Circles - Exercise 15.4 | Extra Documents & Tests for Class 9 PDF Download

 Page 1


    
 
    
   
       
                               
                
         
Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.
 
S o l u t i o n :
It is given that
 And  given
We have to find 
In given triangle 
 Given
 OB = OA               Radiiofthesamecircle
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB      
          ..… 1
                       (Given )
                   
From(1)
So
Again from figure,  is given triangle and 
Now in , 
                  Radiiofthesamecircle
?OAC = ?OCA    
               (Given that )
Page 2


    
 
    
   
       
                               
                
         
Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.
 
S o l u t i o n :
It is given that
 And  given
We have to find 
In given triangle 
 Given
 OB = OA               Radiiofthesamecircle
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB      
          ..… 1
                       (Given )
                   
From(1)
So
Again from figure,  is given triangle and 
Now in , 
                  Radiiofthesamecircle
?OAC = ?OCA    
               (Given that )
Then,
Since
Hence 
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
Page 3


    
 
    
   
       
                               
                
         
Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.
 
S o l u t i o n :
It is given that
 And  given
We have to find 
In given triangle 
 Given
 OB = OA               Radiiofthesamecircle
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB      
          ..… 1
                       (Given )
                   
From(1)
So
Again from figure,  is given triangle and 
Now in , 
                  Radiiofthesamecircle
?OAC = ?OCA    
               (Given that )
Then,
Since
Hence 
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
vii
viii
ix
x
xi
xii
S o l u t i o n :
We have to find  in each figure.
i
It is given that 
?AOC + ?COB = 180°   [Linear pair]
Page 4


    
 
    
   
       
                               
                
         
Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.
 
S o l u t i o n :
It is given that
 And  given
We have to find 
In given triangle 
 Given
 OB = OA               Radiiofthesamecircle
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB      
          ..… 1
                       (Given )
                   
From(1)
So
Again from figure,  is given triangle and 
Now in , 
                  Radiiofthesamecircle
?OAC = ?OCA    
               (Given that )
Then,
Since
Hence 
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
vii
viii
ix
x
xi
xii
S o l u t i o n :
We have to find  in each figure.
i
It is given that 
?AOC + ?COB = 180°   [Linear pair]
 
As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Now ,x =
1
2
?COB = 22
1
2
°
Hence 
ii
As we know that  = x                
Anglesinthesamesegment
line  is diameter passing through centre,
So,
?BCA = 90°       [Angle inscribed in a semicircle is a right angle ]
   
 
?CAB + ?ABC + ?BCA = 180°    [Angle sum property] ? x +40°+90° = 180° ? x = 50°
iii
It is given that
?ABC =
1
2
(Reflex ?AOC)
  
So 
And 
Then 
Hence 
iv
   Linearpair
    
And
x = 
 
 
Hence,
Page 5


    
 
    
   
       
                               
                
         
Q u e s t i o n : 2 0
In the given figure, O is the centre of the circle. If ?APB
= 50°, find ?AOB and ?OAB.
S o l u t i o n :
This question seems to be incorrect.
Q u e s t i o n : 2 1
In the given figure, O is the centre of the circle. Find ?BAC.
 
S o l u t i o n :
It is given that
 And  given
We have to find 
In given triangle 
 Given
 OB = OA               Radiiofthesamecircle
Therefore,  is an isosceles triangle.
So, ?OBA = ?OAB      
          ..… 1
                       (Given )
                   
From(1)
So
Again from figure,  is given triangle and 
Now in , 
                  Radiiofthesamecircle
?OAC = ?OCA    
               (Given that )
Then,
Since
Hence 
Q u e s t i o n : 2 2
If O is the centre of the circle, find the value of x in each of the following figures.
i
ii
iii
iv
v
vi
vii
viii
ix
x
xi
xii
S o l u t i o n :
We have to find  in each figure.
i
It is given that 
?AOC + ?COB = 180°   [Linear pair]
 
As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Now ,x =
1
2
?COB = 22
1
2
°
Hence 
ii
As we know that  = x                
Anglesinthesamesegment
line  is diameter passing through centre,
So,
?BCA = 90°       [Angle inscribed in a semicircle is a right angle ]
   
 
?CAB + ?ABC + ?BCA = 180°    [Angle sum property] ? x +40°+90° = 180° ? x = 50°
iii
It is given that
?ABC =
1
2
(Reflex ?AOC)
  
So 
And 
Then 
Hence 
iv
   Linearpair
    
And
x = 
 
 
Hence,
v
It is given that 
 is an isosceles triangle.
   
Therefore 
And,
In ? AOB, ?AOB + ?OBA + ?BAO = 180° ? 70°+ ?BAO = 180° ? ?BAO = 110°
?AOB = 2(Reflex ?ACB)
                     
Hence,
vi
It is given that 
  
And
?COA + ?AOB = 180° ? ?COA = 180°-60° ? ?COA = 120°
? OCA is an isosceles triangle.
So
Hence, 
vii
                   Angleinthesamesegment
  
In  we have
Hence 
viii
  
As    Radiusofcircle
Therefore,  is an isosceles triangle.