Lines and Angles – Exercise 6.2 – (MATHEMATICS) – 9th Class

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Exercise 6.2      

1. In figure given below, find the values of x and y and then show that AB\parallel CD.


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Sol.    y = 130°     [ Vertically opposite angles ]

Further, 50° + x  = 180°                    [Linear pair]

⇒                        x   = 130°
Hence,              x = y = 130°
Transversal intersects lines  AB  and CD.
Such  that x = y                              [ Alternate interior angles ]

Hence,    AB\parallel CD


2. In figure, if AB\parallel CD , CD\parallel EF and y : z = 3 : 7, find x.


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Sol.   AB\parallel CD and CD\parallel EF                             …(i)  [Given]

∠ 1 = y                                                                                          [Vertically opposite angles ]

∠ 1 + z = 180°                 [ CD\parallel EF   and ∠1 , ∠z are on the same side of the transversal ]

⇒    y + z   = 180°


Given : y : z = 3 : 7  \Rightarrow {y \over z} = {3 \over 7} \Rightarrow y = {{3z} \over 7}
{{3z} \over 7} + z = 180^\circ \Rightarrow {{10z} \over 7} = 180^\circ   [From (ii)]

⇒        z = 126°
∴       y = 180°  – 126°  = 54°                                               [From (i)]

Now, AB\parallel CD  and transversal intersects these  lines.
x + y = 180^\circ \Rightarrow x + 54^\circ = 180^\circ \Rightarrow x = 180^\circ  - 54^\circ = 126^\circ


3. In figure, AB\parallel CD , EF \bot CD   and ∠GED = 126°, find  ∠AGE, ∠GEF and ∠ FGE.


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Sol.   AB\parallel CD  and GE is transversal.
∴    ∠ AGE   =   ∠ GED                                             [Alternate angles ]

⇒  AGE  = 126°
Further,       ∠ GED   = ∠ GEF + ∠FED
⇒                   126°  = ∠GEF  = 126°  – 90°    = 36°
Again,  AB\parallel CD  and GE is transversal.
∴ ∠FGE + ∠GED = 180°            [Sum of interior angles on the same side of transversal is 180°.]

⇒         ∠FGE + 126° = 180°
⇒         ∠FGE = 180° — 126° = 54°.


4. In figure, if PQ\parallel ST , ∠ PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]

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Sol. Construction: Through R draw a line XRY parallel to PQ.
Proof:  PQ\parallel QRY  and QR is transversal.
∴   ∠ PQR =  ∠ QRY = 110°                                         …(i) [Alternate angles]

Also, PQ\parallel ST                                                                          [Given]

and   PQ\parallel RY                                                   [Construction]

∴ ST\parallel XRY  and SR is transversal.
∴  ∠TSR + ∠SRY = 180°
[Sum of interior angles on the same side of transversal]

⇒ 130° + ∠SRY = 180°
⇒ ∠ SRY = 180° — 130° = 50°                                         …(ii)
Also,     ∠QRY = ∠QRS + ∠SRY
⇒    110° =  ∠QRS + 50°                                            [From (i) and (ii)]

⇒      ∠ QRS = 110° — 50° = 60°.


5. In figure, if AB\parallel CD,  ∠ APQ = 50° and  ∠ PRD = 127°, find x and y.

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Sol .
 AB\parallel CD  and PQ is transversal.
∴         ∠ PQR  = ∠APQ  ⇒ x = 50°
Again AB\parallel CD  and PR is transversal.
∴    ∠ APR  =     ∠ PRD                                       [Alternate angles]

⇒     ∠ APQ + ∠ QPR = ∠PRD.
⇒            50° + y = 127° ⇒  y = 127° – 50°    =  77°.


6.  In figure, PQ and RS are two mirrors placed parallel to each other: An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB\parallel CD .

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Sol. Construction: Draw BE perpendicular to PQ and CF perpendicular to RS.
Proof: As BE \bot CF  and CF \bot RS  and PQ\parallel RS .


⇒   BE\parallel CF
Also, we know
Angle of incidence  = angle of reflection.
i.e.,  ∠ ABE = ∠EBC =   x.                                          …..(ii)
and ∠ BCF = ∠FCD = y                                              ….(iii)
From (i),  BE\parallel CF and BC is transversal.
∴  ∠ EBC = ∠BCF                                                                                    [Alternate angles]

      x = y   ⇒        2x = 2y
on ∠ABC  =  ∠BCD                                                                                [From (ii) and (iii)]

But these are alternate angles. Hence, AB\parallel CD.


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