RD Chapter 16 Circles Ex 16.4 Solutions
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 = 2∠APB
∠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 
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
(iv) 
(Linear pair) And
Hence,
(v) It is given that 
is an isosceles triangle. Therefore 
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 haveHence
(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)
Answer - 5 : - We have to prove that 
Since, circumcenter is the intersection of perpendicular bisectors of each side of the triangle.
Now according to figure A, B, C are the vertices of ΔABC
In 
, 
is perpendicular bisector of BC So, BD = CD
OB = OC (Radius of the same circle)
And,
OD = OD (Common)
We know that angle formed any chord of the circle at the center is twice of the angle formed at the circumference by same chord
Therefore,
Therefore,
Hence proved
Question - 6 : - In the given figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.
Answer - 6 : -
It is given that,∠ABC is on circumference of circle BD is passing through centre.
Construction: Join A and C to form AC and extend BO to D such that BD be the perpendicular bisector of AC.
Now in △BDA and △BDC△BDA and △BDC we have
AD = CD (BD is the perpendicular bisector)
So ∠BDA=∠BDC=90°∠BDA=∠BDC=90°
(Common)
△BDA≅△BDC (SAS congruency criterion)△BDA≅△BDC SAS congruency criterion
Hence 
(by cpct)
Question - 7 : - In the given figure, O is the centre of the circle, prove that ∠x = ∠y + ∠z.
Answer - 7 : -
It is given that, O is the center of circle and A, B and C are points on circumference on triangle
We have to prove that ∠x = ∠y + ∠z
∠4 and ∠3 are on same segment
So, ∠4 = ∠3
(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) 
…… (1)
(Exterior angle is equal to the sum of two opposite interior angles) …… (2)∠4=∠z+∠1∠4=∠z+∠1 (Exterior angle is equal to the sum of two opposite interior angles)
…… (3) Adding (2) and (3)
……(4) From equation (1) and (4) we have
Question - 8 : - In the given figure, O and O' are centres of two circles intersecting at B and C. ACD is a straight line, find x.
Answer - 8 : -
It is given that
Two circles having center O and O' and ∠AOB = 130°
And AC is diameter of circle having center O
We have
So
Now, reflex 
So
x°=360°−230°=130°x°=360°-230°=130°
Hence, 
Question - 9 : - In the given figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.
Answer - 9 : -
It is given that O is the centre and ∠ROS=40°∠ROS=40°
We have 
In right angled triangle RQT we have
Question - 10 : - In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
Answer - 10 : -
It is given that ∠ACB = 40° and ∠DPB = 120°
Construction: Join the point A and B
(Angle in the same segment) Now in △BDP△BDP we have