# Class 9 RD Sharma Solutions – Chapter 16 Circles- Exercise 16.4

### Question 1. In Fig., O is the centre of the circle. If âˆ APBâˆ APB= 50Â°, find âˆ AOB and âˆ OAB.

**Solution:**

âˆ APB=50Â°

By degree measure theorem

âˆ AOB=2APB

âˆ APB=2*50Â°=100Â°

since OA=OB [Radius of circle]

Then âˆ OAB=âˆ OBA [Angles opposite to equal sides]

Let âˆ OAB=x

In â–³OAb, by angle sum property

âˆ OAB+âˆ OBA+âˆ AOB=180Â°

x+x+100Â°=180Â°

2x=180Â°-100Â°

2x=80Â°

x=40Â°

âˆ OAB=âˆ OBA=40Â°

### Question 2. In Fig., it is given that O is the centre of the circle and âˆ AOC = 150Â°. Find âˆ ABC.

**Solution:**

âˆ AOC = 150Â°

âˆ AOC +reflex âˆ AOC = 360Â° [complex angle]

150Â°+reflex âˆ AOC = 360Â°

reflex âˆ AOC=210Â°

2âˆ ABC=210Â° [By degree measure theorem]

âˆ ABC=210Â°/2=105Â°

### Question 3. In Fig., O is the centre of the circle. Find âˆ BAC.

**Solution:**

We have âˆ AOB=80Â°

And âˆ AOC=110Â°

Therefore, âˆ AOB+âˆ AOC+âˆ BOC=360Â° [complete angle]

80+100+âˆ BOC=360Â°

âˆ BOC=360Â°-80Â°-110Â°

âˆ BOC=70Â°

By degree measure theorem

âˆ BOC=2âˆ BAC

170=2âˆ BAC

âˆ BAC=170Â°/2=85Â°

### Question 4. If O is the centre of the circle, find the value of x in each of the following figures.

**Solution:**

i)âˆ AOC=135Â°

âˆ AOC+BOC=185Â° [Linear pair of angles]

135Â°+âˆ BOC=180Â°

âˆ BOC=180Â°-135Â°=45Â°

By degree measure theorem

âˆ BOC=2âˆ COB

45=2x

x=45Â°/2=22\frac{1}{2}

ii)We have

âˆ ABC=40Â°

âˆ ACB=90Â° [Angle in semicircle]

In â–³ABC, by angle sum property

âˆ CAB+âˆ ACB+âˆ ABC=180Â°

âˆ CAB+90Â°+40Â°=180Â°

âˆ CAB=180Â°-90Â°-40Â°

âˆ CAB=50Â°

Now,

âˆ CDB=âˆ CAB [Angle is same in segment]

x=50Â°

iii)We have,

âˆ AOC=120Â°

By degree measure theorem

âˆ AOC=2âˆ APC

120Â°=2âˆ APC

âˆ APC=120Â°/2=60

âˆ APC+âˆ ABC=180Â° [Opposite angles of cyclic quadrilaterals]

60Â°+âˆ ABC=180Â°

âˆ ABC=180Â°-60Â°

âˆ ABC=120Â°

âˆ ABC+âˆ DBC=180Â° [Linear pair of angles]

120Â°+x=180Â°

x=180Â°-120Â°=60Â°

iv)We have

âˆ CBD=65Â°

âˆ ABC+âˆ CBD=180Â° [Linear pair of angles]

âˆ ABC=65Â°=180Â°

âˆ ABC=180Â°-65Â°=115Â°

reflex âˆ AOC=2âˆ ABC [By degree measure theorem]

x=2*115Â°

x=230Â°

v)We have,

âˆ OAB=35Â°

Then, âˆ OBA=âˆ OAB=35Â° [Angles opposite to equal radii]

In â–³AOB, by angle sum property

âˆ AOB+âˆ OAB+âˆ OBA=180Â°

âˆ AOB+35Â°+35Â°=180Â°

âˆ AOB=180Â°-35Â°=110Â°

âˆ AOB+reflex âˆ AOB=360Â° [complex angle]

110+reflexâˆ AOB=360Â°

reflexâˆ AOB=360Â°-110Â°=250Â°

By degree measure theorem reflexâˆ AOB=2âˆ ACB

250Â°=2x

x=250Â°/2=125Â°

vi)We have,

âˆ AOB=60

By degree measure theorem reflex

âˆ AOB=2âˆ ACB

60=2âˆ ACB

âˆ ACB=60Â°/2=30Â° [Angle opposite to equal radii]

x=30Â°

vii)We have,

âˆ BAC=50Â° and âˆ DBC=70Â°

âˆ BDC=âˆ BAC=50Â° [Angle in same segment]

In â–³BDC, by angle sum property

âˆ BDC+âˆ BCD+âˆ DBC=180Â°

50Â°+x+70Â°=180Â°

x=180Â°-50Â°-70Â°=60Â°

viii)We have,

âˆ DBO=40Â° and âˆ DBC=90Â° ——-[Angle in a semi circle]

âˆ DBO+âˆ OBC=90Â°

40Â°+âˆ OBC=90Â°

âˆ OBC=90Â°-40Â°=50Â°

By degree measure theorem

âˆ AOC=âˆ OBC

x=2*50Â°=100Â°

ix)In âˆ†DAB, by angle sum property

âˆ ADB+âˆ DAB+âˆ ABD=180Â°

32Â°+âˆ DAB+50Â°=180Â°

âˆ DAB=180Â°-32Â°-50Â°

âˆ DAB=98Â°

Now,

âˆ OAB+âˆ DCB=180Â° [opposite angle of cyclic quadrilateral]

98Â°+x=180Â°

x=180Â°-98Â°=82Â°

x)We have,

âˆ BAC=35Â°

âˆ BDC=âˆ BAC=35Â° [Angle in same segment]

In âˆ†BCD, by angle sum property

âˆ BDC+âˆ BCD+âˆ DBC=180Â°

35Â°+x+65Â°=180Â°

x=180Â°-35Â°-65Â°=80Â°

xi)We have,

âˆ ABD=40Â°

âˆ ACD=âˆ ABD=40Â° [Angle in same segment]

In âˆ†PCD, by angle sum property

âˆ PCD+âˆ CPO+âˆ PDC=180Â°

40Â°+110Â°+x=180Â°

x=180Â°-150Â°

x=30Â°

xii)Given that,

âˆ BAC=52Â°

Then âˆ BDC=âˆ BAC=52Â° [Angle in same segment]

Since OD=OC

Then âˆ ODC=âˆ OCD [ Opposite angle to equal radii]

x=52Â°

### Question 5. O is the circumcenter of the triangle ABC and OD is perpendicular on BC. Prove that âˆ BOD = âˆ A.

**Solution:**

We have to prove that âˆ BOD=âˆ A

since, circum center is the intersection of perpendicular bisector of each side of the triangle. Now according to figure A,B,C are the vertices of âˆ†ABC

In âˆ†BOC, OD is perpendicular bisector of BC.

so, BD=CD

OB=OC ——–(Radius of the same circle)

And,

OD=OD —–[common]

Therefore,

âˆ†BDOâ‰…âˆ†CDO (SSS concurrency criterion )

âˆ BOD=âˆ COD (by cpct)

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,

âˆ BAC=\frac{1}{2} âˆ BOC

âˆ BAC=\frac{1}{2}*2âˆ BOD

âˆ BAC=âˆ BOD

Therefore,

âˆ BOD=âˆ A

### Question 6. In Fig., O is the centre of the circle, BO is the bisector of âˆ ABC. Show that AB = AC.

**Solution:**

Given, BO is the bisector of âˆ ABC

To prove: AB=BC

Proof: Since, BO is the bisector of âˆ ABC.

Then, âˆ ABO=âˆ CBO —-(i)

Since, OB=OA [Radius of circle]

Then, âˆ ABO=âˆ DAB ——–(ii) [opposite angles to equal sides]

Since OB=OC [Radius of circle]

Then, âˆ OAB=âˆ OCB ——–(iii) [opposite angles to equal sides]

compare equations (i), (ii) and (iii)

âˆ OAB=âˆ OCB ——-(iv)

In âˆ†OAB and âˆ†OCB

âˆ OAB=âˆ OCB [From(iv)]

âˆ OBA=âˆ OBC [Given]

OB=OB [common]

Then

âˆ†OABâ‰…âˆ†OCB [By AAS condition]

Therefore, AB=BC [CPCT]

### Question 7. In Fig., O is the centre of the circle, prove that âˆ x = âˆ y + âˆ z.

**Solution:**

We have,

âˆ 3=âˆ 4 [Angles in same segment]

âˆ x=2âˆ 3 [By degree measure theorem]

âˆ x=âˆ 3+âˆ 3â‡’âˆ x=âˆ 3+âˆ 4 ——–(i) [âˆ 3=angle 4]

But âˆ y=âˆ 3+âˆ 1 [By exterior angle property]

â‡’âˆ 3=âˆ y-âˆ 1 —-(ii)

from (i) and (ii)

âˆ x=âˆ y-âˆ 1+âˆ 4

âˆ x=âˆ y+âˆ 4-âˆ 1

âˆ x=âˆ y+âˆ z+âˆ 1-âˆ 1 [By exterior angle property]

âˆ x=âˆ y+âˆ z

### Question 8. In Fig., O and O’ are centres of two circles intersecting at B and C. ACD is a straight line, find x.

**Solution:**

By degree measure theorem

âˆ AOB=2âˆ ACB

130Â°=2âˆ ACBâ‡’âˆ ACB=130Â°/2=65

âˆ ACB+âˆ BCD=180Â° [Linear pair of angles]

65Â°+âˆ BCD=180Â°

âˆ BCD=180-65=115

By degree measure theorem

reflexâˆ BOD=2âˆ BCD

reflexâˆ BOD=2*115Â°=230Â°

Now, reflexâˆ BOD+âˆ BOD=360Â° [complex angle]

230Â°+x=360Â°

x=360Â°-230Â°

x=130Â°

### Question 9. In Fig., O is the centre of a circle and PQ is a diameter. If âˆ ROS = 40Â°, find âˆ RTS.

**Solution:**

Since PQ is diameter

Then,

âˆ PRQ=90Â° [Angle in semi circle]

âˆ PRQ+âˆ TRQ=180Â° [Linear pair of angle]

90+âˆ TRQ=180

âˆ TRQ=180Â°-90Â°=90Â°

By degree measure theorem

âˆ ROS=2âˆ RQS

40=2âˆ RQS

âˆ RQS=40Â°/2=20Â°

In âˆ†RQT, by angle sum property

âˆ RQT+âˆ QRT+âˆ RTS=180Â°

20Â°+90Â°+âˆ RTS=180Â°

### Question 10. In Fig., if âˆ ACB = 40Â°, âˆ DPB = 120Â°, find âˆ CBD.

**Solution:**

We have,

âˆ ACB=40Â°; âˆ DPB=120Â°

âˆ APB=âˆ DCB=40Â° [Angle in same segment]

In âˆ†POB, by angle sum property

âˆ PDB+âˆ PBD+âˆ BPD=180Â°

40+âˆ PBD+120Â°=180Â°

âˆ PBD=180Â°-40Â°-120Â°

âˆ PBD=20Â°

âˆ CBD=20Â°

### Question 11. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

**Solution:**

Construction: O is center and r is radius and given that chord is equal to radius of circle.

Now in âˆ†AOB we have

AO=OB=BA (It is given that chord is equal to radius of circle)

so, âˆ†AOB is an equilateral triangle

âˆ AOB=60Â°

So, âˆ AOB=2âˆ ADB (The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle)

Then âˆ ADB=30Â°

So,

Therefore,

âˆ ADB=30Â° and âˆ AEB=150Â°

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