Geometric constructions kaise karein?

Short mein jawab

Sirf bina nishaan waale ruler aur compass se tum segment ka perpendicular bisector, bindu pe 90° angle, angle ka bisector, angle ki copy aur kisi line ke parallel line bana sakte ho. Har tareeka congruent triangles ki wajah se chalta hai: compass se bani barabar lambaaiyan angles ko barabar hone pe majboor karti hain.

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Sirf ruler aur compass

Compass gol aur arcs banata hai aur ek lambaai ko ek jagah se doosri jagah le jaata hai; bina nishaan ka ruler do binduon ko seedhi line se jodta hai. Sirf in do auzaaron se tum sateek figures bana sakte ho, isliye nahi ki tumne dhyaan se naapa, balki tark ki wajah se. Geometric Twins chapter ke congruent triangles hi har tareeke ke chalne ki wajah hain.

Aankh aur perpendicular bisector

XY ke perpendicular bisector pe centre waale do barabar arcs se bani aankh — Do barabar arcs se aankh

Class 6 mein tumne X aur Y se guzarte do arcs se aankh banaayi thi, unke centre A aur B ka andaaza lagaakar. Suddol aankh ke liye dono arcs ka radius ek hona chahiye, yaani AX = AY = BX = BY. Aise bindu dhoondhne ke liye X aur Y se XY ke upar aur neeche barabar radius ke arcs banaao; wo A aur B pe milte hain. A aur B ko jodne pe XY ke beech ke bindu se right angle pe guzarti line milti hai: perpendicular bisector. Bisection ka matlab hai kisi cheez ko do ek jaise hisson mein baantna.

X aur Y se arcs A aur B pe milte hain; AB, XY ka perpendicular bisector hai — XY ka perpendicular bisector

Ye hamesha kyun chalta hai? SSS se ΔABX ≅ ΔABY, kyunki AX = AY, BX = BY aur AB common hai, isliye ∠XAO = ∠YAO. Phir SAS se ΔAOX ≅ ΔAOY, isliye OX = OY aur ∠AOX = ∠AOY. Ye dono angles milkar 180° banaate hain, isliye har ek 90° hai. Isse ye bhi nikalta hai ki X aur Y se barabar door har bindu isi ek line pe hai, isliye is line ke doosre bindu alag shape ki aankhein dete hain.

Bindu pe right angle

Line ke bindu O pe 90° angle banaane ke liye pehle O ko kisi segment ka beech ka bindu banaao: O ko centre lekar line ko X aur Y pe kaato. Phir X aur Y se bade radius ke arcs banaao jo A pe milein. OA, XY ka perpendicular bisector hai, isliye wo line ke saath 90° banata hai. Sirf ek jodi arcs chahiye, kyunki O pehle se bisector pe hai.

Sulba-Sutras ki rassi

Sulba-Sutras, yagya vedi banaane ke Vedic granth, mein rassi ko compass ki tarah use karke perpendicular banaane ke sateek tareeke the. Katyayana-Sulbasutra ke ek tareeke mein X aur Y pe khoontiyan gaadkar rassi ke sire unse baandhte hain. Rassi ke beech ke bindu ko XY ke upar tana kheenchkar A, aur neeche B lagaate hain. Rassi ke dono aadhe barabar hain, isliye A aur B dono X aur Y se barabar door hain, aur AB perpendicular bisector hai.

Sulba-Sutra ka rassi tareeka: rassi ka beech ka bindu XY ke upar aur neeche kheencha gaya — Rassi compass ki tarah

Angle bisect karna

Aath pankhudi waale phool ko 45° door aath lines chahiye, aur 45°, 90° ka aadha hai. Kisi bhi angle XOY ko bisect karne ke liye O se arc banaao jo arms ko A aur B pe kaate, taaki OA = OB. A aur B se ek hi radius ke arcs banaao jo C pe milein. OA = OB, AC = BC aur OC common hai, isliye SSS se ΔOAC ≅ ΔOBC, yaani ∠AOC = ∠BOC. 90° bisect karne se 45° milta hai, aur phir se karne pe 22.5°.

Angle copy karna aur parallel lines

Angle copy karna: ek arc aur chord BC naye arm pe le jaane se wahi angle milta hai — Angle copy karna

Dohraate designs ko ek angle ki theek copies chahiye. Vertex A se arc banaao jo arms ko B aur C pe kaate. Naye bindu X se ek ray aur usi radius ka arc banaao jo use Z pe kaate. Compass mein lambaai BC lo aur Z se arc pe lagaakar Y pao. Ab SSS se ΔABC ≅ ΔXYZ, isliye ∠X = ∠A. Yahi vichaar parallel lines deta hai: kisi bindu B se, transversal jo angle line m ke saath banata hai use corresponding jagah pe copy karo; nayi line ke corresponding angles barabar hain, isliye wo m ke parallel hai.

Ye kyun zaroori hai

Ye paanch constructions, support lines pe bane arches ke saath, chapter ke har design ka auzaar baksa hain. Aage inhe regular hexagon kaise banayein? mein use karo, aur chapter ka doosra hissa tiling patterns kya hain? mein dekho. Textbook ke sawaalon ke jawab Part 2 Chapter 6 ke NCERT solutions mein hain; poore Part 2 Chapter 6 ke notes dekho.

Practice karo

Perpendicular bisector construction mein bindu A aur B ke liye kya sach hai?
1. (a) AX = AB
2. (b) AX = AY aur BX = BY
3. (c) Sirf AX = BY
4. (d) AB = XY
Jawab dekho

Jawab: (b) AX = AY aur BX = BY
∠XOY bisect karne ke liye pehle arms pe A aur B aise lagaate hain ki:
1. (a) OA = OB
2. (b) OA = AB
3. (c) AB = XY
4. (d) OA = 2 OB
Jawab dekho

Jawab: (a) OA = OB
Kaunsi congruence condition saabit karti hai ki copy kiya angle asli ke barabar hai?
1. (a) SAS
2. (b) SSS
3. (c) ASA
4. (d) RHS
Jawab dekho

Jawab: (b) SSS
Line ke bindu O pe 90° angle kaise banaate hain?

Jawab dekho

Jawab: O ko centre lekar line ko X aur Y pe kaato, taaki OX = OY ho. Bade radius se X aur Y se arcs banaao jo A pe milein. OA, XY ka perpendicular bisector hai, isliye O pe angle 90° hai. Sirf ek jodi arcs chahiye, kyunki O pehle se bisector pe hai.
Angle bisector ka construction kyun chalta hai?

Jawab dekho

Jawab: OA = OB (ek hi arc), AC = BC (ek hi radius) aur OC common hai, isliye SSS se ΔOAC ≅ ΔOBC. Congruent triangles ke milte hisse barabar hote hain, isliye ∠AOC = ∠BOC, aur OC angle ko bisect karta hai.

Students ke sawal

Perpendicular bisector kaise banaate hain?

Compass ko XY ke aadhe se zyaada khol lo. Pehle X se phir Y se segment ke upar arcs banaao; wo A pe milte hain. Usi radius se neeche arcs banaao; wo B pe milte hain. AB jodo. Ye XY ko uske beech ke bindu pe right angle pe kaatti hai.

AB, XY ka perpendicular bisector kyun hai?

Kyunki AX = AY = BX = BY. SSS se ΔABX ≅ ΔABY, isliye ∠XAO = ∠YAO. Phir SAS se ΔAOX ≅ ΔAOY, isliye OX = OY aur ∠AOX = ∠AOY. Ye dono angles milkar seedha angle banaate hain, isliye har ek 90° hai.

Sulba-Sutras ne perpendicular bisector kaise banaaya?

Rassi se. X aur Y pe khoontiyan gaadi jaati hain aur sire pe phande waali rassi unse baandhi jaati hai. Uska beech ka bindu XY ke upar kheench kar tana karke A, aur neeche B lagaate hain. Rassi ke dono aadhe barabar hain, isliye A aur B dono X aur Y se barabar door hain, aur AB perpendicular bisector hai.

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