## Maharashtra State Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2

Question 1.

In the given figure, y = 108° and x = 71°. Are the lines m and n parallel? Justify?

Solution:

y = 108°, x = 71° …[Given]

x + y = 71° + 108°

= 179°

∴ x + y = 180°

∴ The angles x andy are not supplementary.

∴ The angles do not satisfy the interior angles test for parallel lines

∴ line m and line n are not parallel lines.

Question 2.

In the given figure, if ∠a = ∠b then prove that line l || line m.

Given: ∠a ≅ ∠b

To prove: line l| line m

Solution:

Proof:

consider ∠c as shown in the figure ∠a ≅ ∠c …….. (i) [Vertically opposite angles]

But, ∠a ≅ ∠b I (ii) [Given]

∴ ∠b ≅ ∠c [From (i) and (ii)]

But, ∠b and ∠c are corresponding angles on lines l and m when line n is the transversal.

∴ line l || line m. [Corresponding angles test]

Question 3.

In the given figure, if ∠a ≅ ∠b and ∠x ≅ ∠y, then prove that line l | line n.

Given: ∠a ≅ ∠b and ∠x ≅ ∠y

To prove: line l | line n

Solution:

Proof:

∠a = ∠b [Given]

But, ∠a and ∠b are corresponding angles on lines l and m when line k is the transversal.

∴ line l || line m ….(i) [Corresponding angles test]

∠x ≅ ∠y [Given]

But, ∠x and ∠y are alternate angles on lines m and n when seg PQ is the transversal,

∴ line m || line n ……(ii) [Alternate angles test]

∴ From (i) and (ii),

line l || line m || line n

i.e., line l || line n

Question 4.

In the given figure, if ray BA || ray DE, ∠C = 50° and ∠D = 100°. Find the measure of ∠ABC.

(Hint: Draw a line passing through point C and parallel to line AB.)

Solution:

Draw a line FG passing through point

C and parallel to line AB

line FG || ray BA …….(i) [Construction]

Ray BA || ray DE ….(ii) [Given]

line FG || ray BA || ray DE …(iii) [From (i) and (ii)]

line FG||rayDE [From (iii)]

and seg DC is their transvensal

∴ ∠ DCF = ∠ EDC [Alternate angles]

∴ ∠ DCF = 100° [∵ ∠D = 100°]

Now, ∠ DCF = ∠ BCF + ∠ BCD [Angle addition property]

∴ 100° = ∠BCF + 50°

∴ 100° – 50° = ∠BCF

∴ ∠BCF = 50° ….(iv)

Now, line FG || ray BA and seg BC is their transversal.

∴ ∠ABC + ∠BCF = 180° [Interior angles]

∴ ∠ABC + 50° = 180° [From (iv)]

∴ ∠ABC = 180°- 50°

∴ ∠ABC = 130°

Question 5.

In the given figure, ray AE || ray BD, ray AF is the bisector of ∠EAB and ray BC is the bisector of ∠ABD. Prove that line AF || line BC.

Given: Ray AE || ray BD, and

ray AF and ray BC are the bisectors of ∠EAB and ∠ABD respectively.

To prove: line AF || line BC

Solution:

Proof:

Ray AE || ray BD and seg AB is their transversal.

∴ ∠EAB = ∠ABD ….(i) [Alternate angles]

∠FAB = \(\frac { 1 }{ 2 }\)∠EAB [Ray AF bisects ∠EAB]

∴ 2∠FAB = ∠EAB …..(ii)

∠CBA = \(\frac { 1 }{ 2 }\)∠ABD [Ray BC bisects ∠ABD]

∴ 2∠CBA = ∠ABD …(iii)

∴ 2∠FAB = 2∠CBA [From (i), (ii) and (iii)]

∴ ∠FAB = ∠CBA

But, ∠FAB and ∠ABC are alternate angles on lines AF and BC when seg AB is the transversal.

∴ line AF || line BC [Alternate angles test]

Question 6.

A transversal EF of line AB and line CD intersects the lines at points P and Q respectively. Ray PR and ray QS are parallel and bisectors of ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.

Given: Ray PR || ray QS

Ray PR and ray QS are the bisectors of ∠BPQ and ∠PQC respectively.

To prove: line AB || line CD

Solution:

Proof:

Ray PR || ray QS and seg PQ is their transversal.

∠RPQ = ∠SQP ….(i) [Alternate angles]

∠RPQ = \(\frac { 1 }{ 2 }\)∠BPQ …. (ii) [Ray PR bisects ∠BPQ]

∠SQP = \(\frac { 1 }{ 2 }\)∠PQC [Ray QS bisects ∠PQC]

∴ \(\frac { 1 }{ 2 }\)∠BPQ = \(\frac { 1 }{ 2 }\)∠PQC

∴ ∠BPQ = ∠PQC

But, ∠BPQ and ∠PQC are alternate angles on lines AB and CD when line EF is the transversal.

∴ line AB || line CD [Alternate angles test]

**Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.2 Intext Questions and Activities**

Question 1.

In the given figure, how will you decide whether line ¡ and line m are parallel or not? (Textbook pg. no. 19)

Answer:

In the figure, we observe that line I and line m are coplanar and do not intersect each other.

∴ Line l and line m are parallel lines.