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NCERT Class 10 Mathematics Exercise 6.3 Solutions: Triangles

Step-by-step working for every question of Exercise 6.3 in NCERT's Class 10 Mathematics book, chapter “Triangles”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 6.3 Q1

   Which pairs of triangles in Fig. 6.34 (six pairs, (i) to (vi)) are similar? Give the criterion and write the similarity in symbols.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Read the angles and sides marked in each pair of Fig. 6.34 and test the criteria AAA, SSS and SAS.
   2. (i) ∠A = ∠P = 60°, ∠B = ∠Q = 80°, ∠C = ∠R = 40°: all angles match, so △ABC ~ △PQR (AAA).
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   3. (ii) AB = 2, BC = 2.5, CA = 3 and QR = 4, RP = 5, PQ = 6: AB/QR = BC/RP = CA/PQ = 1/2, so △ABC ~ △QRP (SSS).
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   4. (iii) MP = 2, LM = 2.7, LP = 3 and DE = 4, EF = 5, DF = 6: 2/4 = 0.5 but 2.7/5 = 0.54, so the sides are not in one ratio: not similar.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   5. (iv) ∠M = ∠Q = 70°, with MN/QP = 2.5/5 = 1/2 and ML/QR = 5/10 = 1/2: two sides in one ratio with the included angle equal, so △MNL ~ △QPR (SAS).
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   6. (v) In △ABC the 80° angle is at A, but it is not between the two sides that are in proportion with △DEF's sides around its 80° angle, so SAS does not apply: not similar.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   7. (vi) In △DEF, ∠F = 180° − 70° − 80° = 30°; in △PQR, ∠P = 180° − 80° − 30° = 70°. All angles match, so △DEF ~ △PQR (AAA).
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   Answer: (i) △ABC ~ △PQR (AAA); (ii) △ABC ~ △QRP (SSS); (iii) not similar; (iv) △MNL ~ △QPR (SAS); (v) not similar; (vi) △DEF ~ △PQR (AAA)

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2. Ex 6.3 Q2

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

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. D, O, B lie on one straight line, so ∠DOC and ∠BOC form a linear pair: ∠DOC = 180° − 125° = 55°.
   2. In △ODC, the angles add up to 180°: ∠DCO = 180° − 70° − 55° = 55°.
   3. △ODC ~ △OBA matches O ↔ O, D ↔ B and C ↔ A, so ∠OCD corresponds to ∠OAB.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   4. Hence ∠OAB = ∠OCD = ∠DCO = 55°.

   Answer: ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°

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3. Ex 6.3 Q3

   The diagonals AC and BD of trapezium ABCD (AB ∥ DC) meet at O. Use similar triangles to show OA/OC = OB/OD.

   1. Given trapezium ABCD has AB parallel to DC and diagonals AC and BD intersect at O.
   2. In triangles OAB and OCD, angle AOB is equal to angle COD because they are vertically opposite angles.
   3. Angle OAB is equal to angle OCD because AB is parallel to DC and AC is a transversal, making them alternate interior angles.
   4. Therefore, triangle OAB is similar to triangle OCD by the AA similarity criterion.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   5. Since the triangles are similar, the ratio of their corresponding sides is equal.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   6. Thus, OA/OC is equal to OB/OD.

   Answer: OA/OC = OB/OD

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4. Ex 6.3 Q4

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

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. In Fig. 6.36, ∠1 is ∠PQR and ∠2 is ∠PRQ. Since ∠1 = ∠2, the sides opposite them are equal: PR = PQ.
   2. So the given QR/QS = QT/PR becomes QR/QS = QT/PQ, which rearranges to QS/QR = PQ/QT, i.e. PQ/QT = QS/QR.
   3. In △PQS and △TQR, ∠PQS = ∠TQR = ∠1 (the angle at Q is common).
   4. Two sides about the common angle are in the same ratio (PQ/TQ = QS/QR), so △PQS ~ △TQR by the SAS similarity criterion.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   Answer: △PQS ~ △TQR (SAS).

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5. Ex 6.3 Q5

   S and T are on sides PR and QR of △PQR with ∠P = ∠RTS. Show that △RPQ ~ △RTS.

   1. Given that S lies on PR and T lies on QR in triangle PQR.
   2. We are given that angle P equals angle RTS.
   3. In triangle RPQ and triangle RTS, angle P is equal to angle RTS.
   4. Angle R is common to both triangle RPQ and triangle RTS.
   5. Since two angles of triangle RPQ are equal to two angles of triangle RTS, the triangles are similar by the AA similarity criterion.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   6. Therefore, triangle RPQ is similar to triangle RTS.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   Answer: △RPQ ~ △RTS

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6. Ex 6.3 Q6

   In Fig. 6.37, △ABE ≅ △ACD. Show that △ADE ~ △ABC.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. Given that △ABE ≅ △ACD.
   2. From congruence, corresponding sides are equal: AB = AC and AE = AD.
   3. In △ADE and △ABC, we have AD/AB = AE/AC.
   4. The angle ∠DAE is common to both triangles (same as ∠BAC).
   5. By SAS similarity criterion, △ADE ~ △ABC.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   Answer: △ADE ~ △ABC

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7. Ex 6.3 Q7

   Altitudes AD and CE of △ABC meet at P (Fig. 6.38). Show: (i) △AEP ~ △CDP (ii) △ABD ~ △CBE (iii) △AEP ~ △ADB (iv) △PDC ~ △BEC.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) In △AEP and △CDP, ∠APE = ∠CPD (vertically opposite angles).
   2. (i) Also, ∠AEP = ∠CDP = 90° (since AD and CE are altitudes).
   3. (i) Therefore, △AEP ~ △CDP (AA similarity criterion).
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   4. (ii) In △ABD and △CBE, ∠ADB = ∠CEB = 90° (altitudes).
   5. (ii) Also, ∠ABD = ∠CBE (common angle ∠B).
   6. (ii) Therefore, △ABD ~ △CBE (AA similarity criterion).
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   7. (iii) From (i), △AEP ~ △CDP, so ∠EAP = ∠DCP.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   8. (iii) In △AEP and △ADB, ∠AEP = ∠ADB = 90°.
   9. (iii) Also, ∠EAP = ∠DAB (same angle ∠A).
   10. (iii) Therefore, △AEP ~ △ADB (AA similarity criterion).
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   11. (iv) From (ii), △ABD ~ △CBE, so ∠BAD = ∠BCE.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   12. (iv) In △PDC and △BEC, ∠PDC = ∠BEC = 90°.
   13. (iv) Also, ∠PCD = ∠BCE (same angle ∠C).
   14. (iv) Therefore, △PDC ~ △BEC (AA similarity criterion).
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   Answer: (i) △AEP ~ △CDP; (ii) △ABD ~ △CBE; (iii) △AEP ~ △ADB; (iv) △PDC ~ △BEC.

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8. Ex 6.3 Q8

   E is on AD produced of parallelogram ABCD and BE meets CD at F. Show that △ABE ~ △CFB.

   1. Given that ABCD is a parallelogram and E is on AD produced.
   2. Since AB is parallel to CD, the alternate interior angles are equal, so ∠ABE = ∠CFB.
   3. ∠BAE = ∠BAD = ∠BCD = ∠FCB, since opposite angles of a parallelogram are equal.
   4. By the AA similarity criterion, △ABE ~ △CFB.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   Answer: △ABE ~ △CFB

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9. Ex 6.3 Q9

   △ABC and △AMP are right-angled at B and M (Fig. 6.39). Prove (i) △ABC ~ △AMP (ii) CA/PA = BC/MP.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) In △ABC and △AMP, ∠ABC = ∠AMP = 90° (given).
   2. (i) ∠A is common to both triangles.
   3. (i) Therefore, △ABC ~ △AMP by AA similarity criterion.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   4. (ii) Since the triangles are similar, the ratios of corresponding sides are equal.
      why?

      In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

   5. (ii) Thus, CA/PA = BC/MP.

   Answer: (i) △ABC ~ △AMP; (ii) CA/PA = BC/MP

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10. Ex 6.3 Q10

    CD and GH bisect ∠ACB and ∠EGF, with D on AB and H on FE, and △ABC ~ △FEG. Show (i) CD/GH = AC/FG (ii) △DCB ~ △HGE (iii) △DCA ~ △HGF.

    1. Since △ABC ~ △FEG, ∠A = ∠F, ∠B = ∠E, ∠ACB = ∠FGE, and AC/FG = BC/EG.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    2. CD and GH bisect the equal angles ∠ACB and ∠FGE, so ∠ACD = ∠DCB = ∠FGH = ∠HGE.
    3. (i) In △DCA and △HGF: ∠A = ∠F and ∠ACD = ∠FGH, so △DCA ~ △HGF (AA).
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    4. (i) Hence CD/GH = AC/FG (corresponding sides of these similar triangles).
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    5. (ii) In △DCB and △HGE: ∠B = ∠E and ∠DCB = ∠HGE, so △DCB ~ △HGE (AA).
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    6. (iii) This is the similarity △DCA ~ △HGF shown in the third step.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    Answer: (i) CD/GH = AC/FG; (ii) △DCB ~ △HGE (AA); (iii) △DCA ~ △HGF (AA).

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11. Ex 6.3 Q11

    E is on CB produced of isosceles △ABC (AB = AC); AD ⊥ BC and EF ⊥ AC (Fig. 6.40). Prove △ABD ~ △ECF.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Given: △ABC is isosceles with AB = AC, AD ⊥ BC, and E is on CB produced such that EF ⊥ AC.
    2. Since AB = AC, △ABC is isosceles, so ∠ABC = ∠ACB (angles opposite to equal sides are equal).
    3. In △ABD, ∠ADB = 90° (given AD ⊥ BC).
    4. In △ECF, ∠EFC = 90° (given EF ⊥ AC).
    5. Thus, ∠ADB = ∠EFC = 90°.
    6. Also, ∠ABD = ∠ABC and ∠ECF = ∠ACB (same angles as E lies on CB produced and F lies on AC).
    7. Since ∠ABC = ∠ACB, we have ∠ABD = ∠ECF.
    8. Therefore, in △ABD and △ECF: ∠ADB = ∠EFC and ∠ABD = ∠ECF.
    9. By AA similarity criterion, △ABD ~ △ECF.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    Answer: △ABD ~ △ECF (proved by AA similarity criterion)

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12. Ex 6.3 Q12

    Sides AB, BC and median AD of △ABC are proportional to sides PQ, QR and median PM of △PQR (Fig. 6.41). Show △ABC ~ △PQR.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Given that AB/PQ = BC/QR = AD/PM, where AD and PM are medians of △ABC and △PQR respectively.
    2. Since AD is the median, D is the midpoint of BC, so BD = DC = ½BC.
    3. Since PM is the median, M is the midpoint of QR, so QM = MR = ½QR.
    4. From the given proportionality, AB/PQ = BC/QR, so AB/PQ = (2BD)/(2QM) = BD/QM.
    5. Thus, AB/PQ = BD/QM = AD/PM.
    6. By the SSS similarity criterion, △ABD ~ △PQM.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    7. Therefore, ∠ABD = ∠PQM, which means ∠ABC = ∠PQR.
    8. Now, AB/PQ = BC/QR and the included angle ∠ABC = ∠PQR.
    9. By the SAS similarity criterion, △ABC ~ △PQR.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    Answer: △ABC ~ △PQR

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13. Ex 6.3 Q13

    D is on BC of △ABC with ∠ADC = ∠BAC. Show that CA² = CB·CD.

    1. Given that D is a point on BC such that ∠ADC = ∠BAC.
    2. Consider △ADC and △BAC.
    3. We have ∠ADC = ∠BAC (given).
    4. Also, ∠C is common to both triangles.
    5. Therefore, △ADC ~ △BAC by AA similarity criterion.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    6. Since the triangles are similar, the ratios of corresponding sides are equal.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    7. Thus, CA/CB = CD/CA.
    8. Cross-multiplying gives CA × CA = CB × CD.
       why?

       If a/b = c/d (b and d not zero), multiplying both sides by bd gives ad = bc — the same 'do the same to both sides' rule.

    9. Hence, CA² = CB · CD.

    Answer: CA² = CB · CD

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14. Ex 6.3 Q14

    Sides AB, AC and median AD of △ABC are proportional to sides PQ, PR and median PM of △PQR. Show △ABC ~ △PQR.

    1. Given AB/PQ = AC/PR = AD/PM, with D the mid-point of BC and M the mid-point of QR.
    2. Extend AD to E with DE = AD, and PM to L with ML = PM. Then ABEC and PQLR are parallelograms (their diagonals bisect each other), so BE = AC and QL = PR.
    3. So AB/PQ = BE/QL = AE/PL, because BE/QL = AC/PR and AE/PL = 2AD/2PM = AD/PM.
    4. Hence △ABE ~ △PQL (SSS), which gives ∠BAE = ∠QPL.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    5. In the same way, using CE = AB and RL = PQ, △AEC ~ △PLR (SSS), which gives ∠EAC = ∠LPR.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    6. Adding, ∠BAC = ∠BAE + ∠EAC = ∠QPL + ∠LPR = ∠QPR.
    7. Now AB/PQ = AC/PR with the included angles ∠BAC = ∠QPR equal, so △ABC ~ △PQR (SAS).
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    Answer: △ABC ~ △PQR.

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15. Ex 6.3 Q15

    A 6 m pole casts a 4 m shadow while a tower casts a 28 m shadow. Find the tower's height.

    1. Since the sun's rays are parallel, the triangles formed by the pole and its shadow and the tower and its shadow are similar.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    2. The ratio of the height of the pole to its shadow is equal to the ratio of the height of the tower to its shadow.
    3. Let the height of the tower be h meters.
    4. Set up the proportion: 6 / 4 = h / 28.
    5. Solve for h by cross-multiplying: h = (6 × 28) / 4.
       why?

       If a/b = c/d (b and d not zero), multiplying both sides by bd gives ad = bc — the same 'do the same to both sides' rule.

    6. Calculate the value: h = 168 / 4 = 42.

    Answer: 42 m

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16. Ex 6.3 Q16

    AD and PM are medians of △ABC ~ △PQR. Prove AB/PQ = AD/PM.

    1. Since △ABC ~ △PQR, the ratio of corresponding sides is equal, so AB/PQ = BC/QR = AC/PR.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    2. Also, corresponding angles are equal, so ∠B = ∠Q.
    3. AD is a median of △ABC, so BD = BC/2.
    4. PM is a median of △PQR, so QM = QR/2.
    5. Dividing the median lengths by the corresponding side lengths gives BD/BC = 1/2 and QM/QR = 1/2.
    6. Thus BD/BC = QM/QR.
    7. In △ABD and △PQM, we have AB/PQ = BD/QM and the included angle ∠B = ∠Q.
    8. Therefore, △ABD ~ △PQM by SAS similarity criterion.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    9. Since the triangles are similar, the ratio of their corresponding sides is equal, so AB/PQ = AD/PM.
       why?

       In similar triangles the corresponding angles are equal and the corresponding sides are in the same ratio, so one known ratio gives the unknown side. Two equal angles (AA) are enough to show triangles are similar.

    Answer: AB/PQ = AD/PM

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