NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2

 NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2

In this post, you can find a complete explanation of exercise 6.2 class 9 maths which belongs to Chapter 6 Lines and angles. Before you go through maths exercise 6.2 class 9, you should read and study the theorems of chapter 6.

NCERT Maths Class 9 Chapter 6 Exercise 6.2

Here you will find complete solutions to all questions of maths 6.2 class 9 along with the video. You can watch videos related to exercise 6.2 class 9 maths which will help you to understand the topic and questions of class 9 maths lines and angles exercise 6.2.

Solution of Ex- 6.2

Question -1

Question -2

Question -3

Question -4

Question -5

Question -6

Answers to NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.2  

 

Class 9 Maths Chapter 6 exercise 6.2 questions 1

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

Solution Q 1 Ex- 6.2 class 9

We know a linear pair is equal to 180°

x and 50° form a linear pair

∴x+50°=180°

x=180°-50°=130°   …..(i)

Also, y = 130°    (vertically opposite angles)…(ii)

∴x=y

From equations (i) and (ii)

But x and y are alternate angles. Therefore AB||CD

Class 9 Maths Chapter 6 exercise 6.2 questions 2

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

Solution Q 2 Ex- 6.2 class 9

Since CD||EF and transversal PQ intersects them at S and T respectively, therefore

∠CST=∠STF (alternate angles)

180°-y=z

(∵∠y+∠CST=180° being linear pair)

y+z=180°

Given y:z=3:7

Let y=3a ,z=7a

3a+7a=180°

10a=180°

`a=frac{180°}{10}=18°`

a=18°

y=3×18=54°

z=7×18=126°

y=54°,z=126°

Since AB||CD and transversal  PQ intersect them at R and S respectively, therefore 

∠ARS+∠RSC=180° (interior angles on the same side of the transversal are supplementary)

Or     x+y=180°

x=180°-y=180°-54°=126°    (y=54°)

∴x=126°

Class 9 Maths Chapter 6 exercise 6.2 questions 3

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

Solution Q 3 Ex- 6.2 class 9

Here AB||CD and transversal GE cuts them at G and E respectively.

∴         ∠AGE=∠GED  (Alternate angles)

∠AGE=126° [∵∠GED=126°(given)]

∠GEF=∠GED-∠FED=126°-90°=36°

And, ∠FGE=∠GEC (alternate angles)

∠FGE=90°-∠GEF

=90°-36°=54°

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

Class 9 Maths Chapter 6 exercise 6.2 questions 4

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

[Hint: Draw a line parallel to ST through point R]

Solution Q 4 Ex- 6.2 class 9

Produce PQ to intersect SR in a point M.

Now, PM||ST and transversal SM intersect them at M and S respectively.

∴     ∠SMQ=∠TSM (alternate angle)

∠SMQ=130° ∠QMR=180°-130°=50°

(∵∠SMQ+ ∠QMR=180°, linear pair)

Now ray QR stands at Q on PM

∴        ∠PQR+ ∠RQM=180° 

110°+∠RQM=180° 

∠RQM=70° 

∠QRS=180°-(70°+50°)=60

180°-(70°+50°)=130° 

∵   the sum of the angles of a triangle is 180°

Class 9 Maths Chapter 6 exercise 6.2 questions 5

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

 Solution Q 5 Ex- 6.2 class 9

Here, AB||CD and transversal PQ intersects them at P and Q respectively.

∴             ∠PQR= ∠APQ (alternate angles)

               x=50° 

Also, AB||CD and transversal PR intersect them at P and R respectively.

∠APR=∠PRD (alternate angles)

∠APQ+∠QPR=127° (∵∠PRD=127°)

50°+y=127° (∵∠APQ=50°)

y=127°-50°=77° 

x=50° and y=77° 

Class 9 Maths Chapter 6 exercise 6.2 questions 6

In the 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 Q 6 Ex- 6.2 class 9

Two plane mirrors PQ and RS are placed parallel to each other, i.e. PQ||RS. An incident ray AB after reflection takes the paths BC and CD.

Let BN and CM are normals to the plane mirror PQ and RS respectively.

Since BN⊥PQ, CM⊥RS, and PQ∥RS, therefore BN⊥RS⟹BN∥CM.

Thus BN and CM are two parallel lines and a transversal BC cuts them at B  and C respectively.

∠2=∠3(alternate interior angles)

But, ∠1=∠2 and∠3=∠4(by laws of reflection)

∴∠1+∠2=∠2+∠2 

and ∠3+∠4=∠3+∠3 

⟹∠1+∠2=2(∠2) 

and∠3+∠4=2(∠3) 

⇒∠1+∠2=∠3+∠4     [∵∠2=∠3⇒2(∠2)=2(∠3)]

⟹∠ABC=∠BCD 

Thus,lines AB and CD are intersected by transversal BC such that 

∠ABC=∠BCD 

Alternate interior angles are equal.

So AB||CD

Hence proved

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