Triangles: Class 10 Maths Chapter 6 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

75 min
1 simulation
38 practice questions
14 MCQs

Is chapter mein aap kya seekhenge

Similar aur congruent figures mein fark pehchaano, aur matching angles aur matching side ratios se do polygons ki similarity test karo.
Enlargements, reductions, maps aur plans mein lengths nikaalne ke liye scale factor use karo.
Basic Proportionality Theorem aur uska converse state aur prove karo, aur unse lengths nikaalo aur dikhao ki lines parallel hain.
Similarity statements mein vertices ko matching order mein likho.
AA, SSS ya SAS criterion se prove karo ki do triangles similar hain.
Similar triangles se unknown sides, angles aur heights nikaalo, jismein shadow problems aur altitudes, medians aur angle bisectors ke ratios bhi shaamil hain.

NCERT Mathematics Class 10, Chapter 6, pp. 73–98 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

6.1 Introduction

Apne phone par kisi photo ko pinch karke zoom in karo. Picture badi ho jaati hai, phir bhi usmein kuch bhi pichakta nahi: chehre, khidkiyan aur patang apni shape banaye rakhte hain. Class 9 mein tumne congruent figures padhe the, jo shape aur size dono mein match karte hain. Yeh chapter un figures ke baare mein hai jo sirf shape mein match karte hain. Inhe similar figures (समरूप आकृतियाँ) kehte hain.

Similarity ek measuring tool bhi hai. School ke flagpole par koi measuring tape lekar nahi chadhta. Iske bajaye, kisi dhoop wali subah flagpole ko ek metre stick se compare karo: dono ek hi pal mein shadows banaate hain, aur shadows se height pata chal jaati hai (Fig. 6.1).

3/3Dono right triangles ke angles same hain, isliye height ÷ shadow dono ke liye same hai: $h = 7.5 m$ .

Fig. 6.1Parallel sun rays same shape ke do right triangles banaati hain.

Ek flat ground line, left mein ek metre stick aur ek lamba flagpole jis par tiranga hai. Pehle step mein sirf shadows dikhte hain: stick ki 1.6 m aur flagpole ki 12 m. Doosre step mein har object ke top se uski shadow ki tip tak dashed sun rays bani hain; dono rays parallel hain. Teesre step mein har object, uski shadow aur ray se bane do right triangles shade kiye gaye hain, aur flagpole ki height h = 7.5 m di gayi hai.

Har triangle mein height ÷ shadow ek hi number hai. Stick ke liye yeh $\frac{1}{1.6}$ hai, isliye flagpole $\frac{12}{1.6} = 7.5 m$ lamba hai. Surveyors isi idea se pahadiyon ki heights aur nadiyon ki widths nikaalte hain, aur yeh tumhe trigonometry (Chapters 8 aur 9) mein phir milega. Is chapter mein tum seekhoge ki do triangles theek kab similar hote hain aur yeh kaise prove karte hain.

6.2 Similar Figures

Shapes ki kuch families hamesha similar hoti hain. Koi bhi do circles similar hote hain: ₹1 ka sikka aur dinner plate sirf size mein alag hain. Koi bhi do squares similar hote hain, aur koi bhi do equilateral triangles bhi. In har family mein sirf ek hi shape hai; sirf size badal sakta hai.

Congruent figures shape aur size dono mein match karte hain, isliye congruent figures ka har pair similar bhi hota hai. Ulta sach nahi hai: 2 cm ka square aur 5 cm ka square similar hain lekin congruent nahi. Kuch pairs kabhi similar nahi hote, unka size chahe kuch bhi ho. Jaise ek circle aur ek square: koi bhi enlargement ek smooth curve ko chaar corners mein nahi badal sakta.

2/3Har length 1.6 se multiply hoti hai aur shape wahi rehti hai: copy similar hai lekin congruent nahi.

Fig. 6.2Same size, barabar enlargement, ya sirf ek direction mein stretching.NCERT Fig. 6.1 · p. 74

Left mein ghar ki ek chhoti shape (triangular roof wala square, ek door aur ek window) hai, jis par original likha hai. Uske paas ek doosra ghar hai. Pehle step mein doosra ghar same size ki exact copy hai, jis par congruent likha hai. Doosre step mein woh barabar badhta hai taaki har length 1.6 guna lambi ho jaaye, aur us par similar likha hai. Teesre step mein woh iske bajaye sirf side mein khinchta hai, width 2.2 guna par height wahi, isliye roof chapti ho jaati hai; us par not similar likha hai.

Circles aur squares ke saath aankh hi kaafi hai. Do quadrilaterals ya do pentagons ke saath nahi: do four-sided shapes ek jaisi dikh sakti hain aur phir bhi thodi alag ho sakti hain. Humein ek aisa test chahiye jise hum check kar sakein.

Ek photo print karne ke baare mein socho. Ek shop 9 cm × 12 cm ke print ko 15 cm × 20 cm tak enlarge karti hai. Picture ki har length ek hi number se multiply hoti hai, $\frac{15}{9} = \frac{20}{12} = \frac{5}{3} .$ Har angle waisa hi rehta hai: window ka corner ab bhi right angle hai. Ab photo ko kisi editing app mein kholo aur sirf uski width ko 9 cm se 18 cm tak drag karo. Chehre chaude dikhte hain aur har tircha angle badal jaata hai, isliye stretched photo original ke similar nahi hai.

Definition

Similar polygons

Same number of sides wale do polygons tab similar hote hain jab yeh dono baatein sach hon. (i) Ek ka har angle doosre ke matching angle ke barabar ho. (ii) Ek ki har side doosre ki matching side ka same multiple ho.

ExampleEk 9 cm × 12 cm photo aur uska 15 cm × 20 cm enlargement: saare angles $90^{\circ}$ hain aur har side ka multiplier hai $\frac{5}{3} .$

Yeh common multiple scale factor hai, jise representative fraction bhi kehte hain. Maps aur building plans ek scale par banaye jaate hain. 1 : 50 000 scale wale map par paper ka 2 cm ground ke 100 000 cm ko dikhata hai, yaani 1 km.

Activity 6.1NCERT Activity 1 · p. 76

Paper kite ki shadow

Tumhe chahiye

ek phone torch
mote card se kaata gaya ek kite shape
ek plain deewar, ya us par chipki chart paper ki sheet
ek ruler, ek protractor aur ek pencil

Pehle guess karo

Tum card ko deewar ke parallel, torch aur deewar ke theek beech mein pakadte ho. Shadow ka har edge kitna lamba hoga?

card ke edge jitna hicard ke edge ka doublecard ke edge ka aadhasirf kuch edges ke liye lamba

Ek andhere kamre mein torch ko table par aise rakho ki woh seedha deewar par chamke. Bulb ko $O$ kaho.
Card ko deewar ke parallel, torch aur deewar ke theek beech mein pakdo. Uske corners ke naam $A$ , $B$ , $C$ , $D$ rakho.
Kisi dost se chart paper par shadow trace karwao. Matching corners ke naam $A^{'}$ , $B^{'}$ , $C^{'}$ , $D^{'}$ rakho.
Card aur shadow ki sides aur do angles naapo, aur table bharo.

Table 6.1

Parts	Card	Shadow	Ratio
$A B$ aur $A^{'} B^{'}$
$B C$ aur $B^{'} C^{'}$
$C D$ aur $C^{'} D^{'}$
$∠ A$ aur $∠ A^{'}$
$∠ C$ aur $∠ C^{'}$

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 6.4

3/3Shadow ka har corner $O$ se do guna door hai: $O A^{'} = 2 \times O A$ . Har side double hoti hai aur har angle wahi rehta hai.

Fig. 6.3Ek point se enlargement:

O

se har distance double ho jaati hai.NCERT Fig. 6.4 · p. 76

Left mein ek point O hai jise torch mark kiya gaya hai. Uske thoda right mein ek chhota kite ABCD hai. Pehle step mein sirf torch aur chhota kite dikhte hain. Doosre step mein O se dashed rays kite ke har corner se hokar right ki taraf aage jaati hain. Teesre step mein in rays par corners A′, B′, C′ aur D′ wala ek bada kite dikhta hai; har corner O se chhote kite ke matching corner ke mukable do guna door hai, saath mein notes OA′ = 2 × OA, A′B′ = 2 × AB aur angles wahi.

Light seedhi lines mein chalti hai, isliye $A^{'}$ ray $O A$ par hai, $B^{'}$ ray $O B$ par, aur isi tarah aage. Corners jode banaate hain: $A$ ke saath $A^{'}$ , $B$ ke saath $B^{'}$ . Hum $A \leftrightarrow A^{'}$ likhte hain aur aise pairs ko corresponding vertices kehte hain. Corresponding vertices se bani sides aur angles corresponding parts hain: $A B$ ka corresponding $A^{'} B^{'}$ hai aur $∠ B$ ka $∠ B^{'} .$

SimulationSimilar copy banao

Dono axes par 0 se 11 tak ek square grid. Fixed points A (7, 2), B (10, 2) aur C (7, 4) ek right triangle banaate hain jiski sides AB = 3, BC = root 13 aur CA = 2 bani hain aur naapi gayi hain. Fixed points P (1, 6) aur Q (7, 6) ke beech 6 ki doori hai, jo AB ka double hai. Point R (5, 9) se shuru hota hai aur ise whole-number positions par drag kar sakte ho; jaise-jaise tum ise hilaate ho, length QR naapi jaati hai, taaki tum woh jagah dhoondh sako jahan PQR ki har side ABC ki matching side ki double ho.

Fig. 6.4Triangle

A B C

fixed hai, aur

P Q = 2 \times A B

R

ko aise drag karo ki

P Q R

A B C

ka scale factor 2 wala enlargement ban jaaye.

Try karo

$R$ ki woh jagah dhoondo jahan $Q R = 2 \sqrt{13}$ aur $P R = 4$ ho. Uske coordinates kya hain?
Kya $P Q$ ke neeche $R$ ki koi doosri jagah hai jo bhi kaam kare? Kya woh triangle bhi $A B C$ ke similar hai?
Shuru mein $Q R$ , $B C$ ke barabar hai. Isse $P Q R$ , $A B C$ ke similar kyun nahi ban jaata?

Ek aur fact check karna aasaan hai. Agar pehla polygon doosre ke similar hai, aur doosra teesre ke, to pehla teesre ke similar hai. (Dono scale factors ko multiply karo.)

Triangles ke liye, jaisa tum Section 6.4 mein dekhoge, ek condition hamesha doosri ko saath le aati hai. Doosre polygons ke liye aisa nahi hota: tumhe dono conditions check karni padti hain. Fig. 6.5 dikhata hai kyun.

2/33 cm side wala rhombus: har side ratio 1.5 hai, lekin uske angles $60^{\circ}$ aur $120^{\circ}$ hain. Similar nahi.

Fig. 6.5Quadrilaterals ke liye sirf equal angles, ya sirf proportional sides, kaafi nahi hain.NCERT Figs. 6.6–6.7 · p. 77

Left mein 2 cm side ka square hai, right-angle marks ke saath. Right mein ek-ek karke shapes ko usse compare kiya jaata hai. Pehle 3 cm by 2 cm ka rectangle: uske saare angles right angles hain, lekin side ratios 1.5 aur 1 hain, isliye yeh similar nahi hai. Doosra 3 cm side ka rhombus jiska ek angle 60 degree hai: har side ratio 1.5 hai lekin angles alag hain, isliye yeh similar nahi hai. Teesra 3 cm side ka square: saare ratios 1.5 hain aur saare angles 90 degree hain, isliye yeh similar hai.

Quick check

3 mein se 1

Figures ka kaun sa pair hamesha similar hota hai?

do rectanglesdo rhombusesdo regular hexagonsdo isosceles triangles

Answer dekho

Answer (C) do regular hexagons

Har regular hexagon ki chhah equal sides aur chhah

120^{\circ}

angles hote hain, isliye ek hexagon doosre ka enlargement hai. Rectangles lambe ya chaude-chapte ho sakte hain, aur rhombuses patle ya mote, isliye unki shapes badalti rehti hain.

Practice · 4 questions

Practice 6.1: answers ke saath

1
Quick answerEasyNCERT Exercise 6.1, Q1
Har statement poora karo.
1. 3 cm aur 5 cm radii wale do circles hain lekin congruent nahi.
  Answer dekho
  Answer similar
  Same shape, alag-alag sizes.
2. Do squares ke diagonals barabar hain. Unke liye sabse exact word hai:
  sirf similarcongruentdono mein se koi nahi
  Answer dekho
  Answer (B) congruent
  Barabar diagonals se sides barabar hoti hain, isliye squares congruent hain (aur isliye similar bhi).
3. Kaun se triangles hamesha ek doosre ke similar hote hain?
  isoscelesright-angledequilateralscalene
  Answer dekho
  Answer (C) equilateral
  Har equilateral triangle ke teen $60^{\circ}$ angles aur equal sides hote hain. Baaki kinds kai shapes mein aate hain.
4. Do pentagons similar hon, iske liye matching angles ka har pair hona chahiye aur matching sides ka har pair hona chahiye.
  Answer dekho
  Answer (1) equal, (2) proportional
  Polygons ke liye dono conditions zaroori hain.
Short answerEasyNCERT Exercise 6.1, Q2
Roz ki cheezon ke do aise pairs batao jo similar hain aur do aise jo nahi hain. Har ek ka reason do.
Solution dikhao
Similar: ₹1 ka coin aur ₹5 ka coin (dono circles); ek A4 sheet aur ek A5 sheet (A5, A4 ko aadha fold karne se banti hai, aur dono ki sides lagbhag 1 : 1.41 ke ratio mein hain).
Similar nahi: geometry box ke do set squares (ek ke angles $30^{\circ}, 60^{\circ}, 90^{\circ}$ hain, doosre ke $45^{\circ}, 45^{\circ}, 90^{\circ}$ ); ek cricket pitch aur ek carrom board (lambe rectangle aur square ke side ratios alag hote hain).
3
Short answerEasyNCERT Exercise 6.1, Q3
Har pair ke do quadrilaterals similar hain ya nahi, yeh tay karo. Reason bhi do.
1. 2 cm side ka ek square aur 5 cm side ka ek rhombus jiska ek angle $70^{\circ}$ hai.
  Solution dikhao
  Similar nahi. Sides ek hi ratio (2.5) mein hain, lekin square ke angles $90^{\circ}$ hain aur rhombus ke $70^{\circ}$ aur $110^{\circ}$ .
2. Ek 3 cm × 2 cm rectangle aur ek 7.5 cm × 5 cm rectangle.
  Solution dikhao
  Similar. Saare angles $90^{\circ}$ hain, aur $\frac{7.5}{3} = \frac{5}{2} = 2.5$ , isliye sides ek hi ratio mein hain.
Quick answerEasy
Ek 10 cm × 15 cm photo ko aise enlarge kiya jaata hai ki uski chhoti side 25 cm ho jaaye. Lambi side kitni lambi hogi, cm mein?
cm
Answer dekho
Answer 37.5 cm
Scale factor $\frac{25}{10} = 2.5$ , isliye lambi side $15 \times 2.5 = 37.5 c m$ hai.

6.3 Similarity of Triangles

Triangle bhi ek polygon hai, isliye test nahi badalta: matching angles equal hon, aur matching sides sab ek hi ratio mein hon. Jin triangles ke matching angles equal hote hain, unhe equiangular triangles kehte hain.

Thales of Miletus ne equiangular triangles ke baare mein ek khaas baat notice ki: matching sides ke har pair ka ratio same hota hai. Yaani triangles ke liye equal angles apne aap proportional sides le aate hain. Iski chaabi ek theorem hai jo triangle ki ek side ke parallel line ke baare mein hai.

Activity 6.2NCERT Activity 2 · p. 79

Notebook page par parallel lines

Tumhe chahiye

ruled notebook ka ek page
ek nukeeli pencil
millimetre scale wala ek ruler

Pehle guess karo

Ek ruled line kisi triangle ko uske top corner se 4 lines neeche aur base se 3 lines upar kaatti hai. Woh har tirchi side ko kis ratio mein kaatti hai?

3 : 44 : 34 : 71 : 1

Ek ruled line par point $A$ mark karo. Saat lines neeche, do points $B$ aur $C$ ek doosre se door mark karo. $A B$ , $B C$ aur $C A$ ko jodo.
$A$ ke neeche ki chauthi ruled line $A B$ ko $D$ par aur $A C$ ko $E$ par kaatti hai. $D E$ ko pen se gaadha karo: yeh $B C$ ke parallel hai, kyunki ruled lines parallel hoti hain.
$A D$ , $D B$ , $A E$ aur $E C$ ko nearest millimetre tak naapo aur table bharo.
Ek naye triangle ke saath dohrao, is baar $D E$ ke liye $A$ ke neeche ki doosri ruled line lo.

Table 6.2

Length ya ratio	Pehla triangle	Doosra triangle
$A D$
$D B$
$A D \div D B$
$A E$
$E C$
$A E \div E C$

Tumhari entries apne aap save hoti hain.

3/3Har tirchi side par $D E$ ke upar 4 equal gaps hain aur neeche 3: $A D : D B = A E : E C = 4 : 3.$

Fig. 6.6Barabar doori wali parallel lines dono sides ko equal tukdon mein kaatti hain.NCERT Fig. 6.9 · p. 79

Page par aath parallel ruled lines hain. Triangle ABC ka A sabse upar wali line par hai aur B aur C aathvi line par. Pehle step mein sirf lines aur triangle dikhte hain. Doosre step mein A ke neeche ki chauthi line ko AB par D aur AC par E ke beech highlight kiya gaya hai. Teesre step mein dots dikhate hain ki har ruled line dono sides ko kahan kaatti hai, jisse har side par DE ke upar 4 equal gaps aur neeche 3 dikhte hain, saath mein notes AD : DB = 4 : 3 aur AE : EC = 4 : 3.

Yeh fact Basic Proportionality Theorem (आधारभूत समानुपातिकता प्रमेय) hai. Iska proof triangles ke areas ko compare karta hai; Fig. 6.7 woh teen pairs of triangles dikhata hai jo proof mein use hote hain.

2/4 $△ A D E$ aur $△ B D E$ ki height $E N$ same hai, isliye unke areas ka ratio hai $A D : D B .$

Fig. 6.7Area ke teen comparisons se theorem prove hota hai.NCERT Fig. 6.10 · p. 80

Triangle ABC jismein D, AB par aur E, AC par hai, aur DE, BC ke parallel hai, jo arrowheads se mark kiya gaya hai. Pehle step mein sirf yahi dikhta hai. Doosre step mein BE joda gaya hai, triangles ADE aur BDE shade kiye gaye hain aur E se AB par perpendicular EN banaya gaya hai: dono ki height EN same hai. Teesre step mein CD joda gaya hai, triangles ADE aur CDE shade kiye gaye hain aur D se AC par perpendicular DM banaya gaya hai: dono ki height DM same hai. Chauthe step mein triangles BDE aur CDE ek saath shade kiye gaye hain: unka base DE same hai aur woh same parallels ke beech hain, isliye unke areas equal hain.

Theorem 6.1Basic Proportionality Theorem (Thales’ theorem)

Agar triangle ki ek side ke parallel koi line baaki do sides ko alag-alag points par kaatti hai, to woh unhe same ratio mein divide karti hai.

Diya hai: $△ A B C$ mein ek line $A B$ ko $D$ par aur $A C$ ko $E$ par milti hai, aur $D E ∥ B C .$

Prove karna hai: $\frac{A D}{D B} = \frac{A E}{E C} .$

Construction: $B E$ aur $C D$ ko jodo. Do perpendiculars banao: $E N ⊥ A B$ aur $D M ⊥ A C .$

Proof: triangle ka area $= \frac{1}{2} \times$ base $\times$ height. Triangles $A D E$ aur $B D E$ ke bases $A D$ aur $D B$ line $A B$ par hain aur height $E N$ same hai. Isliye $ar (A D E) = \frac{1}{2} A D \cdot E N$ aur $ar (B D E) = \frac{1}{2} D B \cdot E N .$ Divide karne par factor $\frac{1}{2} E N$ cancel ho jaata hai:

\frac{ar (A D E)}{ar (B D E)} = \frac{A D}{D B}

(6.1)

Isi tarah, triangles $A D E$ aur $C D E$ ke bases $A E$ aur $E C$ line $A C$ par hain aur height $D M$ same hai:

\frac{ar (A D E)}{ar (C D E)} = \frac{A E}{E C}

(6.2)

Triangles $B D E$ aur $C D E$ same base $D E$ par khade hain aur same parallels $D E$ aur $B C$ ke beech hain. Class 9 se hum jaante hain ki aise triangles ke areas equal hote hain:

\begin{aligned} ar (B D E) \\ = ar (C D E) \end{aligned}

(6.3)

(6.3) se, (6.1) aur (6.2) ke left sides ek hi fraction hain. Isliye $\frac{A D}{D B} = \frac{A E}{E C} .$ ∎

Kya converse bhi sach hai? Maan lo ek line triangle ki do sides ko proportionally kaatti hai. Kya tab use teesri side ke parallel hona hi padega? Karke dekho.

Activity 6.3NCERT Activity 3 · p. 81

Dono arms par barabar fractions

Tumhe chahiye

ek ruler
do set squares
ek nukeeli pencil

Pehle guess karo

Ek angle ki ek arm par tum har 1.5 cm par points mark karte ho; doosri arm par har 2 cm par. Tum dono arms ka pehla mark jodte ho, aur dono arms ka chautha mark bhi. Kya dono jodne wali lines parallel hongi?

haan, kisi bhi angle ke liyesirf tab jab angle

90^{\circ}

honahi, kyunki spacings alag hainsirf tab jab arms barabar hon

Kisi bhi size ka ek angle $X A Y$ banao.
$A X$ par $A$ se 1.5, 3, 4.5 aur 6 cm par $B_{1}, B_{2}, B_{3}, B$ mark karo. $A Y$ par $A$ se 2, 4, 6 aur 8 cm par $C_{1}, C_{2}, C_{3}, C$ mark karo.
$B C$ ko jodo, phir $B_{1} C_{1}$ , $B_{2} C_{2}$ aur $B_{3} C_{3}$ ko.
Har join ko $B C$ ke against test karo: ek set square $B C$ ke saath rakho, doosra set square uske saath sata kar pakdo, aur pehle ko doosre ke saath slide karo.
Aakhir mein $B_{1} C_{2}$ ko jodo aur use bhi isi tarah test karo.

Table 6.3

Join	$A X$ par ratio	$A Y$ par ratio	Parallel?
$B_{1} C_{1}$
$B_{2} C_{2}$
$B_{3} C_{3}$
$B_{1} C_{2}$

Tumhari entries apne aap save hoti hain.

Theorem 6.2Converse of the Basic Proportionality Theorem

Jo line triangle ki do sides ko proportionally divide karti hai, woh baaki bachi side ke parallel hoti hai.

Diya hai: $A B$ par $D$ aur $A C$ par $E$ , jahan $\frac{A D}{D B} = \frac{A E}{E C} .$ Prove karna hai: $D E ∥ B C .$

Proof (contradiction se): maan lo $D E$ , $B C$ ke parallel nahi hai. $D$ se $B C$ ke parallel line banao; yeh $A C$ ko $E$ se alag ek point $E^{'}$ par milti hai.

$△ A B C$ mein Theorem 6.1 se milta hai: $\frac{A D}{D B} = \frac{A E^{'}}{E^{'} C} .$ Diye gaye ratio ke saath, $\frac{A E}{E C} = \frac{A E^{'}}{E^{'} C} .$

Dono sides mein 1 jodo: $\frac{A E + E C}{E C} = \frac{A E^{'} + E^{'} C}{E^{'} C},$ yaani $\frac{A C}{E C} = \frac{A C}{E^{'} C} .$ Isliye $E C = E^{'} C$ .

$E$ aur $E^{'}$ dono segment $A C$ par $C$ se same distance par hain, isliye woh ek hi point hain. Yeh hamari maanyata ke ulta hai. Isliye $D E ∥ B C$ . ∎

2/3 $D E^{'} ∥ B C$ banao. Theorem 6.1 se $E^{'}$ , $A C$ ko $A D : D B$ ke ratio mein divide karta hai, bilkul $E$ ki tarah.

Fig. 6.8Agar

D

aur

E

sides ko same ratio mein kaatte hain, to

E

ke paas parallel par aane ke alawa koi jagah nahi.NCERT Fig. 6.12 · p. 82

Triangle ABC jismein D, AB ke beech wale hisse mein hai. Pehle step mein ek dashed line D ko AC par ek point E se jodti hai, aise banayi gayi jaise DE, BC ke parallel na ho. Doosre step mein D se BC ke parallel line banayi gayi hai, jo AC ko E ke upar ek doosre point E′ par milti hai. Teesre step mein point E upar khisak kar E′ par aa jaata hai aur dashed line gaayab ho jaati hai, aur DE, BC ke parallel reh jaati hai, saath mein note ki E′ aur E ek hi point hain.

InteractiveBPT lab

Ek triangle ABC jiske across ek line DE hai, aur slider controls. Parallel line mode mein ek slider D ko AB ke saath chalata hai aur E uske saath aise chalta hai ki DE, BC ke parallel rahe; ek table AD, DB, AE aur EC centimetres mein aur ratios AD/DB aur AE/EC dikhati hai, jo hamesha equal rehte hain. Equal ratios mode mein do sliders D ko AB par aur E ko AC par alag-alag set karte hain; DE sirf tab green hoti hai aur parallel marks dikhte hain jab dono ratios equal hon, aur ek doosri table angle ADE ko angle ABC se compare karti hai. New triangle button triangle ki shape badal deta hai.

Fig. 6.9

D

(aur

E

) ko sides ke saath slide karo aur dono ratios compare karo:

\frac{A D}{D B}

aur

\frac{A E}{E C} .

Example 6.1NCERT Example 1 · p. 82

Theorem ka ek aur form

$△ A B C$ mein $D E ∥ B C$ , jahan $D$ , $A B$ par aur $E$ , $A C$ par hai. (a) Dikhao ki $\frac{A D}{A B} = \frac{A E}{A C} .$ (b) Agar $A D = 3$ cm, $A B = 7.5$ cm aur $A C = 10$ cm hai, to $A E$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Table 6.4Theorem 6.1 ke teen forms, sab tab sach jab

D E ∥ B C

Form	Yeh compare karta hai
$\frac{A D}{D B} = \frac{A E}{E C}$	har side ke do tukde
$\frac{A D}{A B} = \frac{A E}{A C}$	upar wala tukda poori side se
$\frac{D B}{A B} = \frac{E C}{A C}$	neeche wala tukda poori side se

Forms ko kabhi mix mat karo: yeh do fractions barabar nahi hain: $\frac{A D}{A B}$ aur $\frac{A E}{E C} .$ Ek form chuno aur dono taraf same tukde rakho.

2/3Diagonal $P R$ banao. $△ P S R$ mein $X Z ∥ S R$ .

Fig. 6.10Ek diagonal trapezium ko do triangles mein baant deta hai.NCERT Figs. 6.14–6.15 · p. 83

Ek trapezium PQRS jiski lambi side PQ neeche hai aur chhoti parallel side SR upar. Uske across ek fence XY dono ke parallel chalti hai, PS par X se QR par Y tak. PX 12 m hai, XS 8 m hai, QY 15 m hai aur YR unknown hai. Pehle step mein sirf khet aur fence dikhte hain. Doosre step mein diagonal PR banaya gaya hai jo fence ko Z par milta hai, aur triangle PSR shade kiya gaya hai. Teesre step mein iske bajaye triangle RPQ shade kiya gaya hai.

Example 6.2NCERT Example 2 · p. 83

Trapezium ki shape wale khet ke across ek fence

Ek khet $P Q R S$ trapezium hai jismein $P Q ∥ S R$ (Fig. 6.10). Ek seedhi fence $X Y$ , jo $P Q$ ke parallel hai, $P S$ par $X$ se $Q R$ par $Y$ tak jaati hai. $P X = 12$ m, $X S = 8$ m aur $Q Y = 15$ m. $Y R$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.3NCERT Example 3 · p. 83

Equal ratios, phir ek isosceles triangle

$△ K L M$ mein $X$ , $K L$ par aur $Y$ , $K M$ par hai, jahan $K X : X L = K Y : Y M = 3 : 4$ . Saath hi $∠ K X Y = 68^{\circ}$ aur $∠ K = 44^{\circ}$ . $∠ K M L$ nikaalo aur dikhao ki $△ K L M$ isosceles hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 10 questions

Practice 6.2: answers ke saath

1
NumericalEasyNCERT Exercise 6.2, Q1
Fig. 6.11 ke dono triangles mein $D E ∥ B C$ .
Fig. 6.11Lengths cm mein; dono mein $D E ∥ B C$ .NCERT Fig. 6.17 · p. 84
Panel (a): triangle ABC jismein D, AB par aur E, AC par hai, DE, BC ke parallel hai; AD 2.4 hai, DB 3.6 hai, AE 2 hai aur EC par question mark hai. Panel (b): ek aur triangle ABC jismein DE, BC ke parallel hai; AD par question mark hai, AE 2.8 hai, EC 4.2 hai aur note mein AB = 9 likha hai. Saari lengths centimetres mein hain.
1. $E C$ nikaalo.
  cm
  Solution dikhao
  $\frac{A D}{D B} = \frac{A E}{E C}$ se milta hai $\frac{2.4}{3.6} = \frac{2}{E C},$ isliye $E C = 2 \times 3.6 \div 2.4 = 3$ cm.
2. $A D$ nikaalo.
  cm
  Solution dikhao
  $A C = 2.8 + 4.2 = 7$ ke saath $\frac{A D}{A B} = \frac{A E}{A C}$ use karo: $A D = 9 \times \frac{2.8}{7} = 3.6$ cm.
2
Quick answerEasyNCERT Exercise 6.2, Q2
$E$ aur $F$ , $△ P Q R$ ki sides $P Q$ aur $P R$ par hain. Har case mein, kya $E F ∥ Q R$ ? (Lengths cm mein.)
1. $P E = 2.4$ , $E Q = 3.2$ , $P F = 1.8$ , $F R = 2.4$
  parallelparallel nahikeh nahi sakte
  Answer dekho
  Answer (A) parallel
  $\frac{2.4}{3.2} = 0.75 = \frac{1.8}{2.4},$ isliye Theorem 6.2 se $E F ∥ Q R$ .
2. $P E = 5$ , $E Q = 4$ , $P F = 6$ , $F R = 5$
  parallelparallel nahikeh nahi sakte
  Answer dekho
  Answer (B) parallel nahi
  $\frac{5}{4} = 1.25$ lekin $\frac{6}{5} = 1.2.$ Ratios alag hain, isliye $E F$ , $Q R$ ke parallel nahi hai.
3. $P Q = 2.25$ , $P E = 0.75$ , $P R = 3.6$ , $P F = 1.2$
  parallelparallel nahikeh nahi sakte
  Answer dekho
  Answer (A) parallel
  $\frac{P E}{P Q} = \frac{1}{3} = \frac{P F}{P R},$ isliye har side $1 : 2$ ke ratio mein kati hai aur $E F ∥ Q R$ .
3
ProofMediumNCERT Exercise 6.2, Q3 aur Q5
$P Q R S$ ek quadrilateral hai aur $X$ uske diagonal $P R$ par ek point hai. $X$ se $R Q$ ke parallel line $P Q$ ko $Y$ par milti hai, aur $X$ se $R S$ ke parallel line $P S$ ko $Z$ par milti hai.
1. Prove karo ki $\frac{P Y}{Y Q} = \frac{P Z}{Z S} .$
  Solution dikhao
  $△ P Q R$ mein $X Y ∥ R Q$ , isliye $\frac{P Y}{Y Q} = \frac{P X}{X R}$ (Theorem 6.1).
  $△ P S R$ mein $X Z ∥ R S$ , isliye $\frac{P Z}{Z S} = \frac{P X}{X R} .$ Dono ek hi ratio ke barabar hain: $\frac{P X}{X R} .$
2. Isse prove karo ki $Y Z ∥ Q S$ .
  Solution dikhao
  $△ P Q S$ mein $Y$ aur $Z$ , $P Q$ aur $P S$ ko same ratio mein divide karte hain (part a), isliye Theorem 6.2 se $Y Z ∥ Q S$ .
NumericalHardNCERT Exercise 6.2, Q4
$△ X Y Z$ mein $L$ , $X Y$ par ek point hai. $L$ se $X Z$ ke parallel line $Y Z$ ko $M$ par milti hai, aur $L$ se $X M$ ke parallel line $Y M$ ko $N$ par milti hai. Agar $Y N = 9$ cm aur $N M = 3$ cm hai, to $M Z$ cm mein nikaalo.
cm
Solution dikhao
$△ Y X M$ mein $L N ∥ X M$ , isliye $\frac{Y N}{N M} = \frac{Y L}{L X} .$
$△ Y X Z$ mein $L M ∥ X Z$ , isliye $\frac{Y L}{L X} = \frac{Y M}{M Z} .$
Isliye $\frac{Y N}{N M} = \frac{Y M}{M Z}$ , jahan $Y M = 9 + 3 = 12$ : $\frac{9}{3} = \frac{12}{M Z},$ aur $M Z = 4$ cm.
ProofMediumNCERT Exercise 6.2, Q6
$O$ , $△ P Q R$ ke andar ek point hai. Points $A$ , $B$ , $C$ kramshah $O P$ , $O Q$ , $O R$ par hain, jahan $A B ∥ P Q$ aur $B C ∥ Q R$ . Prove karo ki $A C ∥ P R$ .
Solution dikhao
$△ O P Q$ mein $A B ∥ P Q$ se milta hai: $\frac{O A}{A P} = \frac{O B}{B Q} .$
$△ O Q R$ mein $B C ∥ Q R$ se milta hai: $\frac{O B}{B Q} = \frac{O C}{C R} .$
Isliye $\frac{O A}{A P} = \frac{O C}{C R},$ aur $△ O P R$ mein Theorem 6.2 se $A C ∥ P R$ milta hai.
ProofEasyNCERT Exercise 6.2, Q7
$△ P Q R$ mein $M$ , $P Q$ ka mid-point hai, aur $M$ se $Q R$ ke parallel line $P R$ ko $N$ par milti hai. Theorem 6.1 use karke dikhao ki $N$ , $P R$ ka mid-point hai.
Solution dikhao
$M N ∥ Q R$ , isliye $\frac{P N}{N R} = \frac{P M}{M Q} = 1$ , kyunki $P M = M Q$ . Isliye $P N = N R$ aur $N$ , $P R$ ka mid-point hai.
ProofEasyNCERT Exercise 6.2, Q8
$△ P Q R$ mein $M$ aur $N$ , $P Q$ aur $P R$ ke mid-points hain. Theorem 6.2 use karke dikhao ki $M N ∥ Q R$ .
Solution dikhao
$\frac{P M}{M Q} = 1 = \frac{P N}{N R},$ isliye $M$ aur $N$ dono sides ko same ratio mein divide karte hain, aur Theorem 6.2 se $M N ∥ Q R$ .
8
Long answerHardNCERT Exercise 6.2, Q9
Trapezium $P Q R S$ mein $P Q ∥ S R$ , aur diagonals $P R$ aur $Q S$ , $O$ par milte hain.
1. Prove karo ki $\frac{P O}{O R} = \frac{Q O}{O S} .$
  Solution dikhao
  $P S$ par $T$ lekar $O T ∥ P Q$ banao; tab $O T ∥ S R$ bhi hai.
  $△ P S R$ mein $T O ∥ S R$ : $\frac{P T}{T S} = \frac{P O}{O R} .$
  $△ S P Q$ mein $T O ∥ P Q$ : $\frac{S T}{T P} = \frac{S O}{O Q},$ isliye $\frac{P T}{T S} = \frac{Q O}{O S} .$
  Isliye $\frac{P O}{O R} = \frac{Q O}{O S} .$
2. Agar $P O = 2 x - 3$ , $O R = 2 x - 2$ , $Q O = 2 x + 4$ aur $O S = 3 x + 1$ hai, to $x$ nikaalo.
  Solution dikhao
  Cross-multiply karo: $(2 x - 3) (3 x + 1)$ $= (2 x + 4) (2 x - 2)$ .
  Isliye $6 x^{2} - 7 x - 3 = 4 x^{2} + 4 x - 8$ , jisse $2 x^{2} - 11 x + 5 = 0$ milta hai, yaani $(2 x - 1) (x - 5) = 0$ .
  $x = \frac{1}{2}$ lene par $P O$ negative ho jaata hai, isliye $x = 5$ . Check: $\frac{7}{8} = \frac{14}{16} .$
ProofHardNCERT Exercise 6.2, Q10
Ek quadrilateral $A B C D$ ke diagonals $O$ par milte hain, aur $\frac{A O}{O C} = \frac{B O}{O D} .$ Prove karo ki $A B C D$ ek trapezium hai.
Solution dikhao
$A D$ par $E$ lekar $O E ∥ D C$ banao. $△ A D C$ mein Theorem 6.1 se milta hai: $\frac{A E}{E D} = \frac{A O}{O C} = \frac{B O}{O D} .$
Isliye $△ D A B$ mein $E$ aur $O$ , $D A$ aur $D B$ ko same ratio mein divide karte hain, aur $E O ∥ A B$ (Theorem 6.2).
Tab $A B ∥ E O ∥ D C$ , isliye $A B C D$ mein parallel sides ka ek pair hai: yeh ek trapezium hai.
NumericalMedium
$△ A B C$ mein $D E ∥ B C$ , jahan $D$ , $A B$ par aur $E$ , $A C$ par hai. $A D = x + 1$ , $D B = 3 x - 1$ , $A E = x + 3$ aur $E C = 3 x + 4$ (cm mein). $x$ nikaalo.
Solution dikhao
$\frac{x + 1}{3 x - 1} = \frac{x + 3}{3 x + 4} .$ Cross-multiply karne par: $3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3$ , isliye $x = 7$ .
Check: $\frac{8}{20} = \frac{10}{25} = 0.4.$

6.4 Criteria for Similarity of Triangles

Definition ke hisaab se, $△ A B C$ aur $△ D E F$ tab similar hain jab angles ke teeno pairs match karein, $∠ A = ∠ D$ , $∠ B = ∠ E$ aur $∠ C = ∠ F$ , aur teeno side ratios barabar hon:

\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D}

(6.4)

Iske liye hum symbol $\sim$ use karte hain, jise "is similar to" padhte hain. Yaani $△ A B C \sim △ D E F$ yeh kehne ka chhota tareeka hai ki dono triangles similar hain, aur $A, B, C$ kramshah $D, E, F$ se match karte hain. Class 9 mein symbol $≅$ congruence ke liye yahi kaam karta tha.

Har baar chhah facts check karna dheema kaam hai. Class 9 mein tumne dekha tha ki teen sahi chune gaye facts congruence tay kar dete hain (SSS, SAS, ASA, RHS). Similarity ke liye bhi kam facts kaafi hain; har aisa shortcut ek similarity criterion hai. Agli teen activities inhe dhoondhti hain.

Activity 6.4NCERT Activity 4 · p. 86

Same angles, alag bases

Tumhe chahiye

ek ruler
ek protractor
ek nukeeli pencil

Pehle guess karo

Do triangles ke bases 4 cm aur 6 cm lambe hain, aur dono ke base ke siron par ek

50^{\circ}

aur ek

70^{\circ}

angle hai. Baaki sides ke liye tum kya expect karte ho?

sirf bases ratio

\frac{2}{3}

mein hainbaaki sides equal hainhar side ratio

\frac{2}{3}

haikoi pattern nahi hai

$B C = 4$ cm banao. $B$ par $B C$ se $50^{\circ}$ par ek ray banao, aur $C$ par $C B$ se $70^{\circ}$ par ek ray, dono same side mein. Rays $A$ par milti hain.
$E F = 6$ cm banao aur $E$ par $50^{\circ}$ aur $F$ par $70^{\circ}$ ke saath dohrao. Rays $D$ par milti hain.
$∠ A$ aur $∠ D$ naapo, phir $A B$ , $C A$ , $D E$ aur $F D$ , aur table bharo.

Table 6.5

Ratio	Value
$B C \div E F$	$4 \div 6 \approx 0.67$
$A B \div D E$
$C A \div F D$

Tumhari entries apne aap save hoti hain.

3/3 $∠ D P Q = ∠ E$ , isliye $P Q ∥ E F$ , aur Theorem 6.1 se milta hai: $\frac{D P}{D E} = \frac{D Q}{D F} .$

Fig. 6.12

△ A B C

ki copy ko corner

D

mein rakhne se problem Theorem 6.1 ban jaati hai.NCERT Fig. 6.24 · p. 87

Left mein ek chhota triangle ABC aur right mein ek bada triangle DEF, jismein B aur E, C aur F, aur A aur D par matching arc marks hain. Pehle step mein dono saath-saath hain. Doosre step mein chhota triangle aise khisakta hai ki A, D par aa jaaye aur uski sides DE aur DF ke saath hon; uske baaki corners DE par P aur DF par Q par aate hain. Teesre step mein PQ ko EF se match karte parallel marks ke saath highlight kiya gaya hai, aur note PQ ∥ EF dikhta hai.

Theorem 6.3AAA similarity criterion

Agar do triangles ke angles pairs mein equal hain, to unki corresponding sides proportional hoti hain, isliye triangles similar hote hain.

Diya hai: $△ A B C$ aur $△ D E F$ , jahan $∠ A = ∠ D$ , $∠ B = ∠ E$ aur $∠ C = ∠ F$ . Maan lo $D E > A B$ . (Agar $D E = A B$ hai, to triangles ASA se congruent hain; agar $D E < A B$ hai, to dono ke roles badal do.)

Construction: $D E$ par $P$ aise mark karo ki $D P = A B$ , aur $D F$ par $Q$ aise ki $D Q = A C$ . $P Q$ ko jodo. Isse $△ A B C$ ki ek copy corner $D$ mein aa jaati hai (Fig. 6.12).

Proof: SAS se $△ A B C ≅ △ D P Q$ ( $A B = D P$ , $∠ A = ∠ D$ , $A C = D Q$ ). Isliye $∠ D P Q = ∠ B = ∠ E$ .

Transversal $D E$ ke saath lines $P Q$ aur $E F$ ke liye $∠ D P Q$ aur $∠ D E F$ corresponding angles hain, aur woh equal hain. Isliye $P Q ∥ E F$ .

Example 6.1 wale form mein Theorem 6.1 se milta hai $\frac{D P}{D E} = \frac{D Q}{D F},$ yaani $\frac{A B}{D E} = \frac{A C}{D F} .$

Copy ko iske bajaye corner $E$ mein rakhne par isi tarah $\frac{A B}{D E} = \frac{B C}{E F}$ milta hai. Isliye teeno ratios equal hain aur $△ A B C \sim △ D E F$ . ∎

CorollaryAA similarity criterion

Jab ek triangle ke do angles doosre triangle ke do angles ke barabar hon, to triangles similar hote hain.

Activity 6.5NCERT Activity 5 · p. 88

Same ratios, phir angles naapo

Tumhe chahiye

ek ruler
ek compass (pair of compasses)
ek protractor

Pehle guess karo

Triangle 1 ki sides 3.5, 4 aur 5 cm hain; triangle 2 ki sides 7, 8 aur 10 cm hain. Angles naapne par tum kya expect karte ho?

bade triangle ke angles bade hainmatching angles equal haintriangle 2 ke angles double hainkoi pattern nahi

$△ A B C$ banao jismein $C A = 5$ cm, $A B = 3.5$ cm aur $B C = 4$ cm (pehle $C A$ banao, phir compass se $B$ dhoondo).
$△ D E F$ banao jismein $F D = 10$ cm, $D E = 7$ cm aur $E F = 8$ cm.
Check karo ki $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D} .$ Phir saare chhah angles naapo.

Table 6.6

Vertices	$△ A B C$ mein	$△ D E F$ mein
$A$ aur $D$
$B$ aur $E$
$C$ aur $F$

Tumhari entries apne aap save hoti hain.

Theorem 6.4SSS similarity criterion

Agar ek triangle ki teeno sides doosre triangle ki teeno sides ke proportional hain, to unke corresponding angles equal hote hain, isliye triangles similar hote hain.

Diya hai: $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D},$ aur yeh common ratio 1 se kam hai.

Construction: $D E$ par $P$ aur $D F$ par $Q$ aise mark karo ki $D P = A B$ aur $D Q = A C$ . $P Q$ ko jodo.

Proof: ab $\frac{D P}{D E} = \frac{D Q}{D F},$ isliye $\frac{D P}{P E} = \frac{D Q}{Q F}$ aur Theorem 6.2 se $P Q ∥ E F$ . Corresponding angles se $∠ D P Q = ∠ E$ aur $∠ D Q P = ∠ F$ milte hain, isliye AA se $△ D P Q \sim △ D E F$ .

Isliye $\frac{P Q}{E F} = \frac{D P}{D E} = \frac{A B}{D E} = \frac{B C}{E F},$ jisse $P Q = B C$ milta hai.

Isliye SSS se $△ A B C ≅ △ D P Q$ . Isliye $∠ A = ∠ D$ , $∠ B = ∠ D P Q = ∠ E$ aur $∠ C = ∠ D Q P = ∠ F$ . ∎

Theorems 6.3 aur 6.4 milkar triangles ke baare mein ek khaas baat kehte hain: equal angles se sides proportional ho jaati hain, aur proportional sides se angles equal ho jaate hain. Quadrilaterals ke liye yeh fail ho gaya tha (Fig. 6.5), lekin triangles ke liye tumhe dono conditions mein se sirf ek check karni hoti hai.

Activity 6.6NCERT Activity 6 · p. 89

Proportional sides ke beech ek equal angle

Tumhe chahiye

ek ruler
ek protractor
ek nukeeli pencil

Pehle guess karo

∠ A = ∠ D = 70^{\circ}

lo, jahan

A B = 2.5

cm,

A C = 4

cm,

D E = 5

cm aur

D F = 8

cm. Dono triangles ke baaki angles bhi match karenge.

SahiGalat

$∠ A = 70^{\circ}$ banao. Uski arms par $B$ aise mark karo ki $A B = 2.5$ cm aur $C$ aise ki $A C = 4$ cm. $B C$ ko jodo.
$∠ D = 70^{\circ}$ banao, jahan $D E = 5$ cm aur $D F = 8$ cm. $E F$ ko jodo.
$∠ B$ , $∠ C$ , $∠ E$ aur $∠ F$ naapo, aur teesri sides $B C$ aur $E F$ bhi.

Table 6.7

Pair	$△ A B C$ mein	$△ D E F$ mein
$∠ B$ aur $∠ E$
$∠ C$ aur $∠ F$
$B C$ aur $E F$

Tumhari entries apne aap save hoti hain.

Theorem 6.5SAS similarity criterion

Do triangles similar hote hain agar ek triangle ka ek angle doosre ke ek angle ke barabar ho aur in angles ko gherne wali sides proportional hon.

Diya hai: $∠ A = ∠ D$ aur $\frac{A B}{D E} = \frac{A C}{D F},$ aur yeh ratio 1 se kam hai. Construction: $D E$ par $P$ aur $D F$ par $Q$ aise mark karo ki $D P = A B$ aur $D Q = A C$ . $P Q$ ko jodo.

Proof: SAS se $△ A B C ≅ △ D P Q$ ( $A B = D P$ , $∠ A = ∠ D$ , $A C = D Q$ ).

Saath hi $\frac{D P}{D E} = \frac{D Q}{D F},$ isliye Theorem 6.2 se $P Q ∥ E F$ , aur corresponding angles se $∠ D P Q = ∠ E$ aur $∠ D Q P = ∠ F$ milte hain.

Isliye $∠ B = ∠ D P Q = ∠ E$ aur $∠ C = ∠ D Q P = ∠ F$ . $∠ A = ∠ D$ ke saath, AAA se triangles similar hain. ∎

3/3SAS: equal angles ka ek pair, aur unhe banane wali sides ek hi ratio mein.

Fig. 6.13Har criterion ko sirf teen facts chahiye.NCERT p. 90

Ek chhota triangle ABC aur ek bada triangle DEF, jo dedh guna bada hai. AA step mein matching arcs B aur E ke equal angles aur C aur F ke equal angles mark karte hain. SSS step mein chhote triangle ki sides par 4, 6 aur 5 aur bade triangle ki sides par 6, 9 aur 7.5 likha hai, har side 1.5 se multiply hui hai. SAS step mein sirf B aur E ke equal angles mark hain, aur unke aas-paas ki sides par 4 aur 6, phir 6 aur 9 likha hai.

Compare

Congruence aur similarity ke tests

Congruent (Class 9)

SSS: teeno sides equal
SAS: do sides aur unke beech ka angle equal
ASA ya AAS: do angles aur ek side equal
RHS: right angle, hypotenuse aur ek side equal

Similar (yeh chapter)

SSS: teeno sides proportional
SAS: proportional sides ke beech ek equal angle
AA: do angles equal (koi side nahi chahiye)
RHS: right angle, hypotenuse aur ek side proportional

Ab criteria use karte hain. Har example mein pehle equal angles ya proportional sides ke pairs dhoondo, phir criterion ka naam likho, phir vertices ko matching order mein rakh kar similarity likho.

2/3Alternate angles equal hain (same marks), aur $C$ par vertically opposite angles bhi.

Fig. 6.14Parallel lines aur vertically opposite angles se do similar triangles bante hain.NCERT Fig. 6.29 · p. 91

Segments AE aur BD, C par ek doosre ko kaatte hain. Left mein AB triangle ABC ki ek vertical side hai; right mein DE triangle EDC ki ek chhoti vertical side hai, jo AB ke parallel hai. AB 9 cm hai, AC 6 cm hai, CE 4 cm hai aur DE unknown hai. Pehle step mein sirf figure aur lengths dikhte hain. Doosre step mein matching arcs A aur E ke, aur B aur D ke equal alternate angles, aur C par vertically opposite angles mark karte hain. Teesre step mein triangle EDC ki ek dashed copy C ke around aadha chakkar ghoomti hai aur triangle ABC ke andar aa jaati hai, uski side AB ke parallel hoti hai.

Example 6.4NCERT Example 4 · p. 91

Bow-tie jaise triangles

Fig. 6.14 mein $A B ∥ E D$ , aur segments $A E$ aur $B D$ , $C$ par ek doosre ko kaatte hain. $A B = 9$ cm, $A C = 6$ cm aur $C E = 4$ cm. Dikhao ki $△ A B C \sim △ E D C$ aur $D E$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.5NCERT Example 5 · p. 91

Matching angle dhoondo

$△ A B C$ mein $A B = 3$ cm, $B C = 6$ cm, $C A = 3 \sqrt{3}$ cm, $∠ A = 90^{\circ}$ aur $∠ B = 60^{\circ}$ . $△ P Q R$ mein $P Q = 12$ cm, $Q R = 6 \sqrt{3}$ cm aur $R P = 6$ cm. $∠ Q$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6.6NCERT Example 6 · p. 92

Product se similar triangles tak

Segments $A C$ aur $B D$ , $O$ par ek doosre ko kaatte hain, jahan $O A = 6$ cm, $O C = 4$ cm, $O B = 3$ cm aur $O D = 8$ cm. (a) Check karo ki $O A \cdot O C = O B \cdot O D$ . (b) Prove karo ki $∠ O A B = ∠ O D C$ . (c) Agar $D C = 10$ cm hai, to $A B$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Jab question mein lengths ka product diya ho, jaise Example 6.6 mein, to divide karke use ratio bana do. Phir un sides ke beech ek equal angle dhoondo.

3/3Right triangles $A B E$ aur $C D E$ mein $∠ E$ common hai, isliye woh similar hain (AA).

Fig. 6.15Bachcha jitna door chalta hai, shadow utni lambi hoti hai.NCERT Fig. 6.32 · p. 92

Left mein ek post AB ke top par street lamp A hai. Pehle step mein ek bachcha post ke paas khada hai aur chhoti shadow banata hai; lamp se ek dashed ray bachche ke sir ko chhooti hui ground par shadow ki tip par milti hai. Doosre step mein bachcha 5 seconds mein post se 6 m door chala gaya hai, D par khada hai aur uska sir C par hai, aur woh lambi shadow DE banata hai. Teesre step mein do right triangles ABE aur CDE shade kiye gaye hain, jahan AB = 4.5 m, CD = 1.5 m, BD = 6 m aur shadow DE = x.

Example 6.7NCERT Example 7 · p. 92

Street lamp ke neeche ek shadow

1.5 m lamba ek bachcha 4.5 m oonche street lamp ke post se 1.2 m/s ki speed se door chalta hai. 5 seconds baad bachche ki shadow kitni lambi hogi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Dhyaan do ki yahan shadow hamesha bachche ki post se distance ki aadhi hai, kyunki $\frac{B D + x}{x} = 3$ se milta hai $x = \frac{B D}{2} .$ Bachcha jaise-jaise door jaata hai, shadow badhti jaati hai.

Example 6.8NCERT Example 8 · p. 93

Similar triangles ke altitudes

$△ A B C \sim △ P Q R$ , aur $A D$ aur $P S$ altitudes hain ( $A D ⊥ B C$ , $P S ⊥ Q R$ ). (a) Prove karo ki $△ A B D \sim △ P Q S$ . (b) Prove karo ki $\frac{A D}{P S} = \frac{A B}{P Q} .$ (c) Agar $A B = 10$ cm, $P Q = 6$ cm aur $A D = 8$ cm hai, to $P S$ nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yahi baat medians aur angle bisectors ke liye bhi sach hai, jaise tum Practice 6.3 mein prove karoge. Similar triangles ke matching segments sides ke ratio mein hote hain.

Quick check

3 mein se 1

Ek hi pal mein 1.2 m lamba ek ladka 0.9 m ki shadow banata hai aur ek neem ka ped 6 m ki shadow. Ped kitna lamba hai, metres mein?

Answer dekho

Answer 8 m

Height ÷ shadow dono ke liye same hai:

\frac{h}{6} = \frac{1.2}{0.9},

isliye

h = 8

6.5 Summary

2/3Corresponding angles equal hain, isliye $△ A D E \sim △ A B C$ (AA).

Fig. 6.17Poora chapter ek triangle mein.

Triangle ABC jismein D, AB par aur E, AC par hai, aur DE, BC ke parallel hai. Pehle step mein DE highlight hai aur note mein AD : DB = AE : EC likha hai, yaani Basic Proportionality Theorem. Doosre step mein chhota triangle ADE shade kiya gaya hai aur D aur B, aur E aur C par matching angle marks dikhate hain ki AA se triangle ADE, triangle ABC ke similar hai. Teesre step mein note mein DE ÷ BC = AD ÷ AB = AE ÷ AC likha hai, yaani parallel segment, BC ka utna hi fraction hai.

Recap

Summary aur key points

Similar figures ki shape same hoti hai; congruent figures ka size bhi same hota hai (Section 6.2).
Same number of sides wale polygons tab similar hote hain jab matching angles equal hon aur matching sides proportional; aam taur par dono zaroori hain (Section 6.2).
Matching sides ka common ratio scale factor hai (Section 6.2).
BPT: triangle ki ek side ke parallel line baaki do sides ko proportionally kaatti hai (Theorem 6.1).
Converse: jo line do sides ko proportionally kaatti hai, woh teesri side ke parallel hoti hai (Theorem 6.2).
AA: equal angles ke do pairs se triangles similar hote hain (Theorem 6.3).
SSS: proportional sides ke teen pairs se triangles similar hote hain (Theorem 6.4).
SAS: proportional sides ke beech ek equal angle se triangles similar hote hain (Theorem 6.5).
Triangles ke liye equal angles aur proportional sides saath-saath aate hain, isliye ek condition kaafi hai (Section 6.4).
Similarity statements mein matching vertices ko order mein likho (Order zaroori hai).
Similar triangles woh cheezein naap lete hain jahan tum pahunch nahi sakte, jaise shadows se heights (Example 6.7).

Flash cards · 6

Khud ko test karo

1 / 6

Board exam ki practice

Yeh questions chapter ke ideas ko board papers wale styles mein milaate hain: multiple choice, assertion aur reason, ek case study, aur lambe proofs.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 6.1, Q1	Practice 6.1 · Q1
NCERT Exercise 6.1, Q2	Practice 6.1 · Q2
NCERT Exercise 6.1, Q3	Practice 6.1 · Q3
NCERT Exercise 6.2, Q1	Practice 6.2 · Q1
NCERT Exercise 6.2, Q2	Practice 6.2 · Q2
NCERT Exercise 6.2, Q3 aur Q5	Practice 6.2 · Q3
NCERT Exercise 6.2, Q4	Practice 6.2 · Q4
NCERT Exercise 6.2, Q6	Practice 6.2 · Q5
NCERT Exercise 6.2, Q7	Practice 6.2 · Q6
NCERT Exercise 6.2, Q8	Practice 6.2 · Q7
NCERT Exercise 6.2, Q9	Practice 6.2 · Q8
NCERT Exercise 6.2, Q10	Practice 6.2 · Q9
NCERT Exercise 6.3, Q1	Practice 6.3 · Q1
NCERT Exercise 6.3, Q2	Practice 6.3 · Q2
NCERT Exercise 6.3, Q3	Practice 6.3 · Q3
NCERT Exercise 6.3, Q4	Practice 6.3 · Q4
NCERT Exercise 6.3, Q5	Practice 6.3 · Q5
NCERT Exercise 6.3, Q6	Practice 6.3 · Q6
NCERT Exercise 6.3, Q7	Practice 6.3 · Q7
NCERT Exercise 6.3, Q8	Practice 6.3 · Q8
NCERT Exercise 6.3, Q9	Practice 6.3 · Q9
NCERT Exercise 6.3, Q10	Practice 6.3 · Q10
NCERT Exercise 6.3, Q11	Practice 6.3 · Q11
NCERT Exercise 6.3, Q12	Practice 6.3 · Q12
NCERT Exercise 6.3, Q13	Practice 6.3 · Q13
NCERT Exercise 6.3, Q14	Practice 6.3 · Q14
NCERT Exercise 6.3, Q15	Practice 6.3 · Q15
NCERT Exercise 6.3, Q16	Practice 6.3 · Q16

Quick check: MCQs

6.2 Similar Figures3 questions
6.3 Similarity of Triangles4 questions
6.4 Criteria for Similarity of Triangles4 questions
6.4 Criteria for Similarity of Triangles3 questions

Glossary · 8

Key terms (English + Hindi)

Basic Proportionality Theoremआधारभूत समानुपातिकता प्रमेयaadhaarbhoot samaanupaatikta pramey: Triangle ki ek side ke parallel line baaki do sides ko same ratio mein divide karti hai.Chapter mein pehli baar yahan aaya →

congruent figuresसर्वांगसम आकृतियाँsarvaangsam aakritiyaan: Same shape aur same size wale figures.Chapter mein pehli baar yahan aaya →
converseविलोमvilom: Woh statement jo kisi theorem mein diye gaye aur nikale gaye hisson ko aapas mein badalne se banta hai.Chapter mein pehli baar yahan aaya →
corresponding partsसंगत भागsangat bhaag: Do figures ke vertices ki di gayi pairing ke hisaab se unke matching vertices, sides aur angles.Chapter mein pehli baar yahan aaya →

equiangular trianglesसमानकोणिक त्रिभुजsamaankonik tribhuj: Aise do triangles jinke angles pairs mein equal hain.Chapter mein pehli baar yahan aaya →

scale factorस्केल गुणकscale gunak: Do similar figures ki matching sides ka common ratio.Chapter mein pehli baar yahan aaya →
similar figuresसमरूप आकृतियाँsamroop aakritiyaan: Aise figures jinki shape ek hi ho, size chahe kuch bhi ho: ek doosre ka exact enlargement ya reduction hota hai.Chapter mein pehli baar yahan aaya →
similarity criterionसमरूपता की कसौटीsamroopta ki kasauti: Facts ka ek chhota set (AA, SSS ya SAS) jo do triangles ko similar prove karne ke liye kaafi hai.Chapter mein pehli baar yahan aaya →

Triangles: Class 10 Maths Chapter 6 Hinglish mein

Is chapter mein aap kya seekhenge

6.1 Introduction

6.2 Similar Figures

Practice 6.1: answers ke saath

6.3 Similarity of Triangles

Practice 6.2: answers ke saath

6.4 Criteria for Similarity of Triangles

Practice 6.3: answers ke saath

6.5 Summary

Summary aur key points

Board exam ki practice

Practice questions answers ke saath (NCERT “Board exam ki practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye