RD Sharma Solutions Class 9 Maths Chapter 16 – Circles (Updated for 2024)

RD Sharma Solutions Class 9 Maths Chapter 16 – Circles: RD Sharma Solutions Class 9 Maths Chapter 16 enables them also to study all topics more effectively & clear their doubts quickly. The students who want to obtain higher marks in the exams must regularly practice mathematics subject exercises with the help of RD Sharma Solutions Class 9 Maths Chapter 16 Circles. This also builds the confidence of the students and helps them to revise each important topic during the exam.

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RD Sharma Class 9 Solutions Chapter 16

Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 16 Circles

RD Sharma class 9 chapter 16 exercise 16a

RD Sharma class 9 chapter 16 exercise 16b

RD Sharma class 9 chapter 16 exercise 16c

RD Sharma class 9 chapter 16 exercise 16d

Access answers of RD Sharma Solutions Class 9 Maths Chapter 16

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

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 the centre to a chord bisects the chord

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

In right ΔOCA:

Using Pythagoras’ theorem,

OA² = AC² + OC²

64 = 36 + OC²

OC² = 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,

OA² = AC² + OC²

100 = AC² + 25

AC² = 100 – 25 = 75

AC = √75 = 8.66

As perpendicular from the centre to the 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,

OA² = AC² + OC²

36 = AC² + 16

AC² = 36 – 16 = 20

AC = √20 = 4.47

Or AC = 4.47cm

As perpendicular from the centre to the 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, and CD of lengths 5 cm, and 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 the centre to the 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,

OC² = OP² + CP²

r² = x² + (11/2) ² …..(1)

In ΔOQA:

By Pythagoras’ theorem,

OA²=OQ²+AQ²

r²= (x+3) ² + (5/2) ² …..(2)

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

(x+3) ² + (5/2) ² = x² + (11/2) ²

Solve the above equation and find the value of x.

x² + 6x + 9 + 25/4 = x² + 121/4

(using identity, (a+b) ² = a² + b² + 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 the radius,

r² = (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 the centre of the circle because we know that, the Perpendicular bisectors of a 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 the 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 the 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 the diameter of the circle which bisects the chord AB at C. (Given)

To Prove: PQ bisects ∠AOB

Now,

In ΔBOC and ΔAOC

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.

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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,

OA²+AR²=OR²

OA²+12²=20²

OA² = 400 – 144 = 256

Or OA = 16 m …(1)

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

So in ΔRSM, ∠RCS = 90⁰ 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 a radius of 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distances 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 triangles pass through the circumcentre and intersect each other at the ratio of 2:1.

Here AD is the median of an 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,

AC² = AD² + DC²

AC² = 60² + (AC/2) ²

AC² = 3600 + AC² / 4

3/4 AC² = 3600

AC² = 4800

or AC = 40√3 m

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

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Exercise 16.4 Page No: 16.60

Question 1: In the figure, O is the centre of the circle. If ∠APB = 50⁰, find ∠AOB and ∠OAB.

Solution:

∠APB = 50⁰ (Given)

By degree measure theorem: ∠AOB = 2∠APB

∠AOB = 2 × 50⁰ = 100⁰

Again, OA = OB [Radius of the 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 figure, it is given that O is the centre of the circle and ∠AOC = 150⁰. Find ∠ABC.

Solution:

∠AOC = 150⁰ (Given)

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

We know, ∠AOC + reflex(∠AOC) = 360⁰ [Complex angle]

150⁰ + reflex∠AOC = 360⁰

or reflex ∠AOC = 360⁰−150⁰ = 210⁰

From (1) => ∠ABC = 210 ^o /2 = 105^o

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

Solution:

Given: ∠AOB = 80⁰ and ∠AOC = 110⁰

Therefore, ∠AOB+∠AOC+∠BOC=360⁰ [Completeangle]

Substitute given values,

80⁰ + 100⁰ + ∠BOC = 360⁰

∠BOC = 360⁰ – 80⁰ – 110⁰ = 170⁰

or ∠BOC = 170⁰

Now, by degree measure theorem

∠BOC = 2∠BAC

170⁰ = 2∠BAC

Or ∠BAC = 170⁰/2 = 85⁰

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

(i)

Solution:

∠AOC = 135⁰ (Given)

From the figure, ∠AOC + ∠BOC = 180⁰ [Linear pair of angles]

135⁰ +∠BOC = 180⁰

or ∠BOC=180⁰−135⁰

or ∠BOC=45⁰

Again, by degree measure theorem

∠BOC = 2∠CPB

45⁰ = 2x

x = 45⁰/2

(ii)

Solution:

∠ABC=40⁰ (given)

∠ACB = 90⁰ [Angle in semicircle]

In ΔABC,

∠CAB+∠ACB+∠ABC=180⁰ [angle sum property]

∠CAB+90⁰+40⁰=180⁰

∠CAB=180⁰−90⁰−40⁰

∠CAB=50⁰

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

This implies, x = 50⁰

(iii)

Solution:

∠AOC = 120⁰ (given)

By degree measure theorem: ∠AOC = 2∠APC

120⁰ = 2∠APC

∠APC = 120⁰/2 = 60⁰

Again, ∠APC + ∠ABC = 180⁰ [Sum of opposite angles of cyclic quadrilaterals = 180 ^o ]

60⁰ + ∠ABC=180⁰

∠ABC=180⁰−60⁰

∠ABC = 120⁰

∠ABC + ∠DBC = 180⁰ [Linear pair of angles]

120⁰ + x = 180⁰

x = 180⁰−120⁰=60⁰

The value of x is 60⁰

(iv)

Solution:

∠CBD = 65⁰ (given)

From figure:

∠ABC + ∠CBD = 180⁰ [ Linear pair of angles]

∠ABC + 65⁰ = 180⁰

∠ABC =180⁰−65⁰=115⁰

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

x=2(115⁰) = 230⁰

The value of x is 230⁰

(v)

Solution:

∠OAB = 35⁰ (Given)

From figure:

∠OBA = ∠OAB = 35⁰ [Angles opposite to equal radii]

InΔAOB:

∠AOB + ∠OAB + ∠OBA = 180⁰ [angle sum property]

∠AOB + 35⁰ + 35⁰ = 180⁰

∠AOB = 180⁰ – 35⁰ – 35⁰ = 110⁰

Now, ∠AOB + reflex∠AOB = 360⁰ [Complex angle]

110⁰ + reflex∠AOB = 360⁰

reflex∠AOB = 360⁰ – 110⁰ = 250⁰

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

250⁰ = 2x

x = 250⁰/2=125⁰

(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 = 30⁰

(vii)

Solution:

∠BAC = 50⁰ and ∠DBC = 70⁰ (given)

From figure:

∠BDC = ∠BAC = 50⁰ [Angle on the same segment]

Now,

In ΔBDC:

Using the angle sum property, we have

∠BDC+∠BCD+∠DBC=180⁰

Substituting given values, we get

50⁰ + x⁰ + 70⁰ = 180⁰

x⁰ = 180⁰−50⁰−70⁰=60⁰

or x = 60^o

(viii)

Solution:

∠DBO = 40⁰ (Given)

Form figure:

∠DBC = 90⁰ [Angle in a semicircle]

∠DBO + ∠OBC = 90⁰

40⁰+∠OBC=90⁰

or ∠OBC=90⁰−40⁰=50⁰

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

or x = 2×50⁰=100⁰

(ix)

Solution:

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

From figure:

In ΔDAB:

Angle sum property: ∠ADB + ∠DAB + ∠ABD = 180⁰

By substituting the given values, we get

32⁰ + ∠DAB + 50⁰ = 180⁰

∠DAB=180⁰−32⁰−50⁰

∠DAB = 98⁰

Now,

∠DAB+∠DCB=180⁰ [Opposite angles of cyclic quadrilateral, their sum = 180 degrees]

98⁰+x=180⁰

or x = 180⁰−98⁰=82⁰

The value of x is 82 degrees.

(x)

Solution:

∠BAC = 35⁰ and ∠DBC = 65⁰

From figure:

∠BDC = ∠BAC = 35⁰ [Angle in the same segment]

In ΔBCD:

Angle sum property, we have

∠BDC + ∠BCD + ∠DBC = 180⁰

35⁰ + x + 65⁰ = 180⁰

or x = 180⁰ – 35⁰ – 65⁰ = 80⁰

(xi)

Solution:

∠ABD = 40⁰, ∠CPD = 110⁰ (Given)

Form figure:

∠ACD = ∠ABD = 40⁰ [Angle in the same segment]

In ΔPCD,

Angle sum property: ∠PCD+∠CPO+∠PDC=180⁰

400 + 110⁰ + x = 180⁰

x=180⁰−150⁰ =30⁰

The value of x is 30 degrees.

(xii)

Solution:

∠BAC = 52⁰ (Given)

From figure:

∠BDC = ∠BAC = 52⁰ [Angle in the same segment]

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

So, x = 52⁰

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 90⁰]

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 the figure, O is the centre of the circle, and 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 the 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.

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Exercise 16.5 Page No: 16.83

Question 1: In the 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 the 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 the 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 the value of x,

y + 80 ^o = 180 ^o

y = 100 ^o

Answer: x = 80 ^oand y = 100 ^o.

Question 4: In the 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 = 80⁰

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)

[Opposite angles of cyclic quadrilateral]

and ∠A + ∠B = 180^o ………(2)

[Co-interior angles]

Forms (1) and (2), we have

∠B = ∠C

Hence proved.

Question 6: In the 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 the figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50⁰, find ∠AOC.

Solution:

Given: ∠OBD = 50⁰

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

∠DBC = 90⁰ ….(i)

[Angle in the semi-circle]

Also, ∠DBC = 50⁰ + ∠OBC

90⁰ = 50⁰ + ∠OBC

or ∠OBC = 40⁰

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

∠AOC = 2∠OBC = 2 x 40⁰ = 80⁰

Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 30⁰. Find m(∠ACB) and m(∠ABC).

Solution:

Given: m(∠CAB)= 30⁰

To Find: m(∠ACB) and m(∠ABC).

Now,

∠ACB = 90⁰ (Angle in a semi-circle)

Now,

In △ABC, by angle sum property: ∠CAB + ∠ACB + ∠ABC = 180⁰

30⁰ + 90⁰ + ∠ABC = 180⁰

∠ABC = 60⁰

Answer: ∠ACB = 90⁰ and ∠ABC = 60⁰

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)

70⁰ + ∠C = 180⁰

∠C = 110⁰

And,

=> ∠B + ∠D = 180⁰ (Opposite angles of Cyclic quadrilateral)

70⁰ + ∠D = 180⁰

∠D = 110⁰

Again, ∠A + ∠C = 180⁰ (Opposite angles of a cyclic quadrilateral)

∠A + 110⁰ = 180⁰

∠A = 70⁰

Answer: ∠A = 70⁰ , ∠C = 110⁰ and ∠D = 110⁰

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

∠A + ∠C = 180^o …..(1)

[Opposite angles of cyclic quadrilateral]

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 the 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( 50⁰) = 100⁰

Since ABCD is a cyclic quadrilateral, we have

∠A + ∠C = 180⁰

50⁰ + y = 180⁰

y = 130⁰

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Exercise VSAQs Page No: 16.89

Question 1: In the 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 the 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, both triangles are congruent which means their corresponding angles are equal.

Therefore, ∠AOB = AO’B = 50°

Now, by the degree measure theorem, we have

∠APB = ∠AOB/2 = 25⁰

Question 3: In the 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 = 58⁰ …(1)

[Angles in the same segment]

ABCD is a cyclic quadrilateral :

The sum of opposite angles = 180 degrees

∠A +∠C = 180⁰

75⁰ + ∠C = 180⁰

∠C = 105⁰

Again, ∠ACB + ∠ACD = 105⁰

∠ACB + 58⁰ = 105⁰

or ∠ACB = 47⁰ …(2)

Now, ∠ACB = ∠ADB = 47⁰

[Angles in the same segment]

Also, ∠D = 77⁰ (Given)

Again From figure, ∠BDC + ∠ADB = 77⁰

∠BDC + 47⁰ = 77⁰

∠BDC = 30⁰

In triangle DPC

∠PDC + ∠DCP + ∠DPC = 180⁰

30⁰ + 58⁰ + ∠DPC = 180⁰

or ∠DPC = 92⁰ 

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

Solution:

Given: ∠AOB = 80⁰ and ∠ABC = 30⁰

To find: ∠CAO

Join OC.

Central angle subtended by arc AC = ∠COA

then ∠COA = 2 x ∠ABC = 2 x 30⁰ = 60⁰ …(1)

In triangle OCA,

OC = OA

[same radii]

∠OCA = ∠CAO …(2)

[Angle opposite to equal sides]

In triangle COA,

∠OCA + ∠CAO + ∠COA = 180⁰

From (1) and (2), we get

2∠CAO + 60⁰ = 180⁰

∠CAO = 60⁰

Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise

• RD Sharma class 9 chapter 16 exercise 16a: This exercise is based on the topics related to the position of a point with respect to a circle, Concentric Circles, Length of an arc, Circular Disc, Minor & Major Arc, & many more. 
• RD Sharma class 9 chapter 16 exercise 16b: This exercise includes topics related to the congruence of circles & arcs. Practising RD Sharma’s class 9 chapter 16 exercise 16b questions enable the students to build an in-depth understanding of the basics of a circle. 
• RD Sharma class 9 chapter 16 exercise 16c: This exercise includes topics based on a few important results on equal chords. The solutions of RD Sharma class 9 chapter 16 exercise 16c are helpful to solve several circle problems. The students will study more circles concepts in detail in order to obtain better marks in the exam.
• RD Sharma class 9 chapter 16 exercise 16d: This exercise includes topics based on arcs & angles subtended by them. Each answer is created by subject experts that assist the students to understand this concept in a better and more fun way. The students can practice RD Sharma class 9 chapter 16 exercise 16d & clear their doubts to secure excellent marks in the final mathematics exam.

Important Topics in the Exercise

RD Sharma Solutions Class 9 Maths Chapter 16 Circles include some important topics that are listed below:

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

This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 16. If you have any doubts regarding the CBSE Class 9 Maths exam, ask in the comments.

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