NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3: Chapter 6 of Class 10 Maths, "Triangles," focuses on the principles of similarity in triangles. Exercise 6.3 emphasizes the application of the Basic Proportionality Theorem (Thales’ Theorem) and criteria for similarity, such as AA, SSS, and SAS.
The chapter is an essential part of the CBSE Class 10 Maths syllabus, covering concepts like similarity criteria, the Pythagoras theorem, and its converse. These solutions help students understand important concepts, practice problem-solving techniques, and prepare effectively for the board exams by following the CBSE exam pattern, which includes theory-based, application-based, and numerical questions. Solving these problems also helps in revising previous years' questions (PYQs) and improving overall mathematical proficiency.
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 Overview
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3, "Triangles," focus on understanding and applying the concepts of similar triangles. This exercise is vital as it introduces the Basic Proportionality Theorem (Thales' Theorem) and similarity criteria like AA, SSS, and SAS, which are fundamental in geometry.
These principles are not only crucial for solving academic problems but also have practical applications in fields like architecture, engineering, and design. By practicing these problems, students develop logical reasoning, proportionality concepts, and problem-solving skills, forming a solid base for higher-level mathematics and competitive exams.
Important Questions for Class 10 Maths Chapter 6
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 PDF
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3, "Triangles," offer a comprehensive understanding of triangle similarity, including the application of the Basic Proportionality Theorem and similarity criteria like AA, SSS, and SAS. These solutions help students enhance their problem-solving skills and conceptual clarity. Below, we have provided a downloadable PDF containing step-by-step solutions to all the questions, making it easy for students to study and revise effectively.
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 PDF
NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 Triangles
Below is the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3 Triangles -
1. State which pairs of triangles in Figure 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) Given, in ΔABC and ΔPQR, ∠A = ∠P = 60° ∠B = ∠Q = 80° ∠C = ∠R = 40° Therefore by AAA similarity criterion, ∴ ΔABC ~ ΔPQR (ii) Given, in ΔABC and ΔPQR, AB/QR = 2/4 = 1/2, BC/RP = 2.5/5 = 1/2, CA/PA = 3/6 = 1/2 By SSS similarity criterion, ΔABC ~ ΔQRP (iii) Given, in ΔLMP and ΔDEF, LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6 MP/DE = 2/4 = 1/2 PL/DF = 3/6 = 1/2 LM/EF = 2.7/5 = 27/50 Here , MP/DE = PL/DF ≠ LM/EF Therefore, ΔLMP and ΔDEF are not similar. (iv) In ΔMNL and ΔQPR, it is given, MN/QP = ML/QR = 1/2 ∠M = ∠Q = 70° Therefore, by SAS similarity criterion ∴ ΔMNL ~ ΔQPR (v) In ΔABC and ΔDEF, given that, AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80° Here , AB/DF = 2.5/5 = 1/2 And, BC/EF = 3/6 = 1/2 ⇒ ∠B ≠ ∠F Hence, ΔABC and ΔDEF are not similar. (vi) In ΔDEF, by sum of angles of triangles, we know that, ∠D + ∠E + ∠F = 180° ⇒ 70° + 80° + ∠F = 180° ⇒ ∠F = 180° – 70° – 80° ⇒ ∠F = 30° Similarly, In ΔPQR, ∠P + ∠Q + ∠R = 180 (Sum of angles of Δ) ⇒ ∠P + 80° + 30° = 180° ⇒ ∠P = 180° – 80° -30° ⇒ ∠P = 70° Now, comparing both the triangles, ΔDEF and ΔPQR, we have ∠D = ∠P = 70° ∠F = ∠Q = 80° ∠F = ∠R = 30° Therefore, by AAA similarity criterion, Hence, ΔDEF ~ ΔPQR
2. In figure 6.35, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
Solution:
As we can see from the figure, DOB is a straight line. Therefore, ∠DOC + ∠ COB = 180° ⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°) = 55° In ΔDOC, sum of the measures of the angles of a triangle is 180º Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180° ⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°) ⇒ ∠DCO = 55° It is given that, ΔODC ~ ΔOBA, Therefore, ΔODC ~ ΔOBA. Hence, Corresponding angles are equal in similar triangles ∠OAB = ∠OCD ⇒ ∠ OAB = 55° ∠OAB = ∠OCD ⇒ ∠OAB = 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 AO/OC = OB/OD
Solution:
In ΔDOC and ΔBOA, AB || CD, thus alternate interior angles will be equal, ∴∠CDO = ∠ABO Similarly, ∠DCO = ∠BAO Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal; ∴∠DOC = ∠BOA Hence, by AAA similarity criterion, ΔDOC ~ ΔBOA Thus, the corresponding sides are proportional. DO/BO = OC/OA ⇒OA/OC = OB/OD Hence, proved.
4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Solution:
In ΔPQR, ∠PQR = ∠PRQ ∴ PQ = PR ……………………… (i) Given, QR/QS = QT/PRUsing equation (i) , we get QR/QS = QT/QP ……………….(ii) In ΔPQS and ΔTQR, by equation (ii), QR/QS = QT/QP ∠Q = ∠Q ∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Solution:
Given, S and T are point on sides PR and QR of ΔPQR And ∠P = ∠RTS. 
In ΔRPQ and ΔRTS, ∠RTS = ∠QPS (Given) ∠R = ∠R (Common angle) ∴ ΔRPQ ~ ΔRTS (AA similarity criterion)
6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Solution:
Given, ΔABE ≅ ΔACD. ∴ AB = AC [By CPCT] ………………………………. (i) And, AD = AE [By CPCT] …………………………… (ii) In ΔADE and ΔABC, dividing eq.(ii) by eq(i), AD/AB = AE/AC ∠A = ∠A [Common angle] ∴ ΔADE ~ ΔABC [SAS similarity criterion]
7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC
Solution:
Given, altitudes AD and CE of ΔABC intersect each other at the point P. (i) In ΔAEP and ΔCDP, ∠AEP = ∠CDP (90° each) ∠APE = ∠CPD (Vertically opposite angles) Hence, by AA similarity criterion, ΔAEP ~ ΔCDP (ii) In ΔABD and ΔCBE, ∠ADB = ∠CEB ( 90° each) ∠ABD = ∠CBE (Common Angles) Hence, by AA similarity criterion, ΔABD ~ ΔCBE (iii) In ΔAEP and ΔADB, ∠AEP = ∠ADB (90° each) ∠PAE = ∠DAB (Common Angles) Hence, by AA similarity criterion, ΔAEP ~ ΔADB (iv) In ΔPDC and ΔBEC, ∠PDC = ∠BEC (90° each) ∠PCD = ∠BCE (Common angles) Hence, by AA similarity criterion, ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Solution:
Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below, 
In ΔABE and ΔCFB, ∠A = ∠C (Opposite angles of a parallelogram) ∠AEB = ∠CBF (Alternate interior angles as AE || BC) ∴ ΔABE ~ ΔCFB (AA similarity criterion)
9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
Solution:
Given, ABC and AMP are two right triangles, right angled at B and M respectively. (i) In ΔABC and ΔAMP, we have, ∠CAB = ∠MAP (common angles) ∠ABC = ∠AMP = 90° (each 90°) ∴ ΔABC ~ ΔAMP (AA similarity criterion) (ii) As, ΔABC ~ ΔAMP (AA similarity criterion) If two triangles are similar then the corresponding sides are always equal, 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 ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF
Solution:
Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. 
(i) From the given condition, ΔABC ~ ΔFEG. ∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE Since, ∠ACB = ∠FGE ∴ ∠ACD = ∠FGH (Angle bisector) And, ∠DCB = ∠HGE (Angle bisector) In ΔACD and ΔFGH, ∠A = ∠F ∠ACD = ∠FGH ∴ ΔACD ~ ΔFGH (AA similarity criterion) ⇒CD/GH = AC/FG (ii) In ΔDCB and ΔHGE, ∠DCB = ∠HGE (Already proved) ∠B = ∠E (Already proved) ∴ ΔDCB ~ ΔHGE (AA similarity criterion) (iii) In ΔDCA and ΔHGF, ∠ACD = ∠FGH (Already proved) ∠A = ∠F (Already proved) ∴ ΔDCA ~ ΔHGF (AA similarity criterion)
11. In the following 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 ΔABD ~ ΔECF.
Solution:
Given, ABC is an isosceles triangle. ∴ AB = AC ⇒ ∠ABD = ∠ECF In ΔABD and ΔECF, ∠ADB = ∠EFC (Each 90°) ∠BAD = ∠CEF (Already proved) ∴ ΔABD ~ ΔECF (using AA similarity criterion)
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Solution:
Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR i.e. AB/PQ = BC/QR = AD/PM We have to prove: ΔABC ~ ΔPQR As we know here, AB/PQ = BC/QR = AD/PM 
⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR) ⇒ ΔABD ~ ΔPQM [SSS similarity criterion] ∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal] ⇒ ∠ABC = ∠PQR In ΔABC and ΔPQR AB/PQ = BC/QR …………………………. (i) ∠ABC = ∠PQR …………………………… (ii) From equation (i) and (ii) , we get, ΔABC ~ ΔPQR [SAS similarity criterion]
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA 2 = CB.CD
Solution:
Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. 
In ΔADC and ΔBAC, ∠ADC = ∠BAC (Already given) ∠ACD = ∠BCA (Common angles) ∴ ΔADC ~ ΔBAC (AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. ∴ CA/CB = CD/CA ⇒ CA 2 = CB.CD. Hence, proved.
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 ΔABC ~ ΔPQR.
Solution:
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that; AB/PQ = AC/PR = AD/PM We have to prove, ΔABC ~ ΔPQR Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN. 
In ΔABD and ΔCDE, we have AD = DE [By Construction.] BD = DC [Since, AP is the median] and, ∠ADB = ∠CDE [Vertically opposite angles] ∴ ΔABD ≅ ΔCDE [SAS criterion of congruence] ⇒ AB = CE [By CPCT] ………………………….. (i) Also, in ΔPQM and ΔMNR, PM = MN [By Construction.] QM = MR [Since, PM is the median] and, ∠PMQ = ∠NMR [Vertically opposite angles] ∴ ΔPQM = ΔMNR [SAS criterion of congruence] ⇒ PQ = RN [CPCT] ……………………………… (ii) Now, AB/PQ = AC/PR = AD/PM From equation (i) and (ii) , ⇒CE/RN = AC/PR = AD/PM ⇒ CE/RN = AC/PR = 2AD/2PM ⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN] ∴ ΔACE ~ ΔPRN [SSS similarity criterion] Therefore, ∠2 = ∠4 Similarly, ∠1 = ∠3 ∴ ∠1 + ∠2 = ∠3 + ∠4 ⇒ ∠A = ∠P ……………………………………………. (iii) Now, in ΔABC and ΔPQR, we have AB/PQ = AC/PR (Already given) From equation (iii), ∠A = ∠P ∴ ΔABC ~ ΔPQR [ SAS similarity criterion]
15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Given, Length of the vertical pole = 6m Shadow of the pole = 4 m Let Height of tower = h m Length of shadow of the tower = 28 m 
In ΔABC and ΔDEF, ∠C = ∠E (angular elevation of sum) ∠B = ∠F = 90° ∴ ΔABC ~ ΔDEF (AA similarity criterion) ∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional) ∴ 6/h = 4/28 ⇒h = (6×28)/4 ⇒ h = 6 × 7 ⇒ h = 42 m Hence, the height of the tower is 42 m.
16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
Solution:
Given, ΔABC ~ ΔPQR 
We know that the corresponding sides of similar triangles are in proportion. ∴AB/PQ = AC/PR = BC/QR ……………………………(i ) Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….….. (ii) Since AD and PM are medians, they will divide their opposite sides. ∴ BD = BC/2 and QM = QR/2 ……………..…………. (iii) From equations (i) and (iii) , we get AB/PQ = BD/QM ……………………….(iv) In ΔABD and ΔPQM, From equation (ii), we have ∠B = ∠Q From equation (iv), we have, AB/PQ = BD/QM ∴ ΔABD ~ ΔPQM (SAS similarity criterion) ⇒AB/PQ = BD/QM = AD/PM
Benefits of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3
Comprehensive Understanding : Provides detailed explanations of theorems like the Basic Proportionality Theorem and similarity criteria (AA, SSS, SAS), ensuring conceptual clarity.
Step-by-Step Solutions : Offers clear, step-by-step answers to all exercise problems, helping students grasp the methodology.
Exam-Oriented : Focuses on important concepts and problem-solving techniques that are frequently tested in board exams.
Time-Saving : Helps students quickly revise and practice, reducing preparation time.
Confidence Boost : Enhances problem-solving skills, building confidence for tackling higher-level geometry and competitive exams.
