9th Mahts 6.3

NCERT Class 9th solution of Exercise 6.1

NCERT Class 9th solution of Exercise 6.2

Exercise 6.3

Q1. In the given figure, sides QP and RQ of PQR are produced to points S and T respectively. If SPR=135° and ∠PQT=110°, find ∠PRQ.
Sol.
∠PQT+∠PQR=180°  [ By linear pair. ]
110°+∠PQR=180°    [ Given ∠PQT=110° ]
∠PQR=180°-110°
∠PQR=70°
∠PRQ+∠PQR=∠SPR  
                            [ By exterior angle property ]
∠PRQ+70°=135°       [ Given ∠SPR=135° ]
∠PRQ=135°-70°
∠PRQ=65°
Answer :
The required value of ∠PRQ=65°.

Q2. In the given figure, ∠X=62°, ∠XYZ=54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of △XYZ, find ∠OZY and ∠YOZ.
Sol. :
In △XYZ,
∠YXZ+∠XYZ+∠XZY=180°  
                                         [ By Angle sum property of ]
62°+54°+∠XZY=180°
∠XZY=180°-62°-54°
∠XZY=180°-116°
∠XZY=64°
∠XZO=∠OZY=frac(1)(2)times∠XZY    
                                                   [ ZO is bisect ∠XZY]
∠XZO=∠OZY=frac(1)(2)times64°         
                                                       [ Given ∠XZY=64°]  
∠XZO=∠OZY=32°
∠XYO=∠OYZ=frac(1)(2)times∠XYZ   
                                                 [ YO is bisect ∠XYZ ]
∠XYO=∠OYZ=frac(1)(2)times54°        
                                                      [ Given ∠XYZ=54° ]
∠XYO=∠OYZ=27°
In △YOZ
∠OYZ+∠OYZ+∠YOZ=180° 
                                      [ By angle sum property of ]
27°+32°+∠YOZ=180°
∠YOZ=180°-27°-32°
∠YOZ=180°-59°
∠YOZ=121°
Answer :
The required ∠OZY=32° and ∠YOZ=121°.

Q3. In the given figure, if AB∥DE, ∠BAC=35° and ∠CDE=53°, find ∠DCE
Sol. :
AB∥DE and AE is transversal.                [ Given ]
∠CED=∠BAC                           [ Alternate angles ]
∠CED=35°
In △CDE
∠DCE+∠CDE+∠CED=180°                 
                               [ By angle sum property of ]
∠DCE+53°+35°=180°                         
                                                [Given ∠CDE=53°]
∠DCE=180°-88°
∠DCE=92°
Answer :
The required value ∠DCE=92°.

Q4. In the given figure, if lines PQ and RS intersect at T, such that ∠PRT=40°, ∠RPT=95°.
Sol. :
In △PRT
∠PTR+∠RPT+∠PRT=180° 
                         [ By angle sum property of ]
∠PRT+95°+40°=180°        
                 [ Given ∠RPT=95° & ∠PRT=40°]
∠PRT=180°-95°-40°
∠PRT=180°-135°
∠PRT=45°
PQ intersect RS at point T
∠QTS=∠PRT             [ Vertically opp. angles]
∠QTS=45°
In △SQT
∠SQT+∠TSQ+∠QTS=180° 
                        [ By angle sum property of ]
∠SQT+75°+45°=180°         
                                        [ Given ∠TSQ=75°]
∠SQT=180°-75°-45°
∠SQT=180°-120°
∠SQT=60°
 Answer :
The required value of ∠SQT=60°.

Q5. In the given figure, if PQ⊥PS, PQ∥SR, ∠SQR=28° and ∠QRT=65°, then find the values of x and y.
Sol. :
PQ∥SR and QR is transversal
∠PQR=∠QRT                         [Alternat angles ]
According to figure
∠PQS+∠SQR=∠QRT
x+28°=65° 
                  [Given ∠SQR=28° & ∠QRT=65°]
x=65°-28°
x=37°
In △PQS
∠QPS+∠PQS+∠PSQ=180 
                         [ By angle sum property of ]
90°+37°+y=180°           [ Given ∠QPS=90°]
y=180°-90°-37°
y=180°-127°
y=53°
Answer :
The required value of x=37° & y=53°.

Q6. In the given 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=frac(1)(2)∠QPR.
 
Sol. :
According to figure 
∠PRS=∠QPS+∠PQR______(1) 
                                   [ By exterior angle property ]
∠TRS=∠QTR+∠TQR_____(2)  
                                   [ By exterior angle property ]
Since QT & RT are bisectors of ∠PQR
& ∠PRS [Given]
∠TQR=frac(1)(2)∠PQR___(3)
∠TRS=frac(1)(2)∠PRS____(4)
∠TRS=frac(1)(2)∠QPR+frac(1)(2)∠PQR___(5)
                                                     [ by eq.(1) &(4)]
∠QTR+∠TQR=frac(1)(2)∠QPR+frac(1)(2)∠PQR 
                                                   [ by eq.(2) & (5) ]
∠QTR+frac(1)(2)∠PQR=frac(1)(2)∠QPR+frac(1)(2)∠PQR
∠QTR=frac(1)(2)∠QPR
Proved.
 

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