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Maths Ace Class 7 Solutions Chapter 9 Lines and Angles
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Maths Ace Prime Class 7 Math Book, Chapter 9, Lines and Angles. Here students can easily find step by step solutions of all the problems for Lines and Angles, Exercise 9.1 and 9.2 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. Here in this post students will get chapter 9 solutions.
Lines and Angles Exercise 9.1 Solution :
Question number – (4)
Solution :
= ∠APC, ∠CPD, ∠CPD and ∠DPB, ∠APD and ∠DPB ∠APC and ∠CPB
Question number – (5)
Solution :
Linear pairs are
= ∠4, ∠5, and ∠1, ∠5
Vertically opposite angles are
= ∠1, ∠4, ∠5, ∠2, ∠3
Question number – (6)
Solution :
(a) Is ∠AOB adjacent to angle ∠BOC
= Yes
(b) Is angle ∠AOC adjacent to angle ∠AOE?
= Yes
(c) Do angle ∠COE and ∠EOF form a linear pair?
= Yes
(d) Are angle ∠BOD and angle ∠BOA supplementary?
= Yes
(e) Is angle ∠COD vertically opposite to angle ∠AOF?
= Yes
(f) What is the vertically opposite angle of angle AOC?
= ∠DOF
Question number – (7)
Solution :
(a) x = 32°
(b) x = 180° – 43
= 137°
(c) 6x + 19 = 180° – 83
or, + 19 = 180 – 137
or, 6x + 19 = 97
or, 6x = 97 – 19 = 78
or, x = 78/6 = 13°
(d) 3x + 2 = 62
or, 3x = 62 – 2 = 60
or, x = 60/3
= 20°
Question number – (8)
Solution :
Y = 30°
X = 180° – 30°
= 150°
Z = 180° – (90 + 30°)
= 180° – 120°
= 60°
Lines and Angles Exercise 9.2 Solution :
Question number – (1)
Solution :
(a) any one pair of corresponding angles.
= ∠1 and ∠3, ∠2 and ∠4, ∠8 and ∠6, ∠7 and ∠5
(b) any one pair of alternate interior angles.
= ∠2 and ∠6, ∠3 and ∠7
(c) any one pair of interior angles on the same side of the transversal
= ∠2 and ∠3, ∠7 and ∠6
(d) any two pairs of vertically opposite angles
= ∠1 and ∠7, ∠2 and ∠8, ∠3 and ∠5, ∠4 and ∠6
Question number – ( 2)
Solution :
= ∠BCF = 45° = ∠D [∴ AB || CD]
∠C = 180° – 45° = 135° = ∠C
∴ ∠D = ∠A = ∠F = 45°
∠C = ∠C = ∠B = ∠G
= 135°
Question number – (3)
Solution :
= ∠AGH = q°
= ∠CHF
∴ x = 180° – ∠CHF
= 180° – q°
Question number – (4)
Solution :
= ∠1 = 75°
= x = 180° – 75°
= 105°
Question number – (5)
Solution :
(a) 17x + 4 = 12 + 15x
or, 17x – 15x
= 12 – 4 = 8
or, 2x = 8
or, x = 8/2
= 4°
(b) x = 100° [∵ a 11b b11l]
Question number – (6)
Solution :
(a) 4z + 6 = 106
or, 4Z = 106 – 6 = 100
or, Z = 100/4 = 25°
= 2y = 180 – 106
or, y = 74/2 = 37°
= x = 37°
(b) 2x + x + 90 = 180°
or, 3x + 90 = 180
or, 3x = 180 – 90 = 90
or, x = 90/3 = 30°
= 2y = 30°
or, y = 30/2 = 15
= 180° – 15°
= 165°
Question number – (7)
Solution :
(a) ∠1 = 180° – 57°
= 123° = ∠7
∴ similarly ∠4 = 57°
∴ l || m
(b) No, the condition is not satisfied the figure.
Next Chapter Solution :
👉 Chapter 1 👈
