NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 6 Lines and Angles Ex 6.3.



Ex 6.3 Class 9 Maths Question 1.
In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If SPR = 135° and PQT = 110°, find PRQ.


Solution:
RPS + RPQ = 180°          [Linear pair of angles]
135° + 
RPQ = 180°           [ RPS = 135° (given)]
 RPQ = 180° – 135° = 45°
Now, 
RPQ + PRQ = PQT      [Exterior angle = sum of interior opposite angles]
 45° + PRQ = 110°
PRQ = 110° – 45° = 65°

 

Ex 6.3 Class 9 Maths Question 2.
In the given figure, 
X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of ΔXYZ, find OZY and YOZ.

Solution:

In ∆XYZ, we have XYZ + YZX + ZXY = 180°     [Angle sum property of a triangle]
But
XYZ = 54° and ZXY = 62°
54° + YZX + 62° = 180°
YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of
XYZ and XZY respectively.
OYZ = ½ XYZ = ½ (54°) = 27°
and
OZY = ½ YZX = ½ (64°) = 32°
Now, in ∆OYZ, we have
YOZ + OYZ + OZY = 180°                   [Angle sum property of a triangle]
YOZ + 27° + 32° = 180°
YOZ = 180° – 27° – 32° = 121°
Thus,
OZY = 32° and YOZ = 121°

 

Ex 6.3 Class 9 Maths Question 3.
In figure, if AB ||
DE, BAC = 35°and CDE = 53°, find DCE.


Solution:
We have, AB || DE
 AED = BAE          [Alternate interior angles]
Now, 
BAE = BAC
BAE = 35°               BAC = 35°Given)]
AED = 35°
In ΔDCE,
DCE + CED + EDC = 180°

[ Sum of all the angles of a triangle is equal to 180°]
 DCE + 35° + 53° = 180°             (∵∠AED = CED = 35°)
 DCE = 180 ° – (35° + 53°) 

 DCE = 92°

 

Ex 6.3 Class 9 Maths Question 4.
In the given figure, if lines PQ and RS intersect at point T such that 
PRT = 40°, RPT = 95° and TSQ = 75°, then find SQT.


Solution:
Given, 
PRT = 40°, RPT = 95° and TSQ = 75°
In ΔPRT, 
PRT + RPT + PTR = 180°    …(i)   [Using angle sum property of a triangle]
On putting 
PRT = 40° and RPT = 95° in eq. (i), we get
40° + 95° + 
PTR = 180°
 135° + PTR = 180° 

 PTR = 180° – 135° 

PTR = 45°
Now, 
PTR = QTS                 [Vertically opposite angles]
 
QTS = 45°

In ΔTQS, QTS + TSQ + SQT = 180°       … (ii)
[Using angle sum property of a triangle]
On putting 
QTS = 45° and TSQ = 75° in eq. (ii), we get
45° + 75° + 
SQT = 180° 

120° + SQT = 180°
SQT = 180° – 120°
 
SQT = 60°

Ex 6.3 Class 9 Maths Question 5.
In the given figure, if PQ
PS, PQ || SR, SQR = 28° and QRT = 65°, then find the values of x and y.


Solution:
For ΔQSR, 
QRT is an exterior angle.
QRT = SQR + QSR           [ Exterior angle = sum of interior opposite angles]
 65° = 28° + QSR               [ QRT = 65° and SQR = 28°    (given)]
QSR = 65° – 28° 

 QSR = 37°
Given PQ || SR and SQ is the transversal which intersects PQ and ST at Q and S, respectively.
 QSR = PQS                      [Alternate interior angles]
 x = 37°
Now, in ΔPQS, 
SPQ + PQS + PSQ = 180°

[Since, sum of all the angles of a triangle is 180°]
 90° + 37° + y = 180°           [ PQ PS  SPQ = 90°]
 127° + y = 180°

 y = 180° – 127° = 53°
Hence, x = 37° and y = 53°.

 

Ex 6.3 Class 9 Maths Question 6.
In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR.


Solution:

In ∆PQR, side QR is produced to S, so by exterior angle property,
PRS = QPR + PQR
 ½ PRS = ½ QPR + ½ PQR
TRS = ½ QPR + TQR                   …(1)
[
QT and RT are bisectors of PQR and PRS respectively.]
Now, in ∆QRT, we have
TRS = TQR + QTR                           …(2)
[Exterior angle property of a triangle]

From (1) and (2), we have ½ QPR + TQR = TQR + QTR
 ½ QPR = QTR
Thus,
QTR = ½ QPR.                Hence proved.


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