NCERT Solutions For Class 9 Maths Chapter 6 Lines And Angles (Ex 6.1)

Exercise 6.1 of Class 9 Maths Chapter 6 Lines and Angles is about pairs of angles. When two lines cross each other, or when a ray stands on a line, the angles formed are linked to each other by fixed rules. Two rules do most of the work in this exercise. First, when a ray stands on a straight line, the two angles on one side of it always add up to 180 degrees. Second, when two lines cross, the angles opposite each other are always equal. Once these are clear, finding a missing angle becomes a two-line job.

This exercise has 6 questions. Some ask you to find a missing angle and some ask you to prove a statement. Every question is solved here in easy steps, with the reason written next to each step, which is how CBSE gives marks. Students can revise on their own, teachers can use the same steps in class, and parents can check homework at home. A free printable PDF of the full exercise is also available.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles: All Exercises

NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles: Exercise 6.1

 

Q.1 In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠ BOD = 40°, find ∠BOE and reflex ∠COE.


Ans

Given: Lines AB and CE itersect at O and∠AOC+∠BOE=70° and ∠BOD=40°To find: ∠BOE and reflex ∠COE.∠AOB is a straight angle, so              ∠AOB=180°∠AOC+∠COE+∠BOE=180°   70°+∠COE=180°⇒    ∠COE=180°−70°=110°∠COD is a straight angle, so              ∠COD=180°∠COE+∠EOB+∠BOD=180°         110°+∠BOE+40°=180°      ∠BOE=180°−150° =30°          Reflex ∠COE=360°−∠COE           =360°−110°           =250°

Q.2 In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.


Ans

Given:​ Lines XY and MN intersect at O. ∠POY=90° and a:b=2:3To find: cSince, a:b = 2:3 so, let a=2x, b=3x              ∠POX+∠POY=180°                      a+b+90°=180°        2x+3x+90°=180°5x=180°   x=180°5=30°   b=2x       =2×30°       =60°MN is a straight line, so              ∠MOX+∠XON=180°  60°+c=180°      c=180°−60°       =120°

Q.3 In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠PQS = ∠PRT.


Ans

Given: In ΔPQR,  ∠PQR=∠PRQTo Prove: ∠PQS=∠PRTProof:​ ∠PQR+∠PQS=180°[Linear pair of angles]         ∠PQR=180°−∠PQS   ...(i)                ∠PRQ+∠PRT=180°[Linear pair of angles]          ∠PRQ=180°−∠PRT   ...(ii)But    ∠PQR=∠PRQ[Given]⇒180°−∠PQS=180°−∠PRT⇒    ∠PQS=∠PRTHence Proved.

Q.4 In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.


Ans

Given: x+y=w+zTo Prove: AOB is a line.Proof:   ∵x+y+w+z=360°             [Complete​​ angle]x+y+x+y=360°            [∵x+y=w+z]        2(x+y)=360°     x+y=360°2      =180°   ∠AOB =180°⇒Since,∠AOB is a straight angle, so AOB is a straight line.

Q.5 In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ.
OS is another ray lying between rays OP and OR. Prove that ∠ROS = (1/2)(∠QOS – ∠POS).


Ans

Given:​ POQ is a line. OR→⊥PQ and OS→ is another ray between       OR→​ and OP→.To Prove: ∠ROS = 12(∠QOS−∠POS)Proof:R.H.S.= 12(∠QOS−∠POS)         =12{(∠QOR+∠ROS)−(∠ROP−∠ROS)}         =12{(90°+∠ROS)−(90°−∠ROS)}      [∠QOR=∠ROP          =90°]        =12(∠ROS+∠ROS)        =12(2∠ROS)         =∠ROS

Q.6 It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information.
If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.


Ans

Given: ∠XYZ=64°, XY is produced to point P.Ray YQ bisects ∠ZYP.To find: ∠XYQ and reflex ∠QYP.​ Since, YQ→ bisects∠ZYP, so ∠ZYQ=∠PYQ=a(let) ∠XYZ+∠ZYP=180° [Linear pair of angles.]            64°+2a=180°   2a=180°−64° =116°      a=116°2 =58°∴       ∠XYQ=64°+a =64°+58° =122°            reflex ∠QYP=360°−∠QYP =360°−58° =302°

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