Isosceles, equilateral triangle ke angles

Isosceles aur equilateral triangle ke angles

Short mein jawab

Isosceles triangle mein barabar sides ke saamne ke angles barabar hote hain. Barabar sides ke beech wale vertex se altitude do right-angled triangles banata hai jo RHS se congruent hain, isliye dono base angles match karte hain. Equilateral triangle mein saari sides barabar hain, isliye saare angles barabar hain, aur har ek 180° ÷ 3 = 60° hai.

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Isosceles aur equilateral triangles

Do barabar sides wala triangle isosceles hota hai. AB = AC wale ΔABC mein angle A barabar sides ke beech hai, aur BC teesri side hai, jise aksar base kehte hain. Teeno sides barabar wala triangle equilateral hota hai. Drawing mein barabar sides pe same nishaan hota hai, jaise har ek pe ek chhoti si lakeer.

Congruence in triangles ke angles ke baare mein naapne se pehle hi kuch bata sakti hai.

Base angles barabar kyun hain

ΔABC lo jisme AB = AC aur ∠A = 80°. A se altitude kheencho, yaani line AD jo BC se right angle pe milti hai.

Isosceles triangle ABC jisme AB = AC aur ∠A = 80°; altitude AD do right triangles banata hai jo RHS se congruent hain, isliye ∠B = ∠C = 50° — Altitude do congruent hisse banata hai

Ab ΔADB aur ΔADC ki tulna karo:

∠ADB = ∠ADC = 90°, kyunki AD altitude hai,
AB = AC, yaani hypotenuse, jaisa diya hai, aur
AD dono triangles ki side hai.

Isliye RHS se ΔADB ≅ ΔADC. Congruent triangles ke corresponding parts barabar hote hain, isliye ∠B = ∠C. Yahi tarq har isosceles triangle pe chalta hai, aur isse niyam milta hai: triangle mein barabar sides ke saamne ke angles barabar hote hain.

Yahan ∠B + ∠C = 180° − 80° = 100°, isliye har ek 50° hai.

Missing angles nikaalna

Niyam aur 180° ke angle sum se do shortcuts milte hain:

Kya pata hai	Kya milta hai
barabar sides ke beech ka angle	har base angle = (180° − wo angle) ÷ 2
ek base angle	doosra base angle utna hi; teesra = 180° − 2 × base angle

Jaise, agar barabar sides ke beech ka angle 40° hai, to har base angle (180° − 40°) ÷ 2 = 70° hai. Agar ek base angle 65° hai, to barabar sides ke beech ka angle 180° − 130° = 50° hai.

Equilateral triangle ke angles

Equilateral triangle ABC mein AB = AC, isliye ∠B = ∠C (in sides ke saamne ke angles). AB = BC bhi hai, isliye ∠C = ∠A. Teeno angles barabar hain, bilkul sides ki tarah. Wo milkar 180° hote hain, isliye har ek 180° ÷ 3 = 60° hai. Ise check karne ke liye compass se equilateral triangle banao aur protractor se uske angles naapo.

Equilateral triangle ABC, saari sides barabar nishaan ke saath aur har angle 60° — Barabar sides, barabar angles

Circle se isosceles triangle

Circle ke koi bhi do radius barabar hote hain, isliye jis triangle ka ek vertex centre pe ho aur baaki do circle pe, wo isosceles hota hai. Agar A centre hai, B aur C circle pe hain aur ∠BAC = 120°, to ∠B = ∠C = (180° − 120°) ÷ 2 = 30°. Isi tarah, agar ∠BAC = 90° ho, to har base angle 45° hoga.

Centre A wala circle aur us par points B aur C; AB = AC radius hain, ∠A = 120° aur ∠B = ∠C = 30° — Radius barabar, isliye base angles barabar

Ye kyun zaroori hai

Ye niyam ek pata angle se teeno angles nikaal deta hai, aur geometry ke sawaalon mein baar-baar kaam aata hai. Ye ye bhi dikhata hai ki congruence kitni kaam ki hai: angles ki ek baat do congruent hisson se nikli. Conditions dobara padho congruent triangles kya hain mein. Textbook ke sawaalon ke jawab Part 2 Chapter 1 ke NCERT solutions mein hain; poore chapter ke liye Part 2 Chapter 1 ke notes.

Practice karo

ΔPQR mein PQ = PR aur ∠P = 50°. ∠Q kitna hai?
1. (a) 50°
2. (b) 65°
3. (c) 130°
4. (d) 40°
Jawab dekho

Jawab: (b) 65° ∠Q = ∠R aur ∠Q + ∠R = 180° − 50° = 130°, isliye har ek 65° hai.
Isosceles triangle ka har base angle 70° hai. Teesra angle kitna hai?
1. (a) 70°
2. (b) 40°
3. (c) 110°
4. (d) 20°
Jawab dekho

Jawab: (b) 40° 180° − 70° − 70° = 40°.
Har equilateral triangle ke liye kaunsi baat sach hai?
1. (a) Uske angles 90°, 45° aur 45° hain
2. (b) Har angle 60° hai
3. (c) Sirf do angles barabar hain
4. (d) Uske angles uske size pe depend karte hain
Jawab dekho

Jawab: (b) Har angle 60° hai
ΔABC mein AB = AC aur ∠B = 35°. ∠C aur ∠A nikaalo.

Jawab dekho

Jawab: ∠C = ∠B = 35°, kyunki ye barabar sides AB aur AC ke saamne hain. Phir ∠A = 180° − 35° − 35° = 110°.
A ek circle ka centre hai, aur B aur C us par points hain jahan ∠BAC = 90°. ∠ABC nikaalo.

Jawab dekho

Jawab: AB aur AC radius hain, isliye AB = AC aur ∠ABC = ∠ACB. Dono milkar 180° − 90° = 90° hain, isliye ∠ABC = 45°.

Students ke sawal

Isosceles triangle ke base angles barabar kyun hote hain?

Barabar sides ke beech wale vertex A se altitude AD kheencho. Triangles ADB aur ADC mein D pe right angle hai, hypotenuse AB aur AC barabar hain, aur AD common side hai, isliye RHS se wo congruent hain. Isliye unke corresponding angles B aur C barabar hain.

Equilateral triangle ka har angle kitna hota hai?

60°. Teeno sides barabar hain, isliye teeno angles barabar hain. Teen barabar angles jo milkar 180° hon, unme se har ek 180° ÷ 3 = 60° hoga.

Ek angle diya ho to isosceles triangle ke angles kaise nikaalein?

Agar diya gaya angle barabar sides ke beech hai, to baaki har angle (180° − diya gaya angle) ÷ 2 hai. Agar wo base angle hai, to doosra base angle bhi utna hi hai aur teesra angle 180° − 2 × base angle hai.

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