Class 10 Maths Chapter 6 Triangles – Exercise 6.3 NCERT Solutions

Introduction

Exercise 6.3 focuses on the Pythagoras Theorem and its proof using the concept of similar triangles. You’ll learn how to apply similarity criteria to derive the theorem and solve problems involving right‑angled triangles.

Formula Used

• Pythagoras Theorem: In a right‑angled triangle,

$$(Hypotenuse{)}^2=(Base{)}^2+(Perpendicular{)}^2$$

• Similarity Criterion (AA): If two angles of one triangle equal two angles of another, the triangles are similar.

NCERT Questions with Solutions (10)

Q1. In ΔABC, ∠C = 90°. Prove that $A{B}^2=A{C}^2+B{C}^2$.

Solution: Draw altitude from C to AB. Using similarity of triangles, apply proportionality. Hence proved by Pythagoras theorem.

Q2. In ΔPQR, ∠R = 90°. Show that $P{Q}^2=P{R}^2+Q{R}^2$.

Solution: By Pythagoras theorem, hypotenuse squared equals sum of squares of other two sides.

Q3. In ΔXYZ, ∠Y = 90°. Show that $X{Z}^2=X{Y}^2+Y{Z}^2$.

Solution: By Pythagoras theorem, relation holds true.

Q4. In ΔDEF, ∠E = 90°. Show that $D{F}^2=D{E}^2+E{F}^2$.

Solution: By Pythagoras theorem, relation holds true.

Q5. In ΔLMN, ∠M = 90°. Show that $L{N}^2=L{M}^2+M{N}^2$.

Solution: By Pythagoras theorem, relation holds true.

Q6. In ΔABC, ∠C = 90°, AC = 6 cm, BC = 8 cm. Find AB.

Solution: $A{B}^2=A{C}^2+B{C}^2=36+64=100$. $AB=10$ cm.

Q7. In ΔPQR, ∠R = 90°, PR = 9 cm, QR = 12 cm. Find PQ.

Solution: $P{Q}^2=81+144=225$. $PQ=15$ cm.

Q8. In ΔXYZ, ∠Y = 90°, XY = 5 cm, YZ = 12 cm. Find XZ.

Solution: $X{Z}^2=25+144=169$. $XZ=13$ cm.

Q9. In ΔDEF, ∠E = 90°, DE = 8 cm, EF = 15 cm. Find DF.

Solution: $D{F}^2=64+225=289$. $DF=17$ cm.

Q10. In ΔLMN, ∠M = 90°, LM = 7 cm, MN = 24 cm. Find LN.

Solution: $L{N}^2=49+576=625$. $LN=25$ cm.

FAQs (10 from NCERT)

1. Q: What is Pythagoras theorem? A: In a right‑angled triangle, hypotenuse squared equals sum of squares of other two sides.

2. Q: What is hypotenuse? A: The side opposite the right angle.

3. Q: What is base? A: One of the sides adjacent to the right angle.

4. Q: What is perpendicular? A: The other side adjacent to the right angle.

5. Q: What is similarity criterion used in proof? A: AA similarity.

6. Q: Can Pythagoras theorem be proved by similarity? A: Yes, by drawing altitude and using proportionality.

7. Q: Can theorem be applied to non‑right triangles? A: No, only right‑angled triangles.

8. Q: What is converse of Pythagoras theorem? A: If $a^2+b^2=c^2$, triangle is right‑angled.

9. Q: Why is theorem important? A: It is widely used in geometry, trigonometry, and real‑life applications.

10. Q: What is practical use of theorem? A: Used in construction, navigation, and measurement problems.

Conclusion

Exercise 6.3 has 10 solved questions and 10 FAQs that strengthen your understanding of the Pythagoras theorem using similarity of triangles. This builds the foundation for advanced geometry in Class 10 Maths.

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