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RD Chapter 16- Circles Ex-16.5 Interview Questions Answers

Question 1 : In the given figure, ΔABC is an equilateral triangle. Find m∠BEC.

Answer 1 :

It is given that,   is an equilateral triangle
 
We have to find  
Since   is an equilateral triangle.
So  
And
  …… (1)
Since, quadrilateral BACE  is a cyclic qualdrilateral

So ,                                          (Sum of opposite angles of cyclic quadrilateral is .)
 
Hence
 

Question 2 : In the given figure, ΔPQR is an isosceles triangle with PQ = PR and m ∠PQR = 35°. Find m ∠QSR and m ∠QTR.

Answer 2 :

Disclaimer: Figure given in the book was showing m∠PQR as m∠SQR. 
It is given that ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°
We have to find the m∠QSR and m∠QTR
Since ΔPQR is an isosceles triangle
So ∠PQR = ∠PRQ = 35° 
Then
 
Since PQTR is a cyclic quadrilateral
So
In cyclic quadrilateral QSRT we have
 
Hence,
 and   

Question 3 : In the given figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.

Answer 3 :

It is given that O is centre of the circle and ∠BOD = 160°
              
We have to find the values of x and y.
As we know that 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.
Therefore,
 
Since, quadrilateral ABCD is a cyclic quadrilateral.
So,
x + y = 180°              (Sum of opposite angles of a cyclic quadrilateral is 180°.)
 
Hence  and  


Question 4 : In the given figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.

Answer 4 :

It is given that ∠BCD = 100° and ∠ABD = 70°
 
We have to find the ∠ADB
We have
∠A + ∠C = 180°                     (Opposite pair of angle of cyclic quadrilateral)
So,
Now in   is   and  
Therefore,
 
Hence,  

Question 5 : If ABCD is a cyclic quadrilateral in which AD || BC (In the given figure). Prove that ∠B = ∠C.

Answer 5 :

It is given that, ABCD is cyclic quadrilateral in which AD || BC
 
We have to prove  
Since, ABCD is a cyclic quadrilateral
So,
 and                ..… (1)
 and                (Sum of pair of consecutive interior angles is 180°) …… (2)
From equation (1) and (2) we have
  …… (3)
  …… (4)
Hence Proved

Question 6 : In the given figure, O is the centre of the circle. Find ∠CBD.

Answer 6 :

It is given that,  
 
We have to find  
Since,      (Given)
So,
(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,

                 (Opposite pair of angle of cyclic quadrilateral)
So,
 
  …… (1)
       (Linear pair)
          ( )
 
Hence  

Question 7 : In the given figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.

Answer 7 :

It is given that, AB and CD are diameter with center O and  
 
We have to find   
Construction: Join the point A and D to form line AD
Clearly arc AD subtends   at B and   at the centre.
Therefore,  ∠AOD=2∠ABD=100°∠AOD=2∠ABD=100° …… (1)
Since CD is a straight line then
        
∠DOA+∠AOC=180°        (Linear pair)∠DOA+∠AOC=180°        Linear pair
      
Hence  

Question 8 : On a semi-circle with AB as diameter, a point C is taken, so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).

Answer 8 :

It is given that,   as diameter,   is centre and  
 
We have to find   and  
Since angle in a semi-circle is a right angle therefore
 
In   we have
 (Given)
     (Angle in semi-circle is right angle)
Now in   we have
                                         
Hence  and   

Question 9 : In a cyclic quadrilateral ABCD if AB || CD and ∠B = 70°, find the remaining angles.

Answer 9 :

It is given that, ABCD is a cyclic quadrilateral such that AB || CD and  
 
Sum of pair of opposite angles of cyclic quadrilateral is 180°.
                   (  given)
So,  
Also AB || CD and BC transversal
So,
 
Now
                   

Question 10 : In a cyclic quadrilateral ABCD, if m ∠A = 3 (m ∠C). Find m ∠A.

Answer 10 :

It is given that
ABCD is cyclic quadrilateral and  
 
We have to find  
Since ABCD is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.
So  
And
Therefore
 
Hence  


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