Class IX Science NCERT Solutions For Lines and Angles 6.2

Exercise 6.2 (Page 103)

1.  In the following figure, find the values of x and y and then show that AB || CD.

Ans. In the figure, we have CD and PQ intersect at Y.

∴ y = 130°          [Vertically opposite angles]

          Again, PQ is a straight line and EA stands on it.

∴ ∠AEP + ∠AEQ = 180°          [Linear pair]

          or 50° + x = 180°

          ⇒ x = 180° – 50° = 130°

...(2)

          From (1) and (2), x = y

          But they are the angles of a pair of interior alternate angles.

          ∴ AB || CD.

2.  In the following Figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Ans. 

          ∴ AB|| EF and PQ is a transversal.

          ∴ Interior alternate angles are equal.

          ∴ ∠x = ∠z

...(1)

          Again, AB || CD,

          ∴ Interior opposite angles are supplementary.

          ⇒ y + z = 180°

          But y : z = 3 : 7

          

          From (1) and (2), we have

                    x = 126°

3.  In the following figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Ans. AB || CD and GE is a transversal.

          ∴ Interior alternate angles are equal.

          ∴ ∠AGE = ∠GED

          But ∠GED = 126°

[Given]

          ∴ ∠AGE = 126°

          Since ∠GED = 126°

          ∴ ∠GEF + ∠FED = ∠GED

          or ∠GEF + 90° = 126°

          or ∠GEF = 126 – 90° = 36°

          Next, AB || CD and GE is a transversal.

          ∴ ∠FGE + ∠GED = 180°

          or ∠FGE + 126° = 180°

          or ∠FGE = 180° – 126° = 54°

          Thus, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°

4.  In the following figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

       Hint: Draw a line parallel to ST through point R.

Ans. ∴ PQ || ST

[Given]

          and EF || ST

[Construction]

          ∴ PQ || EF

          and QR is a transversal,

          ∴ Interior alternate angles are equal i.e ∠PAR = ∠QRF

          But ∠PAR = 110°

[Given]

          ∴ ∠QRF = ∠QRS + ∠SRF = 110°

...(1)

          Again ST || EF [Construction] and RS is a transversal.

          ∴ ∠RST + ∠SRF = 180°

          or 130° + ∠SRF = 180°

          ⇒ ∠SRF = 180° – 130° = 50°

          Now, from (1), we have

          ∠QRS + 50° = 110°

          ⇒ ∠QRS = 110° – 50° = 60°

          Thus, ∠QRS = 60°.

5.  In the following figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Ans. We have AB || CD

[Given]

          and PQ is a transversal.

          ∴ Interior alternate angles are equal.

          ∴ ∠APQ = ∠PQR

          or 50° = x[∴ APQ = 50° (Given)]

...(1)

          Again, AB || CD and PR is a transversal.

          ∴ ∠APR = ∠PRD

[Interior alternate angles]

          ⇒ ∠APR = 127°

[∴ It is given that ∠PRD = 127°]

          But ∠APR = ∠APQ + ∠QPR

          ∴ ∠APQ + ∠QPR + 127°

          ⇒ 50° + y = 127°

[∴ It is given that ∠APQ = 50°]

          ⇒ y = 127° – 50° = 77°

          Thus, x = 50° and y = 77°

6.  In the following 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 || CD.

Ans.

          Remember

          (i) Perpendiculars to the parallel lines are parallel.

          (ii) According to the laws of reflection, angle of incidence = angle of reflection

          Draw ray BL ⊥ PQ and CM ⊥ RS

          ∵ PQ || RS

[Given]

∴ BL || CM [∴ BL ⊥ PQ and CM ⊥ RS]

          and BC is a transversal.

          ∴ ∠LBC = ∠MCB

[Interior alternate angles]

          Since,

          (Angle of incidence) = (Angle of reflection)

          ∴ ∠ABL = ∠LBC and ∠MCD = ∠MCD

          ⇒ ∠ABL = ∠MCD

          ∴ From (1), we have

          ∠LBC + ∠ABL = ∠MCD + ∠MCD

[Equals are added to equals]

          ⇒ ∠ABC = ∠BCD

          i.e. Angles of a pair of interior alternate angles are equal,

          ∴ AB || CD