Class 10 Maths Chapter 6 Triangles – Exercise 6.5 NCERT Solutions

Class 10 Maths Chapter 6 Triangles – Exercise 6.5 NCERT Solutions
Introduction

Exercise 6.5 focuses on the areas of similar triangles. When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. This exercise strengthens your understanding of similarity and its application in geometry.

Formula Used

• Area Ratio of Similar Triangles: If ΔABC ∼ ΔDEF, then

$$\frac{Area(\mathrm{\Delta }ABC)}{Area(\mathrm{\Delta }DEF)}={\left(\frac{AB}{DE}\right)}^2={\left(\frac{BC}{EF}\right)}^2={\left(\frac{AC}{DF}\right)}^2$$

NCERT Questions with Solutions (10)

Q1. ΔABC ∼ ΔDEF, AB = 3 cm, DE = 6 cm. Find ratio of areas.

Solution: $\frac{Area(\mathrm{\Delta }ABC)}{Area(\mathrm{\Delta }DEF)}={\left(\frac{3}{6}\right)}^2=\frac{1}{4}$.

Q2. ΔPQR ∼ ΔXYZ, PQ = 5 cm, XY = 10 cm. Find ratio of areas.

Solution: ${\left(\frac{5}{10}\right)}^2=\frac{1}{4}$.

Q3. ΔLMN ∼ ΔOPQ, LM = 4 cm, OP = 12 cm. Find ratio of areas.

Solution: ${\left(\frac{4}{12}\right)}^2=\frac{1}{9}$.

Q4. ΔABC ∼ ΔDEF, AB = 7 cm, DE = 14 cm. Find ratio of areas.

Solution: ${\left(\frac{7}{14}\right)}^2=\frac{1}{4}$.

Q5. ΔPQR ∼ ΔXYZ, PQ = 8 cm, XY = 16 cm. Find ratio of areas.

Solution: ${\left(\frac{8}{16}\right)}^2=\frac{1}{4}$.

Q6. ΔLMN ∼ ΔOPQ, LM = 9 cm, OP = 27 cm. Find ratio of areas.

Solution: ${\left(\frac{9}{27}\right)}^2=\frac{1}{9}$.

Q7. ΔABC ∼ ΔDEF, AB = 5 cm, DE = 20 cm. Find ratio of areas.

Solution: ${\left(\frac{5}{20}\right)}^2=\frac{1}{16}$.

Q8. ΔPQR ∼ ΔXYZ, PQ = 6 cm, XY = 18 cm. Find ratio of areas.

Solution: ${\left(\frac{6}{18}\right)}^2=\frac{1}{9}$.

Q9. ΔLMN ∼ ΔOPQ, LM = 10 cm, OP = 25 cm. Find ratio of areas.

Solution: ${\left(\frac{10}{25}\right)}^2=\frac{16}{25}$. Correction: Actually ${\left(\frac{10}{25}\right)}^2=\frac{100}{625}=\frac{4}{25}$.

Q10. ΔABC ∼ ΔDEF, AB = 2 cm, DE = 8 cm. Find ratio of areas.

Solution: ${\left(\frac{2}{8}\right)}^2=\frac{1}{16}$.

FAQs (10 from NCERT)

1. Q: What is the relation between areas of similar triangles? A: Ratio of areas = square of ratio of corresponding sides.

2. Q: What is similarity criterion used here? A: AAA, SSS, or SAS similarity.

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

4. Q: Can similar triangles have equal areas? A: Yes, if their corresponding sides are equal.

5. Q: What is the formula for area ratio? A: $\frac{Are{a}_1}{Are{a}_2}={\left(\frac{Sid{e}_1}{Sid{e}_2}\right)}^2$.

6. Q: Why is square taken in formula? A: Because area depends on square of sides.

7. Q: What is practical use of this theorem? A: Indirect measurement of areas.

8. Q: Can theorem be applied to non‑similar triangles? A: No, only similar triangles.

9. Q: What is Thales’ theorem relation here? A: It helps establish similarity, then area ratio applies.

10. Q: Why is this exercise important? A: It builds skills in applying similarity to area problems.

Conclusion

Exercise 6.5 has 10 solved questions and 10 FAQs that strengthen your understanding of the relation between areas of similar triangles. This builds the foundation for advanced geometry in Class 10 Maths.

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