Class 7 Maths Part 2 Chapter 1 NCERT Solutions Hinglish mein: Geometric Twins

Ganita Prakash Grade 7 Part 2 (2026-27) ke Chapter 1, Geometric Twins, ke har “Figure it Out” sawal ka jawab, saath mein text ke beech ke sawaal bhi. Page numbers Part 2 book ke hisaab se hain. Jawabon mein “≅” ka matlab hai “congruent hai”. Short notes: Part 2 Chapter 1 ke notes.

Signboard ka symbol dobara banana (pages 1–3)

Symbol ko doosre board pe kaise banayein? Chhote symbol ko trace kar sakte hain, par bade symbol ke liye naap lekar unse copy banana aasaan hai.

Kya arm lengths AB aur BC figure ko bilkul waisa banane ke liye kaafi hain? Nahi. AB = 4 cm aur BC = 8 cm ke saath arms kai alag angles pe mil sakte hain, aur har angle se alag shape banta hai.

Kya koi aur naap madad karega? Haan, arms ke beech ka angle, ∠ABC. Do arm lengths aur ye angle milkar shape aur size fix karte hain.

AB = 4 cm, BC = 8 cm aur ∠ABC = 80° se symbol banao.

1. BC = 8 cm kheencho.
2. Protractor ko B pe rakho aur BC se 80° mark karo. B se mark ke through ek ray kheencho.
3. Ray pe B se 4 cm door A mark karo. Arms AB aur BC se symbol ban gaya.

Agar do symbols ki arm lengths same hon, to kya wo congruent hain? Hamesha nahi, kyunki arms ke beech ke angles alag ho sakte hain. Agar arm lengths aur angle dono barabar hon, to symbols congruent hain.

Figure it Out (pages 3–4)

1. Check karo ki dono figures congruent hain ya nahi. Ye congruent nahi hain. Inke arms lagbhag barabar lambe hain, par arms ke beech ka angle pehli figure mein lagbhag 90° aur doosri mein lagbhag 70° hai. Ek ko trace karke doosri pe rakho: arms fit nahi hote.

2. Jo jodiyan congruent lagti hain unhe circle karo. Dono boondein congruent lagti hain (ek doosri ki ghumi hui copy hai), aur pattiyon ki dono jodiyan bhi (ek doosri ki palti hui copy hai). Baadal congruent nahi hain, kyunki ek bada hai aur unki outline alag hai, aur star bursts bhi congruent nahi, kyunki ek saaf taur pe bada hai.

3. Diye gaye (a) circle, (b) rectangle ki congruent figure banane ke liye kya naapoge? Do (a) circles, (b) rectangles congruent hain ya nahi, kaise check karoge?

• (a) Circle ke liye sirf uska radius (ya diameter) chahiye. Do circles congruent hain agar unke radius barabar hon.
• (b) Rectangle ke liye uski length aur breadth chahiye, kyunki uske saare angles 90° hote hain. Do rectangles congruent hain agar unki lengths barabar hon aur breadths barabar hon.

4. Dikhayi gayi jaisi do figures (ek lamba segment jisse ek chhota segment milta hai) congruent hain ya nahi, kaise check karein? Har figure mein chaar cheezein naapo: lambe segment ki lambai, chhote segment ki lambai, lambe segment pe kitni door chhota segment milta hai, aur dono ke beech ka angle. Agar chaaron match karein, to figures congruent hain. Book ki jodiyon mein har jodi ki dono figures ke ye chaar naap same aate hain, isliye har jodi congruent hai: ek figure doosri ko bas nayi jagah khiska kar bani hai.

Triangles ki congruence (pages 4–6)

Meera aur Rabia kya kar sakti hain? Frame ke naap lein, jaise uski side lengths, aur cardboard pe unhi naap se triangle banaayein.

Kya tum Meera se sahmat ho ki teen sides kaafi hain? Haan. Di gayi side lengths se triangle sirf ek hi shape aur size mein banta hai, jaisa neeche ka construction dikhata hai.

Kya 4 cm, 6 cm aur 8 cm sides wala triangle dobara bana sakte ho? Haan. AB = 6 cm kheencho. Compass se A se 4 cm radius ka arc aur B se 8 cm radius ka arc lagao. Wo E pe milte hain. AE aur BE jodo.

Kya ΔABE aur ΔABF congruent hain? Haan. Circles AB ke neeche F pe bhi milte hain, par ΔABF, AB ke aar-paar ΔABE ki mirror image hai, line ke neeche usi tarah bana hua. Ek ka cut-out palatne ke baad doosre pe fit ho jaata hai. Dono ki sides 4 cm, 6 cm aur 8 cm hain, isliye aise saare triangles congruent hain (SSS condition).

Congruence likhne ka tareeka (pages 6–8)

ΔABC aur ΔXYZ ke kaunse vertices ek doosre pe aane chahiye? A, X pe, B, Y pe aur C, Z pe, taaki barabar sides ek doosre pe aayein.

Kya vertices ko milaane ke aur tareeke hain? In triangles ke liye nahi, kyunki inki teeno sides alag lambai ki hain aur har side sirf usi lambai wali side pe aa sakti hai. Do ya teen barabar sides wale triangle ko ek se zyaada tareekon se milaaya ja sakta hai (page 21 ka sawal 4 dekho).

Fig. 1.1 (diagonal BD wala rectangle ABCD) mein congruent triangles ki jodi pehchaano. ΔABD aur ΔCDB. AB = CD aur AD = CB, kyunki ye rectangle ki aamne-saamne ki sides hain, aur BD common hai, isliye ye SSS se congruent hain.

Kya correspondence A–C, B–B, D–D ho sakta hai? Nahi. Isse ΔABD ki AB, ΔCDB ki CB pe aayegi, aur ye sides barabar hon, zaroori nahi.

Sahi correspondence batao. AB ko CD pe aana hai, AD ko CB pe aur BD ko DB pe. Isliye A ka match C, B ka D aur D ka B: ΔABD ≅ ΔCDB.

Figure it Out (pages 8–9)

1. ΔHEN ≅ ΔBIG. Is congruence ko likhne ke baaki sahi tareeke batao. H ka match B, E ka I aur N ka G hai, isliye letters jodiyon mein hi badalne chahiye:

• ΔHNE ≅ ΔBGI
• ΔEHN ≅ ΔIBG
• ΔENH ≅ ΔIGB
• ΔNHE ≅ ΔGBI
• ΔNEH ≅ ΔGIB

Inme se har ek ulta bhi likh sakte ho, jaise ΔBIG ≅ ΔHEN.

2. Kya triangles RED aur JAM congruent hain? Haan, SSS se: RE = JA = 3.5 cm, ED = AM = 5 cm aur RD = JM = 6 cm. R, 3.5 cm aur 6 cm sides ke beech hai, aur J bhi; E, 3.5 cm aur 5 cm sides ke beech hai, aur A bhi. Isliye ΔRED ≅ ΔJAM.

3. AB = AD aur CB = CD. Kya congruent triangles ki koi jodi hai? Kya AC, ∠BAD aur ∠BCD ko do barabar hisson mein baantta hai? Haan. ΔABC aur ΔADC mein AB = AD, CB = CD aur AC common hai, isliye SSS se ΔABC ≅ ΔADC. Corresponding angles barabar hain, isliye ∠BAC = ∠DAC aur ∠BCA = ∠DCA. Isliye AC, ∠BAD aur ∠BCD dono ko do barabar hisson mein baantta hai.

4. DF = DG aur FE = GE. Kya ΔDFE aur ΔGED congruent hain? Figure ke dono triangles, DFE aur DGE, SSS se congruent hain: DF = DG, FE = GE, aur DE common hai. Sahi statement hai ΔDFE ≅ ΔDGE (D ka D, F ka G, E ka E). ΔDFE ≅ ΔGED likhna galat hoga, kyunki ye side DF ko GE se jodega.

Angles naapna (pages 9–10)

Kya Meera aur Rabia teen angles se triangle bilkul waisa bana sakti hain? Nahi.

30°, 70° aur 80° angles se kya bilkul copy ban sakti hai? Nahi. Koi bhi base kheencho, ek sire pe 30° aur doosre pe 70° banao; teesra angle apne aap 80° ho jaata hai. Lamba base lene se same angles wala bada triangle banta hai. In sab triangles ka shape same hai par size nahi, isliye ye sab congruent nahi hain.

Do sides aur unke beech ka angle (page 10)

AB = 6 cm, AC = 5 cm aur ∠A = 30° wala triangle banao aur classmates se milaao. AB = 6 cm kheencho, A pe 30° ka angle banao, aur nayi arm pe A se 5 cm door C mark karo. BC jodo. Sabko wahi triangle milta hai, kyunki AB kheenchne ke baad angle AC ki disha fix kar deta hai aur lambai C ko fix kar deti hai; tab BC ki sirf ek hi lambai ho sakti hai. Isliye do barabar sides aur barabar included angle wale triangles congruent hain: SAS condition.

Do sides aur non-included angle (pages 10–12)

AB = XY = 6 cm, AC = XZ = 4 cm aur ∠B = ∠Y = 30°. Kya triangles congruent hain? Zaroori nahi. PQ = 6 cm kheencho aur P se PQ ke saath 30° pe line l. Q se 4 cm radius ka arc l ko do points, R aur S, pe kaatta hai. ΔPQR aur ΔPQS dono ke naap diye gaye naap jaise hain, par dono alag triangles hain. Isliye SSA condition se triangles congruent nahi hote.

Do angles aur unke beech ki side (pages 12–13)

BC = YZ = 5 cm, ∠B = ∠Y = 50° aur ∠C = ∠Z = 30°. Kya triangles congruent hain? Haan. BC = 5 cm kheencho, B pe 50° aur C pe 30° ka angle banao. Dono arms sirf ek point A pe milti hain, isliye in naap wala har triangle ek hi hai. Ye ASA condition hai, aur ΔABC ≅ ΔXYZ.

O, AD aur BC ka midpoint hai. AB aur CD ke baare mein kya keh sakte ho? AO = OD aur BO = OC, aur ∠AOB = ∠DOC kyunki ye vertically opposite angles hain. Isliye SAS se ΔAOB ≅ ΔDOC, jahan A ka match D, O ka O aur B ka C hai. AB aur DC corresponding sides hain, isliye AB = DC.

Figure it Out (pages 13–14)

1. Kya triangles ABC aur XYZ congruent hain? Haan, SAS se. AB = XZ = 7 cm, BC = ZY = 5 cm, aur in sides ke beech ke angles barabar hain: ∠B = ∠Z = 47°. B ka match Z, A ka X aur C ka Y hai, isliye ΔABC ≅ ΔXZY.

2. CD aur AB parallel hain aur AB = CD. Baaki barabar parts kaunse hain? DB aur CA, O pe katte hain. DB parallel lines ki transversal hai, isliye alternate angles ∠CDO aur ∠ABO barabar hain. CA bhi transversal hai, isliye ∠DCO = ∠BAO. In angles ke beech DC = BA hai, isliye ASA se ΔDOC ≅ ΔBOA. Isliye OD = OB aur OC = OA, yaani O, DB aur CA dono ka midpoint hai. Saath hi ∠DOC = ∠BOA (vertically opposite angles).

3. ∠ABC = ∠DBC aur ∠ACB = ∠DCB. Dikhao ki ∠BAC = ∠BDC. Kya triangles congruent hain? ΔABC aur ΔDBC mein B pe angles barabar hain, C pe angles barabar hain, aur BC common hai, isliye ASA se ΔABC ≅ ΔDBC. Isliye corresponding angles hone se ∠BAC = ∠BDC. (Angle sum se bhi yahi milta hai: ∠BAC = 180° − ∠ABC − ∠ACB = 180° − ∠DBC − ∠DCB = ∠BDC.)

4. ∠ABD = ∠DCA aur ∠ACB = ∠DBC. Barabar parts pehchaano. Diye gaye angles jodo: ∠ABC = ∠ABD + ∠DBC = ∠DCA + ∠ACB = ∠DCB. ΔABC aur ΔDCB mein ∠ABC = ∠DCB, BC common hai, aur ∠ACB = ∠DBC, isliye ASA se ΔABC ≅ ΔDCB. Barabar parts hain AB = DC, AC = DB, ∠BAC = ∠CDB aur ∠ABC = ∠DCB, aur diye gaye angles bhi.

Do angles aur non-included side (pages 14–16)

∠A = ∠X = 35°, ∠C = ∠Z = 75° aur BC = YZ = 4 cm. Kya triangles congruent hain? Haan. Angle sum se ∠B = 180° − 35° − 75° = 70°, aur isi tarah ∠Y = 70°. Ab ∠B = ∠Y, BC = YZ aur ∠C = ∠Z, isliye ASA se ΔABC ≅ ΔXYZ. Diye gaye facts do angles aur ek aisi side hain jo unke beech nahi: ye AAS condition hai, jo congruence ke liye kaafi hai.

Right triangle mein do sides (pages 16–17)

BC = YZ = 4 cm, ∠B = ∠Y = 90° aur AC = XZ = 5 cm. Kya triangles congruent hain? Banao: QR = 4 cm kheencho, Q pe QR ke perpendicular line l kheencho, aur R se 5 cm radius ke arc se l ko kaato. Us point ko P kaho aur PR jodo.

Kya arc QR ke neeche bhi l se milta hai? Kya wo triangle alag hai? Haan, wo l ke neeche wale hisse ko ek doosre point pe milta hai. Wo triangle QR ke aar-paar ΔPQR ki mirror image hai, isliye palatne ke baad ΔPQR ke congruent hai. In naap wale saare triangles congruent hain, isliye ΔABC ≅ ΔXYZ. Naapoge to dono mein PQ 3 cm aata hai. Ye RHS condition hai: right angle, hypotenuse (right angle ke saamne ki side) aur ek aur side.

Isosceles aur equilateral triangles ke angles (pages 17–19)

ΔABC isosceles hai jisme AB = AC aur ∠A = 80°. ∠B aur ∠C ke baare mein kya keh sakte hain? Altitude AD kheencho. RHS se ΔADB ≅ ΔADC (D pe right angles, AB = AC, AD common), isliye ∠B = ∠C. Dono milkar 180° − 80° = 100° hain, isliye ∠B = ∠C = 50°.

Equilateral triangle ke angles kya hain? Teeno barabar hain, kyunki har jodi ki sides barabar hai. Teen barabar angles jo milkar 180° hon, har ek 60° hai.

Construction se check karo. BC = 5 cm kheencho. Compass ko 5 cm kholkar B aur C se arcs lagao; wo A pe milte hain. AB aur AC jodo. Protractor se teeno angles naapo: har ek 60° hai.

Asli zindagi mein congruent triangles (pages 19–20)

Har picture mein dikhne waale congruent triangles ke baare mein batao.

• Louvre Museum, Paris: glass pyramid ke chaar dhalaan waale faces congruent triangles hain, aur uska metal frame har face ko ek hi shape aur size ke bahut se chhote tukdon mein baantta hai.
• Pyramid of Giza, Egypt: iske chaar dhalaan waale faces congruent triangles hain jo square base pe khade hain.
• Dome design: frame bahut se chhote triangles se bana hai; dome ke har ghere ke triangles congruent hain.
• Rangoli design: beech ka star do rangon ke congruent triangles se bana hai jo centre ke chaaron taraf lage hain, aur background mein bhi congruent shapes dohraaye gaye hain.
• Rabindra Setu (Howrah Bridge): steel frame poore bridge pe wahi triangles dohraata hai, aur bridge ke dono aadhe hisse ek doosre ki mirror image hain.

Figure it Out (pages 20–21)

1. ΔAIR ≅ ΔFLY. Corresponding vertices, sides aur angles batao.

Corresponding	Jodiyan
vertices	A aur F, I aur L, R aur Y
sides	AI aur FL, IR aur LY, AR aur FY
angles	∠A aur ∠F, ∠I aur ∠L, ∠R aur ∠Y

2. Kaunsi jodiyon ke triangles congruent hain?

• (a) AB = DE, BC = EF, CA = DF: SSS se congruent, ΔABC ≅ ΔDEF.
• (b) AB = EF, ∠A = ∠E, AC = ED: ∠A, AB aur AC ke beech hai, aur ∠E, EF aur ED ke beech, isliye SAS se congruent, ΔABC ≅ ΔEFD.
• (c) AB = DF, ∠B = ∠D = 90°, AC = FE: dono right triangles hain, AC aur FE hypotenuse hain, aur AB = DF ek aur side hai, isliye RHS se congruent, ΔABC ≅ ΔFDE.
• (d) ∠A = ∠D, ∠B = ∠E, AC = DF: do angles aur ek aisi side jo unke beech nahi, isliye AAS se congruent, ΔABC ≅ ΔDEF.
• (e) AB = DF, ∠B = ∠F, AC = DE: ∠B, AB aur AC ke beech nahi hai, isliye ye SSA hai. In naap se triangles congruent nahi hote; do alag triangles ho sakte hain.

3. OB = OC aur OA = OD. Dikhao ki AB, CD ke parallel hai. ΔAOB aur ΔDOC mein OA = OD, OB = OC, aur ∠AOB = ∠DOC (vertically opposite angles), isliye SAS se ΔAOB ≅ ΔDOC. Isliye ∠OAB = ∠ODC, yaani ∠BAD = ∠CDA. Ye transversal AD dwara lines AB aur CD ke saath bane alternate angles hain. Alternate angles barabar hain, isliye AB ∥ CD.

4. ABCD ek square hai. Dikhao ki ΔABC ≅ ΔADC. Kya ΔABC, ΔCDA ke bhi congruent hai? Square ki saari sides barabar hain aur AC common hai. AB = AD, BC = DC aur AC = AC, isliye SSS se ΔABC ≅ ΔADC. Saath hi AB = CD, BC = DA aur AC = CA, isliye ΔABC ≅ ΔCDA bhi. Aisa isliye hota hai kyunki ΔABC ki do sides barabar hain, AB = BC, isliye use ΔADC pe do tareekon se rakh sakte hain.

Aise aur examples do jahan triangles do tareekon se congruent hon, aur chhe tareekon se. Koi bhi do congruent isosceles triangles: agar PQ = PR aur XY = XZ, same lambaaiyon aur P aur X pe same angle ke saath, to ΔPQR ≅ ΔXYZ aur ΔPQR ≅ ΔXZY bhi. Same side wale do equilateral triangles chhe tareekon se congruent hain: ΔABC ≅ ΔXYZ, ΔXZY, ΔYXZ, ΔYZX, ΔZXY aur ΔZYX.

5. Agar A circle ka centre hai, to ∠B aur ∠C nikaalo. AB aur AC radius hain, isliye AB = AC aur ∠B = ∠C. ∠A = 120° hai, isliye ∠B + ∠C = 60°, aur ∠B = ∠C = 30°.

6. Missing angles nikaalo. Book andar ke chaar points ka naam nahi deti, isliye hum unhe G, H, O aur M kehte hain, jaisa picture mein dikhaya hai. Figure ek rectangle hai, isliye uske corners 90° ke hain, aur seedhi line pe angles milkar 180° hote hain.

Triangles ko isi kram mein hal karo, har ek pichhle waalon ka istemaal karta hai:

Triangle	Missing angles aur reason
ΔCRU	CR = CU aur ∠C = 90°, isliye ∠CRU = ∠CUR = 45°
ΔRVO	RV = VO aur ∠RVO = 68°, isliye ∠VRO = ∠VOR = 56°
ΔVDO	∠OVD = 180° − 68° = 112°; VO = VD, isliye ∠VOD = ∠VDO = 34°
ΔUHA	UA = UH, isliye ∠UHA = ∠UAH = 56° aur ∠AUH = 68°
ΔUGH	UG = UH aur ∠GUH = 34°, isliye ∠UGH = ∠UHG = 73°
ΔRGO	∠RGO = 180° − 34° − 44° = 102°
ΔURG	R pe seedhi line: ∠URG = 180° − 45° − 34° − 56° = 45°; U pe seedhi line: ∠RUG = 180° − 45° − 34° − 68° = 33°; isliye ∠RGU = 102°
ΔGOH	G ke chaaron taraf: ∠OGH = 360° − 102° − 102° − 73° = 83°; isliye ∠GHO = 180° − 46° − 83° = 51°
ΔAHK	∠AKH = 180° − 34° − 44° = 102°
ΔHOM	∠OHM = 180° − 90° − 56° = 34°
ΔHKM	H ke chaaron taraf: ∠KHM = 360° − 73° − 51° − 34° − 44° − 56° = 102°; isliye ∠HKM = 180° − 102° − 30° = 48°
ΔKML	K pe seedhi line: ∠MKL = 180° − 102° − 48° = 30°; ∠MLK = 90°, isliye ∠KML = 60°
ΔFMB	MF = FB = BM, isliye ye equilateral hai: 60°, 60° aur 60°
ΔMLB	∠LBM = 90° − 60° = 30° aur ∠MLB = 90°, isliye ∠LMB = 60°
ΔOFM	F pe seedhi line: ∠OFM = 180° − 98° − 60° = 22°; M ke chaaron taraf: ∠OMF = 360° − 56° − 30° − 60° − 60° − 60° = 94°; isliye ∠MOF = 64°
ΔDOF	O ke chaaron taraf: ∠DOF = 360° − 34° − 56° − 44° − 46° − 90° − 64° = 26°; isliye ∠ODF = 180° − 98° − 26° = 56°

Do checks: ∠VDO + ∠ODF = 34° + 56° = 90°, jo rectangle ka corner D hai. Aur teen nishaan wali sides bhi match karti hain: ASA (34°, RO = AH, 44°) se ΔRGO ≅ ΔAKH, aur sach mein ∠RGO = ∠AKH = 102°.

Puzzle time (page 23)

Safed squares ko 6 chhote congruent hisson mein baanto. 24 safed squares hain, isliye har hisse mein 4 squares honge. 4 squares ke chhe L-shapes se ye ho jaata hai, kuch ghume hue aur kuch palte hue: