Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 2 Parallel Lines Solutions

Question 1. Select the correct alternative and fill in the blanks in the following statements.

(i) If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is _____.
(A) 0°
(B) 90°
(C) 180°
(D) 360°
Answer: (C) 180°
In simple words: When a transversal cuts two parallel lines, the two interior angles on the same side of the transversal always add up to 180 degrees. This is a fundamental property of parallel lines.

🎯 Exam Tip: Remember the interior angles on the same side property (consecutive interior angles) for parallel lines; it's a common concept tested in geometry.

Question 1.
(ii) The number of angles formed by a transversal of two lines is _____.
(A) 2
(B) 4
(C) 8
(D) 16
Answer: (C) 8
In simple words: When a straight line (transversal) cuts across two other straight lines, it creates eight distinct angles at the points of intersection. These angles include corresponding, alternate interior, alternate exterior, and interior angles.

🎯 Exam Tip: Visualizing the intersections helps; count the angles formed at each intersection point to easily identify the total number.

Question 1.
(iii) A transversal intersects two parallel lines. If the measure of one of the angles is 40°, then the measure of its corresponding angle is _____.
(A) 40°
(B) 140°
(C) 50°
(D) 180°
Answer: (A) 40°
In simple words: Corresponding angles are angles in the same relative position at each intersection when a transversal crosses two parallel lines, and they are always equal in measure. If one is 40°, its corresponding angle is also 40°.

🎯 Exam Tip: The property of corresponding angles being equal is crucial for solving problems involving parallel lines; ensure you can identify corresponding angles correctly.

Question 1.
(iv) In ΔABC, ∠A = 76°, ∠B = 48°, then ∠C = _____.
(A) 66°
(B) 56°
(C) 124°
(D) 28°
Answer: (B) 56° In ΔABC, ∠A + ∠B + ∠C = 180°
∴ ∠C = 180° - 76° - 48° = 56°
(B) 56°
In simple words: The sum of all three interior angles in any triangle is always 180 degrees. By subtracting the given two angles from 180°, we can find the measure of the third angle.

🎯 Exam Tip: The angle sum property of a triangle (sum of angles = 180°) is fundamental and frequently used in geometry proofs and calculations.

Question 1.
(v) Two parallel lines are intersected by a transversal. If measure of one of the alternate interior angles is 75° then the measure of the other angle is _____.
(A) 105°
(B) 15°
(C) 75°
(D) 45°
Answer: (C) 75°
In simple words: When a transversal intersects two parallel lines, alternate interior angles are equal in measure. If one alternate interior angle is 75°, the other one will also be 75°.

🎯 Exam Tip: Knowing that alternate interior angles are congruent when lines are parallel is a key theorem; practice identifying these pairs.

Question 2. Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of (i) A pair of complementary angles
(ii) A pair of supplementary angles
(iii) A pair of congruent angles.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र किरण PQ और किरण PR को दर्शाता है जो एक दूसरे पर लंबवत हैं, जिससे ∠RPQ 90° का कोण बनता है। इसमें बिंदु B और A क्रमशः इस कोण के आंतरिक और बाहरी भाग में स्थित हैं। किरण PB और PA भी एक दूसरे पर लंबवत दिखाई गई हैं।
(i) Complementary angles:
∠RPQ = 90° [Ray PQ ⊥ ray PR]
∴ ∠RPB + ∠BPQ = 90° [Angle addition property]
∠RPB and ∠BPQ are pair of complementary angles
∠APB = 90° [Ray PA ⊥ ray PB]
∴ ∠APR + ∠RPB = 90°
∠APR and ∠RPB are pair of complementary angles.
(ii) Supplementary angles:
∠APB + ∠RPQ = 90° + 90° = 180°
∴ ∠APB and ∠RPQ are a pair of supplementary angles.
(iii) Congruent angles:
a. ∠APB = ∠RPQ [Each is of 90°]
b. ∠APB = ∠RPQ
∴ ∠APR + ∠RPB = ∠RPB + ∠BPQ [Angle addition property]
∴ ∠APR = ∠BPQ
∴ ∠APR ≅ ∠BPQ
In simple words: This problem involves identifying different angle relationships (complementary, supplementary, congruent) based on perpendicular rays and angle addition properties. Complementary angles add up to 90°, supplementary angles add up to 180°, and congruent angles have the same measure.

🎯 Exam Tip: Clearly understanding definitions of complementary, supplementary, and congruent angles, along with the angle addition postulate, is essential for these types of multi-part geometry questions.

Question 3. Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समानांतर रेखाओं AB और CD को दर्शाता है, जिन्हें एक तिर्यक रेखा EF बिंदु P और Q पर प्रतिच्छेद करती है। रेखा EF, रेखा AB पर लंबवत है, जैसा कि बिंदु P पर 90° का कोण दर्शाया गया है। Given: line AB || line CD and line EF intersects them at P and Q respectively. line EF ⊥ line AB To prove: line EF ⊥ line CD Solution: Proof: line EF ⊥ line AB [Given]
∴ ∠APR = 90° ....(i) line AB || line CD and line EF is their transversal.
∴ ∠EPB = ∠PQD .....(ii) [Corresponding angles]
∴ ∠PQD = 90° [From (i) and (ii)]
∴ line EF ⊥ line CD
In simple words: This proof shows that if a line crosses two parallel lines and forms a 90-degree angle with one, it must also form a 90-degree angle with the other. This is because corresponding angles between parallel lines are equal.

🎯 Exam Tip: This is a standard proof involving properties of parallel lines and transversals. Mastering the concept of corresponding angles and their equality is key.

Question 4. In the given figure, measures of some angles are shown. Using the measures find the measures of ∠x and ∠y and hence show that line l || line m.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में दो रेखाएँ l और m एक तिर्यक रेखा n द्वारा प्रतिच्छेदित की गई हैं। तिर्यक रेखा के एक ही तरफ के आंतरिक कोण 130° और 50° दर्शाए गए हैं, और साथ ही अज्ञात कोण x और y भी दिखाए गए हैं। Solution: Proof: ∠x = 130° ∠y = 50° [Vertically opposite angles] Here, m∠PQT + m∠QTS = 130° + 50° = 180° But, ∠PQT and ∠QTS are a pair of interior angles on lines l and m when line n is the transversal,
∴ line l || line m [Interior angles test]
In simple words: We find the unknown angles using properties like vertically opposite angles. Then, by showing that the sum of interior angles on the same side of the transversal is 180°, we prove that the lines are parallel.

🎯 Exam Tip: This question combines several angle properties. Remember that if the sum of consecutive interior angles is 180°, then the lines are parallel; this is the converse of the interior angles theorem.

Question 5. In the given figure, Line AB || line CD || line EF and line QP is their transversal. If y : z = 3 : 7 then find the measure of ∠x.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन समानांतर रेखाओं AB, CD और EF को दिखाता है, जिन्हें एक तिर्यक रेखा QP प्रतिच्छेद करती है। इसमें कोण x, y और z को चिह्नित किया गया है, जिनका उपयोग रेखाओं के समानांतर होने के गुणों को समझने के लिए किया जाता है। Solution: y : z = 3 : 7 [Given] Let the common multiple be m
∴ ∠y = 3m and ∠z = 7m ....(i) line AB || line EF and line PQ is their transversal [Given] ∠x = ∠z
∴ ∠x = 7m .....(ii) [From (i)] line AB || line CD and line PQ is their transversal [Given] ∠x + ∠y = 180°
∴ 7m + 3m = 180°
∴ 10m = 180°
∴ m = 18°
∴ ∠x = 7m = 7(18°) [From (ii)]
∴ ∠x = 126°
In simple words: We use the properties of parallel lines (corresponding angles and interior angles) along with the given ratio to set up equations. Solving these equations allows us to find the common multiple and subsequently the measure of angle x.

🎯 Exam Tip: When multiple parallel lines are involved, apply the angle properties (like corresponding or interior angles) step-by-step for each pair of parallel lines with the transversal. Ratios often require using a common multiple.

Question 6. In the given figure, if line q || line r, line p is their transversal and if a = 80°, find the values of f and g.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में दो समानांतर रेखाएँ q और r एक तिर्यक रेखा p द्वारा प्रतिच्छेदित की गई हैं। विभिन्न कोणों, जैसे a, b, c, d, e, f, g, h को चिह्नित किया गया है, जो समानांतर रेखाओं और तिर्यक रेखा के गुणों का प्रदर्शन करते हैं। Solution: i. ∠a = 80° [Given] ∠g = ∠a [Alternate exterior angles]
∴ ∠g = 80° ....(i) ii. Now, line q || line r and line p is their transversal.
∴ ∠f + ∠g = 180°
∴ ∠f + 80° = 180° [Interior angles]
∴ ∠f = 180° - 80° [From (i)]
∴ ∠f = 100°
In simple words: Given parallel lines and a transversal, we use the property of alternate exterior angles to find angle g, and then the property of consecutive interior angles (which sum to 180°) to find angle f.

🎯 Exam Tip: Accurately identifying the relationship between angle pairs (alternate exterior, interior, corresponding, vertically opposite) is crucial. Always state the reason for each step in your solution.

Question 7. In the given figure, if line AB || line CF and line BC || line ED then prove that ∠ABC = ∠FDE.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समानांतर रेखाओं AB और CF को दिखाता है, और साथ ही एक और जोड़ी समानांतर रेखाएँ BC और ED को भी दर्शाता है। इसमें कोण ABC और कोण FDE को दर्शाया गया है, जिनका उपयोग रेखाओं के समानांतर होने के गुणों को सिद्ध करने में किया जाता है। Given: line AB || line CF and line BC || line ED To prove: ∠ABC = ∠FDE Solution: Proof: line AB || line PF and line BC is their transversal.
∴ ∠ABC = ∠BCD ....(i) [Alternate angles] line BC || line ED and line CD is their transversal.
∴ ∠BCD = ∠FDE ....(ii) [Corresponding angles]
∴ ∠ABC = ∠FDE [From (i) and (ii)]
In simple words: To prove ∠ABC = ∠FDE, we use a chain of angle relationships. First, since AB || CF, alternate interior angles ∠ABC and ∠BCD are equal. Then, since BC || ED, corresponding angles ∠BCD and ∠FDE are equal. By transitivity, ∠ABC equals ∠FDE.

🎯 Exam Tip: For proofs involving multiple pairs of parallel lines, break down the problem into steps, using one pair of parallel lines and its transversal at a time. Connect the relationships to form a logical chain.

Question 8. In the given figure, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that □ QXRY is a rectangle.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में समानांतर रेखाएँ AB और CD एक तिर्यक रेखा PS द्वारा प्रतिच्छेदित की गई हैं। कोण AQR, QRC, BQR और QRD के समद्विभाजक (angle bisectors) QX, RX, QY, RY खींचे गए हैं, जो चतुर्भुज QXRY बनाते हैं। Given: line AB || line CD Rays QX, RX, QY, RY are the bisectors of ∠AQR, ∠QRC, ∠BQR and ∠QRD respectively. To prove: □QXRY is a rectangle. Proof: ∠XQA = ∠XQR = x° ......(i) [Ray QX bisects ∠AQR] ∠YQR = ∠YQB =y° ......(ii) [Ray QY bisects ∠BQR] ∠XRQ = ∠XRC = u° ........ (iii) [Ray RX bisects ∠CRQ] ∠YRQ = ∠YRD = v° ......(iv) [Ray RY bisects ∠DRQ] line AB || line CD and line PS is their transversal.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समानांतर रेखाओं AB और CD को दिखाता है जो तिर्यक रेखा PS द्वारा काटी जाती हैं। इसमें कोण AQR, QRC, BQR और QRD के कोण समद्विभाजकों QX, RX, QY, RY को दिखाया गया है, और साथ ही विभिन्न कोणों x, u, y, v को भी चिह्नित किया गया है। ∠AQR+ ∠CRQ= 180° [Interior angles] (∠XQA + ∠XQR) + (∠XRQ + ∠XRC) = 180° [Angle addition property]
∴ (x + x) + (u + u) = 180° [From (i) and (ii)]
∴ 2x + 2u = 180°
∴ 2(x + u) = 180°
∴ x + u = 90° ........(v) In ΔXQR ∠XQR + ∠XRQ + ∠QXR = 180° [Sum of the measures of the angles of triangle is 180°]
∴ x + u + ∠QXR = 180° [From (i) and (iii)]
∴ 90 + ∠QXR = 180° [From (v)]
∴ ∠QXR = 180°- 90°
∴ ∠QXR = 90° .....(vi) Similarly we can prove that,
∴ y + v = 90° Hence ∠QYR = 90° ......(vii) Now, ∠AQR + ∠BQR = 180° [Angles is linear pair] (∠XQA + ∠XQR) + (∠YQR + ∠YQB) = 180° [Angle addition property]
∴ (x + x) + (y + y) = 180° [From (i) and (ii)]
∴ 2x + 2y = 180°
∴ 2(x+y) = 180°
∴ x +y = 90° i.e. ∠XQR + ∠YQR = 90° [From (i) and (ii)]
∴ ∠XQY = 90° ....(viii) [Angle addition property] Similarly we can prove that, ∠XRY = 90° ...(ix) In □QXRY ∠QXR = ∠QYR = ∠XQY = ∠XRY = 90° [From (vi), (vii), (viii) and (ix)]
∴ □ QXRY is a rectangle.
In simple words: This proof shows that when angle bisectors of angles formed by a transversal intersecting parallel lines meet, they form a rectangle. The key steps involve proving that the interior angles formed by these bisectors are 90 degrees, using angle sum properties of triangles and parallel lines.

🎯 Exam Tip: This is a challenging proof often requiring multiple steps. Focus on using properties of parallel lines (interior angles sum to 180°), angle bisectors (divide angles into two equal halves), and the angle sum property of triangles (sum of angles = 180°) to show that all angles of the quadrilateral are 90 degrees.

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Problem Set 2 Intext Questions and Activities

Question 1. To verify the properties of angles formed by a transversal of two parallel lines. (Textbook pg. no. 14) Take a piece of thick coloured paper. Draw a pair of parallel lines and a transversal on it. Paste straight sticks on the lines. Eight angles will be formed. Cut pieces of coloured paper, as shown in the figure, which will just fit at the corners of ∠1 and ∠2. Place the pieces near different pairs of corresponding angles, alternate angles and interior angles and verify their properties.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र समानांतर रेखाओं और एक तिर्यक रेखा द्वारा बनने वाले कोणों के गुणों को समझने के लिए एक गतिविधि को दर्शाता है। इसमें विभिन्न कोणों को चिह्नित किया गया है, जैसे 1 और 2, और कुछ कोणों को छायांकित किया गया है, जिससे छात्र संगत कोणों, एकांतर कोणों और आंतरिक कोणों के गुणों का सत्यापन कर सकें।
In simple words: This activity provides a hands-on way to visually confirm the properties of angles formed when a transversal intersects parallel lines, such as corresponding angles being equal or interior angles being supplementary, by using physical cut-outs.

🎯 Exam Tip: Practical activities like this reinforce theoretical concepts. Understanding the properties through visual verification can help you remember and apply them more effectively in problem-solving.