Class 7 Maths Part 2 Chapter 6 NCERT Solutions Hinglish mein: Constructions and Tilings

Ganita Prakash Grade 7 (2026-27) ke Part 2 Chapter 6, Constructions and Tilings, ke har sawaal ka jawab: andar ke sawaal, har Figure it Out, Math Talk aur Try This, arch aur star designs, tangram figures aur saari tiling paheliyan. Page numbers book se milte hain (pages 136–163). Constructions mein sirf bina nishaan ka ruler aur compass use hota hai. Chhote notes: Part 2 Chapter 6 ke notes.

Aankhein aur perpendicular bisector (pages 136–139)

Isse aankh banaao. Support line XY banaao. XY ke aadhe se zyaada kisi bhi radius se X aur Y se XY ke upar aur neeche arcs banaao; wo A aur B pe milte hain. A centre aur AX radius lekar XY ke neeche X se Y tak arc banaao; B centre aur BX radius lekar XY ke upar X se Y tak arc banaao. Support lines mita do.

AB, XY ko kahan aur kis angle pe milti hai? XY ke beech ke bindu pe, 90° pe.

Kya lambaaiyan kuch bhi hon, do milne ke binduon se guzarti line hamesha perpendicular bisector hai? Haan. Kaunse do triangles congruent hone chahiye? ΔAOX aur ΔAOY. AX = AY aur AO common hai; saath hi SSS se ΔABX ≅ ΔABY (AX = AY, BX = BY, AB common), isliye ∠XAO = ∠YAO. To SAS se ΔAOX ≅ ΔAOY, jisse OX = OY aur ∠AOX = ∠AOY. Dono angles ka jod 180° hai, isliye har ek 90° hai.

Alag shape ki aankhein; kya C aur D, AB pe honge? Aise aur bindu C aur D chuno jinke liye CX = CY = DX = DY ho, aur unhe centre banaao. Haan, wo AB pe honge: X aur Y se barabar door kinhi do binduon ko jodne se perpendicular bisector milta hai, aur XY ka perpendicular bisector ek hi hai.

Saabit karo: X aur Y se barabar door har bindu XY ke perpendicular bisector pe hai. Maan lo P aisa bindu hai, aur XY ke neeche B bhi X aur Y se barabar door hai. Jo humne saabit kiya, uske hisaab se line PB, XY ka perpendicular bisector hai. Aisi line ek hi hai, isliye P usi pe hai. (Agar P khud XY pe ho, to PX = PY se P beech ka bindu hai, jo bisector pe bhi hai.)

Figure it Out (page 140)

1. Kya XY ke upar aur neeche ke arcs ka radius ek hi hona chahiye? Nahi. Upar waali jodi ke liye ek radius aur neeche waali ke liye doosra lo. Har milne ka bindu phir bhi X aur Y se barabar door hai, isliye dono perpendicular bisector pe hain, aur do bindu poori line tay kar dete hain.

2. Kya arcs ki dono jodiyan XY ki ek hi taraf ho sakti hain? Haan, do alag radius se, taaki milne ke bindu alag hon. Dono bindu X aur Y se barabar door hain, isliye unse guzarti line perpendicular bisector hai. Sateek line ke liye binduon ko kaafi door rakho.

3. Ek jodi ke andar, kya dono arcs ka radius ek hona chahiye? Haan. Milne ka bindu X aur Y se barabar door hona chahiye. Agar X waale arc ka radius Y waale se alag ho, to milne ka bindu ek sire ke paas hoga aur perpendicular bisector pe nahi hoga.

4. Chaar pankhudi ka design. Ek segment aur uska perpendicular bisector banaao, taaki do lines O pe right angle pe kaatein. Chaaron arms pe O se ek hi doori pe bindu lagao. Har pankhudi ek aadhe arm pe bani aankh hai: us aadhe arm ke perpendicular bisector pe, uske dono taraf barabar doori pe, do centre dhoondho aur dono arcs banaao. Phir outline rang se banaao.

Ek bindu pe 90° angle (page 141)

O pe 90° angle banaao. Kya arcs ki do jodiyan chahiye? O centre lekar line ko X aur Y pe kaato, taaki O, XY ka beech ka bindu ho. X aur Y se bade radius ke arcs banaao jo A pe milein, aur OA jodo. Nahi, ek jodi kaafi hai, kyunki O pehle se perpendicular bisector ka ek bindu hai.

Sulba-Sutras ke rassi constructions (pages 141–142)

1. Fig. 6.4 mein AB perpendicular bisector kyun hai? Rassi ke dono aadhe barabar hain aur dono poore tane rehte hain, isliye AX = AY aur BX = BY, har ek rassi ka aadha. A aur B dono X aur Y se barabar door hain, isliye line AB, XY ka perpendicular bisector hai.

2. Bindu O pe 90° ke liye rassi ke aur tareeke. (i) Rassi ke chhote tukde se O ke dono taraf line pe X aur Y lagao, taaki OX = OY ho; phir modi hui rassi ke tareeke se A dhoondho jahan AX = AY; OA line pe perpendicular hai. (ii) (Extra) Sulba-Sutras 3 : 4 : 5 ke anupaat ki bhujaon waale rassi triangle bhi use karte the: 3, 4 aur 5 barabar hisson pe gaanth lagaakar rassi ko triangle mein taana jaaye, to 3 aur 4 waali bhujaon ke beech ka angle right angle hota hai.

Angle bisection (pages 143–145)

Fig. 6.5 ki padosi support lines ke beech ka angle, aur use kaise banaayein. 360° ko 8 lines barabar baantein to 45°. 90° banaao aur use bisect karo.

Figure it Out (pages 144–145)

1. Chaar alag angles bisect karo. Har angle ke liye vertex O se arc banaao jo arms ko A aur B pe kaate; A aur B se AB ke aadhe se zyaada ek hi radius ke arcs banaao jo C pe milein; OC jodo.

2. Aath pankhudi waali figure. O se do perpendicular lines banaao aur chaaron right angles bisect karo, jisse 45° door aath lines milti hain. Har line pe O se ek hi doori pe bindu lagao aur O se us bindu tak aankh ki tarah, do barabar arcs se, pankhudi banaao.

3. Arcs doosri taraf, O ki taraf banaaye jaayein: kya OC phir bhi bisector hai? Haan, jab tak arcs O se alag kisi bindu C pe milein. OA = OB, AC = BC aur OC common hai, isliye SSS se ΔOAC ≅ ΔOBC, aur ∠AOC = ∠BOC.

4. Bisection se aur angles; kya 65.5° bana sakte ho? 90° se 45°, 22.5°, 11.25° vagairah milte hain; jodne se 67.5° = 45° + 22.5° aur 135° = 90° + 45°. 65.5° is tarah nahi ban sakta, kyunki wo aise aadhon ka jod nahi hai. (Book shayad 67.5° kehna chahti hai, jo ban sakta hai.)

5. Rassi se angle bisect karo. O pe rassi ki ek lambaai pakadkar dono arms pe A aur B lagao, taaki OA = OB. Rassi ke sire A aur B pe baandho, uska beech ka bindu taanke C lagao, jahan CA = CB, aur OC jodo.

6. Square mein jitni badi ho sakein utni chaar pankhudiyan. Perpendicular bisectors se chaaron bhujaon ke beech ke bindu lagao. Har beech ke bindu ko centre aur aadhi bhuja ko radius lekar square ke andar arc banaao. Har arc centre aur do konon se guzarta hai, aur chaaron arcs centre se konon tak chaar pankhudiyan banaate hain.

Dohraate units aur angle copy karna (pages 145–147)

Ek angle banaao aur uski copy karo. Vertex A se arc banaao jo arms ko B aur C pe kaate. X se ek ray aur usi radius ka arc banaao jo use Z pe kaate. Compass ko BC jitna kholo aur naye arc pe Y lagao, jahan ZY = BC. XY jodo. SSS se ΔABC ≅ ΔXYZ, isliye ∠X = ∠A.

Figure it Out (page 147)

1. Chaar angles ki alag disha mein copy karo. Har angle ke liye wahi steps, aur nayi ray apni pasand ki kisi bhi disha mein.

2. Fig. 6.6 banaao. Har unit ek pankha hai: do barabar arms aur unke siron ke beech vertex centre waala arc. Ek unit banaao. Ek arm ke door waale sire pe wahi angle ulti taraf copy karo, usi arm lambaai aur radius se, taaki padosi units ek arm share karein. Aise hi aage badho, phir har doosre unit mein rang bharo.

Parallel lines (pages 147–148)

Figure it Out (page 148)

1. Parallel lines ki chaar jodiyan. Har ek ke liye line m aur use A pe kaatti transversal l banaao; l pe B chuno; A ka angle B pe corresponding jagah copy karo; B se guzarti nayi line m ke parallel hai.

2. Aath point waala star. Centre O ka circle banaao aur right angles bisect karke O se 45° door aath lines banaao; wo circle ko A se H tak kaatti hain. Star ka har point ek parallelogram poora karta hai: S ke liye A se OH ke parallel line aur H se OA ke parallel line banaao; wo S pe milti hain. T, U, V, W, X, Y aur Z ke liye bhi yahi karo, har bahari bindu ko O se jodo, aur har parallelogram ka ek aadha hissa rang do.

Arch designs (pages 149–151)

Arches kaise banaaye gaye? Pehle unhe kaagaz ya patthar jaisi samtal sateh pe support lines aur arcs se banaakar.

Trefoil arch: support lines aur arch. AD banaao. Ek angle copy karke A aur D pe barabar angles banaao, aur AB = DC lekar B aur C lagao; BC jodo. Perpendicular bisectors se AB, BC aur CD ke beech ke bindu nikaalo, aur AB, BC aur CD pe bahar ki taraf semicircles banaao. Arch zyaada sundar lage to radius badal sakte ho.

Pointed arch: kaun si support lines? Kya beech ke binduon se ban sakta hai? Upar milte do barabar segments, jinke beech ke bindu lage hon. Haan: har segment pe neeche ka aadha hissa bahar ubhra arc aur upar ka aadha andar dabaa arc banaao, Class 6 ki wavy wave ki tarah. Dono upar ke arcs ek nok pe milte hain.

Figure it Out (page 151): 1. Alag shape ke pointed arches; 2. Apne arches. Arcs ke radius, ya support lines ke beech ka angle, badalkar patle ya chaude arches banaao; apne designs ke liye semicircles aur pointed arcs milao.

Regular hexagon (pages 151–154)

Try This: barabar bhujaon waala pentagon aur hexagon. Ek lambaai chuno, ek bhuja banaao aur compass se har agli bhuja usi lambaai ki rakho. Bhujaayein barabar aati hain, par angles aksar nahi, isliye shapes regular nahi hoti.

Kya chhe ke chhe triangle equilateral honge? Kya chhe triangle ek bindu ke chaaron taraf fit honge? Haan. Ek bindu ke chaaron taraf angles ka jod 360° hai, aur chhe 60° angles theek 360° banaate hain, isliye chhe congruent equilateral triangle bina gap ya overlap fit hote hain. Unse bane hexagon ki bhujaayein barabar hain, aur uska har angle 60° + 60° = 120° hai, isliye wo regular hai.

Kya 70° ka angle gap mein fit hoga? 40° + 60° + 50° + 30° + 40° + 90° = 310°, isliye gap ∠AOI = 360° − 310° = 50° hai. 70° ka angle fit nahi hoga; wo 20° overlap karega.

AOD, BOE aur COF seedhi lines kyun hain? Jaise ∠AOB + ∠BOC + ∠COD = 60° + 60° + 60° = 180°, jo seedha angle hai. BOE aur COF ke liye bhi yahi sach hai.

4 cm aur 5 cm side ke regular hexagon banaao. 4 cm (ya 5 cm) radius ka circle banaao. Usi khulaav se circle pe ek bindu lagao aur circle pe aage badhte hue chhe bindu lagao; unhe kram se jodo. Har side radius ke barabar hai, kyunki har do padosi bindu centre ke saath equilateral triangle banaate hain.

120° kaise banaayein? ∠CAX = 60° kyun hai? 60° banaao aur seedhi line pe uske bagal ka angle lo: 180° − 60° = 120°. 60° ke construction mein AB = AC (pehla arc) aur BC = AB (doosra arc, wahi radius), isliye ΔABC equilateral hai aur uska har angle 60° hai.

Jude hue constructions (pages 154–155)

Math Talk: 30° aur 15°. 60° bisect karke 30°, aur 30° bisect karke 15°.

6-pointed star; kya uske chhe point triangle equilateral hain? Regular hexagon GHIJKL banaao aur uski bhujaayein aage badhaao; ya ek bade regular hexagon ABCDEF ke ek chhod ek konon ko jodkar do bade equilateral triangle banaao. Haan, har point triangle equilateral hai: uska har base angle seedhi line pe hexagon ke 120° angle ke bagal mein hai, isliye 60° hai, aur teesra angle 180° − 120° = 60° hai.

Figure it Out (pages 154–155)

1. (a) Inflexed arc. Ek hi oonchaai ke do khade khambhe banaao aur unke sire jodo; is line ka beech ka bindu M lagao. Har khambhe ke sire ke theek upar, chaudaai ke aadhe jitni oonchaai pe, ek bindu lagao. In binduon ko centre lekar har khambhe ke sire se M ke theek upar waale bindu tak arcs banaao; dono arcs upar ek nok pe milte hain.

(b) Sirf compass se chhe pankhudi ka phool. Ek circle banaao. Usi radius se circle pe aage badhte hue chhe bindu lagao. Har bindu ko centre aur wahi radius lekar, circle ka wo hissa banaao jo pehle circle ke bahar hai, ek padosi bindu se doosre tak. Chhe arcs nokdaar siron waali chhe pankhudiyan banaate hain.

(c) Circle mein hexagon. Circle banaao, radius ko circle pe chhe baar aage badhaao aur chhe binduon ko jodo.

(d) Ring mein chhe circle. Radius r ka circle banaao aur r ko circle pe aage badhaakar chhe bindu pao. Ek radius bisect karke r/2 nikaalo, aur chhe binduon pe r/2 radius ke circle banaao; padosi circle bas ek doosre ko chhoote hain.

(e) Triangles waala hexagon. Ek bada regular hexagon banaao aur centre se use chhe equilateral triangles mein baanto. Saari bhujaon ke beech ke bindu lagao aur unhe bhujaon ke parallel lines se jodo, taaki har triangle chaar chhote equilateral triangles mein kat jaaye. Har chhote triangle mein uske konon ko uske centre se jodo, jo do angle bisectors ke milne pe milta hai.

2. Optical illusion. Hume ek chamakta safed triangle doosre triangle ke upar dikhta hai, jabki uski koi bhuja bani hi nahi. Hamara dimaag teen kaale circles ke kate konon ko jodkar triangle bana leta hai. Ise dobara banaane ke liye halke se ek equilateral triangle banaao, har kone pe kaala circle banaao, ek doosre, ulte triangle ke kone banaao, aur circles aur triangle ke andar ki halki lines mita do.

3. Hexagon mein star. Regular hexagon banaao aur ek chhod ek konon ko jodkar do equilateral triangle banaao. Unki outline chhe point waala star hai jiske point hexagon ke kone hain. Hexagon aur star ki outline rakho aur andar ki lines mita do.

4. Line l ke bahar ke bindu P se l pe perpendicular. P centre lekar arc banaao jo l ko X aur Y pe kaate, taaki PX = PY ho. Ab P, XY ke perpendicular bisector pe hai. l ki doosri taraf X aur Y se barabar radius ke arcs banaao jo Q pe milein. Line PQ, l pe perpendicular hai.

Tangram (pages 155–156)

Tangram ke tukde har figure kaise banaate hain? Har figure saaton tukde use karti hai aur uska area asli square jitna hai. Pehle do bade triangle, jo aadha area dhakte hain, figure ke sabse bade triangle jaise hisson mein rakho; phir medium triangle, square aur parallelogram fit karo; do chhote triangle aksar kaan, pankh aur pair jaisi patli nokon ko bharte hain. Chaar figures hal karke:

• C shape: baayin taraf neeche tak ek bada triangle, baayin taraf nok waala medium triangle, aur doosra bada triangle; upar ek chhota triangle aur square; neeche parallelogram aur doosra chhota triangle.
• Machhli: sir ek bada triangle; ek patti jisme ek chhota triangle (upar ka pankh), square, doosra chhota triangle aur parallelogram (neeche ka pankh); sharir doosra bada triangle; poonchh medium triangle.
• Chamgaadad: pankh do bade triangle; unke beech parallelogram, diamond ki tarah tircha square jiske neeche ek chhota triangle, aur medium triangle jiske neeche doosra chhota triangle.
• Billi: chehra square, jiske kaan do chhote triangle; sharir do bade triangle aur unke beech medium triangle; poonchh parallelogram.

2 × 1 tiles se grid dhakna (pages 157–160)

Kya 4 × 7 grid dhak sakta hai? 5 × 7 grid? 4 × 7: haan, 7 columns mein se har ek mein do khadi tiles. 5 × 7: nahi.

Wajah poori karo. Har tile theek 2 squares dhakta hai, isliye kitni bhi tiles even ginti ke squares hi dhakengi. 5 × 7 grid mein 35 squares hain, jo odd hai, isliye tiles ka koi bhi set use theek se nahi dhak sakta.

m × n grid jisme m aur n dono even hain: haan, khadi tiles har column bhar deti hain, kyunki rows ki ginti even hai.

m aur n mein ek even, doosra odd: haan. Rows ki ginti even ho to har column khadi tiles se bharo; columns ki ginti even ho to har row leti tiles se.

m aur n dono odd: nahi, kyunki tab m × n odd hai.

5 × 3 grid jiska ek kona hata hai: haan. Chhoti row mein ek leti tile rakho, phir neeche ke 4 × 3 block ko khadi tiles se bharo, har column mein do.

5 × 3 grid jiski doosri row ka beech ka square hata hai: haan. Rows ko upar se 1 se 5 aur columns ko 1 se 3 ginte hue: column 1 mein khadi tiles (rows 1–2 aur 3–4), column 2 mein (rows 3–4) aur column 3 mein (rows 2–3 aur 4–5), aur row 1 mein leti tile (columns 2–3) aur row 5 mein (columns 1–2).

Fig. 6.13 (upar ka beech ka square hata hai): kya dhak sakta hai? Pakka kaise jaanein? Nahi. Squares ko chessboard ki tarah rango. Har 2 × 1 tile ek kaala aur ek safed square dhakta hai, isliye dhake gaye hisse mein dono barabar chahiye. Is hisse mein 8 safed aur 6 kaale squares hain.

Agar saada grid dhak sakta hai, to kya kaala-safed grid bhi? Aur ulta? Haan, dono. Har tile do padosi squares dhakta hai, jinke rang hamesha alag hote hain, isliye saadi tiling apne aap har tile ko ek kaale aur ek safed square pe rakhti hai. Aur rang bhool jaao to kaali-safed tiling ek saadi tiling hi hai.

Kya Fig. 6.14 ka kaala-safed hissa dhak sakta hai? Nahi: 8 safed squares, 6 kaale.

Aisa aur square jise hataane se 5 × 3 grid na dhak sake. Koi bhi kaala square: neeche ka beech waala square, centre ka square, ya doosri ya chauthi row ka pehla ya aakhiri square. 8 safed squares (kone, rows 2 aur 4 ke beech ke squares, ya row 3 ke sire ke squares) mein se koi bhi hataane pe 7 aur 7 bachte hain, aur ye saare grid dhak sakte hain.

Figure it Out (page 160)

1. L-shape tile waala hissa: ho sakta hai. Hisse mein 12 squares hain, isliye 4 tiles chahiye. Ek tareeka, rows ko upar se 1 se 4 aur columns ko 1 se 4 ginte hue: tile 1 row 1 (columns 1, 2) aur row 2 column 1 dhakti hai; tile 2 row 2 column 2 aur row 3 (columns 2, 3); tile 3 row 3 column 1 aur row 4 (columns 1, 2); tile 4 row 3 column 4 aur row 4 (columns 3, 4).

2. 8 × 8 grid jiske do aamne saamne ke kone hate hain: nahi ho sakta. Isme 62 squares hain. Chessboard ki tarah rangne pe aamne saamne ke kone ek hi rang ke hote hain, isliye ek rang ke 32 aur doosre ke 30 squares bachte hain, jabki har 2 × 1 tile har rang ka ek square dhakta hai.

Plane ki tiling (pages 160–162)

Aisi shape jiski copies plane dhakein; aur kaun se regular polygons? Square, aur equilateral triangle aur regular hexagon bhi. (Extra) Koi bhi rectangle, parallelogram ya triangle bhi chalta hai. Ye teen hi aise regular polygons hain jo akele plane dhakte hain: har bindu pe angles ka jod 360° hona chahiye, jo 6 × 60°, 4 × 90° aur 3 × 120° se hota hai, par pentagon ka 108° aur zyaada bhujaon waale polygons ke bade angles 360° theek nahi bhar sakte.

Roz ki zindagi mein tilings. Farsh aur deewar ki tiles, eenton ki deewarein, paving stones, chessboard, madhumakkhi ke chhatte aur tataiyon ke chhatton ke hexagon khaane.