Looking for accurate SEBA Class 9 Maths Chapter 6 Exercise 6.2 (Lines and Angles) textual solutions for the 2026–27 academic year? Access step-by-step solutions for parallel lines, transversal properties, alternate interior angles, corresponding angles, and co-interior angle proofs designed to help you score top marks in your annual exams.
EXERCISE 6.2
1. In Fig. 6.24, find the values of x and y and then show that AB ∥ CD.
Solution 1:
From the figure, 50° + x = 180° (Linear pair)
⇒ x = 180° – 50°
⇒ x = 130°
Also, y = 130° (Vertically opposite angles)
Since x = y = 130°, the alternate interior angles are equal.
Hence, AB ∥ CD.
2. In Fig. 6.25, if AB ∥ CD, CD ∥ EF and y : z = 3 : 7, find x.
Solution 2:
Given: AB ∥ CD, CD ∥ EF, so AB ∥ EF.
Therefore, x = z (Alternate interior angles) — (i)
Also, x + y = 180° (Co-interior angles)
⇒ z + y = 180° [Using (i)]
Given ratio y : z = 3 : 7.
Let y = 3a and z = 7a.
3a + 7a = 180°
⇒ 10a = 180°
⇒ a = 18°
z = 7 × 18° = 126°
Since x = z,
x = 126°
Class 9 Maths Chapter 6 Exercise 6.2 Answers
3. In Fig. 6.26, if AB ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution 3:
Given: AB ∥ CD, EF ⊥ CD (∠FED = 90°), and ∠GED = 126°.
∠AGE = ∠GED (Alternate interior angles)
⇒ ∠AGE = 126°
∠GED = ∠GEF + ∠FED
⇒ 126° = ∠GEF + 90°
⇒ ∠GEF = 126° – 90°
⇒ ∠GEF = 36°
∠AGE + ∠FGE = 180° (Linear pair)
⇒ 126° + ∠FGE = 180°
⇒ ∠FGE = 180° – 126°
⇒ ∠FGE = 54°
4. In Fig. 6.27, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]
Solution 4:
Construction: Draw a line RU parallel to ST through point R, extending towards the left (so RU ∥ ST ∥ PQ).
Since ST ∥ RU:
∠RST + ∠SRU = 180° (Co-interior angles)
⇒ 130° + ∠SRU = 180°
⇒ ∠SRU = 50°
Since PQ ∥ RU:
∠PQR = ∠QRU (Alternate interior angles)
⇒ ∠QRU = 110°
∠QRU = ∠QRS + ∠SRU
⇒ 110° = ∠QRS + 50°
⇒ ∠QRS = 110° – 50°
⇒ ∠QRS = 60°
SEBA Class 9 Maths Chapter 6 Ex 6.2 Notes
5. In Fig. 6.28, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Solution 5:
Given: AB ∥ CD.
∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
∠APR = ∠PRD (Alternate interior angles)
⇒ ∠APQ + ∠QPR = ∠PRD
⇒ 50° + y = 127°
⇒ y = 127° – 50°
⇒ y = 77°
6. In Fig. 6.29, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB ∥ CD.
Solution 6:
Construction: Draw normals BM ⊥ PQ and CN ⊥ RS.
Since PQ ∥ RS, their perpendiculars are also parallel, so BM ∥ CN.
By Laws of Reflection (Angle of incidence = Angle of reflection):
Let ∠1 = ∠2 at point B and ∠3 = ∠4 at point C.
Since BM ∥ CN and BC is a transversal:
∠2 = ∠3 (Alternate interior angles)
Multiplying by 2:
2(∠2) = 2(∠3)
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠BCD
Since alternate interior angles are equal, AB ∥ CD.
Lines and Angles Class 9 SEBA Ex 6.2 Solutions
7. This question consists of two statements namely Assertion (A) and Reason (R).
Assertion (A) : If the side BC of a Δ ABC is produced to D, then ∠ACD = ∠A + ∠B.
Reason (R) : The sum of the angles of a triangle is 180°.
Select the answer from the following options:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Solution 7:
Assertion (A): True. (By exterior angle theorem, an exterior angle of a triangle equals the sum of its two opposite interior angles: ∠ACD = ∠A + ∠B).
Reason (R): True. (The sum of angles in a triangle is 180°).
Explanation: Reason (R) is used to prove Assertion (A) (∠A + ∠B + ∠ACB = 180° and ∠ACD + ∠ACB = 180°).
Correct Option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
8. Match Column-I with Column-II
| Column-I | Column-II |
| (A) If x° and y° be the measures of two complementary angles such that 2x = 3y, then x = | (P) 90° |
| (B) If an angle is the complement of itself, then the measure of the angle is…. | (Q) 54° |
| (C) If x° and y° be the angles forming a linear pair such that x – y = 60°, then y = …. | (R) 60° |
| (D) If an angle is the suplement of itself, then the measure of the angle is …. | (S) 45° |
Choose the correct alternative–
(d) A → R, B → S, C → Q, D → P
(a) A → Q, B → S, C → R, D → P
(b) A → Q, B → P, C → S, D → R
(c) A → P, B → Q, C → R, D → S
Solution 8:
(A) Complementary angles: x + y = 90°.
Given 2x = 3y ⇒ y = (2/3)x.
x + (2/3)x = 90° ⇒ (5/3)x = 90° ⇒ x = 54° → (Q)
(B) Angle equal to its complement: x + x = 90° ⇒ 2x = 90° ⇒ x = 45° → (S)
(C) Linear pair: x + y = 180° and x – y = 60°.
Subtracting equations: 2y = 120° ⇒ y = 60° → (R)
(D) Angle equal to its supplement: x + x = 180° ⇒ 2x = 180° ⇒ x = 90° → (P)
Matching pairs: A → Q, B → S, C → R, D → P
Correct Option: (a)
Class 9 Maths Chapter 6 Exercise 6.2 Question Answer SEBA 2026-27
📐 Updated Exercise 6.2 Solutions Notice: This page features complete, step-by-step SEBA Class 9 Maths Chapter 6 (Lines and Angles) Exercise 6.2 Solutions updated for the 2026–27 academic session. All geometry proofs, transversal line properties, and angle calculations strictly follow the latest revised Assam Board (SEBA) textbook.
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