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Ncert Solution | Ncert Solution for class 9th | Ncert Solution for Class 9 Maths | NCERT SOLUTIONS MATHS CHAPTER 6 LINES AND ANGLES
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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles are prepared by CBSE student eCARE expert to score good marks in class 9. This chapter contain many topics which are very important to score good in class 9. There are some important topic/Activity mentioned below:

Question 1
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Q1
Solution:
Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ?∠AOC + ∠BOE = 70° (Given)]
or ∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
Also, AB and CD intersect at O.
∴∠COA = ∠BOD [Vertically opposite angles]
But ∠BOD = 40° [Given]
∴ ∠COA = 40°
Also, ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250°.



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Some questions are asked by class 9 students related to Maths

Question 1.
In figure, find the values of x and y and then show that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Q1
Solution:
In the figure, we have CD and PQ intersect at F.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Q1.1
∴ y = 130° …(1)
[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
As they are pair of alternate interior angles.
∴ AB || CD

Ex 6.2 Class 9 Maths Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Q2
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
z = 73 y = 73(180°- z) [By (2)]
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°.

 

Ex 6.2 Class 9 Maths Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Q3
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°.

 

Ex 6.2 Class 9 Maths Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Q4.1
Solution:
Draw a line EF parallel to ST through R.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2 Q4
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°.