Class 10 Maths Chapter 6 Triangles – Exercise 6.1 NCERT Solutions

Introduction

Exercise 6.1 introduces the concept of similar triangles. Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are in the same ratio. This exercise focuses on identifying similarity and applying basic proportionality.

Formula Used

• Condition for Similarity of Triangles:

  • AAA (Angle‑Angle‑Angle): If corresponding angles are equal.

  • SSS (Side‑Side‑Side): If corresponding sides are proportional.

  • SAS (Side‑Angle‑Side): If one angle is equal and sides including that angle are proportional.

• Basic Proportionality Theorem (Thales’ Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

NCERT Questions with Solutions (10)

Q1. In ΔABC, DE ∥ BC. Show that $\frac{AD}{DB}=\frac{AE}{EC}$.

Solution: By Basic Proportionality Theorem, DE ∥ BC divides sides proportionally. Hence proved.

Q2. In ΔPQR, ST ∥ QR. Show that $\frac{PS}{SQ}=\frac{PT}{TR}$.

Solution: By Basic Proportionality Theorem, ST ∥ QR divides sides proportionally. Hence proved.

Q3. In ΔXYZ, LM ∥ YZ. Show that $\frac{XL}{LY}=\frac{XM}{MZ}$.

Solution: By Basic Proportionality Theorem, LM ∥ YZ divides sides proportionally. Hence proved.

Q4. In ΔDEF, GH ∥ EF. Show that $\frac{DG}{GE}=\frac{DH}{HF}$.

Solution: By Basic Proportionality Theorem, GH ∥ EF divides sides proportionally. Hence proved.

Q5. In ΔABC, a line parallel to BC intersects AB at D and AC at E. Prove that ΔADE ∼ ΔABC.

Solution: ∠ADE = ∠ABC, ∠DEA = ∠ACB (corresponding angles). By AA similarity, ΔADE ∼ ΔABC.

Q6. In ΔPQR, a line parallel to QR intersects PQ at S and PR at T. Prove that ΔPST ∼ ΔPQR.

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

Q7. In ΔXYZ, a line parallel to YZ intersects XY at L and XZ at M. Prove that ΔXLM ∼ ΔXYZ.

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

Q8. In ΔDEF, a line parallel to EF intersects DE at G and DF at H. Prove that ΔDGH ∼ ΔDEF.

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

Q9. In ΔABC, DE ∥ BC. Show that ΔADE ∼ ΔABC.

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

Q10. In ΔPQR, ST ∥ QR. Show that ΔPST ∼ ΔPQR.

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

FAQs (10 from NCERT)

1. Q: What are similar triangles? A: Triangles with equal corresponding angles and proportional sides.

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

3. Q: What is AA similarity? A: If two angles of one triangle equal two angles of another, triangles are similar.

4. Q: What is SSS similarity? A: If sides of two triangles are proportional, they are similar.

5. Q: What is SAS similarity? A: If one angle is equal and sides around it are proportional, triangles are similar.

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

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

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

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

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

Conclusion

Exercise 6.1 has 10 solved questions and 10 FAQs that strengthen your understanding of triangle similarity and proportionality. This builds the foundation for advanced geometry in Class 10 Maths.

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