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# RD Chapter 16- Circles Ex-16.5 Interview Questions Answers

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

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.

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.

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 Todays Deals  