Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 2 Parallel Lines.
9th Standard Maths 2 Practice Set 2.1 Chapter 2 Parallel Lines Textbook Answers Maharashtra Board
Class 9 Maths Part 2 Practice Set 2.1 Chapter 2 Parallel Lines Questions With Answers Maharashtra Board
   Question :
   
   In the given figure, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.
   
    
   
   i. ∠RHD
   
   ii. ∠PHG
   
   iii. ∠HGS
   
   iv. ∠MGK
   
   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)]
  
   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.
   
    
   
   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 lis 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°
  
   Question 3.
   
   In the given figure, line 11| line m and line n || line p. Find ∠a, ∠b, ∠c from the given measure of an angle.
   
    
   
   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)]
  
   Question 4.
   
   In the given figure, sides of ∠PQR and ∠XYZ are parallel to each other. Prove that, ∠PQR ≅ ∠XYZ.
   
    
   
   Given: Ray YZ || ray QRandray 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)]
  
   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.
   
    
   
   i. ∠ART
   
   ii. ∠CTQ
   
   iii. ∠DTQ
   
   iv. ∠PRB
   
   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)]
  
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:
   
    
   
   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°
   
    
   
   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
   
    
  
   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°.