LINES AND ANGLES — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 6  ·  Q–
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Class 9 Mathematics Exercise-6.2 NCERT Solutions Olympiad Board Exam
Chapter 6

LINES AND ANGLES

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

5 Questions
10–15 min Ideal time
Q1 Now at
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Q1
NUMERIC2 marks
In Fig. 6.23, if \(\small AB \parallel CD\), \(\small CD \parallel EF\) and \(\small y : z = 3 : 7\), find \(\small x\).
📘 Concept & Theory Concept used ›
  • If two parallel lines are cut by a transversal, then interior angles on the same side of the transversal are supplementary.
  • Supplementary angles add up to: \[\small 180^\circ \]
  • If three lines are parallel to one another, then corresponding angle relationships remain the same throughout the figure.
  • Ratio method: If two angles are in the ratio \(\small3:7\), then total parts: \[\small 3+7=10 \]
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Use supplementary angle property to form equation: \[\small y+z=180^\circ \]

  2. Use ratio \(\small 3:7\) to calculate values of \(\small y\) and \(\small z\).

  3. Again apply supplementary angle property between \(\small x\) and \(\small y\).

  4. Solve for \(\small x\).

📊 Graph / Figure Graph / Figure ›
A B C D E F x y z ∥ ∥
Fig. 6.23
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  17 steps
  1. Given
  2. \[\small AB \parallel CD \] \[\small CD \parallel EF \] Since both \(AB\) and \(EF\) are parallel to \(CD\), therefore: \[\small AB \parallel EF \] Also, \[\small y:z=3:7 \]

    We know that angles \(\small y\) and \(\small z\) are interior angles on the same side of the transversal.

    Therefore, they are supplementary.

    \[\small y+z=180^\circ \]
  3. Let
  4. \[\small y=3k \] and \[\small z=7k \]
  5. Substituting into supplementary equation:
  6. \[\small \begin{aligned} 3k+7k&=180^\circ\\ 10k&=180^\circ\\ k&=\frac{180^\circ}{10}\\ k&=18^\circ \end{aligned} \]
  7. Therefore,
  8. \[\small \begin{aligned} y&=3k\\ y&=3\times18^\circ\\ y&=54^\circ \end{aligned} \]
  9. Also,
  10. \[\small \begin{aligned} z&=7k\\ z&=7\times18^\circ\\ z&=126^\circ \end{aligned} \]
  11. Now angles \(\small x\) and \(\small y\) are also interior angles on the same side of the transversal.
  12. Therefore,
  13. \[\small x+y=180^\circ\]
  14. Substituting \(\small y=54^\circ\),
  15. \[\small x+54^\circ=180^\circ\]
  16. Subtracting \(\small 54^\circ\) from both sides:
  17. \[ \begin{aligned} x&=180^\circ-54^\circ\\ x&=126^\circ \end{aligned} \]
💡 Answer Final Answer ›
Final Answer : \( \boxed{x=126^\circ} \)
🎯 Exam Significance Exam Significance ›
  • Very important for CBSE Board examinations because it tests understanding of parallel lines and transversal properties.
  • Frequently asked in objective questions, short answer questions, and case-study based problems.
  • Strengthens angle-property fundamentals required in higher geometry chapters.
  • Helpful for preparation of competitive examinations such as NTSE, Olympiads, Polytechnic entrance tests, and other foundation-level engineering aptitude exams.
  • Develops logical reasoning and multi-step angle-solving skills.
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1 / 5  ·  20%
Q2 →
Q2
NUMERIC3 marks
In Fig. 6.24, if \(\small AB \parallel CD\), \(\small EF \perp CD\) and \(\small \angle GED = 126^\circ\), find \(\small \angle AGE\), \(\small \angle GEF\) and \(\small angle FGE\).
📘 Concept & Theory Concept Used ›
  • Perpendicular lines form a right angle: \[\small 90^\circ \]
  • Sum of angles on a straight line: \[\small 180^\circ \]
  • Sum of interior angles of a triangle: \[\small 180^\circ \]
  • Adjacent angles forming a straight line are called a linear pair.
  • Parallel line properties are frequently used with transversals in geometry problems.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Use perpendicular property to identify the right angle at point \(\small E\).

  2. Split \(\small \angle GED\) into two parts and calculate \(\small \angle GEF\).

  3. Apply angle sum property of triangle \(\small GEF\) to find \(\small \angle FGE\).

  4. Use linear pair property to calculate \(\small \angle AGE\).

📊 Graph / Figure Graph / Figure ›
A B C D F E G ∥ ∥
Fig. 6.24
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  29 steps
  1. Given
  2. \[\small AB \parallel CD \] \[\small EF \perp CD \] \[\small \angle GED = 126^\circ \]
  3. Since \(\small EF \perp CD\), the angle between \(\small EF\) and \(\small ED\) is a right angle.
  4. Therefore,\[\small \angle FED = 90^\circ \]
  5. Now angle \(\small GED\) is made up of two angles:
  6. \[\small \angle GED = \angle GEF + \angle FED\]
  7. Substituting the known values:
  8. \[\small 126^\circ = \angle GEF + 90^\circ\]
  9. Subtracting \(\small 90^\circ\) from both sides:
  10. \[\small\begin{aligned} \angle GEF &= 126^\circ - 90^\circ\\ \angle GEF &= 36^\circ\end{aligned}\]
  11. Thus, \[ \boxed{\angle GEF = 36^\circ} \]
  12. In triangle \(\small GEF\),
  13. \[\small \angle GEF + \angle GFE + \angle FGE = 180^\circ\]
  14. Since \(\small EF\) is perpendicular to \(\small CD\), angle \(\small GFE\) is a right angle.
  15. Therefore, \[\small \angle GFE = 90^\circ \]
  16. Substituting the values:
  17. \[\small 36^\circ + 90^\circ + \angle FGE = 180^\circ\]
  18. Adding known angles:
  19. \[\small 126^\circ + \angle FGE = 180^\circ\]
  20. Subtracting \(126^\circ\) from both sides:
  21. \[\small \begin{aligned} \angle FGE &= 180^\circ - 126^\circ\\ \angle FGE &= 54^\circ\end{aligned}\]
  22. Therefore, \[\small \boxed{\angle FGE = 54^\circ} \]
  23. Now \(\small\angle AGE\) and \(\small\angle FGE\) form a linear pair.
  24. Therefore, \[\small \angle AGE + \angle FGE = 180^\circ \]
  25. Substituting \(\angle FGE = 54^\circ\),
  26. \[\small \angle AGE + 54^\circ = 180^\circ\]
  27. Subtracting \(\small 54^\circ\) from both sides:
  28. \[\small \angle AGE = 180^\circ - 54^\circ\]
  29. \[\small \angle AGE = 126^\circ\]
💡 Answer Final Answer ›
Final Answer: \[ \boxed{\angle GEF = 36^\circ} \] \[ \boxed{\angle FGE = 54^\circ} \]
🎯 Exam Significance Exam Significance ›
  • This question is highly important for CBSE Board exams because it combines multiple angle properties in a single problem.
  • Frequently asked in objective questions and competency-based problems.
  • Builds strong understanding of:
    • Parallel lines
    • Perpendicular lines
    • Triangle angle sum property
    • Linear pair concept
  • Important for Olympiads, NTSE, and other foundation-level competitive examinations where logical geometry is tested.
  • Improves multi-step reasoning and diagram interpretation skills.
← Q1
2 / 5  ·  40%
Q3 →
Q3
NUMERIC3 marks
In Fig. 6.25, if \(\small PQ \parallel ST\), \(\small \angle PQR = 110^\circ\) and \(\small \angle RST = 130^\circ\), find \(\small \angle QRS\).
📘 Concept & Theory Concept Used ›
  • If two parallel lines are cut by a transversal, alternate interior angles are equal.
  • Interior angles on the same side of a transversal are supplementary.
  • Construction of an auxiliary parallel line is a very important geometry technique used in higher mathematics.
  • Angles on a straight line add up to: \[\small 180^\circ \]
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Draw an auxiliary line through point \(\small R\) parallel to \(\small ST\).

  2. Use alternate angle property to transfer \(\small \angle RST\) to point \(\small R\).

  3. Use same side interior angle property to calculate \(\small \angle ARQ\).

  4. Subtract the obtained angle from \(\small 130^\circ\) to find \(\small \angle QRS\).

📊 Graph / Figure Graph / Figure ›
P Q S T R A B 110° 130°
Fig. 6.25
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  18 steps

  1. Given:

    \[\small PQ \parallel ST \] \[\small \angle PQR = 110^\circ \] \[\small \angle RST = 130^\circ \]
  2. Construction

    Draw a line \(\small AB\) through point \(\small R\) such that:

    \[\small AB \parallel ST \]
  3. Since \(\small AB \parallel ST\), and \(\small RS\) acts as a transversal, alternate interior angles are equal.
  4. Therefore, \[\small \begin{aligned} \angle SRA &= \angle RST\\ \angle SRA &= \angle ARQ + \angle QRS \end{aligned} \]
  5. Angle \(\small \angle SRA\) is made up of two angles:
  6. \[\small \angle SRA = \angle ARQ + \angle QRS\]
  7. Therefore, \[\small \angle ARQ + \angle QRS = 130^\circ \tag{1}\]
  8. Now \(\small PQ \parallel AB\), and \(\small QR\) is a transversal.
  9. Hence interior angles on the same side of the transversal are supplementary.
  10. Therefore, \[\small \angle ARQ + \angle PQR = 180^\circ \]
  11. Substituting \(\small \angle PQR = 110^\circ\),
  12. \[\small \angle ARQ + 110^\circ = 180^\circ\]
  13. Subtracting \(\small 110^\circ\) from both sides:
  14. \[\small \begin{aligned} \angle ARQ &= 180^\circ - 110^\circ\\ \angle ARQ &= 70^\circ \end{aligned} \]
  15. Substitute \(\small \angle ARQ = 70^\circ\) into Equation (1):
  16. \[\small 70^\circ + \angle QRS = 130^\circ\]
  17. Subtracting \(\small 70^\circ\) from both sides:
  18. \[\small \begin{aligned} \angle QRS &= 130^\circ - 70^\circ\\ \angle QRS &= 60^\circ \end{aligned} \]
💡 Answer Final Answer ›
Final Answer: \(\boxed{\angle QRS = 60^\circ}\)
🎯 Exam Significance Exam Significance ›
  • This is an excellent example of applying auxiliary line construction in geometry.
  • Frequently asked in CBSE Board examinations and competency-based questions.
  • Develops understanding of:
    • Alternate interior angles
    • Same side interior angles
    • Parallel line properties
    • Geometrical constructions
  • Important for Olympiads, NTSE, and other competitive entrance examinations involving logical geometry.
  • Strengthens multi-step deductive reasoning, which is very important in advanced mathematics.
← Q2
3 / 5  ·  60%
Q4 →
Q4
NUMERIC3 marks
In Fig. 6.26, if \(\small AB \parallel CD\), \(\small \angle APQ = 50^\circ\) and \(\small \angle PRD = 127^\circ\), find \(\small x\) and \(\small y\).
📘 Concept & Theory Concept Used ›
  • If two parallel lines are cut by a transversal, alternate interior angles are equal.
  • Angles forming a linear pair add up to: \[\small 180^\circ \]
  • Sum of interior angles of a triangle is: \[\small 180^\circ \]
  • Multi-step geometry problems often require combining several angle properties together.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Use alternate interior angle property to find angle \(\small x\).

  2. Use linear pair property at point \(\small R\) to calculate \(\small \angle QRP\).

  3. Apply angle sum property of triangle \(\small PQR\) to find angle \(\small y\).

  4. Write the final values of both unknown angles.

📊 Graph / Figure Graph / Figure ›
A P B C Q R D 50° y x 127°
Fig. 6.26
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  18 steps
  1. Given
  2. \[\small AB \parallel CD \] \[\small \angle APQ = 50^\circ \] \[\small \angle PRD = 127^\circ \]
  3. Since \(\small AB \parallel CD\), and \(\small PQ\) is a transversal, alternate interior angles are equal.
  4. Therefore, \[\small \angle APQ = \angle PQR \]
  5. Since \(\small \angle APQ = 50^\circ\),
  6. \[\small \angle PQR = 50^\circ\]
  7. But \[\small \angle PQR = x\]
  8. Therefore, \[\small x = 50^\circ \]
  9. Now line \(\small CD\) is a straight line, therefore \(\small \angle QRP\) and \(\small\angle PRD\) form a linear pair.
  10. Hence, \[\small \angle QRP + \angle PRD = 180^\circ \]
  11. Substituting \(\small \angle PRD = 127^\circ\), \[\small \angle QRP + 127^\circ = 180^\circ \]
  12. Subtracting \(127^\circ\) from both sides:
  13. \[\small \begin{aligned} \angle QRP &= 180^\circ - 127^\circ\\ \angle QRP &= 53^\circ \end{aligned} \]
  14. In triangle \(\small PQR\), the sum of all interior angles is \(\small 180^\circ\).
  15. Therefore, \[\small \angle PQR + \angle QRP + \angle RPQ = 180^\circ \]
  16. Substituting known values: \[\small 50^\circ + 53^\circ + y = 180^\circ \]
  17. Adding known angles: \[\small 103^\circ + y = 180^\circ \]
  18. Subtracting \(103^\circ\) from both sides: \[ \begin{aligned} y &= 180^\circ - 103^\circ\\ y &= 77^\circ \end{aligned} \]
💡 Answer Final Answer ›
Final Answer: \[ \boxed{x = 50^\circ} \] \[ \boxed{y = 77^\circ} \]
🎯 Exam Significance Exam Significance ›
  • This question combines three important geometry concepts: alternate interior angles, linear pair, and triangle angle sum property.
  • Frequently asked in CBSE Board examinations as competency-based and reasoning-type questions.
  • Helps students understand how multiple angle properties work together in one figure.
  • Important for NTSE, Olympiads, and foundation-level engineering entrance preparation.
  • Strengthens geometrical visualization and step-by-step logical deduction skills.
← Q3
4 / 5  ·  80%
Q5 →
Q5
NUMERIC3 marks
In Fig. 6.27, \(\small PQ\) and \(\small RS\) are two mirrors placed parallel to each other. An incident ray \(\small AB\) strikes the mirror \(\small PQ\) at \(\small B\), the reflected ray moves along the path \(\small BC\) and strikes the mirror \(\small RS\) at \(\small C\) and again reflects back along \(\small CD\). Prove that: \[\small AB \parallel CD \]
📘 Concept & Theory Concept Used ›
  • Law of reflection:
    • Angle of incidence = Angle of reflection
  • If two lines are perpendicular to the same line, then they are parallel to each other.
  • If alternate interior angles are equal, then the two lines are parallel.
  • Reflection problems combine concepts of geometry and light rays from physics.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Draw perpendiculars from the reflection points to the mirrors.

  2. Use the law of reflection at both mirrors.

  3. Show that corresponding angles formed with transversal \(\small BC\) are equal.

  4. Prove that alternate interior angles are equal.

  5. Conclude that: \[\small AB \parallel CD \]

📊 Graph / Figure Graph / Figure ›
P B E F Q R C S A D
Fig. 6.27
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  20 steps
  1. Given \[\small PQ \parallel RS \]

    Ray \(\small AB\) is reflected from point \(\small B\).

    Ray \(\small BC\) is reflected from point \(\small C\).

  2. To Prove \[\small AB \parallel CD\]
  3. Conctruction
    Draw line \(\small BE\) through point \(\small B\) such that: \[\small BE \perp PQ\]
  4. Draw line \(CF\) through point \(C\) such that: \[\small CF \perp RS\]
  5. Proof
  6. By the law of reflection, angle of incidence equals angle of reflection.
  7. Therefore, \[\small \angle ABE = \angle EBC \tag{1}\]
  8. Similarly, at point \(\small C\),
  9. \[\small \angle BCF = \angle FCD \tag{2}\]
  10. Since:\[\small BE \perp PQ \] and \[\small CF \perp RS \] also, \[\small PQ \parallel RS \] therefore perpendiculars drawn to parallel lines are parallel.
  11. Hence, \[\small BE \parallel CF \]
  12. Now \(\small BC\) acts as a transversal.
  13. Therefore alternate interior angles are equal. \[\small \angle EBC = \angle BCF \tag{3} \]
  14. From Equations (1) and (3):
  15. \[\small \angle ABE = \angle BCF \]
  16. Adding equal angles on both sides: \[\small \angle ABE + \angle EBC = \angle BCF + \angle FCD \]
  17. Therefore, \[\small \angle ABC = \angle BCD \tag{4} \]
  18. Now \(\small BC\) is a transversal and alternate interior angles are equal.
  19. Therefore, \[\small AB \parallel CD \]
  20. Hence Proved
🎯 Exam Significance Exam Significance ›
  • This is a very important proof-based geometry question combining mathematics and physics concepts.
  • Frequently asked in CBSE Board examinations as a reasoning and proof-oriented question.
  • Strengthens understanding of:
    • Law of reflection
    • Parallel lines
    • Alternate interior angles
    • Geometrical proof writing
  • Helpful for Olympiads, NTSE, and foundation-level competitive examinations.
  • Develops analytical thinking and structured proof presentation skills.
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NCERT Class 9 Maths Exercise 6.2 Solutions
NCERT Class 9 Maths Exercise 6.2 Solutions — Complete Notes & Solutions · academia-aeternum.com
Geometry is one of the most fascinating branches of Mathematics, and Chapter 6 “Lines and Angles” marks an important step in exploring it. In this chapter, students learn about the fundamental concepts of points, lines, rays, and angles — the very foundation upon which the entire structure of geometry is built. Through clear definitions, diagrams, and logical reasoning, the chapter explains types of angles such as acute, obtuse, right, straight, and reflex angles, along with their…
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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