NCERT Solutions for Class 10 Maths Chapter 6: Triangles (NCERT 2026–27)

These Class 10 Maths Chapter 6 solutions cover Triangles from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 6.1, 6.2 and 6.3 is solved step by step — including the Basic Proportionality Theorem (Thales) proofs, the AAA, SSS and SAS similarity criteria, and the standard height-and-shadow applications — so you can write exam-ready answers and revise quickly.

Class: 10 Subject: Mathematics Chapter: 6 – Triangles Exercises: 6.1, 6.2, 6.3 Total questions: 29 Session: 2026–27
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Chapter 6 Overview

Chapter 6 of Class 10 Maths, Triangles, studies figures that have the same shape but not necessarily the same size — that is, similar figures. After defining similarity of polygons (equal corresponding angles and proportional corresponding sides), the chapter focuses on the similarity of triangles. It proves the Basic Proportionality Theorem (Thales Theorem) and its converse, and develops three working criteria for triangle similarity — AAA/AA, SSS and SAS. These tools are then used to prove relations between sides, to solve indirect-measurement problems such as the heights of towers and the lengths of shadows, and to handle medians and bisectors of similar triangles. The Class 10 Maths Chapter 6 solutions below work through every question of Exercises 6.1, 6.2 and 6.3 with complete reasoning.

Key Concepts & Definitions

Similar figures: two figures with the same shape but not necessarily the same size. All congruent figures are similar, but similar figures need not be congruent.

Similar polygons: two polygons of the same number of sides are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (proportion). Either condition alone is not enough.

Similar triangles: ▵ABC ~ ▵DEF means ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD. The order of letters shows the correspondence of vertices.

Scale factor: the common ratio of corresponding sides of two similar figures (also called the representative fraction).

Equiangular triangles: triangles whose corresponding angles are equal. Thales found that the ratio of any two corresponding sides of equiangular triangles is constant.

Indirect measurement: using similarity (equal angles + proportional sides) to find heights and distances that cannot be measured directly, e.g. the height of a tower from its shadow.

Important Theorems & Formulas (Chapter 6)

Theorem 6.1 – Basic Proportionality Theorem (BPT / Thales): if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In ▵ABC with DE || BC, AD/DB = AE/EC.

Theorem 6.2 – Converse of BPT: if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Theorem 6.3 – AAA similarity: if corresponding angles of two triangles are equal, their corresponding sides are proportional and the triangles are similar. (Two pairs of equal angles is enough → AA criterion.)

Theorem 6.4 – SSS similarity: if the sides of one triangle are proportional to the sides of another, their corresponding angles are equal and the triangles are similar.

Theorem 6.5 – SAS similarity: if one angle of a triangle equals one angle of another and the sides including these angles are proportional, the triangles are similar.

RHS similarity (Note to the Reader): if the hypotenuse and one side of a right triangle are proportional to those of another right triangle, the triangles are similar.

Exercise 6.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original.

1. Fill in the blanks using the correct word given in brackets: (i) All circles are ______. (congruent, similar) (ii) All squares are ______. (similar, congruent) (iii) All ______ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ______ and (b) their corresponding sides are ______. (equal, proportional)

SOLUTION (i) All circles are similar (they have the same shape but vary in radius). (ii) All squares are similar (all angles 90° and all sides in equal ratio). (iii) All equilateral triangles are similar (each angle is 60° and the sides are in the same ratio). (iv) (a) equal  and  (b) proportional.

2. Give two different examples of pair of (i) similar figures.   (ii) non-similar figures.

SOLUTION (i) Similar figures: (a) any two circles; (b) any two equilateral triangles. (Other valid pairs: two squares, or a small and an enlarged photograph from the same negative.) (ii) Non-similar figures: (a) a square and a rectangle (equal angles but sides not in the same ratio); (b) a square and a rhombus (sides in the same ratio but angles unequal). (Also valid: a scalene triangle and an equilateral triangle.)

3. State whether the following quadrilaterals are similar or not:

SOLUTION In Fig. 6.8 the two quadrilaterals are a square and a rhombus. Their corresponding sides are in the same ratio (1 : 2), but their corresponding angles are not equal — the square has all 90° angles, while the rhombus does not. Since both conditions for similarity must hold and the angle condition fails, the two quadrilaterals are not similar.

Exercise 6.2 Solutions

1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

SOLUTION By the Basic Proportionality Theorem (DE || BC ⇒ AD/DB = AE/EC). (i) Given AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Then 1.5/3 = 1/EC ⇒ EC = (1 × 3)/1.5 = 2 cm. (ii) Given DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm. Then AD/7.2 = 1.8/5.4 ⇒ AD = (1.8 × 7.2)/5.4 = 12.96/5.4 = 2.4 cm.

2. E and F are points on the sides PQ and PR respectively of a ▵PQR. For each of the following cases, state whether EF || QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

SOLUTION By the converse of BPT, EF || QR ⇔ PE/EQ = PF/FR. (i) PE/EQ = 3.9/3 = 1.3; PF/FR = 3.6/2.4 = 1.5. Since 1.3 ≠ 1.5, EF is not parallel to QR. (ii) PE/QE = 4/4.5 = 8/9; PF/RF = 8/9. The ratios are equal, so EF || QR. (iii) EQ = PQ − PE = 1.28 − 0.18 = 1.10 cm; FR = PR − PF = 2.56 − 0.36 = 2.20 cm. Then PE/EQ = 0.18/1.10 = 9/55 and PF/FR = 0.36/2.20 = 9/55. The ratios are equal, so EF || QR.

3. In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

SOLUTION In ▵ABC, LM || CB. By BPT, AM/MB = AL/LC, i.e. AB/AM = (AM + MB)/AM = 1 + MB/AM = 1 + LC/AL = AC/AL. So AM/AB = AL/AC.   …(1) In ▵ACD, LN || CD. By BPT, AN/ND = AL/LC, and by the same working AN/AD = AL/AC.   …(2) From (1) and (2), both AM/AB and AN/AD equal AL/AC. Therefore AM/AB = AN/AD. ∎

4. In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

SOLUTION In ▵ABC, DE || AC. By BPT, BD/DA = BE/EC.   …(1) In ▵ABE, DF || AE. By BPT, BD/DA = BF/FE.   …(2) From (1) and (2), the left sides are equal, so the right sides are equal: BF/FE = BE/EC. ∎

5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

SOLUTION In ▵PQO, DE || OQ. By BPT, PE/EQ = PD/DO.   …(1) In ▵PRO, DF || OR. By BPT, PF/FR = PD/DO.   …(2) From (1) and (2), PE/EQ = PF/FR. In ▵PQR this means the line EF divides PQ and PR in the same ratio, so by the converse of BPT, EF || QR. ∎

6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

SOLUTION In ▵OPQ, AB || PQ. By BPT, OA/AP = OB/BQ.   …(1) In ▵OPR, AC || PR. By BPT, OA/AP = OC/CR.   …(2) From (1) and (2), OB/BQ = OC/CR. In ▵OQR this means BC divides OQ and OR in the same ratio, so by the converse of BPT, BC || QR. ∎

7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

SOLUTION In ▵ABC, let D be the mid-point of AB, so AD = DB, and let a line through D parallel to BC meet AC at E. Since DE || BC, by BPT (Theorem 6.1): AD/DB = AE/EC. But AD = DB ⇒ AD/DB = 1, so AE/EC = 1, giving AE = EC. Hence E is the mid-point of AC — the line bisects the third side. ∎

8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

SOLUTION In ▵ABC, let D and E be the mid-points of AB and AC, so AD = DB and AE = EC. Then AD/DB = 1 and AE/EC = 1, so AD/DB = AE/EC. Thus the line DE divides sides AB and AC in the same ratio. By the converse of BPT (Theorem 6.2), DE || BC. ∎

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

SOLUTION Through O draw a line EO || AB || DC, meeting AD at E (E on AD). In ▵ADC, EO || DC. By BPT, AE/ED = AO/OC.   …(1) In ▵ABD, EO || AB. By BPT, AE/ED = BO/OD.   …(2) From (1) and (2), AO/OC = BO/OD, which rearranges to AO/BO = CO/DO. ∎

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

SOLUTION Given AO/BO = CO/DO, which rearranges to AO/OC = BO/OD.   …(1) Through O draw OE || AB, with E on AD. In ▵ABD, OE || AB gives (by BPT) AE/ED = BO/OD.   …(2) From (1) and (2), AE/ED = AO/OC. In ▵ADC the line EO divides AD and AC in the same ratio, so by the converse of BPT, EO || DC. Since OE || AB and OE || DC, we get AB || DC. A quadrilateral with one pair of parallel sides is a trapezium. ∎

Exercise 6.3 Solutions

1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

SOLUTION (i) In ▵ABC and ▵PQR all three angles match (60°, 80°, 40°). By AAA: ▵ABC ~ ▵PQR. (ii) Sides are proportional: AB/QR = BC/RP = CA/PQ = 2/4 = 1/2. By SSS: ▵ABC ~ ▵QRP. (iii) The two sides of one are not proportional to the corresponding sides including the equal angle of the other, so the triangles are not similar. (iv) ∠M = ∠Q = 70° and the including sides are proportional (ML/QP = MN/QR = 1/2). By SAS: ▵MNL ~ ▵QPR. (v) Only one pair of equal angles (80°) is known and the sides including it are not proportional, so the triangles are not similar. (vi) In ▵DEF, ∠D = 70°, ∠E = 80°, so ∠F = 30°. In ▵PQR, ∠Q = 80°, ∠R = 30°, so ∠P = 70°. Two pairs of angles are equal. By AA: ▵DEF ~ ▵PQR.

2. In Fig. 6.35, ▵ODC ~ ▵OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

SOLUTION DOB is a straight line, so ∠DOC + ∠BOC = 180° ⇒ ∠DOC = 180° − 125° = 55°. In ▵DOC: ∠DCO = 180° − ∠CDO − ∠DOC = 180° − 70° − 55° = 55°. Since ▵ODC ~ ▵OBA, ∠OAB corresponds to ∠OCD. Therefore ∠OAB = ∠DCO = 55°.

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

SOLUTION In ▵OAB and ▵OCD: ∠AOB = ∠COD (vertically opposite angles). Since AB || DC and AC is a transversal, ∠OAB = ∠OCD (alternate angles). By the AA criterion, ▵OAB ~ ▵OCD. Corresponding sides are proportional: OA/OC = OB/OD. ∎

4. In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ▵PQS ~ ▵TQR.

SOLUTION ∠1 = ∠2 means ∠PQR = ∠PRQ, so in ▵PQR the sides opposite these angles are equal: PR = PQ. Given QR/QS = QT/PR. Replacing PR by PQ: QR/QS = QT/PQ, i.e. QT/QR = PQ/QS (rearranging). In ▵PQS and ▵TQR: the included angle ∠PQS = ∠TQR (it is the same angle ∠Q), and the sides about it are proportional (QT/QR = QP/QS). By the SAS similarity criterion, ▵PQS ~ ▵TQR. ∎

5. S and T are points on sides PR and QR of ▵PQR such that ∠P = ∠RTS. Show that ▵RPQ ~ ▵RTS.

SOLUTION In ▵RPQ and ▵RTS: ∠RPQ = ∠RTS (given). ∠R = ∠R (common angle to both triangles). By the AA criterion, ▵RPQ ~ ▵RTS. ∎

6. In Fig. 6.37, if ▵ABE ≅ ▵ACD, show that ▵ADE ~ ▵ABC.

SOLUTION Since ▵ABE ≅ ▵ACD, corresponding parts are equal: AB = AC and AE = AD. So AD/AB = AE/AC (both equal the same ratio, since AD = AE and AB = AC give AD/AB = AE/AC). In ▵ADE and ▵ABC: ∠A = ∠A (common), and the sides including it are proportional (AD/AB = AE/AC). By the SAS similarity criterion, ▵ADE ~ ▵ABC. ∎

7. In Fig. 6.38, altitudes AD and CE of ▵ABC intersect each other at the point P. Show that: (i) ▵AEP ~ ▵CDP (ii) ▵ABD ~ ▵CBE (iii) ▵AEP ~ ▵ADB (iv) ▵PDC ~ ▵BEC

SOLUTION (i) In ▵AEP and ▵CDP: ∠AEP = ∠CDP = 90° (CE ⊥ AB, AD ⊥ BC) and ∠APE = ∠CPD (vertically opposite). By AA, ▵AEP ~ ▵CDP. (ii) In ▵ABD and ▵CBE: ∠ADB = ∠CEB = 90° and ∠B = ∠B (common). By AA, ▵ABD ~ ▵CBE. (iii) In ▵AEP and ▵ADB: ∠AEP = ∠ADB = 90° and ∠A = ∠A (common ∠EAP = ∠DAB). By AA, ▵AEP ~ ▵ADB. (iv) In ▵PDC and ▵BEC: ∠PDC = ∠BEC = 90° and ∠C = ∠C (common ∠DCP = ∠ECB). By AA, ▵PDC ~ ▵BEC. ∎

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ▵ABE ~ ▵CFB.

SOLUTION In ▵ABE and ▵CFB: ∠A = ∠C (opposite angles of parallelogram ABCD are equal). Since AE || BC (AD produced is parallel to BC) and BE is a transversal, ∠AEB = ∠CBF (alternate angles). By the AA criterion, ▵ABE ~ ▵CFB. ∎

9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ▵ABC ~ ▵AMP (ii) CA/PA = BC/MP

SOLUTION (i) In ▵ABC and ▵AMP: ∠ABC = ∠AMP = 90° (given) and ∠A = ∠A (common). By AA, ▵ABC ~ ▵AMP. (ii) Since the triangles are similar, corresponding sides are proportional: CA/PA = BC/MP (= AB/AM). Hence CA/PA = BC/MP. ∎

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ▵ABC and ▵EFG respectively. If ▵ABC ~ ▵FEG, show that: (i) CD/GH = AC/FG (ii) ▵DCB ~ ▵HGE (iii) ▵DCA ~ ▵HGF

SOLUTION Since ▵ABC ~ ▵FEG: ∠A = ∠F, ∠B = ∠E, ∠ACB = ∠FGE. As CD and GH bisect ∠ACB and ∠EGF, half-angles are equal: ∠ACD = ∠DCB = ½∠ACB and ∠FGH = ∠HGE = ½∠FGE, so ∠ACD = ∠FGH and ∠DCB = ∠HGE. (iii) In ▵DCA and ▵HGF: ∠A = ∠F and ∠ACD = ∠FGH. By AA, ▵DCA ~ ▵HGF. (i) From ▵DCA ~ ▵HGF, corresponding sides give CD/GH = AC/FG. (ii) In ▵DCB and ▵HGE: ∠B = ∠E and ∠DCB = ∠HGE. By AA, ▵DCB ~ ▵HGE. ∎

11. In Fig. 6.40, 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.

SOLUTION Since AB = AC, ▵ABC is isosceles, so the base angles are equal: ∠ABC = ∠ACB, i.e. ∠ABD = ∠ECF (as ∠ECF and ∠ACB are the same angle). Also ∠ADB = ∠EFC = 90° (AD ⊥ BC and EF ⊥ AC). By the AA criterion, ▵ABD ~ ▵ECF. ∎

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ▵PQR (see Fig. 6.41). Show that ▵ABC ~ ▵PQR.

SOLUTION Given AB/PQ = BC/QR = AD/PM. Since AD and PM are medians, D and M are mid-points of BC and QR, so BD = ½BC and QM = ½QR. Then BD/QM = (½BC)/(½QR) = BC/QR. So AB/PQ = BD/QM = AD/PM. In ▵ABD and ▵PQM all three pairs of sides are proportional, so by SSS, ▵ABD ~ ▵PQM. Hence ∠B = ∠Q (corresponding angles). Now in ▵ABC and ▵PQR: ∠B = ∠Q, and AB/PQ = BC/QR (given). By the SAS criterion, ▵ABC ~ ▵PQR. ∎

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB·CD.

SOLUTION In ▵ADC and ▵BAC: ∠ADC = ∠BAC (given) and ∠C = ∠C (common angle ∠ACD = ∠BCA). By the AA criterion, ▵ADC ~ ▵BAC. Corresponding sides are proportional: CA/CB = CD/CA. Cross-multiplying: CA × CA = CB × CD, i.e. CA2 = CB·CD. ∎

14. 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.

SOLUTION Given AB/PQ = AC/PR = AD/PM. Produce AD to E so that AD = DE, and join BE, CE; similarly produce PM to N so that PM = MN, and join QN, RN. Then ABEC and PQNR are parallelograms (diagonals bisect each other), so BE = AC and QN = PR. Now AE = 2AD and PN = 2PM, so AB/PQ = AC/PR = (2AD)/(2PM) = AE/PN. Replacing AC by BE and PR by QN: AB/PQ = BE/QN = AE/PN. In ▵ABE and ▵PQN all three pairs of sides are proportional, so by SSS, ▵ABE ~ ▵PQN. Hence ∠BAE = ∠QPN. Similarly ∠CAE = ∠RPN, so ∠BAC = ∠QPR. Now in ▵ABC and ▵PQR: ∠BAC = ∠QPR and AB/PQ = AC/PR. By the SAS criterion, ▵ABC ~ ▵PQR. ∎

15. 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.

SOLUTION The pole and the tower are vertical, and the sun’s rays make the same angle at the same time, so the pole–shadow triangle and the tower–shadow triangle are similar (AA). Therefore (pole height)/(pole shadow) = (tower height)/(tower shadow): 6/4 = h/28. h = (6 × 28)/4 = 168/4 = 42 m. The height of the tower is 42 m.

16. If AD and PM are medians of triangles ABC and PQR, respectively where ▵ABC ~ ▵PQR, prove that AB/PQ = AD/PM.

SOLUTION Since ▵ABC ~ ▵PQR: AB/PQ = BC/QR and ∠B = ∠Q.   …(1) AD and PM are medians, so D and M are mid-points of BC and QR: BD = ½BC and QM = ½QR. Hence BD/QM = BC/QR = AB/PQ.   …(2) In ▵ABD and ▵PQM: ∠B = ∠Q (from 1) and AB/PQ = BD/QM (from 2). By the SAS criterion, ▵ABD ~ ▵PQM. Corresponding sides give AB/PQ = AD/PM. ∎

Common Mistakes to Avoid

Watch out for these

  • Writing the similarity in the wrong vertex order — ▵ABC ~ ▵DEF must list corresponding vertices in order, or your side ratios will be wrong.
  • Calling two figures similar when only one condition holds — a square and a rhombus, or a square and a rectangle, are not similar.
  • Applying BPT (Theorem 6.1) without first confirming the line is parallel to a side, or applying its converse without showing the two ratios are equal.
  • Mixing up the criteria: AA needs two equal angles, SAS needs the angle to be between the two proportional sides, SSS needs all three side ratios equal.
  • For median problems, forgetting that a median goes to the mid-point, so BD = ½BC must be substituted before comparing ratios.
  • In shadow/tower problems, pairing the height of one object with the shadow of the other — keep height with its own shadow in the ratio.

Practice MCQs & Assertion–Reason

1. If in ▵ABC, DE || BC with D on AB and E on AC, and AD = 2 cm, DB = 3 cm, AE = 4 cm, then EC is:

(a) 5 cm    (b) 6 cm    (c) 8 cm    (d) 2.67 cm

2. The Basic Proportionality Theorem is also known as:

(a) Pythagoras Theorem    (b) Thales Theorem    (c) Midpoint Theorem    (d) Heron’s Theorem

3. All ______ triangles are similar.

(a) isosceles    (b) right-angled    (c) equilateral    (d) scalene

4. Two triangles are similar if their corresponding angles are equal. This is the:

(a) SSS criterion    (b) SAS criterion    (c) AAA criterion    (d) RHS criterion

5. A vertical pole 5 m high casts a shadow 3 m long when a tower casts a shadow 24 m long. The height of the tower is:

(a) 30 m    (b) 40 m    (c) 14.4 m    (d) 45 m

6. In ▵PQR, E on PQ and F on PR. EF || QR if and only if:

(a) PE = PF    (b) PE/EQ = PF/FR    (c) EQ = FR    (d) PE/PF = EQ/FR

7. For the SAS similarity criterion, the equal angle must be:

(a) opposite the larger side    (b) any angle    (c) included between the two proportional sides    (d) a right angle

8. D is on BC of ▵ABC with ∠ADC = ∠BAC. Then CA2 equals:

(a) CB·BD    (b) CB·CD    (c) BD·CD    (d) AB·AC

9. If ▵ABC ~ ▵DEF and AB = 4 cm, DE = 6 cm, BC = 5 cm, then EF is:

(a) 7.5 cm    (b) 3.33 cm    (c) 6 cm    (d) 10 cm

10. Which of the following pairs of figures is always similar?

(a) two rectangles    (b) two rhombuses    (c) two circles    (d) two isosceles triangles

Answer key: 1-(b), 2-(b), 3-(c), 4-(c), 5-(b), 6-(b), 7-(c), 8-(b), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: All congruent figures are similar.

Reason: Congruent figures have the same shape and the same size, so corresponding angles are equal and corresponding sides are in the ratio 1 : 1.

A-R 2. Assertion: A square and a rhombus are always similar.

Reason: For two polygons to be similar, both the corresponding angles must be equal and the corresponding sides must be in the same ratio.

A-R 3. Assertion: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, it divides those sides in the same ratio.

Reason: This is the Basic Proportionality (Thales) Theorem.

A-R 4. Assertion: Two triangles with two pairs of equal angles are similar.

Reason: By the angle sum property, if two angles are equal the third pair is also equal, giving AAA similarity.

A-R 5. Assertion: In ▵ABC ~ ▵PQR, the medians AD and PM satisfy AB/PQ = AD/PM.

Reason: Corresponding medians of similar triangles are in the same ratio as their corresponding sides.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • Similar figures have the same shape but not necessarily the same size; all congruent figures are similar, but not conversely.
  • Two polygons are similar when corresponding angles are equal and corresponding sides are proportional — one condition alone is not enough.
  • BPT (Theorem 6.1): DE || BC in ▵ABC ⇒ AD/DB = AE/EC; its converse (Theorem 6.2) is used to prove lines parallel.
  • AA / AAA: equal corresponding angles ⇒ similar triangles (sides proportional).
  • SSS: all three pairs of sides proportional ⇒ similar; SAS: one equal angle between two proportional sides ⇒ similar.
  • Similar triangles have corresponding medians, altitudes and angle bisectors all in the same ratio as their sides.
  • Similarity gives indirect measurement: pole/shadow = tower/shadow to find unknown heights.

How to score full marks in this chapter

Always state the criterion you use (AA, SSS, SAS, BPT or its converse) and quote the reason for every equal angle (common, vertically opposite, alternate, or right angle). Write the similarity with vertices in the correct correspondence before reading off side ratios. In proof questions, number your statements and end with the required result; in numerical questions, substitute carefully into a single proportion and simplify. Neat, justified steps earn the method marks even if the final arithmetic slips.

Frequently Asked Questions

What is Class 10 Maths Chapter 6 Triangles about?

Chapter 6 deals with similar figures and, in particular, the similarity of triangles. It proves the Basic Proportionality Theorem (Thales) and its converse, establishes the AAA/AA, SSS and SAS similarity criteria, and applies them to prove side relations and to find heights and distances by indirect measurement.

How many exercises are there in Class 10 Maths Chapter 6?

There are three exercises in the latest NCERT textbook: Exercise 6.1 (3 questions), Exercise 6.2 (10 questions) and Exercise 6.3 (16 questions) — 29 questions in total, all solved on this page.

What is the Basic Proportionality Theorem (BPT)?

The BPT, also called the Thales Theorem, states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then those two sides are divided in the same ratio. In ▵ABC with DE || BC, AD/DB = AE/EC.

What are the three criteria for similarity of triangles?

The three criteria are AAA/AA (equal corresponding angles), SSS (all corresponding sides proportional) and SAS (one equal angle included between two proportional sides). For right triangles there is also an RHS similarity criterion mentioned in the Note to the Reader.

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