Class 10 Maths Chapter 6 Triangles – Exercise 6.4 NCERT Solutions

Introduction

Exercise 6.4 focuses on applying the Basic Proportionality Theorem (Thales’ theorem) and properties of similar triangles to solve numerical problems. You’ll learn how to use ratios of sides and similarity criteria to calculate unknown lengths and prove geometric relations.

Formula Used

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

$$\frac{AD}{DB}=\frac{AE}{EC}$$

• Similarity Property: If ΔABC ∼ ΔDEF, then

$$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$$

NCERT Questions with Solutions (10)

Q1. In ΔABC, DE ∥ BC intersects AB at D and AC at E. If AD = 3 cm, DB = 2 cm, AE = 4.5 cm, find EC.

Solution: By BPT: $\frac{AD}{DB}=\frac{AE}{EC}$. $\frac{3}{2}=\frac{4.5}{EC}$. $EC=3$.

Q2. In ΔPQR, ST ∥ QR intersects PQ at S and PR at T. If PS = 4 cm, SQ = 6 cm, PT = 5 cm, find TR.

Solution: By BPT: $\frac{PS}{SQ}=\frac{PT}{TR}$. $\frac{4}{6}=\frac{5}{TR}$. $TR=7.5$.

Q3. In ΔXYZ, LM ∥ YZ intersects XY at L and XZ at M. If XL = 2 cm, LY = 3 cm, XM = 4 cm, find MZ.

Solution: By BPT: $\frac{XL}{LY}=\frac{XM}{MZ}$. $\frac{2}{3}=\frac{4}{MZ}$. $MZ=6$.

Q4. In ΔDEF, GH ∥ EF intersects DE at G and DF at H. If DG = 2 cm, GE = 3 cm, DH = 4 cm, find HF.

Solution: By BPT: $\frac{DG}{GE}=\frac{DH}{HF}$. $\frac{2}{3}=\frac{4}{HF}$. $HF=6$.

Q5. In ΔABC, DE ∥ BC intersects AB at D and AC at E. If AD = 2 cm, DB = 4 cm, AE = 3 cm, find EC.

Solution: $\frac{AD}{DB}=\frac{AE}{EC}$. $\frac{2}{4}=\frac{3}{EC}$. $EC=6$.

Q6. In ΔPQR, ST ∥ QR intersects PQ at S and PR at T. If PS = 3 cm, SQ = 9 cm, PT = 2 cm, find TR.

Solution: $\frac{PS}{SQ}=\frac{PT}{TR}$. $\frac{3}{9}=\frac{2}{TR}$. $TR=6$.

Q7. In ΔXYZ, LM ∥ YZ intersects XY at L and XZ at M. If XL = 5 cm, LY = 10 cm, XM = 6 cm, find MZ.

Solution: $\frac{XL}{LY}=\frac{XM}{MZ}$. $\frac{5}{10}=\frac{6}{MZ}$. $MZ=12$.

Q8. In ΔDEF, GH ∥ EF intersects DE at G and DF at H. If DG = 3 cm, GE = 6 cm, DH = 5 cm, find HF.

Solution: $\frac{DG}{GE}=\frac{DH}{HF}$. $\frac{3}{6}=\frac{5}{HF}$. $HF=10$.

Q9. In ΔABC, DE ∥ BC intersects AB at D and AC at E. If AD = 4 cm, DB = 6 cm, AE = 5 cm, find EC.

Solution: $\frac{AD}{DB}=\frac{AE}{EC}$. $\frac{4}{6}=\frac{5}{EC}$. $EC=7.5$.

Q10. In ΔPQR, ST ∥ QR intersects PQ at S and PR at T. If PS = 2 cm, SQ = 8 cm, PT = 3 cm, find TR.

Solution: $\frac{PS}{SQ}=\frac{PT}{TR}$. $\frac{2}{8}=\frac{3}{TR}$. $TR=12$.

FAQs (10 from NCERT)

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

2. Q: What is AA similarity criterion? A: If two angles are equal, triangles are similar.

3. Q: What is SSS similarity criterion? A: If sides are proportional, triangles are similar.

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

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

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

7. Q: Why is BPT important? A: It helps in solving geometric problems involving ratios.

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

9. Q: What is practical use of similarity? A: Indirect measurement and geometry proofs.

10. Q: Why is this exercise important? A: It builds skills in proportionality and similarity applications.

Conclusion

Exercise 6.4 has 10 solved questions and 10 FAQs that strengthen your understanding of proportionality and triangle similarity. This builds the foundation for advanced geometry in Class 10 Maths.

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