Question

write true or false. (a) hcf of two no.'s is always a factor of their lcm. (b) every even integer is of the form 2, where m is an integer. (c)every odd integer is of the form 2m-1,where m is an integer. (d)there can be two no. having 12 as their hcf and 350 as their lcm​

Answer 1

Answer:

Question 1.

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

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 1

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.

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

Given: ∠a ≅ ∠b

To prove: line l| line m

Solution:

Proof:

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 3

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.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 4

Given: ∠a ≅ ∠b and ∠x ≅ ∠y

To prove: line l | line n

Solution:

Proof:

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 5

∠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,

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 6

∴ 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.)

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 7

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.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 8

∴ ∠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.

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

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 = 12∠EAB [Ray AF bisects ∠EAB]

∴ 2∠FAB = ∠EAB …..(ii)

∠CBA = 12∠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.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 10

∴ 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.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 11

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:

Maharashtra Board Class 9 Maths Solutions Chapter 2 Parallel Lines Practice Set 2.2 12

Proof:

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

∠RPQ = ∠SQP ….(i) [Alternate angles]

∠RPQ = 12∠BPQ …. (ii) [Ray PR bisects ∠BPQ]

∠SQP = 12∠PQC [Ray QS bisects ∠PQC]

∴ 12∠BPQ = 12∠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]

Step-by-step explanation:

Answer 2

Step-by-step explanation:

1. True
2. False
3. True
4. False

All attached answers are here