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RD Chapter 16- Circles Ex-16.4 Interview Questions Answers

Question 1 : In the given figure, O is the centre of the circle. If ∠APB∠APB= 50°, find ∠AOB and ∠OAB.

Answer 1 :

APB = 500 (Given)

By degree measure theorem: AOB = 2APB

AOB = 2× 500 = 1000

Again, OA = OB [Radius of circle]

Then OAB = OBA [Angles opposite to equal sides]

Let OAB = m

In ΔOAB,

By angle sum property: OAB+OBA+AOB=1800

=> m + m + 1000 =1800

=>2m = 1800 –1000 = 800

=>m = 800/2 = 400

OAB = OBA = 400

Question 2 : In the given figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.

Answer 2 :

It is given that O is the centre of circle and A, B and C are points on circumference.
(Given)
 
We have to find ∠ABC
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.
Hence, 

Question 3 : In the given figure, O is the centre of the circle. Find ∠BAC.

Answer 3 :

It is given that
  And   (given)
 
We have to find  
In given triangle  
  (Given)
  OB = OA               (Radii of the same circle)
Therefore,   is an isosceles triangle.
So, ∠OBA=∠OAB      ∠OBA=∠OAB                ..… (1)
 
                        (Given  )
                     [From (1)]
So
Again from figure,   is given triangle and  
Now in  , 

                   (Radii of the same circle)

∠OAC=∠OCA    ∠OAC=∠OCA    
 
                (Given that  )
 
Then,
Since
 
Hence  

Question 4 :
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)

Answer 4 :

We have to find   in each figure.
(i) It is given that  
∠AOC+∠COB=180°   [Linear pair]∠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.

Hence  
(ii) As we know that   = x                 [Angles in the same segment]
line   is diameter passing through centre,
So,

∠BCA= 90°       [Angle inscribed in a semicircle is a right angle ]∠BCA= 90°       Angle inscribed in a semicircle is a right angle 
(iii) It is given that
 
So  
And  
 
Then  
Hence  
(iv)  
    (Linear pair)
 
And
Hence,
 (v) It is given that  
  is an isosceles triangle.
    
Therefore  

And,

Hence, 
(vi) It is given that  
And
ΔOCA is an isosceles triangle.∆OCA is an isosceles triangle.
So
 
Hence,  
(vii)               (Angle in the same segment)
In   we have
 
Hence  
(viii)
   
As     (Radius of circle)

Therefore,   is an isosceles triangle.
So            (Vertically opposite angles)
Hence,  
(ix) It is given that  
  …… (1)          (Angle in the same segment)
∠ADB=∠ACB=32°∠ADB=∠ACB=32°  ......(2)           (Angle in the same segment)
Because   and   are on the same segment   of the circle.
Now from equation (1) and (2) we have
 
Hence,  
(x) It is given that  
∠BAC=∠BDC=35°∠BAC=∠BDC=35°                (Angle in the same segment)
 Now in ΔBDC∆BDC we have
∠BDC+∠DCB+∠CBD=180°⇒35°+65°+∠CBD=180°⇒∠CBD=180°−100°=80°∠BDC+∠DCB+∠CBD=180°⇒35°+65°+∠CBD=180°⇒∠CBD=180°-100°=80°
Hence,  
(xi)