Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 2 Set 2.1 Parallel Lines Solutions

Question 1. In the given figure, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाओं RP और MS को दर्शाता है, जिन्हें एक तिर्यक रेखा DK काट रही है। बिंदु H रेखा RP पर है और बिंदु G रेखा MS पर है, जहां तिर्यक रेखा इन समांतर रेखाओं को काटती है। कोण DHP 85 डिग्री दर्शाया गया है।
(i) ∠RHD
(ii) ∠PHG
(iii) ∠HGS
(iv) ∠MGK
Answer:
Solution:
i. ∠DHP = 85° .....(i)
∠DHP + ∠RHD = 180° [Angles in a linear pair]
85° + ∠RHD = 180°
∴ ∠RHD = 180°- 85°
∴ ∠RHD = 95° .....(ii)
ii. ∠PHG = ∠RHD [Vertically opposite angles]
∴ ∠PHG = 95° [From (ii)]
iii. line RP || line MS and line DK is their transversal. [Corresponding angles]
∴ ∠HGS = ∠DHP .....(iii) [From (i)]
iv. ∠HGS = 85° [Vertically opposite angles]
∴ ∠MGK = ∠HGS
∴ ∠MGK = 85° [From (iii)] In simple words: We used properties of angles on a straight line (linear pair), vertically opposite angles, and corresponding angles formed by parallel lines and a transversal to find the measures of all requested angles, starting from the given angle DHP.

🎯 Exam Tip: Remember to clearly state the geometric reason for each step (e.g., linear pair, vertically opposite, corresponding angles) to score full marks.

Question 2. In the given figure line p line q and line l and line m are tranversals. Measures of some angles are shown. Hence find the measures of ∠a, ∠b, ∠c, ∠d.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाओं p और q को दर्शाता है, जिन्हें दो तिर्यक रेखाएँ l और m काट रही हैं। रेखा l पर एक कोण 110 डिग्री और रेखा m पर एक कोण 115 डिग्री दर्शाया गया है। हमें कोण a, b, c, और d के मान ज्ञात करने हैं।
Answer:
Solution:
i. 110° + ∠a = 180° [Angles in a linear pair]
∴ ∠a = 180° – 110°
∴ ∠a = 70°
ii. consider ∠e as shown in the figure
line p || line q, and line l is their transversal.
∠e + 110° = 180° [Interior angles]
∴ ∠e = 180° – 110°
∴ ∠e = 70°
But, ∠b = ∠e [Vertically opposite angles]
∴ ∠b = 70°
iii. line p || line q, and line m is their transversal.
∴ ∠c = 115° [Corresponding angles]
iv. 115° + ∠d = 180° [Angles in a linear pair]
∴ ∠d = 180° – 115°
∴ ∠d = 65° In simple words: We used properties of linear pairs, interior angles, vertically opposite angles, and corresponding angles formed by parallel lines and transversals to find the unknown angles a, b, c, and d.

🎯 Exam Tip: Always clearly identify which pair of parallel lines and which transversal you are referring to when applying angle properties, especially in complex diagrams.

Question 3. In the given figure, line l || line m and line n || line p. Find ∠a, ∠b, ∠c from the given measure of an angle.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में, दो समांतर रेखाएँ l और m हैं जिन्हें एक तिर्यक रेखा n काट रही है, और दो अन्य समांतर रेखाएँ n और p हैं जिन्हें एक तिर्यक रेखा l काट रही है। एक कोण 45 डिग्री दर्शाया गया है। हमें कोण a, b, और c के मान ज्ञात करने हैं।
Answer:
Solution:
i. consider ∠d as shown in the figure
line l || line m, and line p is their transversal.
∴ ∠d = 45° [Corresponding angles]
Now, ∠d + ∠b = 180° [Angles in a linear pair]
∴ 45° + ∠b = 180°
∴ ∠b = 180°- 45°
∴ ∠b = 135° .....(i)
ii. ∠a = ∠b [Vertically opposite angles]
∴ ∠a = 135° [From (i)]
iii. line n || line p, and line m is their transversal.
∴ ∠c = ∠b [Corresponding angles]
∴ ∠c = 135° [From (i)] In simple words: By identifying parallel lines and transversals, we applied the properties of corresponding angles and angles in a linear pair to determine the unknown angles a, b, and c based on the given 45-degree angle.

🎯 Exam Tip: When multiple pairs of parallel lines and transversals are present, carefully trace the specific lines involved for each angle relationship to avoid errors.

Question 4. In the given figure, sides of ∠PQR and ∠XYZ are parallel to each other. Prove that, ∠PQR = ∠XYZ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो कोणों ∠PQR और ∠XYZ को दर्शाता है, जहाँ कोण PQR की भुजाएँ (QP और QR) और कोण XYZ की भुजाएँ (YX और YZ) एक-दूसरे के समांतर हैं। हमें यह सिद्ध करना है कि ∠PQR = ∠XYZ।
Answer:
Given: Ray YZ || ray QR and ray YX || ray QP
To prove: ∠PQR = ∠XYZ
Construction: Extend ray YZ in the opposite direction. It intersects ray QP at point S.
Solution:
Proof:
Ray YX || ray QP [Given]
Ray YX || ray SP and seg SY is their transversal [P-S-Q]
∴ ∠XYZ = ∠PSY ......(i) [Corresponding angles]
ray YZ || ray QR [Given]
ray SZ || ray QR and seg PQ is their transversal. [S-Y-Z]
∴ ∠PSY ≈ ∠SQR [Corresponding angles]
∴ ∠PSY ≃ ∠PQR ......(ii) [P-S-Q]
∴ ∠PQR = ∠XYZ [From (i) and (ii)] In simple words: We proved the equality of two angles whose sides are parallel by extending one ray and using the property of corresponding angles twice. This establishes a transitive relationship showing that if two angles have parallel sides, they are equal.

🎯 Exam Tip: For proof-based questions, clearly state Given, To Prove, Construction (if any), and provide reasons for each step in the proof for logical flow and clarity.

Question 5. In the given figure, line AB || line CD and line PQ is transversal. Measure of one of the angles is given. Hence find the measures of the following angles.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाओं AB और CD को दर्शाता है, जिन्हें एक तिर्यक रेखा PQ काट रही है। कोण BRT का मान 105 डिग्री दर्शाया गया है। हमें निम्नलिखित कोणों के माप ज्ञात करने हैं।
(i) ∠ART
(ii) ∠CTQ
(iii) ∠DTQ
(iv) ∠PRB
Answer:
Solution:
i. ∠BRT = 105° ....(i)
∠ART + ∠BRT = 180° [Angles in a linear pair]
∴ ∠ART + 105° = 180°
∴ ∠ART = 180°- 105°
∴ ∠ART = 75° ...(ii)
ii. line AB || line CD and line PQ is their transversal.
∴ ∠CTQ = ∠ART [Corresponding angles]
∴ ∠CTQ = 75° [From (ii)]
iii. line AB || line CD and line PQ is their transversal.
∴ ∠DTQ = ∠BRT [Corresponding angles]
∴ ∠DTQ = 105° [From (i)]
iv. ∠PRB = ∠ART [Vertically opposite angles]
∴ ∠PRB = 75° [From (ii)] In simple words: By using the properties of angles in a linear pair, corresponding angles, and vertically opposite angles, we determined the measures of ∠ART, ∠CTQ, ∠DTQ, and ∠PRB based on the given angle ∠BRT = 105°.

🎯 Exam Tip: When given one angle in a parallel line-transversal setup, systematically apply all related angle properties (linear pairs, vertically opposite, corresponding, alternate interior/exterior, interior angles on same side) to find all other angles.

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.1 Intext Questions and Activities

Question 1. Angles formed by two lines and their transversal. (Textbook pg. no. 13) When a transversal (line n) intersects two lines (line l and m) in two distinct points, 8 angles are formed as shown in the figure. Pairs of angles formed out of these angles are as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो रेखाओं l और m को दर्शाता है जिन्हें एक तिर्यक रेखा n काट रही है। इस प्रतिच्छेदन से 8 कोण बनते हैं, जिन्हें d, a, c, b, h, e, g, f के रूप में दर्शाया गया है।
Answer:
Pairs of corresponding angles
i. ∠d, ∠h
ii. ∠a, ∠e
iii. ∠c, ∠g
iv. ∠b, ∠f
Pairs of alternate interior angles
i. ∠c, ∠e
ii. ∠b, ∠h
Pairs of alternate exterior angles
i. ∠d, ∠f
ii. ∠a, ∠g
Pairs of interior angles on the same side of the transversal
i. ∠c, ∠h
ii. ∠b, ∠e
Some important properties:
1. When two lines intersect, the pairs of vertically opposite angles formed are congruent.
Example:
In the given diagram,
line l and m intersect at point P.
The pairs of vertically opposite angles that are congruent are:
i. ∠a ≈ ∠b
ii. ∠c ≃ ∠d
2. The angles in a linear pair are supplementary.
Example:
For the given diagram,
∠a and ∠c are in linear pair
∴ ∠a + ∠c = 180°
Also, ∠d and ∠b are in linear pair
∴ ∠d + ∠b = 180°
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो प्रतिच्छेदी रेखाओं l और m को दर्शाता है जो बिंदु P पर एक दूसरे को काटती हैं। कोणों को a, b, c, d के रूप में दर्शाया गया है।
3. When one pair of corresponding angles is congruent, then all the remaining pairs of corresponding angles are congruent.
Example:
In the given diagram,
If ∠a ≈ ∠b
then ∠e = ∠f, ∠c = ∠d and ∠g = ∠h
4. When one pair of alternate angles is congruent, then all the remaining pairs of alternate angles are congruent.
Example:
For the given diagram,
If ∠e = ∠d, then ∠g = ∠b
Also, ∠a ≈ ∠h and ∠c ≃ ∠f
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाओं l और m को दर्शाता है जिन्हें एक तिर्यक रेखा n काट रही है। इस प्रतिच्छेदन से 8 कोण बनते हैं, जिन्हें a, e, g, b, d, f, h के रूप में दर्शाया गया है।
5. When one pair of interior angles on one side of the transversal is supplementary, then the other pair of interior angles is also supplementary.
Example:
For the given diagram,
If ∠e + ∠b = 180°, then ∠g + ∠d = 180°. In simple words: This section defines and illustrates the different types of angle pairs formed by two lines and a transversal (corresponding, alternate interior, alternate exterior, interior on same side) and summarizes key properties related to these angles when the lines are parallel.

🎯 Exam Tip: Memorize all angle pair definitions and their properties (congruent or supplementary) for parallel lines. A clear understanding of these basics is fundamental for solving geometry problems.