASP (matching angles), isliye ∠APS+∠BPQ=∠APS+∠ASP=90∘, jisse ∠SPQ=90∘ bachta hai. Isliye PQRS square hai, aur khaas taur par ek parallelogram.

    Theorem 9 ka converse, agar PQRS parallelogram hai to P, Q, R, S midpoints hain, galat hai.

  15. ProofMediumNCERT End-of-chapter Exercises, Q15

    Midpoint Theorem ka doosra proof. △ABC mein M aur N, AB aur AC ke midpoints hain.

    1. C se BA ke parallel line draw karo, jo line MN (N ke aage badhayi gayi) ko S par mile. △AMN≅△CSN prove karo, aur iska use karke dikhao ki MN∥BC aur MN=12BC.

      Solution dikhao

      AN=CN, ∠ANM=∠CNS (vertically opposite) aur ∠AMN=∠CSN (alternate angles, kyunki AM∥SC), isliye △AMN≅△CSN (AAS). Isliye MN=NS aur CS=AM=MB.

      Isliye CS aur MB equal aur parallel hain, aur MBCS parallelogram hai (Theorem 5). Tab MS∥BC aur MS=BC, isliye MN=12MS=12BC.

    2. Theorem 7 ko bhi isi tarah prove karo. Maan lo M se BC ke parallel line AC ko Q par milti hai, aur line MQ par S aise chuno ki CS∥AB.

      Solution dikhao

      MBCS ki opposite sides ke dono pairs parallel hain, isliye woh parallelogram hai aur CS=MB=AM.

      △AMQ aur △CSQ mein: AM=CS, ∠AQM=∠CQS (vertically opposite) aur ∠MAQ=∠SCQ (alternate angles). Triangles congruent hain (AAS), isliye AQ=QC.

  16. Short answerMediumNCERT End-of-chapter Exercises, Q16

    Neeche har statement kisi special parallelogram ki property ka converse hai. Tay karo ki woh sach hai ya nahi.

    1. (Rectangle: har angle 90∘ hai.) Jis quadrilateral ke chaaron angles equal hain, woh rectangle hai.

      Answer dekho

      Answer Sahi

      Chaar equal angles jinka total 360∘ hai, har ek 90∘ hota hai.

    2. (Rhombus: diagonals perpendicular hain.) Perpendicular diagonals wala quadrilateral rhombus hai.

      Answer dekho

      Answer Galat

      Kite ke diagonals perpendicular hote hain lekin woh rhombus nahi hai jab tak chaaron sides equal na hon.

    3. (Square: diagonals equal, perpendicular hain aur ek doosre ko bisect karte hain.) Jis quadrilateral ke diagonals equal, perpendicular hain aur ek doosre ko bisect karte hain, woh square hai.

      Answer dekho

      Answer Sahi

      Bisect karne wale diagonals use parallelogram banate hain, equal diagonals rectangle (Practice 12.2, Q1(c)) aur perpendicular diagonals rhombus, isliye woh square hai.

    4. (Rectangle: iska Varignon parallelogram rhombus hai.) Agar kisi quadrilateral ka Varignon parallelogram rhombus hai, to woh quadrilateral rectangle hai.

      Answer dekho

      Answer Galat

      Varignon parallelogram tab bhi rhombus hota hai jab diagonals equal hon, jaise isosceles trapezium mein, jo rectangle nahi hai.
  17. Short answerHardNCERT End-of-chapter Exercises, Q17

    Card par ek convex quadrilateral ABCD draw karo aur use diagonal AC ke saath fold karo, taaki △ABC aur △ADC ab ek plane mein na rahein.

    1. Fold karne par 4-gon ke kaun se angles badalte hain, aur kaun se wahi rehte hain?

      Solution dikhao

      B aur D wale angles rigid triangles ke andar hain, isliye woh nahi badalte. A wala angle AB aur AD ke beech hai, jo ab alag planes mein hain, isliye woh chhota ho jaata hai: ∠A<∠BAC+∠CAD. C par bhi yahi hota hai.

    2. Samjhao ki fold kiye gaye (non-planar) 4-gon ka angle sum 360∘ se kam kyun hai.

      Solution dikhao
      ∠A+∠B+∠C+∠D<(∠BAC+∠B+∠BCA)+(∠CAD+∠D+∠ACD)

      Right-hand side do triangles ka angle sum hai, 180∘+180∘=360∘.

    3. Tum angle sum ko 10∘ se kam kaise bana sakte ho?

      Solution dikhao

      Dono triangles ko lamba aur patla banao, jinke B aur D par angles bahut chhote hon, aur lagbhag poora band hone tak fold karo taaki D, B ke paas aa jaaye. Tab A aur C wale angles bhi bahut chhote ho jaate hain, aur sum ko jitna chaho utna chhota banaya ja sakta hai.

  18. ProofHardNCERT End-of-chapter Exercises, Q18

    Do half-turns ek slide banate hain. Tile ABCD mein maan lo X aur Y, AB aur BC ke midpoints hain. Ek point Z ko X ke about half-turn diya jaata hai, jo Z1 par aata hai, aur phir Y ke about half-turn, jo Z2 par aata hai.

    1. Dikhao ki ZZ2, XY ke parallel hai aur usse do guna lamba hai.

      Solution dikhao

      △ZZ1Z2 mein X, Z1Z ka midpoint hai aur Y, Z1Z2 ka midpoint hai. Midpoint Theorem se XY∥ZZ2 aur XY=12ZZ2.

    2. Nateeja nikaalo ki har point same direction mein same doori tak move karta hai, jo AC ke barabar hai: do half-turns diagonal AC ke saath ek slide banate hain.

      Solution dikhao

      △ABC mein XY∥AC aur XY=12AC. Isliye har point Z ke liye ZZ2, AC ke parallel hai aur length mein uske barabar hai: har point A se C ki direction mein AC jitna slide karta hai.

    3. Iska use karke samjhao ki half-turn tiling mein ek tile ko AC ke saath uski length jitna slide karne par woh bilkul doosri tile par kyun aa jaati hai.

      Solution dikhao

      Sides ke midpoints ke about half-turns aise moves hain jo tiling ko apne aap par le jaate hain (Method 1). Unmein se do karna, pehle X ke about aur phir Y ke about, bhi tiling ko apne aap par le jaata hai, aur part (b) dikhata hai ki result AC ke saath slide hai. BC aur CD ke midpoints ke about half-turns isi tarah BD ke saath slide dete hain. Tiler mein blue copies bilkul yahi slides hain.

  19. Short answerMediumNCERT End-of-chapter Exercises, Q19

    Chaar lengths kab 4-gon banati hain?

    1. Kya 3 cm, 4 cm aur 8 cm ki sticks triangle bana sakti hain? Jo rule use kar rahe ho, woh batao.

      Solution dikhao

      AnswerNahi.

      Triangle tabhi banta hai jab do chhoti lengths ka total sabse lambi length se zyada ho. Yahan 3+4=7, jo 8 se kam hai.

    2. Ek 4-gon ki teen sides 3 cm, 4 cm aur 12 cm hain. Kya chauthi side 20 cm ho sakti hai? 9 cm? 5 cm?

      Solution dikhao

      Answer20 cm: nahi. 9 cm: haan. 5 cm: nahi.

      Kisi bhi 4-gon mein har side baaki teeno ke total se chhoti hoti hai: AB ke liye AB<AC+CB aur AC<AD+DC, isliye AB<AD+DC+CB.

      Chauthi side x ke saath: agar x sabse lambi hai, to x<3+4+12=19; aur 12<3+4+x ke liye x>5 chahiye. Isliye 5<x<19. 20 fail hota hai, 9 chalta hai, aur 5 fail hota hai kyunki woh 4-gon ko chapta karke segment bana dega.

    3. Ek rule batao jisse tay ho ki lengths a, b, c, d kisi 4-gon ki sides ho sakti hain ya nahi.

      Solution dikhao

      Sabse lambi length baaki teen ke total se kam honi chahiye. Part (b) dikhata hai ki yeh zaroori hai. Jab yeh sach ho, to tum hamesha ek bana sakte ho: sticks ko lagbhag seedha bichhao aur joints ko hinge ki tarah kholo.

  20. NumericalMediumNCERT End-of-chapter Exercises, Q20

    n-gon ka diagonal do aise vertices ko jodta hai jo adjacent nahi hain.

    1. 9-gon ke kitne diagonals hote hain?

      Answer dekho

      Answer 27

      Har vertex se 9−3=6 diagonals shuru hote hain (khud tak ya apne do padosiyon tak nahi). 9×6=54 har diagonal ko do baar ginta hai, isliye 27 hain.

    2. n-gon ke diagonals ki ginti ka formula nikaalo, aur use n=3,4,5,6 ke liye check karo.

      Solution dikhao

      Answern(n−3)2

      n vertices mein se har ek se n−3 diagonals shuru hote hain, aur har diagonal do baar gina jaata hai, jisse n(n−3)2 milta hai. n=3,4,5,6 ke liye yeh 0,2,5,9 hai.

    3. Jab n-gon (n+1)-gon banta hai, to kitne diagonals judte hain? Check karo ki tumhara formula match karta hai.

      Solution dikhao

      Answern−1

      Padosi U aur W ke beech ek naya vertex V daalo. V ko n−2 diagonals milte hain (U aur W ke alawa har purane vertex tak), aur purani side UW diagonal ban jaati hai: kul n−1.

      (n+1)(n−2)2−n(n−3)2=2n−22=n−1
  21. NumericalMediumNCERT End-of-chapter Exercises, Q21

    n sides wale polygon (flat, sides cross na karein) ka angle sum kitna hai?

    1. Convex pentagon aur convex hexagon ka angle sum nikaalo.

      Solution dikhao

      Answer540∘ aur 720∘.

      Ek vertex se diagonals pentagon ko 3 triangles mein aur hexagon ko 4 mein kaatte hain, isliye sums 3×180∘=540∘ aur 4×180∘=720∘ hain.

    2. Ek formula nikaalo aur convex n-gon ke liye use prove karo.

      Solution dikhao

      Answer(n−2)×180∘

      Ek vertex se n−3 diagonals convex n-gon ko n−2 triangles mein kaatte hain, jinke angles milkar bilkul polygon ke angles banate hain. Isliye sum (n−2)×180∘ hai. Formula non-convex polygons ke liye bhi sach hai, lekin proof mein zyada dhyaan chahiye kyunki diagonal bahar bhi ho sakta hai.

    3. 10-gon ka angle sum degrees mein nikaalo.

      °
      Answer dekho

      Answer 1440 °

      (10−2)×180∘=1440∘.
  22. ProofHardNCERT End-of-chapter Exercises, Q22

    Kai converses. △ABC mein M, AB ka ek point hai aur N, AC ka ek point. Chaar conditions ke naam rakho: (M1) M, AB ka midpoint hai; (M2) N, AC ka midpoint hai; (Par) MN∥BC; (Half) MN=12BC. Midpoint Theorem kehta hai ki (M1) aur (M2) milkar (Par) aur (Half) dete hain.

    1. Prove karo ki (Par) aur (Half) milkar (M1) aur (M2) dete hain.

      Solution dikhao

      N se AB ke parallel line draw karo, jo BC ko K par mile. MNKB ki opposite sides ke dono pairs parallel hain, isliye woh parallelogram hai: BK=MN=12BC, isliye K, BC ka midpoint hai; aur NK=MB.

      △CBA mein CB ke midpoint K se BA ke parallel line CA ko bisect karti hai (Theorem 7), isliye N, AC ka midpoint hai aur NK=12AB. Tab MB=12AB, isliye M, AB ka midpoint hai.

    2. Theorem 7 kaun si conditions di hui maanta hai, aur kaun si conclude karta hai?

      Solution dikhao

      Answer(M1) aur (Par) diye hue hain; yeh (M2) aur (Half) conclude karta hai.

      Theorem 7, AB ke midpoint M (M1) aur usse BC ke parallel line (Par) se shuru hota hai. Yeh prove karta hai ki yeh line AC ko uske midpoint par milti hai (M2) aur triangle ke andar wala hissa 12BC hai (Half).

    3. Ek triangle mein ∠B=90∘, AB=8 cm, BC=6 cm aur AC=10 cm hai; maan lo M, AB ka midpoint hai, aur N, AC ka woh point hai jahan AN=1.4 cm. Dikhao ki (M1) aur (Half) sach hain lekin (M2) nahi. Isse kya pata chalta hai?

      Solution dikhao

      B ko origin par rakho, A=(0,8) aur C=(6,0), taaki M=(0,4). N, A se C ki taraf 1.410=0.14 hisse par hai, isliye N=(0.84,6.88).

      MN2=0.842+2.882=0.7056+8.2944=9, isliye MN=3=12BC: (Half) sach hai. Lekin AN=1.4≠5, isliye (M2) fail hota hai, aur MN, BC ke parallel nahi hai.

      Isliye (M1) aur (Half) se (M2) zaroori nahi hota: M ke around 12BC radius wala circle AC ko do points par mil sakta hai, aur unmein se sirf ek midpoint hai. In converses mein Theorem 7 sabse useful hai, isliye chapter mein usi ko jagah di gayi hai.

  23. ProofHardNCERT End-of-chapter Exercises, Q23

    Samjhao ki Section 12.4 ka half-turn method poore plane ko bina gaps aur bina overlaps ke kyun dhakta hai.

    Solution dikhao

    Tile ABCD ko BC ke midpoint ke about uske half-turn ke saath jodo, jiske corners A′, C, B, D′ hain (A′ aur D′, A aur D ki images hain). Dono milkar hexagon A B D′ A′ C D banate hain. Iski opposite sides (AB aur A′C, BD′ aur CD, D′A′ aur DA) ek doosre ki half-turn images hain, isliye equal aur parallel hain.

    Is hexagon ko AC ke saath length AC jitna slide karo: side AB bilkul side CA′ par aa jaati hai. Ise BD ke saath length BD jitna slide karo: side BD′, side DC par aa jaati hai. Isliye slide ki gayi copies original ke saath poore edges par fit hoti hain, madhumakhi ke chhatte ke cells ki tarah. Hexagons ki rows ek strip bharti hain, aur strips aapas mein fit hoti hain, isliye hexagons plane ko bina gaps ya overlaps ke dhak lete hain.

    Har hexagon bilkul do tiles hai, isliye tiles bhi plane ko dhak leti hain. Yeh wahi slides hain jo Q18 mein mile the, isi wajah se tiler ki blue copies pehli tile ki slides hain.

  24. NumericalHardNCERT End-of-chapter Exercises, Q24

    Ek square tile ABCD ki side 10 cm hai. A se BC ke midpoint M tak, B se CD ke midpoint P tak, C se DA ke midpoint N tak, aur D se AB ke midpoint Q tak lines khichi gayi hain. Yeh beech mein ek chhota square gherti hain.

    1. Chhota square tile ka kitna hissa hai? (Decimal mein do.)

      Solution dikhao

      Tile ke centre ke about quarter-turn har line ko agli line par le jaata hai, isliye chaaron lines right angles par milti hain aur ek square gherti hain; uski side s kaho. Maan lo AM, BP ko E par aur DQ ko F par milti hai; tab F aur E chhote square ke corners hain, isliye FE=s.

      DPBQ mein DP aur QB equal aur parallel hain, isliye woh parallelogram hai aur DQ∥BP. Right △ABE mein line DQ, AB ke midpoint Q se guzarti hai aur BE ke parallel hai, isliye woh AE ko F par bisect karti hai (Theorem 7): AF=FE=s aur AE=2s.

      Quarter-turn △DAF ko △ABE par le jaata hai, isliye BE=AF=s. Isliye △ABE ka area 12×s×2s=s2 hai. Tile chhota square plus aise chaar triangles hai, isliye uska area 5s2 hai aur chhota square uska 15=0.2 hissa hai.

    2. Chhote square ka area aur uski side ki length nikaalo.

      Solution dikhao

      Answer20 cm²; side 20=25≈4.47 cm hai.

      5s2=100, isliye s2=20 cm² aur s=20=25≈4.47 cm.

  25. Quick answerEasy

    Parallelogram ABCD mein ∠A=(4x+7)∘ aur ∠B=(x+18)∘. ∠C nikaalo.

    Answer dekho

    Answer (A) 131∘

    (4x+7)+(x+18)=180 se 5x=155 milta hai, isliye x=31, ∠A=131∘ aur ∠C=∠A=131∘. 49∘, ∠B hai, aur 31 sirf x hai.
  26. Quick answerEasy

    △ABC ki sides ke midpoints D, E, F ko jodne se △DEF banta hai, jiska perimeter 14.5 cm hai. △ABC ka perimeter cm mein nikaalo.

    cm
    Answer dekho

    Answer 29 cm

    △DEF ki har side △ABC ki ek side ki aadhi hai, isliye perimeter double hota hai: 2×14.5=29 cm.
  27. Quick answerMedium

    True ya false: agar quadrilateral ke diagonals equal hain aur ek doosre ko bisect karte hain, to quadrilateral rectangle hai.

    Answer dekho

    Answer Sahi

    Ek doosre ko bisect karne wale diagonals use parallelogram banate hain (Theorem 4), aur equal diagonals wala parallelogram rectangle hota hai (Practice 12.2, Q1(c)).
  28. Quick answerEasy

    Kisi bhi triangle mein centroid har median ko vertex se ginte hue ratio mein baantta hai.

    Answer dekho

    Answer 2:1

    Vertex se centroid tak ka hissa centroid se side ke midpoint tak ke hisse ka do guna hai.
  29. ProofHard

    △ABC mein D, BC ka midpoint hai aur E, AD ka midpoint hai. Line BE ko badhane par woh AC ko F par milti hai. Prove karo ki AF=13AC.

    Solution dikhao

    D se DG∥BF draw karo, jo AC ko G par mile.

    △BCF mein D, BC ka midpoint hai aur DG∥BF, isliye G, CF ka midpoint hai (Theorem 7): FG=GC.

    △ADG mein E, AD ka midpoint hai aur EF∥DG, isliye F, AG ka midpoint hai: AF=FG.

    Isliye AF=FG=GC, aur AF=13AC.

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 meinYahan practice question
NCERT Exercise Set 12.1, Q1Practice 12.1 · Q1
NCERT Exercise Set 12.1, Q2Practice 12.1 · Q2
NCERT Exercise Set 12.1, Q3Practice 12.1 · Q3
NCERT Exercise Set 12.1, Q4Practice 12.1 · Q4
NCERT Exercise Set 12.1, Q5Practice 12.1 · Q5
NCERT Exercise Set 12.2, Q1Practice 12.2 · Q1
NCERT Exercise Set 12.2, Q2Practice 12.2 · Q2
NCERT Exercise Set 12.2, Q3Practice 12.2 · Q3
NCERT Exercise Set 12.2, Q4Practice 12.2 · Q4
NCERT Exercise Set 12.3, Q1Practice 12.3 · Q1
NCERT Exercise Set 12.3, Q2Practice 12.3 · Q2
NCERT Exercise Set 12.3, Q3Practice 12.3 · Q3
NCERT Exercise Set 12.3, Q4Practice 12.3 · Q4
NCERT Exercise Set 12.3, Q5Practice 12.3 · Q5
NCERT Exercise Set 12.4, Q1Practice 12.4 · Q1
NCERT Exercise Set 12.4, Q2Practice 12.4 · Q2
NCERT End-of-chapter Exercises, Q1Chapter ke end ki exercises · Q1
NCERT End-of-chapter Exercises, Q2Chapter ke end ki exercises · Q2
NCERT End-of-chapter Exercises, Q3Chapter ke end ki exercises · Q3
NCERT End-of-chapter Exercises, Q4Chapter ke end ki exercises · Q4
NCERT End-of-chapter Exercises, Q5Chapter ke end ki exercises · Q5
NCERT End-of-chapter Exercises, Q6Chapter ke end ki exercises · Q6
NCERT End-of-chapter Exercises, Q7Chapter ke end ki exercises · Q7
NCERT End-of-chapter Exercises, Q8Chapter ke end ki exercises · Q8
NCERT End-of-chapter Exercises, Q9Chapter ke end ki exercises · Q9
NCERT End-of-chapter Exercises, Q10Chapter ke end ki exercises · Q10
NCERT End-of-chapter Exercises, Q11Chapter ke end ki exercises · Q11
NCERT End-of-chapter Exercises, Q12Chapter ke end ki exercises · Q12
NCERT End-of-chapter Exercises, Q13Chapter ke end ki exercises · Q13
NCERT End-of-chapter Exercises, Q14Chapter ke end ki exercises · Q14
NCERT End-of-chapter Exercises, Q15Chapter ke end ki exercises · Q15
NCERT End-of-chapter Exercises, Q16Chapter ke end ki exercises · Q16
NCERT End-of-chapter Exercises, Q17Chapter ke end ki exercises · Q17
NCERT End-of-chapter Exercises, Q18Chapter ke end ki exercises · Q18
NCERT End-of-chapter Exercises, Q19Chapter ke end ki exercises · Q19
NCERT End-of-chapter Exercises, Q20Chapter ke end ki exercises · Q20
NCERT End-of-chapter Exercises, Q21Chapter ke end ki exercises · Q21
NCERT End-of-chapter Exercises, Q22Chapter ke end ki exercises · Q22
NCERT End-of-chapter Exercises, Q23Chapter ke end ki exercises · Q23
NCERT End-of-chapter Exercises, Q24Chapter ke end ki exercises · Q24

Quick check: MCQs

  1. Convex quadrilaterals4 questions
  2. 12.2 Parallelograms4 questions
  3. 12.3.1 Triangles ki Ek Important Property4 questions
  4. Quadrilateral ke midpoints4 questions
  5. 12.4 Kisi Bhi 4-gon se Plane ko Tile Karna3 questions
Glossary · 12

Key terms (English + Hindi)

centroidकेंद्रकkendrak
Woh point jahan triangle ke teeno medians milte hain, har median par vertex se do-tihaai doori par.Chapter mein pehli baar yahan aaya →
concurrentसंगामीsangami
Aisi lines jo sab ek common point se guzarti hain.Chapter mein pehli baar yahan aaya →
converseविलोमvilom
Agar A to B se banaya gaya statement agar B to A. Iska apna alag proof chahiye.Chapter mein pehli baar yahan aaya →
convex quadrilateralउत्तल चतुर्भुजuttal chaturbhuj
Aisa quadrilateral jiske saare internal angles 180∘ se kam hon. Non-convex quadrilateral mein ek dent hota hai, jahan ek angle 180∘ se zyada hota hai.Chapter mein pehli baar yahan aaya →
diagonalविकर्णvikarn
Polygon ke do aise vertices ko jodne wala segment jo ek hi side ke ends nahi hain.Chapter mein pehli baar yahan aaya →
medianमाध्यिकाmadhyika
Triangle ke vertex ko opposite side ke midpoint se jodne wala segment.Chapter mein pehli baar yahan aaya →
non-planar 4-gon
Chaar sides wali aisi figure jiske chaaron corners ek plane mein nahi hote, jaise diagonal ke saath moda gaya card.Chapter mein pehli baar yahan aaya →
parallelogramसमांतर चतुर्भुजsamantar chaturbhuj
Aisa quadrilateral jiski opposite sides parallel hon.Chapter mein pehli baar yahan aaya →
quadrilateralचतुर्भुजchaturbhuj
Chaar points ko jodne wale chaar segments se bani flat figure, jismein koi teen points ek line par nahi hote aur sides sirf shared corners par milti hain.Chapter mein pehli baar yahan aaya →
self-intersecting 4-gon
Chaar sides wali aisi figure jismein do sides ek doosre ko cross karti hain; yahan ise quadrilateral nahi gina jaata.Chapter mein pehli baar yahan aaya →
tiling
Ek ya zyada shapes ki copies se plane ko bina gaps aur bina overlaps ke dhakna.Chapter mein pehli baar yahan aaya →
Varignon parallelogram
Quadrilateral ki sides ke midpoints ko order mein jodne se bana parallelogram.Chapter mein pehli baar yahan aaya →

Tum chapter ke end tak pahunch gaye