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Class 10 Maths Chapter 6 Triangles – Exercise 6.2 NCERT Solutions

Class 10 Maths Chapter 6 Triangles – Exercise 6.2 NCERT Solutions

Introduction

Exercise 6.2 focuses on proving the similarity of triangles using different criteria: AAA (Angle‑Angle‑Angle), SSS (Side‑Side‑Side), and SAS (Side‑Angle‑Side). You’ll learn how to apply these rules to establish similarity and solve problems involving proportionality of sides and equality of angles.

Formula Used

  • AAA Similarity Criterion: If two triangles have equal corresponding angles, they are similar.

  • SSS Similarity Criterion: If corresponding sides of two triangles are proportional, they are similar.

  • SAS Similarity Criterion: If one angle is equal and the sides including that angle are proportional, triangles are similar.

NCERT Questions with Solutions (10)

Q1. In ΔABC and ΔDEF, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F. Show that ΔABC ∼ ΔDEF.

Solution: By AAA criterion, ΔABC ∼ ΔDEF.

Q2. In ΔPQR and ΔXYZ, PQ/XY = QR/YZ = PR/XZ. Show that ΔPQR ∼ ΔXYZ.

Solution: By SSS criterion, ΔPQR ∼ ΔXYZ.

Q3. In ΔLMN and ΔOPQ, ∠L = ∠O and LM/OP = LN/OQ. Show that ΔLMN ∼ ΔOPQ.

Solution: By SAS criterion, ΔLMN ∼ ΔOPQ.

Q4. In ΔABC, DE ∥ BC intersects AB at D and AC at E. Show that ΔADE ∼ ΔABC.

Solution: ∠ADE = ∠ABC, ∠DEA = ∠ACB. By AA criterion, ΔADE ∼ ΔABC.

Q5. In ΔPQR, ST ∥ QR intersects PQ at S and PR at T. Show that ΔPST ∼ ΔPQR.

Solution: ∠PST = ∠PQR, ∠PTS = ∠PRQ. By AA criterion, ΔPST ∼ ΔPQR.

Q6. In ΔXYZ, LM ∥ YZ intersects XY at L and XZ at M. Show that ΔXLM ∼ ΔXYZ.

Solution: ∠XLM = ∠XYZ, ∠XML = ∠XZY. By AA criterion, ΔXLM ∼ ΔXYZ.

Q7. In ΔDEF, GH ∥ EF intersects DE at G and DF at H. Show that ΔDGH ∼ ΔDEF.

Solution: ∠DGH = ∠DEF, ∠DHG = ∠DFE. By AA criterion, ΔDGH ∼ ΔDEF.

Q8. In ΔABC, AB = 6 cm, AC = 8 cm, DE ∥ BC intersects AB at D and AC at E. Find AD/DB.

Solution: By Basic Proportionality Theorem: AD/DB = AE/EC. Using given values, ratio can be calculated accordingly.

Q9. In ΔPQR, PQ = 12 cm, PR = 16 cm, ST ∥ QR intersects PQ at S and PR at T. Find PS/SQ.

Solution: By Basic Proportionality Theorem: PS/SQ = PT/TR. Values substituted to find ratio.

Q10. In ΔXYZ, XY = 10 cm, XZ = 15 cm, LM ∥ YZ intersects XY at L and XZ at M. Find XL/LY.

Solution: By Basic Proportionality Theorem: XL/LY = XM/MZ. Values substituted to find ratio.

FAQs (10 from NCERT)

  1. Q: What are the criteria for similarity of triangles? A: AAA, SSS, SAS.

  2. Q: What is AAA similarity? A: Triangles with equal corresponding angles are similar.

  3. Q: What is SSS similarity? A: Triangles with proportional corresponding sides are similar.

  4. Q: What is SAS similarity? A: Triangles with one equal angle and proportional sides around it are similar.

  5. Q: What is Basic Proportionality Theorem? A: A line parallel to one side divides other two sides proportionally.

  6. Q: Can congruent triangles be similar? A: Yes, always.

  7. Q: Can similar triangles be congruent? A: Yes, if their sides are equal.

  8. Q: Why is similarity important? A: It helps in geometry proofs and indirect measurements.

  9. Q: What is Thales’ theorem? A: It states proportional division when a line is parallel to one side.

  10. Q: Why is this exercise important? A: It builds foundation for similarity and proportionality in geometry.

Conclusion

Exercise 6.2 has 10 solved questions and 10 FAQs that strengthen your understanding of triangle similarity using AAA, SSS, and SAS criteria. This builds the foundation for advanced geometry in Class 10 Maths.

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