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Triangles

Lesson 106 of 113

1. Ye topic hai kya?

Do triangles ka same shape (angles same, sides scale pe) — similar. Parallel line beech mein kaat de to ratios lock (BPT). Right triangle pe a²+b²=c². Paper: corresponding side, DE ∥ BC pe AD/DB, area 9:16 se side, ladder wall se kitni door.

Yahan lock

Correspondence (order of vertices), AA, SAS, SSS, BPT + converse, area ∝ (side)², Pythagoras + converse, scale k pe perimeter k, area k².

Yahan kya nahi

sin/cos, 30°–45° tower. Circle tangent. Compass se triangle banana. Coordinate √[(x₂−x₁)²+(y₂−y₁)²]. AP 6,10,14. Reasoning wala F/L mirror.

2. Similar — same shape, scale alag

Congruent = same shape aur same size (ek ke upar dusra baith jaye). Similar = same shape, size zoom in/out ho sakta. Har congruent pair similar bhi hai, scale k=1. Har similar congruent nahi.

B A C Q P R

ΔABC ~ ΔPQR — shape same, PQR zoomed. Angles A=P, B=Q, C=R.

ΔABC ~ ΔPQR likha = vertices uss order mein match: A↔P, B↔Q, C↔R. Sides: AB/PQ = BC/QR = CA/RP. Angle A ke saamne wali side BC, angle P ke saamne QR — corresponding.

Scale factor k = AB/PQ (chhote/bade — sawal ke hisaab se kisko numerator). Perimeter ratio bhi wahi k. Area nahi — area k².

Order galat = pairing galat. ΔABC ~ ΔPRQ matlab A↔P, B↔R, C↔Q — tab AB/PR, BC/RQ. Paper pe “ABC ~ DEF” padho, phir sides jodo. Random AB/DF nahi.

3. Criteria — kab similar?

Do triangles similar tab, jab ek inme se pack ho. Teesra angle khud 180 se nikalta — isliye AA kaafi (NCERT AA likhta, AAA nahi alag se).

NaamKya chahiyeSocho
AADo corresponding angles equalTeesra automatic. Right triangles: ek acute equal → similar.
SASDo sides proportional aur included angle equalBeech wala angle. Adjacent nahi, included.
SSSTeen corresponding sides proportionalAngles khud equal ho jaate.

Example — SSS. Sides 8, 10, 12 aur 12, 15, 18. Ratios: 8/12=2/3, 10/15=2/3, 12/18=2/3. Same — similar, k=2/3.

Example — AA. Ek triangle 70° aur 50°. Dusre mein 70° aur 60°. Pehle ka teesra 60°. Dusre ka teesra 50°. Dono 70°, 60°, 50° — AA, similar.

SAS trap. Included angle = un do sides ke beech. Agar equal angle included nahi (kisi aur vertex pe), SAS nahi lagta.

SSS similarity sides ke ratio pe. SSS congruence (Class IX) sides equal. Yahan zoom allowed. SAS yahan ratio + included; congruence wala SAS equal sides + included.

4. BPT — Thales, parallel kaat

ΔABC. Point D on AB, E on AC. DE ∥ BC. Phir DE do sides ko same ratio mein kaat-ti hai:

Basic Proportionality Theorem: AD/DB = AE/EC. (Equally: AD/AB = AE/AC = DE/BC — kyunki ΔADE ~ ΔABC, AA: corresponding angles, DE ∥ BC.)
A B C D E DE ∥ BC

DE ∥ BC. AD:DB = AE:EC. Figure se andaza nahi — numbers se ratio.

Example

DE ∥ BC. AD = 2, DB = 3, AE = 4. EC?

Solution

  1. BPT

    AD/DB = AE/EC → 2/3 = 4/EC.

  2. Cross

    2·EC = 12 → EC = 6.

Answer6

Converse: agar D, E aise hain ke AD/DB = AE/EC, to DE ∥ BC. Ratio check → parallel prove. Parallel given → ratio.

AD/DB segments of same side. AD/AB alag fraction (whole side). Mix mat karo. AD=2, AB=5 → AD/AB=2/5, AD/DB=2/3 — dono sahi, sawal kya maangta.

5. Areas of similar triangles

ΔABC ~ ΔPQR, scale k = AB/PQ. Heights bhi k (same angles). Area = ½ × base × height → dono factors k → area k².

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)². Sides 2:3 → areas 4:9. Areas 9:16 → sides 3:4 (positive square root). Perimeter ratio = side ratio, area nahi.
Example

Similar, corresponding sides 3 : 5. Chhote ka area 36. Bade ka area?

Solution

  1. k²

    Sides 3/5 → areas 9/25.

  2. Bade

    36 / (chhota) = 9/25 of bada → bada = 36 × 25/9 = 100.

Answer100

Agar DE ∥ BC aur AD/AB = 2/5, to ΔADE ~ ΔABC, DE/BC = 2/5, ar(ADE)/ar(ABC) = 4/25. Trapezium DECB ka area = bada minus chhota = 21/25 of ABC.

6. Pythagoras — right triangle

Right angle C pe. Legs a, b; hypotenuse c (right ke saamne, sabse lamba).

Pythagoras: a² + b² = c². Converse: teen sides pe a²+b²=c² (sabse badi ko c) → triangle right-angled, right us vertex pe jo c ke opposite.
A B C a b c

∠C = 90°. a² + b² = c². Square-on-sides picture yahan zaroorat nahi — formula sides pe.

LDC triples (scale bhi): 3-4-5 → 6-8-10, 9-12-15. 5-12-13. 8-15-17. 7-24-25. Check: 8²+15² = 64+225 = 289 = 17².

Example

Right triangle, legs 8 aur 15. Hypotenuse?

Solution

  1. Square

    64 + 225 = 289.

  2. Root

    √289 = 17. (9-12-15 nahi — ye 8-15 pair.)

Answer17

Converse example: 9, 12, 15. 81+144=225=15² → right, hypotenuse 15. 6, 8, 11: 36+64=100 ≠ 121 → right nahi.

Altitude to hypotenuse — teen similar

Right ΔABC, right C pe. CD ⊥ hypotenuse AB. Phir ΔACD ~ ΔABC ~ ΔCBD (AA: dono right, common acute). Isse: AC² = AD·AB, BC² = BD·AB, CD² = AD·BD, aur AC·BC = AB·CD (area do tarah).

5-12-13: AB=13, CD = (5×12)/13 = 60/13. Unique — quadratic product 56 nahi, AP nahi.

7. Word — shadow, ladder — angle nahi

Dhoop ki kirnein parallel → ground pe do objects ke triangles similar (AA: dono right ground se, sun-ray same slope). Height / shadow = dusre ka height / shadow. Angle 30°–60° mat lao — woh Applications of trigonometry.

Example

6 m pole ki shadow 4 m. Us waqt tree ki shadow 10 m. Tree ki height?

Solution

  1. Similar

    h/10 = 6/4 = 3/2.

  2. h

    h = 10 × 3/2 = 15 m.

Answer15 m
Example

13 m sihi, deewar se 5 m door. Deewar pe kitni unchi pahunchti? (ground level, wall vertical.)

Solution

  1. Right

    Hypotenuse 13, base 5. Height h: 5² + h² = 13².

  2. h

    h² = 169−25 = 144 → h = 12 m. (5-12-13.)

Answer12 m

8. Sawal — basic se pro

Q1 · Basic

Kaun sa hamesha true?

  1. Similar triangles congruent hote hain
  2. Congruent triangles similar hote hain
  3. Equal area ⇒ similar
  4. SSS sides equal ⇒ sirf similar, congruent nahi

Solution

  1. k=1

    Congruent = similar with scale 1. B.

  2. A/C/D

    A ulta. C: 3-4-5 aur isosceles area same ho sakta, shape nahi. D: equal sides ⇒ congruent.

AnswerB
Q2 · Basic

ΔABC ~ ΔPQR. AB = 6, PQ = 9, BC = 8. QR?

  1. 12
  2. 5.33
  3. 11
  4. 16

Solution

  1. Pair

    AB↔PQ, BC↔QR. 6/9 = 8/QR.

  2. QR

    QR = 8 × 9/6 = 12. B: 8×6/9. D: 8×2.

Answer12
Q3 · Basic

DE ∥ BC, AD = 2, DB = 3, AE = 4. EC?

  1. 6
  2. 5
  3. 8
  4. 3

Solution

  1. BPT

    2/3 = 4/EC → EC = 6.

  2. Trap

    C: AD+AE. D: DB copy.

Answer6
Q4 · Medium

Right triangle, legs 8 aur 15. Hypotenuse?

  1. 17
  2. 23
  3. 13
  4. 7

Solution

  1. 8²+15²

    64+225=289=17². C 5-12-13 mix. B 8+15.

Answer17
Q5 · Medium

Sides 9, 12, 15. Triangle?

  1. Obtuse, Pythagoras fail
  2. Right-angled, hypotenuse 15
  3. Right-angled, hypotenuse 12
  4. Equilateral nahi isliye right nahi

Solution

  1. Converse

    81+144=225=15². 3-4-5 ×3. Right, badi side hypotenuse.

AnswerB
Q6 · Medium

Similar, sides 2 : 3. Areas ka ratio?

  1. 2 : 3
  2. 4 : 9
  3. 8 : 27
  4. 2 : 9

Solution

  1. k²

    4:9. A perimeter/side. C volume (3D, yahan nahi).

AnswerB
Q7 · Medium

6 m pole, shadow 4 m. Tree shadow 10 m. Tree height?

  1. 15 m
  2. 24 m
  3. 12 m
  4. 8 m

Solution

  1. h/10=6/4

    h=15. B: 6×4. C: 6+ something. Angle nahi diya — trig nahi.

Answer15 m
Q8 · Medium

DE ∥ BC. AD/AB = 2/5. BC = 20. DE?

  1. 8
  2. 50
  3. 10
  4. 4

Solution

  1. ~

    ΔADE ~ ΔABC → DE/BC = AD/AB = 2/5.

  2. DE

    20 × 2/5 = 8. C: AD/DB=2/3 se 20×2/3 galat pairing. B: 20×5/2.

Answer8
Q9 · Pro

ΔABC ~ ΔDEF. AB=8, BC=10, CA=12, DE=12. EF + FD?

  1. 33
  2. 18
  3. 15
  4. 24

Solution

  1. k

    AB/DE=8/12=2/3. ABC chhota. Corresponding: BC/EF=CA/FD=2/3.

  2. EF, FD

    EF = 10 × 12/8 = 15. FD = 12 × 12/8 = 18. Sum 33. C sirf EF. D sirf FD×kuch.

Answer33
Q10 · Pro

13 m sihi, deewar se 5 m door. Deewar pe height?

  1. 12 m
  2. 8 m
  3. 18 m
  4. √194 m

Solution

  1. 13 hyp

    5²+h²=13² → h=12. D: 13²+5² galat (hyp ko leg bana diya).

Answer12 m

Practice

Pehle khud, phir neeche kholo.

12345 678910

P1

Do triangles, angles 80° aur 40° dono mein. Similar?

  1. Haan, AA (teesra 60° dono)
  2. Nahi, teesra alag ho sakta
  3. Sirf SSS se
  4. Sirf jab sides equal

Solution

  1. 180

    Teesra 60° dono. AA. B galat — sum lock.

AnswerA
P2

Right, legs 5 aur 12. Hypotenuse?

  1. 13
  2. 17
  3. 7
  4. 60

Solution

  1. 25+144

    169=13². B 8-15 mix. D product.

Answer13
P3

DE ∥ BC. AD = 3, AB = 8, AE = 6. EC?

  1. 10
  2. 16
  3. 5
  4. 11

Solution

  1. DB

    AB=8, AD=3 → DB=5. AD/DB=3/5=AE/EC=6/EC → EC=10.

  2. Ya

    AD/AB=AE/AC → 3/8=6/(6+EC) → 3(6+EC)=48 → 6+EC=16 → EC=10.

Answer10
P4

Similar, areas 9 : 16. Corresponding sides?

  1. 9 : 16
  2. 3 : 4
  3. 81 : 256
  4. 4.5 : 8

Solution

  1. √

    Sides √9 : √16 = 3:4.

AnswerB
P5

6, 8, 10. Right-angled?

  1. Haan, hypotenuse 10
  2. Nahi, 6+8≠10
  3. Haan, hypotenuse 8
  4. Sirf 3,4,5 pe chalta

Solution

  1. 36+64

    100=10². 3-4-5 ×2. B triangle-inequality mix (6+8>10, equality nahi chahiye).

AnswerA
P6

ΔABC ~ ΔDEF. AB corresponding kis se?

  1. DE
  2. EF
  3. DF
  4. Jo bhi longest

Solution

  1. Order

    A↔D, B↔E → AB↔DE. C: AC↔DF.

AnswerA
P7

SAS similarity ke liye equal angle kahan?

  1. Do proportional sides ke beech (included)
  2. Koi bhi ek angle
  3. Sirf right angle
  4. Opposite the longer side, hamesha

Solution

  1. Included

    Beech wala. AA alag criterion. Right special case AA/HL congruence IX, yahan SAS included.

AnswerA
P8

Similar, k = 2/5 (chhote/bade). Perimeter ratio chhote : bade?

  1. 2 : 5
  2. 4 : 25
  3. 5 : 2
  4. 8 : 125

Solution

  1. Linear

    Perimeter ~ side. 2:5. B area.

AnswerA
P9

7, 24, 25. Check Pythagoras.

  1. Right, hypotenuse 25
  2. 49+576 ≠ 625
  3. Right, hypotenuse 24
  4. Isosceles

Solution

  1. 49+576

    625=25². B arithmetic fail.

AnswerA
P10

DE ∥ BC, AD = DB. AE : EC?

  1. 1 : 1
  2. 1 : 2
  3. 2 : 1
  4. Nahi nikal sakte

Solution

  1. BPT

    AD/DB=1 → AE/EC=1. D midpoint of AC bhi. Midpoint theorem (IX) isi family — yahan BPT ka special.

Answer1 : 1

← Index · Arithmetic progressions

Agle topic: Coordinate geometry.