RD Sharma Class 9 Solutions of Chapter 16 consist of various topics such as Circles explained in a step by step and detailed way for an easy understanding. In brief, a circle is a closed plane curve consisting of all points at a given distance from a fixed point within it, which is called the center of the circle. The center of a circle always lies in the interior of the circle. The distance between any points of the circumference of the circle and the center of the circle is called the radius. Also, the longest chord of a circle is its diameter. RD Sharma Solutions help students to practice questions based on different concepts such as radius, segment, sector and diameter of circle. Check the detailed RD Sharma solutions for class 9 for Chapter 16 and score good marks in exams. The comprehensive solutions to the exercise questions can be checked from the following link.

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### Access Answers to Maths RD Sharma Chapter 16 Circles

## Exercise 16.1 Page No: 16.5

**Question 1: Fill in the blanks:**

**(i) All points lying inside/outside a circle are called ______ points/_______ points.**

**(ii) Circles having the same centre and different radii are called _____ circles.**

**(iii) A point whose distance from the center of a circle is greater than its radius lies in _________ of the circle.**

**(iv) A continuous piece of a circle is _______ of the circle.**

**(v) The longest chord of a circle is a ____________ of the circle.**

**(vi) An arc is a __________ when its ends are the ends of a diameter.**

**(vii) Segment of a circle is a region between an arc and _______ of the circle.**

**(viii) A circle divides the plane, on which it lies, in _________ parts.**

**Solution:**

**(i)** Interior/Exterior

**(ii)** Concentric

**(iii)** The Exterior

**(iv)** Arc

**(v)** Diameter

**(vi)** Semi-circle

**(vii)** Center

**(viii)** Three

**Question 2: Write the truth value (T/F) of the following with suitable reasons:**

**(i) A circle is a plane figure.**

**(ii) Line segment joining the center to any point on the circle is a radius of the circle,**

**(iii) If a circle is divided into three equal arcs each is a major arc.**

**(iv) A circle has only finite number of equal chords.**

**(v) A chord of a circle, which is twice as long as its radius is the diameter of the circle.**

**(vi) Sector is the region between the chord and its corresponding arc.**

**(vii) The degree measure of an arc is the complement of the central angle containing the arc.**

**(viii) The degree measure of a semi-circle is 180 ^{0}.**

**Solution:**

**(i)** T

**(ii)** T

**(iii)** T

**(iv)** F

**(v) **T

**(vi)** T

**(vii)** F

**(viii)** T

## Exercise 16.2 Page No: 16.24

**Question 1: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.**

**Solution:**

Radius of circle (OA) = 8 cm (Given)

Chord (AB) = 12cm (Given)

Draw a perpendicular OC on AB.

We know, perpendicular from centre to chord bisects the chord

Which implies, AC = BC = 12/2 = 6 cm

In right ΔOCA:

Using Pythagoras theorem,

OA^{2} = AC^{2} + OC^{2}

64 = 36 + OC^{2}

OC^{2} = 64 – 36 = 28

or OC = √28 = 5.291 (approx.)

The distance of the chord from the centre is 5.291 cm.

**Question 2: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.**

**Solution:**

Distance of the chord from the centre = OC = 5 cm (Given)

Radius of the circle = OA = 10 cm (Given)

In ΔOCA:

Using Pythagoras theorem,

OA^{2} = AC^{2} + OC^{2}

100 = AC^{2 }+ 25

AC^{2} = 100 – 25 = 75

AC = √75 = 8.66

As, perpendicular from the centre to chord bisects the chord.

Therefore, AC = BC = 8.66 cm

=> AB = AC + BC = 8.66 + 8.66 = 17.32

Answer: AB = 17.32 cm

**Question 3: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.**

**Solution:**

Distance of the chord from the centre = OC = 4 cm (Given)

Radius of the circle = OA = 6 cm (Given)

In ΔOCA:

Using Pythagoras theorem,

OA^{2} = AC^{2} + OC^{2}

36 = AC^{2 }+ 16

AC^{2} = 36 – 16 = 20

AC = √20 = 4.47

Or AC = 4.47cm

As, perpendicular from the centre to chord bisects the chord.

Therefore, AC = BC = 4.47 cm

=> AB = AC + BC = 4.47 + 4.47 = 8.94

Answer: AB = 8.94 cm

**Question 4: Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.**

**Solution:**

Given: AB = 5 cm, CD = 11 cm, PQ = 3 cm

Draw perpendiculars OP on CD and OQ on AB

Let OP = x cm and OC = OA = r cm

We know, perpendicular from centre to chord bisects it.

Since OP⊥CD, we have

CP = PD = 11/2 cm

And OQ⊥AB

AQ = BQ = 5/2 cm

In ΔOCP:

By Pythagoras theorem,

OC^{2} = OP^{2} + CP^{2}

r^{2 }= x^{2 }+ (11/2)^{ 2 }…..(1)

In ΔOQA:

By Pythagoras theorem,

OA^{2}=OQ^{2}+AQ^{2}

r^{2}= (x+3)^{ 2 }+ (5/2)^{ 2} …..(2)

From equations (1) and (2), we get

(x+3)^{ 2 }+ (5/2)^{ 2} = x^{2 }+ (11/2)^{ 2}

Solve above equation and find the value of x.

x^{2} + 6x + 9 + 25/4 = x^{2} + 121/4

(using identity, (a+b)^{ 2} = a^{2} + b^{2} + 2ab )

6x = 121/4 – 25/4 − 9

6x = 15

or x = 15/6 = 5/2

Substitute the value of x in equation (1), and find the length of radius,

r^{2 }= (5/2)^{2 }+ (11/2)^{ 2}

= 25/4 + 121/4

= 146/4

or r = √146/4 cm

**Question 5: Give a method to find the centre of a given circle.**

**Solution:**

Steps of Construction:

Step 1: Consider three points A, B and C on a circle.

Step 2: Join AB and BC.

Step 3: Draw perpendicular bisectors of chord AB and BC which intersect each other at a point, say O.

Step 4: This point O is a centre of the circle, because we know that, the Perpendicular bisectors of chord always pass through the centre.

**Question 6: Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.**

**Solution:**

From figure, Let C is the mid-point of chord AB.

To prove: D is the mid-point of arc AB.

Now, In ΔOAC and ΔOBC

OA = OB [Radius of circle]

OC = OC [Common]

AC = BC [C is the mid-point of chord AB (given)]

So, by SSS condition: ΔOAC ≅ ΔOBC

So, ∠AOC = ∠BOC (BY CPCT)

Therefore, D is the mid-point of arc AB. Hence Proved.

**Question 7: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.**

**Solution:**

Form figure: PQ is a diameter of circle which bisects the chord AB at C. (Given)

To Prove: PQ bisects ∠AOB

Now,

In ∠BOC and ∠AOC

OA = OB [Radius]

OC = OC [Common side]

AC = BC [Given]

Then, by SSS condition: ΔAOC ≅ ΔBOC

So, ∠AOC = ∠BOC [By c.p.c.t.]

Therefore, PQ bisects ∠AOB. Hence proved.

## Exercise 16.3 Page No: 16.40

**Question 1: Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.**

**Solution:**

Let R, S and M be the position of Ishita, Isha and Nisha respectively.

Since OA is a perpendicular bisector on RS, so AR = AS = 24/2 = 12 cm

Radii of circle = OR = OS = OM = 20 cm (Given)

In ΔOAR:

By Pythagoras theorem,

OA^{2}+AR^{2}=OR^{2}

OA^{2}+12^{2}=20^{2}

OA^{2 }= 400 – 144 = 256

Or OA = 16 m …(1)

From figure, OABC is a kite since OA = OC and AB = BC. We know that, diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

So in ΔRSM, ∠RCS = 90^{0} and RC = CM …(2)

Now, Area of ΔORS = Area of ΔORS

=>1/2×OA×RS = 1/2 x RC x OS

=> OA ×RS = RC x OS

=> 16 x 24 = RC x 20

=> RC = 19.2

Since RC = CM (from (2), we have

RM = 2(19.2) = 38.4

So, the distance between Ishita and Nisha is 38.4 m.

**Question 2: A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.**

**Solution:**

Since, AB = BC = CA. So, ABC is an equilateral triangle

Radius = OA = 40 m (Given)

We know, medians of equilateral triangle pass through the circumcentre and intersect each other at the ratio 2 : 1.

Here AD is the median of equilateral triangle ABC, we can write:

OA/OD = 2/1

or 40/OD = 2/1

or OD = 20 m

Therefore, AD = OA + OD = (40 + 20) m = 60 m

Now, In ΔADC:

By Pythagoras theorem,

AC^{2} = AD^{2} + DC^{2}

AC^{2} = 60^{2} + (AC/2)^{ 2}

AC^{2 }= 3600 + AC^{2} / 4

3/4 AC^{2 }= 3600

AC^{2 }= 4800

or AC = 40√3 m

Therefore, length of string of each phone will be 40√3 m.

## Exercise 16.4 Page No: 16.60

**Question 1: In figure, O is the centre of the circle. If ∠APB = 50 ^{0}, find ∠AOB and ∠OAB.**

**Solution:**

∠APB = 50^{0} (Given)

By degree measure theorem: ∠AOB = 2∠APB

∠AOB = 2 × 50^{0} = 100^{0}

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=180^{0}

=> m + m + 100^{0} = 180^{0}

=>2m = 180^{0} – 100^{0} = 80^{0}

=>m = 80^{0}/2 = 40^{0}

∠OAB = ∠OBA = 40^{0}

**Question 2: In figure, it is given that O is the centre of the circle and ∠AOC = 150 ^{0}. Find ∠ABC.**

**Solution:**

∠AOC = 150^{0} (Given)

By degree measure theorem: ∠ABC = (reflex∠AOC)/2 …(1)

We know, ∠AOC + reflex(∠AOC) = 360^{0} [Complex angle]

150^{0} + reflex∠AOC = 360^{0}

or reflex ∠AOC = 360^{0}−150^{0} = 210^{0}

From (1) => ∠ABC = 210^{ o} /2 = 105^{o}

**Question 3: In figure, O is the centre of the circle. Find ∠BAC. **

** **

** Solution:**

Given: ∠AOB = 80^{0} and ∠AOC = 110^{0}

Therefore, ∠AOB+∠AOC+∠BOC=360^{0} [Completeangle]

Substitute given values,

80^{0} + 100^{0} + ∠BOC = 360^{0}

∠BOC = 360^{0} – 80^{0} – 110^{0} = 170^{0}

or ∠BOC = 170^{0}

Now, by degree measure theorem

∠BOC = 2∠BAC

170^{0} = 2∠BAC

Or ∠BAC = 170^{0}/2 = 85^{0}

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

**(i)**

**Solution:**

∠AOC = 135^{0} (Given)

From figure, ∠AOC + ∠BOC = 180^{0} [Linear pair of angles]

135^{0} +∠BOC = 180^{0}

or ∠BOC=180^{0}−135^{0}

or ∠BOC=45^{0}

Again, by degree measure theorem

∠BOC = 2∠CPB

45^{0} = 2x

x = 45^{0}/2

**(ii)**

**Solution:**

∠ABC=40^{0} (given)

∠ACB = 90^{0} [Angle in semicircle]

In ΔABC,

∠CAB+∠ACB+∠ABC=180^{0} [angle sum property]

∠CAB+90^{0}+40^{0}=180^{0}

∠CAB=180^{0}−90^{0}−40^{0}

∠CAB=50^{0}

Now, ∠CDB = ∠CAB [Angle is on same segment]

This implies, x = 50^{0}

**(iii)**

**Solution:**

∠AOC = 120^{0} (given)

By degree measure theorem: ∠AOC = 2∠APC

120^{0 }= 2∠APC

∠APC = 120^{0}/2 = 60^{0}

Again, ∠APC + ∠ABC = 180^{0} [Sum of opposite angles of cyclic quadrilaterals = 180 ^{o} ]

60^{0} + ∠ABC=180^{0}

∠ABC=180^{0}−600

∠ABC = 120^{0}

∠ABC + ∠DBC = 180^{0} [Linear pair of angles]

120^{0 }+ x = 180^{0}

x = 180^{0}−120^{0}=60^{0}

The value of x is 60^{0}

**(iv) **

**Solution:**

∠CBD = 65^{0} (given)

From figure:

∠ABC + ∠CBD = 180^{0} [ Linear pair of angles]

∠ABC + 65^{0} = 180^{0}

∠ABC =180^{0}−65^{0}=1150

Again, reflex ∠AOC = 2∠ABC [Degree measure theorem]

x=2(115^{0}) = 230^{0}

The value of x is 230^{0}

**(v)**

**Solution:**

∠OAB = 35^{0} (Given)

From figure:

∠OBA = ∠OAB = 35^{0} [Angles opposite to equal radii]

InΔAOB:

∠AOB + ∠OAB + ∠OBA = 180^{0} [angle sum property]

∠AOB + 35^{0} + 35^{0} = 180^{0}

∠AOB = 180^{0} – 35^{0} – 35^{0} = 110^{0}

Now, ∠AOB + reflex∠AOB = 360^{0} [Complex angle]

110^{0} + reflex∠AOB = 360^{0}

reflex∠AOB = 360^{0} – 110^{0} = 250^{0}

By degree measure theorem: reflex ∠AOB = 2∠ACB

250^{0} = 2x

x = 250^{0}/2=125^{0}

**(vi)**

**Solution:**

∠AOB = 60^{o} (given)

By degree measure theorem: reflex∠AOB = 2∠OAC

60^{ o} = 2∠ OAC

∠OAC = 60^{ o} / 2 = 30^{ o} [Angles opposite to equal radii]

Or x = 30^{0}

**(vii)**

**Solution:**

∠BAC = 50^{0} and ∠DBC = 70^{0} (given)

From figure:

∠BDC = ∠BAC = 50^{0 }[Angle on same segment]

Now,

In ΔBDC:

Using angle sum property, we have

∠BDC+∠BCD+∠DBC=180^{0}

Substituting given values, we get

50^{0} + x^{0} + 70^{0} = 180^{0}

x^{0 }= 180^{0}−50^{0}−70^{0}=60^{0}

or x = 60^{o} . Answer!!

**(viii) **

**Solution:**

∠DBO = 40^{0} (Given)

Form figure:

∠DBC = 90^{0} [Angle in a semicircle]

∠DBO + ∠OBC = 90^{0}

40^{0}+∠OBC=90^{0}

or ∠OBC=90^{0}−40^{0}=50^{0}

Again, By degree measure theorem: ∠AOC = 2∠OBC

or x = 2×50^{0}=100^{0}

**(ix)**

**Solution:**

∠CAD = 28, ∠ADB = 32 and ∠ABC = 50 (Given)

From figure:

In ΔDAB:

Angle sum property: ∠ADB + ∠DAB + ∠ABD = 180^{0}

By substituting the given values, we get

32^{0} + ∠DAB + 50^{0} = 180^{0}

∠DAB=180^{0}−32^{0}−50^{0}

∠DAB = 98^{0}

Now,

∠DAB+∠DCB=180^{0 } [Opposite angles of cyclic quadrilateral, their sum = 180 degrees]

98^{0}+x=180^{0}

or x = 180^{0}−98^{0}=82^{0}

The value of x is 82 degrees.

**(x)**

**Solution:**

∠BAC = 35^{0} and ∠DBC = 65^{0}

From figure:

∠BDC = ∠BAC = 35^{0} [Angle in same segment]

In ΔBCD:

Angle sum property, we have

∠BDC + ∠BCD + ∠DBC = 180^{0}

35^{0} + x + 65^{0} = 180^{0}

or x = 180^{0} – 35^{0} – 65^{0} = 80^{0}

**(xi)**

**Solution:**

∠ABD = 40^{0}, ∠CPD = 110^{0} (Given)

Form figure:

∠ACD = ∠ABD = 40^{0} [Angle in same segment]

In ΔPCD,

Angle sum property: ∠PCD+∠CPO+∠PDC=180^{0}

400 + 110^{0} + x = 180^{0}

x=180^{0}−150^{0 }=30^{0}

The value of x is 30 degrees.

**(xii)**

**Solution:**

∠BAC = 52^{0} (Given)

From figure:

∠BDC = ∠BAC = 52^{0} [Angle in same segment]

Since OD = OC (radii), then ∠ODC = ∠OCD [Opposite angle to equal radii]

So, x = 52^{0}

**Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.**

**Solution:**

In ΔOBD and ΔOCD:

OB = OC [Radius]

∠ODB = ∠ODC [Each 90^{0}]

OD = OD [Common]

Therefore, By RHS Condition

ΔOBD ≅ ΔOCD

So, ∠BOD = ∠COD…..(i)[By CPCT]

Again,

By degree measure theorem: ∠BOC = 2∠BAC

2∠BOD = 2∠BAC [Using(i)]

∠BOD = ∠BAC

Hence proved.

**Question 6: In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC. **

** **

**Solution:**

Since, BO is the bisector of ∠ABC, then,

∠ABO = ∠CBO …..(i)

From figure:

Radius of circle = OB = OA = OB = OC

∠OAB = ∠OCB …..(ii) [opposite angles to equal sides]

∠ABO = ∠DAB …..(iii) [opposite angles to equal sides]

From equations (i), (ii) and (iii), we get

∠OAB = ∠OCB …..(iv)

In ΔOAB and ΔOCB:

∠OAB = ∠OCB [From (iv)]

OB = OB [Common]

∠OBA = ∠OBC [Given]

Then, By AAS condition : ΔOAB ≅ ΔOCB

So, AB = BC [By CPCT]

**Question 7: In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.**

**Solution: **

From the figure:

∠3 = ∠4 ….(i) [Angles in same segment]

∠x = 2∠3 [By degree measure theorem]

∠x = ∠3 + ∠3

∠x = ∠3 + ∠4 (Using (i) ) …..(ii)

Again, ∠y = ∠3 + ∠1 [By exterior angle property]

or ∠3 = ∠y − ∠1 …..(iii)

∠4 = ∠z + ∠1 …. (iv) [By exterior angle property]

Now, from equations (ii) , (iii) and (iv), we get

∠x = ∠y − ∠1 + ∠z + ∠1

or ∠x = ∠y + ∠z + ∠1 − ∠1

or x = ∠y + ∠z

Hence proved.

## Exercise 16.5 Page No: 16.83

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

**Solution:**

ΔABC is an equilateral triangle. (Given)

Each angle of an equilateral triangle is 60 degrees.

In quadrilateral ABEC:

∠BAC + ∠BEC = 180^{o} (Opposite angles of quadrilateral)

60^{o} + ∠BEC = 180^{ o}

∠BEC = 180^{ o} – 60^{ o}

∠BEC = 120^{ o}

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

**Solution:**

Given: ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°

In ΔPQR:

∠PQR = ∠PRQ = 35^{o} (Angle opposite to equal sides)

Again, by angle sum property

∠P + ∠Q + ∠R = 180^{ o}

∠P + 35^{ o} + 35^{ o} = 180^{ o}

∠P + 70^{ o} = 180^{ o}

∠P = 180^{ o} – 70^{ o}

∠P = 110^{ o}

Now, in quadrilateral SQTR,

∠QSR + ∠QTR = 180^{ o} (Opposite angles of quadrilateral)

110^{ o} + ∠QTR = 180^{ o}

∠QTR = 70^{ o}

** Question 3: In figure, O is the centre of the circle. If ∠BOD = 160 ^{o}, find the values of x and y.**

**Solution: **

From figure: ∠BOD = 160^{ o}

By degree measure theorem: ∠BOD = 2 ∠BCD

160** ^{ o}** = 2x

or x = 80^{ o}

Now, in quadrilateral ABCD,

∠BAD + ∠BCD = 180** ^{ o}** (Opposite angles of Cyclic quadrilateral)

y + x = 180^{ o}

Putting value of x,

y + 80** ^{ o}** = 180

^{ o}y = 100^{ o}

Answer: x = 80** ^{ o} **and y = 100

**.**

^{ o}**Question 4: In figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100 ^{o} and ∠ABD = 70^{o}, find ∠ADB.**

**Solution: **

From figure:

In quadrilateral ABCD,

∠DCB + ∠BAD = 180** ^{o}** (Opposite angles of Cyclic quadrilateral)

100** ^{ o}** + ∠BAD = 180

^{o}∠BAD = 80^{0}

In Δ BAD:

By angle sum property: ∠ADB + ∠DAB + ∠ABD = 180^{ o}

∠ADB + 80^{o} + 70^{ o} = 180^{ o}

∠ADB = 30^{o}

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

**Solution:**

Given: ABCD is a cyclic quadrilateral with AD ‖ BC

=> ∠A + ∠C = 180^{o} ………(1)

and ∠A + ∠B = 180^{o} ………(2)

Form (1) and (2), we have

∠B = ∠C

Hence proved.

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

**Solution: **

Given: ∠BOC = 100^{o}

By degree measure theorem: ∠AOC = 2 ∠APC

100^{ o} = 2 ∠APC

or ∠APC = 50^{ o}

Again,

∠APC + ∠ABC = 180^{ o} (Opposite angles of a cyclic quadrilateral)

50^{o} + ∠ABC = 180^{ o}

or ∠ABC = 130^{ o}

Now, ∠ABC + ∠CBD = 180^{ o} (Linear pair)

130^{o} + ∠CBD = 180^{ o}

or ∠CBD = 50^{ o}

**Question 7: In figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50 ^{0}, find ∠AOC.**

**Solution: **

Given: ∠OBD = 50^{0}

Here, AB and CD are the diameters of the circles with centre O.

∠DBC = 90** ^{0}** ….(i)

Also, ∠DBC = 50** ^{0}** + ∠OBC

90** ^{0}** = 50

**+ ∠OBC**

^{0}or ∠OBC = 40^{0}

Again, By degree measure theorem: ∠AOC = 2 ∠ABC

∠AOC = 2∠OBC = 2 x 40** ^{0}** = 80

^{0}**Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 30 ^{0}. Find m(∠ACB) and m(∠ABC).**

**Solution: **

Given: m(∠CAB)= 30^{0}

To Find: m(∠ACB) and m(∠ABC).

Now,

∠ACB = 90** ^{0}** (Angle in semi-circle)

Now,

In △ABC, by angle sum property: ∠CAB + ∠ACB + ∠ABC = 180^{0}

30** ^{0}** + 90

**+ ∠ABC = 180**

^{0}

^{0}∠ABC = 60^{0}

Answer: ∠ACB = 90** ^{0}** and ∠ABC = 60

^{0}**Question 9: In a cyclic quadrilateral ABCD if AB||CD and B = 70 ^{o} , find the remaining angles.**

**Solution: **

A cyclic quadrilateral ABCD with AB||CD and B = 70^{o}.

∠B + ∠C = 180^{o} (Co-interior angle)

70^{0} + ∠C = 180^{0}

∠C = 110^{0}

And,

=> ∠B + ∠D = 180^{0} (Opposite angles of Cyclic quadrilateral)

70^{0} + ∠D = 180^{0}

∠D = 110^{0}

Again, ∠A + ∠C = 180^{0} (Opposite angles of cyclic quadrilateral)

∠A + 110^{0} = 180^{0}

∠A = 70^{0}

Answer: ∠A = 70^{0} , ∠C = 110^{0 }and ∠D = 110^{0}

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

∠A + ∠C = 180^{o} …..(1)

Since m ∠A = 3(m∠C) (given)

=> ∠A = 3∠C …(2)

Equation (1) => 3∠C + ∠C = 180^{ o}

or 4∠C = 180^{o}

or ∠C = 45^{o}

From equation (2)

∠A = 3 x 45^{o} = 135^{o}

**Question 11: In figure, O is the centre of the circle ∠DAB = 50°. Calculate the values of x and y.**

**Solution: **

Given : ∠DAB = 50^{o}

By degree measure theorem: ∠BOD = 2 ∠BAD

so, x = 2( 50^{0}) = 100^{0}

Since, ABCD is a cyclic quadrilateral, we have

∠A + ∠C = 180^{0}

50^{0} + y = 180^{0}

y = 130^{0}

## Exercise VSAQs Page No: 16.89

**Question 1: In figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.**

**Solution: **

By degree measure theorem: ∠AOB = 2 ∠APB

so, ∠AOB = 2 × 70° = 140°

Since AOBC is a cyclic quadrilateral, we have

∠ACB + ∠AOB = 180°

∠ACB + 140° = 180°

∠ACB = 40°

**Question 2: In figure, two congruent circles with centres O and O’ intersect at A and B. If ∠AO’B = 50°, then find ∠APB.**

**Solution: **

As we are given that, both the triangle are congruent which means their corresponding angles are equal.

Therefore, ∠AOB = AO’B = 50°

Now, by degree measure theorem, we have

∠APB = ∠AOB/2 = 25^{0}

**Question 3: In figure, ABCD is a cyclic quadrilateral in which ∠BAD=75°, ∠ABD=58° and ∠ADC=77°, AC and BD intersect at P. Then, find ∠DPC.**

**Solution**:

∠DBA = ∠DCA = 58^{0} …(1)

ABCD is a cyclic quadrilateral :

Sum of opposite angles = 180 degrees

∠A +∠C = 180^{0}

75^{0} + ∠C = 180^{0}

∠C = 105^{0}

Again, ∠ACB + ∠ACD = 105^{0}

∠ACB + 58^{0} = 105^{0}

or ∠ACB = 47^{0} …(2)

Now, ∠ACB = ∠ADB = 47^{0}

Also, ∠D = 77^{0} (Given)

Again From figure, ∠BDC + ∠ADB = 77^{0}

∠BDC + 47^{0} = 77^{0}

∠BDC = 30^{0}

In triangle DPC

∠PDC + ∠DCP + ∠DPC = 180^{0}

30^{0} + 58^{0} + ∠DPC = 180^{0}

or ∠DPC = 92^{0 }. Answer!!

**Question 4: In figure, if ∠AOB = 80° and ∠ABC=30°, then find ∠CAO.**

**Solution: **

Given: ∠AOB = 80^{0} and ∠ABC = 30^{0}

To find: ∠CAO

Join OC.

Central angle subtended by arc AC = ∠COA

then ∠COA = 2 x ∠ABC = 2 x 30^{0} = 60^{0} …(1)

In triangle OCA,

OC = OA

[same radii]∠OCA = ∠CAO …(2)

[Angle opposite to equal sides]In triangle COA,

∠OCA + ∠CAO + ∠COA = 180^{0}

From (1) and (2), we get

2∠CAO + 60^{0 }= 180^{0}

∠CAO = 60^{0}

## RD Sharma Solutions For Class 9 Maths Chapter 16 Exercises:

Get detailed solutions for all the questions listed under below exercises:

## RD Sharma Solutions for Chapter 16 Circles

In the 16th chapter of Class 9, RD Sharma Solutions students will study important concepts. Some are listed below:

- Circle Introduction
- Position of a point with respect to a circle
- Circular Disc
- Concentric Circles
- Degree measure of an arc
- Chord and segment of a circle
- Congruence of circles and arcs