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Brilliant’s Composite Mathematics Class 6 Solutions Chapter 13 Parallel Lines
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 13, Parallel Lines. Here students can easily find step by step solutions of all the problems for Parallel Lines, Exercise 13.1, 13.2, 13.3 and 13.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Parallel Lines Exercise 13.1 Solution
Question no – (1)
Solution :
Parallel line segments in Figure –
(i) ST || QR
Parallel line segments in Figure –
(ii) AD || BC , AB || DC
Parallel line segments in Figure –
(iii) PQ || BC, AC || PR , AB || RQ
Parallel line segments in Figure –
(iv) AB || CD, AB || EH , AB || GF , BE || CF , BF || GD , GH || EF ; GH || EF , GH || AD, DC || GF
Parallel line segments in Figure –
(v) AB || CD, AD || BC , AE || FC
Parallel line segments in Figure –
(vi) AB || CF, AB || ED , AF || DC, BC || EF , AE || BD
Parallel line segments in Figure –
(vii) AB || CD, AB || EF, CD || EF
Question no – (4)
Solution :
Yes, Because their corresponding lines are Parallel.
Question no – (5)
Solution :
(i) If two rays do not intersect, even when produced, then they are Parallel.
(ii) Two parallel lines are every where Equal distance apart.
(iii) Two lines are Parallel if they do not meet, even when produced.
(iv) Lines l and m are parallel, they are denoted by l || m.
(v) The symbol l ∦ m stands for Line l is not parallel to the line m.
Question no – (6)
Solution :
In the figure – (i) Parallel
In the figure – (ii) Parallel
In the figure – (iii) not parallel.
Parallel Lines Exercise 13.2 Solution
Question no – (1)
Solution :
(i) Alternate angle of ∠PQR is ∠ARQ
Alternate angle of ∠PQE is ∠QPR
(ii) Corresponding angle of ∠ PQE is ∠ARQ
Question no – (2)
Solution :
(i) Pairs of alternate exterior angles
∠b and ∠g , ∠c and ∠h
(ii) pairs of corresponding angles
∠b and ∠e , ∠a and ∠h , ∠c and ∠f ∠d and ∠g
(iii) pairs of alternate interior angles
∠a and ∠f , ∠d and ∠e
(iv) pair of interior angles on the same side of the transversal.
∠a and ∠e , ∠d and ∠f
Parallel Lines Exercise 13.3 Solution
Question no – (1)
Solution :
∠a = 120° and ∠b = 65 ° [ vertically opposite sides]
∠a = ∠c = 120° [corresponding angles]
∠b = ∠d = 65° [alternate angles]
Question no – (2)
Solution :
∠1 = 50
∴ ∠2 = 75° [alternate angles]
Question no – (3)
Solution :
According to the question,
∠1 = 75° than ∠2
= 180° – 75°
= 105°
∴ ∠1 = ∠3 = 75° [opposite angles]
∠3 = ∠5 = 75 ° [alternate angles]
∠5 = ∠7 = 75° [opposite angles]
and, ∠2 = ∠4 = 105° [opposite angles]
∠4 = ∠6 = 105° [opposite angles]
∠6 = ∠8 = 105° [opposite angles]
Question no – (4)
Solution :
No, l || m
Because, ∠1 = ∠7 [corresponding angles]
But hare ∠1 ≠ ∠7
Question no – (5)
Solution :
Yes, ∠3 = ∠7 = 65° [corresponding angles]
∴ l || m
Question no – (6)
Solution :
∠d = 70° [vertically opposite angle]
∠d = ∠c = 70 °
∠d = ∠b = 70°
∠b = ∠a = 70°
Question no – (7)
Solution :
∠x = 105° [vertically opposite]
∠x = ∠α = 105°[alternative]
∠β = 45°[corresponding]
Question no – (8)
Solution :
∠1 = ∠3 = 45° [opposite]
∠3 = ∠2 = 45° [alternative]
∠5 = 180° – 45° = 135°
∠4 = ∠5 = 135° [alternative]
Question no – (9)
Solution :
∠PEA = 110° = ∠EFC [corresponding]
∠AEF = 180° – 100°
= 70°
∠AEF = ∠EFD = 70° [alternative]
and, ∠AEF + ∠EFC = 70° + 110°
= 180°
(The sum of the interior angles on the same side of the transversal is equal to 180°)
∴ AB || CD
= ∠X = 150° [alternative angle]
Next Chapter Solution :
👉 Chapter 15 👈
