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BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.2 Book Answers |
Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 Textbooks Solutions PDF
Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.2 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.2 Textbooks. These Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 Books Solutions
Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 9th |
Subject | Maths Chapter 6 Lines and Angles Ex 6.2 |
Chapters | All |
Provider | Hsslive |
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BSEB Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 9th Maths Chapter 6 Lines and Angles Ex 6.2 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2
Question 1.
In figure, find the values of x and y and then show that AB || CD.
Solution:
Since AB || CD and transversal PQ intersects them at R and S respectively.
∴ ∠ARS = ∠RSD [Alternate angles]
⇒ x = y
But ∠RSD = ∠CSQ
[ Vertically opp. angles]
⇒ y = 130° [∵∠CSQ = 130°]
Hence, x = y = 130°.
Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Solution:
Since CD || EF and transversal PQ intersects them at S and T respectively.
∠CST = ∠STF [Alternate Angles]
⇒ 180° – y = z
[∵ ∠y + ∠CST = 180° being linear pair]
⇒ y + z = 180°
Given y : z = 3 : 7.
So, the sum of ratios = 3 + 7 = 10
∴ y = 310 x 180°
= 3 x 18° = 54°
and, z = 710 x 180°
= 7 x 18° = 126°
Since AB || CD and transversal PQ intersects them at R and S respectively.
∴ ∠ARS + ∠RSC = 180° [Consecutive interior angles are supplementary]
⇒ x + y = 180°.
⇒ x = 180° – y
= 180° – 54° = 126° [∵ y = 54°]
Hence x = 126°.
Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
Since AB || CD and transversal GE cuts them at G and E respectively.
∠AGE = ∠GED [Alternate angles]
∠AGE = 126° [∵ ∠GED = 1260 (given)]
∠GEF = ∠GED – ∠FED = 126° – 90° = 36°
∠FGE = ∠GEC [Alternate angles]
∠FGE = 90° – ∠GEF
= 90° – 36° = 54°
Hence, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°.
Question 4.
In figure, if PQ || ST, ∠PQR = 110° and ∠RST =130°, find ∠QRS.
Solution:
Produce PQ to intersect SR in a point M.
Now, PM || ST and transversal SM intersects them at M and R respectively.
∴ ∠SMQ = ∠TSM [Alternate angles]
⇒ ∠SMQ = 130°
⇒ ∠QMR = 180° – 130° = 50° [∵ ∠SMQ + ∠QMR = 180° linear pairs]
Since, ray RQ stands at Q on PM.
∴ ∠PQR + ∠RQM = 180°
⇒ 110° + ∠RQM = 180°
⇒ ∠RQM = 70°
∴ ∠QRS = 180° – (70° + 50°) = 60°
[∵ Sum of the angles of a triangle is 180°]
Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Solution:
∵ AB || CD and transversal PQ intersects them at P and Q respectively.
∴ ∠PQR = ∠APQ [Alternate angles]
⇒ x = 50° [∵ ∠APQ = 50° (given)]
∵ AB || CD and transversal PR intersects them at P and R respectively.
∴ ∠APR = ∠PRD [Alternate angles]
⇒ ∠APQ + ∠QPR = 127° [∵ ∠PRD = 127°]
⇒ 50° + y = 127° [∵∠APQ = 50°]
⇒ y = 127° – 50° = 77°
Hence, x = 50° and y = 11°.
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:
Two-plane mirrors PQ and RS are placed parallel to each other i.e., PQ || RS. An incident ray AB after reflections takes the path BC and CD.
BN and CM are the normals to the plane mirrors PQ and RS respective.
Since BN ⊥ PQ, CM ⊥ RS and PQ || RS
∴ 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.
i.e., alternate interior angles are equal.
Therefore, AB || CD.
BSEB Textbook Solutions PDF for Class 9th
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- BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Book Answers
- BSEB Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 Book Answers
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- BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.1 Book Answers
- BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.2 Book Answers
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- BSEB Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.3 Book Answers
- BSEB Class 9 Maths Chapter 7 Triangles Ex 7.4 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.4 Book Answers
- BSEB Class 9 Maths Chapter 7 Triangles Ex 7.5 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.5 Book Answers
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- BSEB Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.2 Book Answers
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- BSEB Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Book Answers
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