Triangles | Class 10 Maths

🔍 Similar vs Congruent — Farq kya hai?

📸 Photo copy ki tarah:

Congruent: Same shape, same size — bilkul identical (≅). Jaise photocopy 100%.
Similar: Same shape, alag size — proportional (∼). Jaise photo ko zoom karo — shape same, size alag.
Class 10 mein focus hai similarity par!

Similar Triangles (∼): △ABC ∼ △PQR iff
1. Corresponding angles equal: ∠A=∠P, ∠B=∠Q, ∠C=∠R
2. Corresponding sides proportional: AB/PQ = BC/QR = CA/RP

📐 BPT — Basic Proportionality Theorem

Theorem (Thales Theorem): Ek triangle mein agar ek line ek side ke parallel ho aur doosri do sides ko alag-alag points par intersect kare, to woh dono sides ko same ratio mein divide karegi.

If DE ∥ BC in △ABC, then: AD/DB = AE/EC

Example 1: DE ∥ BC. AD=3, DB=5, AE=6. EC nikalo.

AD/DB = AE/EC → 3/5 = 6/EC → EC = 10

Example 2: DE ∥ BC. AD=x, DB=x−2, AE=x+2, EC=x−1. x nikalo.

x/(x−2) = (x+2)/(x−1)
x(x−1) = (x+2)(x−2)
x²−x = x²−4
−x = −4 → x = 4

🔗 Criteria for Similarity

AA (Angle-Angle)

Do angles equal ho to triangles similar

∠A=∠P, ∠B=∠Q → △∼△

SSS (Side-Side-Side)

Teen sides ka ratio equal ho

AB/PQ = BC/QR = CA/RP

SAS (Side-Angle-Side)

Do sides proportional aur included angle equal

AB/PQ = AC/PR, ∠A=∠P

AA Example: △ABC mein ∠ACB = 90°. CD ⊥ AB. Prove karo △ACD ∼ △ABC.

In △ACD aur △ABC:
∠A = ∠A (common)
∠ACD = ∠ABC (kyunki ∠ACD = 90°−∠A = ∠ABC... wait, more precisely ∠ADC=∠ACB=90°)
∠ADC = ∠ACB = 90°
By AA: △ACD ∼ △ABC ■

📊 Area of Similar Triangles

Theorem

△ABC ∼ △PQR to:
Area(△ABC) / Area(△PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²

Areas ka ratio = corresponding sides ke squares ka ratio

Example: Do similar triangles ke corresponding sides 4:9 ratio mein hain. Areas ka ratio?

Area ratio = 4²:9² = 16:81

Example: △ABC ∼ △DEF. AB=4, DE=6, Area(△ABC)=16 cm². Area(△DEF)?

16/Area(DEF) = (4/6)² = 16/36
Area(DEF) = 16 × 36/16 = 36 cm²

📐 Pythagoras Theorem

Theorem: Right angle triangle mein, hypotenuse ka square = doosri do sides ke squares ka sum.
AB² = BC² + CA² (jahan ∠C = 90°)

Proof using Similar Triangles

△ABC mein ∠A=90°. Altitude AD ⊥ BC.
△ABD ∼ △CAB (AA criterion)
→ AB/CB = DB/AB → AB² = CB × DB ...(1)
△ACD ∼ △CAB (AA criterion)
→ AC/BC = DC/AC → AC² = BC × DC ...(2)
(1)+(2): AB² + AC² = CB(DB+DC) = CB×BC = BC²
∴ BC² = AB² + AC² ■

Converse of Pythagoras Theorem

Agar triangle mein AB² = BC² + CA², to ∠C = 90°

Common Pythagorean triplets:
3,4,5 | 5,12,13 | 8,15,17 | 7,24,25 | 6,8,10 | 9,40,41

Example: Triangle sides 6, 8, 10. Right angle hai kya?

10² = 100. 6²+8² = 36+64 = 100. ✓ Yes, right angle triangle! ∠ opposite to 10 = 90°

📝 Practice Questions

Q1. BPT: △ABC mein DE∥BC. AD=1.5, DB=3, AE=1. EC nikalo.

AD/DB = AE/EC → 1.5/3 = 1/EC → EC = 2

Q2. △ABC ∼ △DEF. ∠A=47°, ∠E=83°. ∠C nikalo.

△ABC∼△DEF → ∠B=∠E=83°. ∠C=180°−47°−83° = 50°

Q3. Do similar triangles ke areas 81 cm² aur 49 cm² hain. In ke corresponding sides ka ratio nikalo.

Area ratio = 81:49. Side ratio = √81:√49 = 9:7

Q4. Ek ladder 15 m lambi hai. Neeche se 9 m door wall se lagayi. Kitni uchhai tak pahunchegi?

h² + 9² = 15² → h² = 225−81 = 144 → h = 12 m

Q5. △ABC ∼ △PQR. PQ = 12, AB = 8. Area(△ABC) = 48 cm². Area(△PQR)?

Area ratio = (8/12)² = 64/144. 48/Area(PQR) = 64/144 → Area(PQR) = 48×144/64 = 108 cm²

🎮 Geometry Lab → Next: Circles →

Triangles — Similarity 📐