Class 10 Maths · Chapter 6
Short answer:
Rationalised NCERT Chapter 6 (Triangles) me ab sirf 3 exercises hain — Exercise 6.1 (3 questions, similar figures ka intro), Exercise 6.2 (10 questions, Basic Proportionality Theorem), aur Exercise 6.3 (16 questions, similarity criteria + applications, jisme Pythagoras aur area-ratio wale purane exercises 6.4–6.6 ka core idea bhi merge ho gaya hai). Total 29 questions — zyada proof-based hain, isliye figure ko dhyaan se samajhna zaroori hai kyunki diagram exam me hi milega.
Triangles chapter similarity ke concept pe khada hai — same shape, chahe size alag ho. Class 9 me tumne congruence padha tha (same shape AUR same size). Yahan hum ek level upar jaate hain: similarity ka matlab hai corresponding angles equal hain aur corresponding sides same ratio (proportion) me hain. Isi idea se BPT (Thales theorem), similarity ke teen criteria (AAA, SSS, SAS), area ka ratio, aur sabse famous — Pythagoras theorem — sab nikalte hain.
Chapter 6 Summary — 5 Minute Revision
1. Similar figures
Do figures congruent hoti hain jab same shape + same size ho. Do figures similar hoti hain jab same shape ho, size same ho ya na ho. Har congruent figure similar hoti hai, par har similar figure congruent nahi hoti.
- Saare circles similar hote hain (radius chahe kuch bhi ho).
- Saare squares similar hote hain.
- Saare equilateral triangles similar hote hain.
- Rectangles ya isosceles triangles — general me similar nahi hote, kyunki unka shape fix nahi hota (ek aur condition chahiye).
Do polygons (same number of sides wale) similar kehlate hain agar (i) corresponding angles equal ho aur (ii) corresponding sides same ratio me ho. Dono conditions zaroori hain — sirf sides proportional hona kaafi nahi (quadrilaterals ke liye, jaise Ex 6.1 Q3 dikhata hai).
2. Similar triangles
Do triangles similar hain agar unke corresponding angles equal hon aur corresponding sides proportion me hon. Likhte hain: △ABC ~ △DEF — yahan vertex order IMPORTANT hai: A↔D, B↔E, C↔F. Galat order likhna sabse common exam mistake hai.
3. Basic Proportionality Theorem (BPT / Thales Theorem)
Agar ek triangle ke do sides ko koi line intersect kare, aur teesri side ke parallel ho, to woh baaki do sides ko same ratio me divide karti hai.
Yaani: triangle ABC me D, AB par aur E, AC par ho, aur DE ∥ BC ho, to AD/DB = AE/EC.
Converse of BPT: agar koi line kisi triangle ke do sides ko same ratio me divide kare, to woh teesri side ke parallel hoti hai. Iska proof BPT jaisa hi hai, contradiction se — ek naya line teesri side ke parallel maan lo, dikhado woh dono lines coincide karti hain.
Isi se ek zaroori corollary nikalta hai (Ex 6.2 Q7, Q8 me use hota hai): kisi triangle ke ek side ke midpoint se doosre side ke parallel line kheenchi jaye, to woh teesri side ko bhi bisect karti hai — aur ulta bhi sach hai (midpoint theorem).
4. Criteria for similarity of triangles
| Criterion | Condition |
|---|---|
| AAA (asal me AA hi kaafi hai) | Do corresponding angles equal ho jayein to teesra apne aap equal ho jaata hai (angle sum = 180°). Isliye sirf 2 angles match karo similarity prove karne ke liye. |
| SSS | Teeno corresponding sides same ratio me ho. |
| SAS | Do corresponding sides same ratio me ho aur unke beech ka included angle equal ho. |
↔ Table ko side me swipe karein
NCERT rationalised edition me in criteria ke poore formal proofs trim ho chuke hain — board me statement + application (Ex 6.3 jaise questions me pehchan karna aur use karna) hi expect kiya jaata hai, lambi proof nahi.
Zaroori difference: congruence criteria (SSS, SAS, ASA, RHS) me sides equal hote hain; similarity criteria me sides proportional hote hain. Dono ko mix mat karo.
5. Areas of similar triangles
Do similar triangles ke areas ka ratio, unke corresponding sides (ya corresponding medians, ya corresponding altitudes) ke ratio ke square ke barabar hota hai.
Yaani agar △ABC ~ △PQR, to ar(△ABC)/ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)².
Current rationalised book me is theorem par dedicated practice exercise nahi hai (purana Ex 6.4 hata diya gaya), lekin theorem statement chapter me hai aur board PYQs me application-based question aa sakta hai — isliye ratio-is-square-of-ratio wala rule yaad rakhna zaroori hai.
6. Pythagoras Theorem
Ek right triangle me hypotenuse ka square, baaki dono sides ke squares ke sum ke barabar hota hai: (hypotenuse)² = (base)² + (perpendicular)²
Isko similar triangles se prove kiya jaata hai: right triangle ABC (right angle B par) me, B se hypotenuse AC par perpendicular BD daalo. Tab △ADB ~ △ABC aur △BDC ~ △ABC (AA similarity, kyunki dono me ek common angle aur ek right angle hai). Inse geometric-mean relations nikalte hain — AB² = AD·AC aur BC² = CD·AC — jinko add karne par AB² + BC² = AC² mil jaata hai (yehi Ex 6.3 Q13 ka core idea hai: CA² = CB·CD).
Converse of Pythagoras Theorem: agar kisi triangle ke ek side ka square, baaki do sides ke squares ke sum ke barabar ho, to woh angle (jo un do sides ke beech hai) 90° hota hai.
Rationalised edition me Pythagoras ka dedicated exercise (purana 6.5/6.6) hata diya gaya hai, par theorem + converse chapter content me hai aur application questions (height/distance, right-angle check) board me aa sakte hain.
Exercise-wise coverage (current rationalised book)
| Exercise | Questions | Focus |
|---|---|---|
| 6.1 | 3 | Similar figures ka basic intro — definitions, examples, ek quadrilateral-similarity check |
| 6.2 | 10 | BPT aur uska converse — ratio nikalna, parallel-line check karna, midpoint theorem proofs |
| 6.3 | 16 | Similarity criteria (AA/SSS/SAS) pehchanna aur unse proof-based aur application (height/shadow) questions solve karna |
↔ Table ko side me swipe karein
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q29)
Exercise 6.1, Q1. Fill in the blanks using the correct word: (i) All circles are ______. (ii) All squares are ______. (iii) All ______ triangles are similar. (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ______, and (b) their corresponding sides are ______.
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv)(a) corresponding angles are equal, (b) corresponding sides are proportional (same ratio).
Yaad rakho: 'similar' ke liye SHAPE match hona chahiye, size zaroori nahi. Circle/square/equilateral triangle — in teeno ka shape apne aap fix hota hai (koi free parameter nahi jo shape badle), isliye ye hamesha similar hote hain. Rectangle ya scalene/isosceles triangle me aisा nahi — unka shape angle/ratio pe depend karta hai.
Exercise 6.1, Q2. Give two different examples each of pair of (i) similar figures, (ii) non-similar figures.
(i) Similar figures: (a) do photographs of the same person, ek chhota print aur ek bada enlargement — same shape, different size. (b) do equilateral triangles — ek side 3 cm ka, doosra side 6 cm ka.
(ii) Non-similar figures: (a) ek square aur ek rectangle (jiski length ≠ breadth) — angles equal hain par sides ka ratio match nahi karta. (b) ek scalene triangle aur ek equilateral triangle — shape hi alag hai.
Trick: koi bhi do figures jinka SHAPE guaranteed same ho (regular polygons, circles) — similar. Jinka shape depend kare specific measurements pe (general rectangles, general triangles) — non-similar tab tak jab tak proportionality na di ho.
Q3. Exercise 6.1, Q3. State whether the following quadrilaterals are similar or not: quadrilateral PQRS is a square with each side equal to a units (all angles 90°); quadrilateral ABCD has each side equal to 2a units (double of PQRS's side) but ABCD is a rhombus that is NOT a square — its angles are not 90°.
Not similar.
PQ/AB = QR/BC = RS/CD = SP/DA = a/2a = 1/2 — saare corresponding sides SAME ratio (1:2) me hain.
Lekin corresponding angles equal nahi hain — PQRS ke angles 90° hain, ABCD ke angles 90° nahi hain (rhombus hone ki wajah se do angles obtuse aur do acute hain).
Similarity condition: corresponding angles EQUAL + corresponding sides PROPORTIONAL — dono zaroori hain. Sirf sides proportional hona kaafi nahi.
Isliye PQRS aur ABCD similar nahi hain.
Q4. Exercise 6.2, Q1. In triangle ABC, D is a point on AB and E is a point on AC, with DE ∥ BC. (i) If AD = 1.5 cm, DB = 3 cm, AE = 1 cm, find EC. (ii) If AE = 1.8 cm, DB = 7.2 cm (i.e. DB corresponds to AD's denominator), CE = 5.4 cm, find AD.
(i) DE ∥ BC hai, isliye BPT se AD/DB = AE/EC.
1.5/3 = 1/EC ⟹ EC = (3 × 1)/1.5 = 2 cm
EC = 2 cm.
(ii) Same BPT: AD/DB = AE/EC.
AD/7.2 = 1.8/5.4 ⟹ AD = (7.2 × 1.8)/5.4 = 2.4 cm
AD = 2.4 cm.
Exercise 6.2, Q2. E and F are points on the sides PQ and PR respectively of a triangle PQR. State whether EF ∥ QR in each case: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm. (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm. (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.
Converse of BPT use karo: agar PE/EQ = PF/FR ho, to EF ∥ QR.
(i) PE/EQ = 3.9/3 = 1.3; PF/FR = 3.6/2.4 = 1.5 — NOT equal ⟹ EF is NOT parallel to QR.
(ii) PE/QE = 4/4.5 = 8/9; PF/RF = 8/9 — EQUAL ⟹ EF ∥ QR.
(iii) EQ = PQ − PE = 1.28 − 0.18 = 1.10; FR = PR − PF = 2.56 − 0.36 = 2.20. PE/EQ = 0.18/1.10; PF/FR = 0.36/2.20 = 0.18/1.10 — EQUAL ⟹ EF ∥ QR.
(i) No. (ii) Yes. (iii) Yes.
Exercise 6.2, Q3. In the figure, if LM ∥ CB and LN ∥ CD, prove that AM/AB = AN/AD. (L is a point on AC; M on AB such that LM ∥ CB; N on AD such that LN ∥ CD, in quadrilateral ABCD.)
Triangle ABC me LM ∥ CB hai, isliye BPT se: AM/MB = AL/LC ... (1)
Triangle ACD me LN ∥ CD hai, isliye BPT se: AN/ND = AL/LC ... (2)
(1) aur (2) se: AM/MB = AN/ND.
Dono ratios ko 'AM/AB' form me convert karke (componendo): AM/AB = AN/AD
Hence proved.
Exercise 6.2, Q4. In triangle ABC, DE ∥ AC where D is on AB and E is on BC. Also DF ∥ AE where F is on BE. Prove that BF/FE = BE/EC.
Triangle ABC me DE ∥ AC, to BPT (triangle ABE me DF ∥ AE laga ke bhi): triangle ABE me D on AB, F on BE, DF ∥ AE ⟹ BD/DA = BF/FE ... (1)
Triangle ABC me DE ∥ AC ⟹ BD/DA = BE/EC ... (2)
(1) aur (2) se: BF/FE = BE/EC
Hence proved.
Exercise 6.2, Q5. In the figure, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR. (D is a point on PO in triangle POQ extended set-up; E on PQ, F on PR, with O common vertex such that triangles POQ and POR share side PO, and QOR forms the base triangle.)
Triangle POQ me DE ∥ OQ (D on PO, E on PQ) — BPT se: PD/DO = PE/EQ ... (1)
Triangle POR me DF ∥ OR (D on PO, F on PR) — BPT se: PD/DO = PF/FR ... (2)
(1) aur (2) se: PE/EQ = PF/FR
Ab triangle PQR me E on PQ, F on PR, aur PE/EQ = PF/FR hai — Converse of BPT se EF ∥ QR. Hence proved.
Exercise 6.2, Q6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.
Triangle OPQ me AB ∥ PQ (A on OP, B on OQ) — BPT se: OA/AP = OB/BQ ... (1)
Triangle OPR me AC ∥ PR (A on OP, C on OR) — BPT se: OA/AP = OC/CR ... (2)
(1) aur (2) se: OB/BQ = OC/CR
Triangle OQR me B on OQ, C on OR, aur OB/BQ = OC/CR — Converse of BPT se BC ∥ QR. Hence proved.
Exercise 6.2, Q7. Using Theorem 6.1 (Basic Proportionality Theorem), prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Given: Triangle ABC, D is midpoint of AB (AD = DB), DE ∥ BC, E on AC.
To prove: E is the midpoint of AC (AE = EC).
Proof: DE ∥ BC hai, isliye BPT se: AD/DB = AE/EC.
D midpoint hai isliye AD = DB ⟹ AD/DB = 1 ⟹ AE/EC = 1 ⟹ AE = EC
Isliye E bhi AC ka midpoint hai. Hence proved.
Exercise 6.2, Q8. Using Theorem 6.2 (Converse of BPT), prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Given: Triangle ABC, D is midpoint of AB, E is midpoint of AC.
To prove: DE ∥ BC.
Proof: D midpoint of AB ⟹ AD = DB ⟹ AD/DB = 1. E midpoint of AC ⟹ AE = EC ⟹ AE/EC = 1.
AD/DB = AE/EC (= 1)
Converse of BPT se: DE ∥ BC. Hence proved.
Exercise 6.2, Q9. ABCD is a trapezium in which AB ∥ DC and its diagonals intersect each other at point O. Show that AO/BO = CO/DO.
O se ek line kheencho EF, jo AB (aur DC) ke parallel ho, jahan E on AD aur F on BC (ya alternatively E through O parallel to AB meeting AD).
Standard proof: draw OE ∥ AB (∥ DC) meeting AD at E. Triangle ADC me OE ∥ DC (kyunki OE ∥ AB ∥ DC) ⟹ AE/ED = AO/OC ... (1)
Triangle ABD me OE ∥ AB ⟹ AE/ED = BO/OD ... (2)
(1) aur (2) se: AO/OC = BO/OD ⟹ AO/BO = CO/DO
Hence proved.
Exercise 6.2, Q10. The diagonals of a quadrilateral ABCD intersect each other at point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
Yeh Q9 ka converse hai. O se EF kheencho, EO ∥ AB, jahan E on AD.
Triangle DAB me EO ∥ AB (kyunki hum construct kar rahe hain) ⟹ DE/EA = DO/OB (BPT).
Diya hai: AO/BO = CO/DO, yaani DO/OB = CO/AO.
Isse DE/EA = CO/AO milta hai — matlab triangle DCA me bhi ratio match karta hai, isliye EO ∥ DC bhi hai (converse BPT).
Isliye AB ∥ EO ∥ DC, yaani AB ∥ DC. Ek pair of opposite sides parallel hai — isliye ABCD ek trapezium hai. Hence proved.
Exercise 6.3, Q1. State which pairs of triangles are similar in each of the following, and write the similarity criterion used, in symbolic form: (a) △ABC and △PQR with ∠A=∠P, ∠B=∠Q, ∠C=∠R. (b) △ABC and △QRP with AB/QR = AC/QP = BC/RP = 1/2. (c) △LMP and △FED with MP/ED = PL/DF = 1/2 but LM/FE ≠ 1/2. (d) △MNL and △QPR with ∠M=∠Q=70°, MN/QP = ML/QR = 1/2. (e) △ACB and △FDE with ∠A=∠F=80° but sides not proportional. (f) △DEF and △PQR where remaining angles are found using angle-sum property.
(a) ∠A=∠P, ∠B=∠Q, ∠C=∠R ⟹ △ABC ~ △PQR (AA similarity).
(b) AB/QR = AC/QP = BC/RP = 1/2, teeno sides same ratio me ⟹ △ABC ~ △QRP (SSS similarity). Vertex order dhyaan se: A↔Q, B↔R, C↔P.
(c) MP/ED = PL/DF = 1/2 hai par LM/FE ≠ 1/2 — teesra ratio match nahi karta ⟹ similar NAHI hain.
(d) ∠M = ∠Q (included angle) aur MN/QP = ML/QR = 1/2 (do sides proportional) ⟹ △MNL ~ △QPR (SAS similarity).
(e) ∠A = ∠F hai par sides proportional nahi hain — sirf ek angle match karna kaafi nahi (AA ke liye do angles chahiye, aur SAS ke liye angle ke saath sides bhi chahiye) ⟹ similar NAHI hain.
(f) Angle sum property (180°) se baaki angles nikaalo — jab dono triangles ke teeno angles match ho jaate hain to AA similarity se similar hain.
Har part me pehle criterion check karo (AA / SSS / SAS), phir vertex correspondence sahi order me likho — yehi sabse zyada marks katne wali galti hai.
Exercise 6.3, Q2. In the figure, △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
∠BOC = 125° aur DOB ek straight line hai (diagonals kaट रहे hain), isliye:
∠DOC = 180° − ∠BOC = 180° − 125° = 55°
Triangle DOC me angle sum property:
∠DCO = 180° − ∠DOC − ∠CDO = 180° − 55° − 70° = 55°
△ODC ~ △OBA hai, isliye corresponding angles equal: ∠OAB corresponds to ∠OCD (O↔O, D↔B, C↔A).
∠OAB = ∠OCD = 55°
∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.
Exercise 6.3, Q3. Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at point O. Show that AO/OC = OB/OD, using a similarity criterion for two triangles.
Triangle AOB aur triangle COD lo. AB ∥ DC hai, aur AC, BD transversals hain.
∠OAB = ∠OCD (alternate angles, AB ∥ DC, transversal AC) aur ∠OBA = ∠ODC (alternate angles, AB ∥ DC, transversal BD).
Isliye △AOB ~ △COD (AA similarity criterion)
Similar triangles ke corresponding sides proportional hote hain:
AO/CO = OB/OD ⟹ AO/OC = OB/OD
Hence proved.
Exercise 6.3, Q4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that △PQS ~ △TQR.
Diya hai ∠1 = ∠2, yaani ∠PQS = ∠TQR (in fact ∠PQR = ∠TQR wale angles, jisse ∠PQS ho jaata hai common setup me PQ=PR type isosceles-adjacent construction).
Diya hai: QR/QS = QT/PR — yeh ratio △PQS aur △TQR ke corresponding sides ka hai (QR corresponds to QT ke against, PR/QS ka pairing).
∠PQS = ∠TQR (common/given angle) aur QP/QT = QR/QS wala ratio set karke — SAS similarity criterion apply hota hai.
Isliye △PQS ~ △TQR (SAS similarity). Hence proved.
Note: is figure ka exact labeling text-only format me thoda ambiguous hai — core method yeh hai ki given ek common angle aur ek proportional-sides pair se SAS milaake similarity establish karni hai.
Exercise 6.3, Q5. S and T are points on sides PR and QR of triangle PQR such that ∠P = ∠RTS. Show that △RPQ ~ △RTS.
Triangle RPQ aur triangle RTS lo.
∠RPQ = ∠P = ∠RTS = ∠RTS (diya hai) — yeh ek common-type angle match hai.
∠R common hai dono triangles me (same vertex R, angle PRQ = angle TRS, kyunki S on PR aur T on QR, so angle at R is shared).
∠R = ∠R (common) aur ∠RPQ = ∠RTS (given) ⟹ AA similarity criterion
Isliye △RPQ ~ △RTS. Hence proved.
Exercise 6.3, Q6. If △ABE ≅ △ACD, show that △ADE ~ △ABC.
△ABE ≅ △ACD diya hai. Congruent triangles ke corresponding sides equal hote hain:
AB = AC aur AE = AD
In dono equations ko rearrange karo:
AD/AB = AE/AC (kyunki AB=AC aur AE=AD)
Ab triangle ADE aur triangle ABC dekho: ∠A common hai (∠DAE = ∠BAC, same angle), aur AD/AB = AE/AC.
∠A = ∠A (common), AD/AB = AE/AC ⟹ SAS similarity criterion
Isliye △ADE ~ △ABC. Hence proved.
Exercise 6.3, Q7. Altitudes AD and CE of triangle ABC intersect each other at point P. Show that: (i) △AEP ~ △CDP, (ii) △ABD ~ △CBE, (iii) △AEP ~ △ADB, (iv) △PDC ~ △BEC.
AD ⊥ BC aur CE ⊥ AB hai (altitudes), P inka intersection point (orthocentre) hai.
(i) Triangle AEP aur CDP me: ∠AEP = ∠CDP = 90° (dono altitude ke foot par right angle), aur ∠APE = ∠CPD (vertically opposite angles).
AA similarity ⟹ △AEP ~ △CDP
(ii) Triangle ABD aur CBE me: ∠ADB = ∠CEB = 90°, aur ∠B common hai.
AA similarity ⟹ △ABD ~ △CBE
(iii) Triangle AEP aur ADB me: ∠AEP = ∠ADB = 90°, aur ∠A common hai (∠EAP = ∠DAB, same angle at A).
AA similarity ⟹ △AEP ~ △ADB
(iv) Triangle PDC aur BEC me: ∠PDC = ∠BEC = 90°, aur ∠C common hai.
AA similarity ⟹ △PDC ~ △BEC
Sab hence proved (har part me AA criterion — right angle + ek common/vertically-opposite angle).
Exercise 6.3, Q8. E is a point on side AD produced of a parallelogram ABCD, and BE intersects CD at F. Show that △ABE ~ △CFB.
ABCD parallelogram hai, isliye AB ∥ DC (aur AD ∥ BC).
AB ∥ CF (kyunki CF, DC ka hi hissa hai, aur DC ∥ AB) — transversal BE ke saath: ∠AEB = ∠CBF (alternate angles, AB ∥ DC, transversal BE)... in fact ∠ABE = ∠CFB aur ∠A = ∠BCF (opposite angles ke through AA milta hai).
∠A = ∠BCF (ABCD parallelogram ke opposite angles ka property, extended) aur ∠AEB = ∠CBF (alternate angles, AB ∥ CD) ⟹ AA similarity
Isliye △ABE ~ △CFB. Hence proved.
Exercise 6.3, Q9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) △ABC ~ △AMP, (ii) CA/PA = BC/MP.
(i) Triangle ABC aur AMP me: ∠ABC = ∠AMP = 90° (dono right angle par diya hai), aur ∠A common hai (∠BAC = ∠MAP, same angle at A, kyunki M, P respectively AB, AC ki lines pe hote hain aisi typical figure me).
∠A = ∠A (common), ∠ABC = ∠AMP = 90° ⟹ AA similarity ⟹ △ABC ~ △AMP
(ii) Similar triangles ke corresponding sides proportional hote hain (correspondence A↔A, B↔M, C↔P):
AB/AM = BC/MP = CA/PA ⟹ CA/PA = BC/MP
Hence proved.
Exercise 6.3, Q10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of triangles ABC and FEG respectively. If △ABC ~ △FEG, show that: (i) CD/GH = AC/FG, (ii) △DCB ~ △HGE, (iii) △DCA ~ △HGF.
△ABC ~ △FEG diya hai, isliye ∠A = ∠F, ∠B = ∠E, ∠C = ∠G, aur AC/FG = BC/EG = AB/FE.
CD, ∠ACB ko bisect karta hai aur GH, ∠EGF ko bisect karta hai, aur ∠ACB = ∠FGE, isliye ∠ACD = ∠FGH = ∠DCB = ∠HGE (aadha-aadha equal).
(ii) Triangle DCB aur HGE me: ∠B = ∠E (given similarity se) aur ∠DCB = ∠HGE (abhi dikhaya).
AA similarity ⟹ △DCB ~ △HGE
Isse CD/GH = CB/GE milta hai.
(i) Chunki CB/GE = AC/FG hai hi (original similarity se), isliye:
CD/GH = CB/GE = AC/FG ⟹ CD/GH = AC/FG
(iii) Triangle DCA aur HGF me: ∠A = ∠F (given) aur ∠DCA = ∠HGF (bisected halves equal).
AA similarity ⟹ △DCA ~ △HGF
Sab hence proved.
Exercise 6.3, Q11. In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that △ABD ~ △ECF.
AB = AC hai (isosceles), isliye ∠ABD = ∠ACB (base angles equal). Aur E, CB ke produced part par hai, so ∠ACB aur ∠ECF wahi angle hai jo ∠ABD ke corresponding hai kyunki ∠ECF, ∠ACB ka hi supplementary-linked angle ban ke bhi effectively equal nikalta hai (isosceles ke exterior angle property se).
AD ⊥ BC ⟹ ∠ADB = 90°. EF ⊥ AC ⟹ ∠EFC = 90°.
∠ADB = ∠EFC = 90°, aur ∠ABD = ∠ECF (isosceles triangle base-angle property se) ⟹ AA similarity
Isliye △ABD ~ △ECF. Hence proved.
Exercise 6.3, Q12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of another triangle PQR. Show that △ABC ~ △PQR.
Diya hai: AB/PQ = BC/QR = AD/PM (sides aur medians same ratio me).
D, BC ka midpoint hai aur M, QR ka midpoint hai (kyunki AD, PM medians hain), isliye BD = BC/2 aur QM = QR/2.
BC/QR = BD/QM (dono halves same ratio me hain) — isliye AB/PQ = BD/QM = AD/PM
Triangle ABD aur PQM me teeno sides same ratio me hain ⟹ SSS similarity ⟹ △ABD ~ △PQM.
Isse ∠ABD = ∠PQM, yaani ∠B = ∠Q
Ab triangle ABC aur PQR me: AB/PQ = BC/QR (diya hai) aur included angle ∠B = ∠Q (abhi dikhaya).
SAS similarity ⟹ △ABC ~ △PQR
Hence proved.
Exercise 6.3, Q13. D is a point on side BC of triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB·CD.
Triangle ADC aur triangle BAC lo. ∠ADC = ∠BAC (diya hai) aur ∠C common hai (∠ACD = ∠BCA, same angle).
∠ADC = ∠BAC, ∠C = ∠C (common) ⟹ AA similarity ⟹ △ADC ~ △BAC
Corresponding sides proportional (correspondence A↔B, D↔A, C↔C):
CA/CB = CD/CA
Cross-multiply karo:
CA² = CB × CD
Hence proved. (Yeh bilkul wahi geometric-mean relation hai jo Pythagoras theorem prove karne me use hoti hai.)
Exercise 6.3, Q14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that △ABC ~ △PQR.
Diya hai: AB/PQ = AC/PR = AD/PM. D, BC ka midpoint hai; M, QR ka midpoint hai.
Construction: AD ko E tak extend karo taaki AD = DE, aur PM ko N tak extend karo taaki PM = MN. Phir BDEC aur QMNR parallelograms ban jaate hain (diagonals bisect each other).
AE = 2·AD aur PN = 2·PM, aur BE = AC, QN = PR (parallelogram property)
Isliye AB/PQ = AE/PN = AC/PR — teeno sides proportional hain triangle ABE aur PQN me.
SSS similarity ⟹ △ABE ~ △PQN ⟹ ∠BAE = ∠QPN aur similarly ∠DAC = ∠MPR
In dono angles ko add karke: ∠BAC = ∠QPR.
AB/PQ = AC/PR (given), aur included angle ∠BAC = ∠QPR ⟹ SAS similarity ⟹ △ABC ~ △PQR
Hence proved.
Exercise 6.3, Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground, and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Pole aur tower dono same time par shadow bana rahe hain, isliye sun ka angle of elevation dono ke liye same hai. Dono right triangles similar honge (AA similarity: right angle + common elevation angle).
Pole ki height / Pole ki shadow = Tower ki height / Tower ki shadow
6/4 = h/28 ⟹ h = (6 × 28)/4 = 42
Tower ki height = 42 m.
Exercise 6.3, Q16. If AD and PM are medians of triangles ABC and PQR respectively, where △ABC ~ △PQR, prove that AB/PQ = AD/PM.
△ABC ~ △PQR diya hai, isliye ∠B = ∠Q aur AB/PQ = BC/QR ... (1)
D, BC ka midpoint hai (AD median), M, QR ka midpoint hai (PM median), isliye BD = BC/2 aur QM = QR/2.
BC/QR = BD/QM, isliye (1) se: AB/PQ = BD/QM
Triangle ABD aur PQM me: AB/PQ = BD/QM aur included angle ∠B = ∠Q.
SAS similarity ⟹ △ABD ~ △PQM ⟹ AB/PQ = BD/QM = AD/PM
Isliye AB/PQ = AD/PM. Hence proved.
Important Equations — Ek Nazar Me
| Theorem / Rule | Statement | Kab use karo |
|---|---|---|
| Basic Proportionality Theorem (BPT / Thales) | Triangle ke ek side ke parallel line, baaki do sides ko same ratio me divide karti hai: AD/DB = AE/EC | Jab DE ∥ BC diya ho aur ek unknown length nikalni ho |
| Converse of BPT | Agar line do sides ko same ratio me divide kare, to woh teesri side ke parallel hai | Jab check karna ho ki EF ∥ QR hai ya nahi (ratio milaake) |
| Midpoint theorem (BPT ka corollary) | Midpoint se doosre side ke parallel line teesra side bhi bisect karti hai (aur ulta) | Midpoint-based proofs |
| AA similarity | Do corresponding angles equal ⟹ triangles similar (teesra apne aap equal) | Jab do angles common/equal diye ho |
| SSS similarity | Teeno corresponding sides same ratio me ⟹ triangles similar | Jab sirf sides ke ratios diye ho, koi angle nahi |
| SAS similarity | Do sides same ratio me + unke beech ka included angle equal ⟹ similar | Jab ek angle + do adjacent sides ka ratio diya ho |
| Area of similar triangles | ar(△1)/ar(△2) = (side ratio)² = (median ratio)² = (altitude ratio)² | Jab area ka ratio poochha jaye — RATIO KA SQUARE lena hai, ratio nahi |
| Pythagoras theorem | (hypotenuse)² = (base)² + (perpendicular)² — sirf right triangle me | Right angle diya ho ya construct kiya ho |
| Converse of Pythagoras | Agar ek side² = baaki do sides² ka sum, to opposite angle 90° hai | Right angle CHECK karna ho (koi angle diya na ho) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Vertex correspondence galat order me likhna. △ABC ~ △PQR likhne ka matlab hai A↔P, B↔Q, C↔R strictly. Agar actual correspondence A↔Q hai to △ABC ~ △QPR likhna padega, △ABC ~ △PQR nahi — galat order se corresponding sides/angles ka ratio bhi galat nikal jaayega.
- Congruence criteria aur similarity criteria confuse karna. SSS/SAS congruence me sides EQUAL hote hain; SSS/SAS similarity me sides PROPORTIONAL (same ratio) hote hain. Dono naam same hain isliye exam me mix ho jaata hai.
- Area ka ratio bhool jaana ki woh sides ke ratio ka SQUARE hai. Agar sides ka ratio 2:3 hai to area ka ratio 4:9 hoga, 2:3 nahi. Yeh sabse common numerical mistake hai.
- Pythagoras theorem non-right-angled triangle pe laga dena. a² + b² = c² SIRF right triangle me valid hai — koi bhi triangle me directly nahi laga sakte jab tak 90° angle diya ya proved na ho.
- BPT ka direction galat samajhna — 'DE parallel to BC' na hokar kisi aur side ke parallel maan lena. Jis side ke parallel line hai, ratio hamesha baaki DO sides ka hota hai, teesri (parallel wali) side ka nahi.
- AAA ke liye teeno angles check karna zaroori samajhna. Sirf DO corresponding angles equal dikhana kaafi hai — teesra angle sum property (180°) se apne aap equal ho jaata hai, waqt zaya mat karo teesra bhi prove karne me.
Board-Style Important Questions
- 1 mark: Agar △ABC ~ △DEF aur ∠A = 47°, ∠E = 63°, to ∠C ka value nikalo.
- 2 marks: Ek triangle ABC me D aur E respectively AB aur AC par is tarah hain ki DE ∥ BC. Agar AD/DB = 2/3 aur AC = 15 cm, to AE nikalo.
- 3 marks: Prove karo ki agar ek line kisi triangle ke do sides ko same ratio me divide kare, to woh teesri side ke parallel hoti hai (Converse of BPT).
- 3 marks: Do similar triangles ke areas ka ratio unke corresponding sides ke ratio ke square ke barabar hota hai — is theorem ko state karke, ek numerical example se verify karo.
- 5 marks: Ek right triangle me right angle wale vertex se hypotenuse par perpendicular daala jaata hai — prove karo ki jo do triangles banti hain woh dono original triangle ke similar hoti hain, aur isse Pythagoras theorem derive karo.
Quick Quiz — Score Check Karein
Q1. Do figures similar kab kehlate hain?
Q2. Konsi shapes hamesha similar hoti hain, chahe unka size kuch bhi ho?
Q3. Exercise 6.1 Q3 me PQRS square hai (side a) aur ABCD rhombus hai (side 2a, angles 90° nahi). In dono ke baare me sahi statement kya hai?
Q4. Triangle ABC me D, AB par aur E, AC par hai. Agar AD = 1.5 cm, DB = 3 cm, AE = 1 cm aur DE ∥ BC ho, to EC kitna hoga?
Q5. Triangle PQR me E on PQ aur F on PR hain. PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm diya hai. EF, QR ke parallel hai ya nahi?
Q6. Congruence criteria (SSS, SAS) aur similarity criteria (SSS, SAS) me kya bada farak hai?
Q7. AAA similarity criterion me actually kitne angles check karne zaroori hain?
Q8. Agar do similar triangles ke corresponding sides ka ratio 2:3 hai, to unke areas ka ratio kya hoga?
Q9. Ek vertical pole (6 m) ki shadow 4 m lambi hai, aur usi waqt ek tower ki shadow 28 m lambi hai. Tower ki height kya hogi?
Q10. D, side BC par is tarah hai ki ∠ADC = ∠BAC. Similarity se kaunsa result prove hota hai?
Aksar Poochhe Jaane Wale Sawaal
Similar aur congruent triangles me kya farak hai?
Congruent triangles same shape AUR same size ke hote hain (sab sides aur angles exactly equal). Similar triangles sirf same shape ke hote hain — angles equal, sides proportional, par size same hona zaroori nahi. Har congruent pair similar hota hai, par har similar pair congruent nahi hota.
BPT sirf triangles me hi apply hota hai kya?
BPT ka original statement triangle ke liye hai, lekin isko trapeziums aur quadrilaterals ke diagonal-intersection problems me bhi apply karte hain (jaise Ex 6.2 Q9, Q10) — usually ek helper line/construction kheenchke ek triangle bana lete hain, phir BPT lagate hain.
AAA similarity aur AA similarity me kya difference hai?
Koi practical difference nahi — agar triangle ke do angles equal hain to teesra apne aap equal ho jaata hai (angle sum 180° fixed hai). Isliye NCERT me isko 'AAA' bolte hain but proof/application me sirf do angles check karna kaafi hota hai, isi liye ise 'AA criterion' bhi kaha jaata hai.
Area of similar triangles wala exercise current book me kyun nahi hai?
Rationalisation (2023) me NCERT ne purane exercises 6.4, 6.5, 6.6 (jisme area-ratio aur Pythagoras ke dedicated practice questions the) hata diye. Lekin theorem statements chapter ke text me abhi bhi hain, aur board application questions (jaise height/distance ya area comparison) in results ko use kar sakte hain — isliye inhe skip mat karo.
Pythagoras theorem ka converse kaise use hota hai?
Jab kisi triangle ke teeno sides ki length di ho aur poochha jaye ki woh right triangle hai ya nahi — sabse badi side ka square nikaalo aur check karo ki woh baaki do sides ke squares ke sum ke barabar hai ya nahi. Barabar hai to right triangle hai, aur right angle sabse badi side ke opposite wale vertex par hoga.
Similarity criteria (AA/SSS/SAS) me se kaunsa criterion kab use karna chahiye?
Agar sirf angles diye hain — AA use karo. Agar sirf sides ke ratios diye hain (koi angle nahi) — SSS use karo. Agar do sides ka ratio + unke BEECH ka angle diya hai — SAS use karo. Question me kya diya hai usi se criterion decide hota hai, guess mat karo.
Class 10 Maths — Saare Chapters
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Class 10 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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