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

Maharashtra State Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1

Question :
In the given figure, line RP || line MS and line DK is their transversal. ∠DHP = 85°. Find the measures of following angles.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 1
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.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 2
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 3
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.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 4
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 5
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.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 6
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.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 7
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:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 8
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°
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 9
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
Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.1 10

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°.

Maharashtra Board Class 9 Maths Solutions

 

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