NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.2

NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles

TextbookNCERT
Class9th
SubjectMathematics
Chapter6th
Chapter NameLines and Angles
CategoryClass 9th Mathematics
Medium English
SourceLast Doubt

Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.2 In This Chapter We Will Learn About Lines and Angles, Line-segment, Collinear points, Angle, Arms, Vertex, Ray, Acute, Right angle, Obtuse angle, Straight angle, Reflex angle, Complementry angle, Supplementry angle, Adjacent, Vertically opposite angle,  with Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.2.

NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles

Chapter -6

Lines and Angles

Exercise – 6.2

Question 1. In figure, find the values of x and y and then show that AB || CD.
ch 6 6.2 vk
Solution: In the figure, we have CD and PQ intersect at F.
ch 6 6.2 vk
∴ y = 130° …(1) [Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∠AEP + ∠AEQ = 180° [Linear pair]
50° + x = 180°
⇒ x = 180° – 50° = 130° …(2)
From (1) and (2), x = y
As they are pair of alternate interior angles.
So, AB || CD
Question 2. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Class 9th Maths Chapter - 6 Lines and Angles Exercise - 6.2
Solution:
AB || CD, and CD || EF [Given]
∴ AB || EF
∴ x = z [Alternate interior angles] ….(1)
Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
⇒ z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
Let 3x + 7x = 10x
10x = 180°
x = 180°/10
x = 18
Now, 3x = 3 × 18 = 54°
7x = 7 × 18 = 126°
x = 126°.
Question 3. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Class 9th Maths Chapter - 6 Lines and Angles Exercise - 6.2
Solution:
AB || CD and GE is a transversal.
∴ ∠AGE = ∠GED [Alternate interior angles]
But ∠GED = 126° [Given]
∴∠AGE = 126°
Also, ∠GEF + ∠FED = ∠GED
or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]
x = z [Alternate interior angles]… (1) Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
∠GEF = 126° -90° = 36°
Now, AB || CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180° [Co-interior angles]
or ∠FGE + 126° = 180°
or ∠FGE = 180° – 126° = 54°
Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°.
Question 4. In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.

Solution: Draw a line EF parallel to ST through R.
Class 9th Maths Chapter - 6 Lines and Angles Exercise - 6.2
Since PQ || ST [Given]
and EF || ST [Construction]
∴ PQ || EF and QR is a transversal
⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given]
∴∠QRF = ∠QRS + ∠SRF = 110° …(1)
Again ST || EF and RS is a transversal
∴ ∠RST + ∠SRF = 180° [Co-interior angles] or 130° + ∠SRF = 180°
⇒ ∠SRF = 180° – 130° = 50°
Now, from (1), we have ∠QRS + 50° = 110°
⇒ ∠QRS = 110° – 50° = 60°
Thus, ∠QRS = 60°.
Question 5. In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
ch 6 6.2 vk
Solution:
We have AB || CD and PQ is a transversal.
∴ ∠APQ = ∠PQR
[Alternate interior angles]
⇒ 50° = x [ ∵ ∠APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD [Alternate interior angles]
⇒ ∠APR = 127° [ ∵ ∠PRD = 127° (given)]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]
⇒ y = 127°- 50° = 77°
Thus, x = 50° and y = 77°.
Question 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 || CD.

Solution: Draw ray BL ⊥PQ and CM ⊥ RS
ch 6 6.2 vk
∵ PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABL = ∠LBC and ∠MCB = ∠MCD
⇒ ∠ABL = ∠MCD …(2) [By (1)]
Adding (1) and (2), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.

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