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Exercise 6.2

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

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.

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.

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.]

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.

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 $AB\parallel CD$ .

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, $AB\parallel CD$.