RD Chapter 16 Circles Ex 16.5 Solutions
Question - 11 : - In the given figure, O is the centre of the circle and ∠DAB = 50° . Calculate the values of x and y.
Answer - 11 : - It is given that, O is the centre of the circle and ∠DAB=50°∠DAB=50°.
We have to find the values of x and y.
ABCD is a cyclic quadrilateral and 
So,
50° + y = 180°
y = 180° − 50°
y = 130°
Clearly 
is an isosceles triangle with OA = OB and 
Then, 
(Since ) So, 
x + ∠AOB∠AOB = 180° (Linear pair)
Therefore, x = 180° − 80° = 100°
Hence,
Answer - 12 : -
It is given that, 
and 
We have to find the 
In given 
we have
Question - 13 : - In the given figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
Answer - 13 : -
It is given that, ABC is an equilateral triangle
We have to find 
and 
Since is an equilateral triangle So, 
And 
is cyclic quadrilateral So 
(Sum of opposite pair of angles of a cyclic quadrilateral is 180°.) Then,
Similarly BECD is also cyclic quadrilateral
so
Hence,
and 
.
Question - 14 : - In the given figure, O is the centre of the circle. If ∠CEA = 30°, Find the values of x, y and z.
Answer - 14 : -
It is given that, O is the centre of the circle and 
We have to find the value of x, y and z.
Since, angle in the same segment are equal
So 
And z = 30°
As 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.
Since 
Then,
y = 2z
= 2 × 30°
= 60°
Since, the sum of opposite pair of angles of a cyclic quadrilateral is 180°.
z + x = 180°
x = 180° − 30°
= 150°
Hence,
x = 150°, y = 60° and z = 30°
Question - 15 : - In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
Answer - 15 : -
It is given that, 
and 
, 
are cyclic quadrilateral We have to find the value of x and y.
Since , is a cyclic quadrilateral So 
(Opposite angle of a cyclic quadrilateral are supplementary) 
( ) 
..… (1) x = 180° − 102°
= 78°
Now in cyclic quadrilateral DCFE
x + y = 180° (Opposite angles of a cyclic quadrilateral are supplementary)
y = 180° − 78°
= 102°
Hence, x = 78° and y = 102°
Question - 16 : - In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
Answer - 16 : -
It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.
We have to prove that smaller of two is 60°
Since ABCD is a cyclic quadrilateral
So ∠A + ∠C = 180° (Sum of opposite pair of angles of cyclic quadrilateral is 180°) ..… (1)
And,
∠A – ∠C = 60° (Given) ..… (2)
Adding equation (1) and (2) we have
So, ∠C = 60°
Hence, smaller of two is 60°.
Question - 17 : - In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.
Answer - 17 : -
Here, ABCD is a cyclic quadrilateral, we need to find x.
In cyclic quadrilateral the sum of opposite angles is equal to 180°.
Therefore,
Hence, the value of x is 100°.
Question - 18 : - ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
(iii) ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Answer - 18 : -
(i) It is given that 
, 
and 
We have to find 
In cyclic quadrilateral ABCD
..… (1) 
..… (2) Since, So,
Hence, 
Question - 19 : - Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Answer - 19 : -
To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.
Proof:
We know that the perpendicular bisector of every chord of a circle always passes through the centre.
Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.
Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.
Question - 20 : - Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Answer - 20 : -
Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.
Let O be the centre of the circle.
We know that the angle formed in the semicircle is 90°.
Since, ABCD is a rectangle, So
Therefore, AC and BD are diameter of the circle.
We also know that the intersection of any two diameter is the centre of the circle.
Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.