Parents can use ICSE Class 9 Maths Solutions S Chand Chapter 13 Circle Ex 13(B) to provide additional support to their children.

## S Chand Class 9 ICSE Maths Solutions Chapter 13 Circle Ex 13(B)

Question 1.

Find:

Solution:

(i) ∵ Angles at a point = 360°

∴ 4x – 2 + 6x + 6 + 7x – 18 = 360°

⇒ 17x – 14 = 360°

⇒ 17x = 360° + 14

⇒ 17x = 374°

⇒ x = \(\frac { 374° }{ 17 }\) = 22°

Now, m\(\overparen{Q R}\) = 7x – 18 = 7 x 22 – 18°

= 154°- 18° = 136°

(ii) ∵ AD is the diameter of the circle with centre O

∴ m∠AOD = 180°

∴ 10y = 180° ⇒ y = \(\frac { 180° }{ 10 }\)= 18°

Now, m∠BOC = 6y = 6 x 18 = 108°

(iii) In the figure,

\(\overparen{A C}\) = \(\overparen{B C}\)

∴ ∠AOC = ∠BOC

∴ 8y – 8 = 6y

⇒ 8y – 6y = 8

⇒ 2y = 8

⇒ y = 4

∴ m∠AOC = 8y – 8 = 8 x 4 – 8 = 32 – 8 = 24°

∴ m∠AOC = 24°

(iv) In the given figure,

\(\odot\) A = \(\odot\) B

⇒ Circles with centres A and B are equal

and \(\overparen{\mathrm{CD}}=\overparen{\mathrm{EF}}\)

45 – 6x = 9x ⇒ 45 = 9x + 6x

⇒ 15x = 45 ⇒ x = \(\frac { 45 }{ 15 }\) = 3

Now, m∠EBF = 9x = 9 x 3 = 27°

OR

If m∠CAD = 45 + 6x, then

45 + 6x = 9x ⇒ 45 = 9x – 6x

⇒ 3x = 45 ⇒ x = \(\frac { 45 }{ 15 }\) = 15

m(∠EBF) = 9x = 9 x 15 = 135°

(v) In the given figure, a circle with centre O and QT and PS are diameters

\(m \overparen{\mathrm{PR}}=\overparen{\mathrm{PQ}}+m \overparen{\mathrm{QR}}\)

= m\(\overparen{S T}\) + mQR

(∵ ∠POQ = ∠SOT vertically opposite angles)

= 55° + 100° = 155°

and m\(\overparen{PRT}\) = m∠PQ + m∠QRT

= 55° + 180° = 235°

(vi) In the figure, chord AB = chord CD

∴ 4y + y = y + 68°

(Angles at the centre by two equal chords are equal)

5y = y + 68° ⇒ 5y – y = 68°

⇒ 4y = 68°

⇒ y = \(\frac { 68° }{ 4 }\) = 17°

m\(\overparen{A B}\) = 4y + y = 5y

= 5 x 17° = 85°

Question 2.

∆PQR is inscribed in a circle. ∠P = ∠Q. Prove that PR = QR.

Solution:

∆ABC is inscribed in a circle and ∠P = ∠Q

∴ QR = PR

∵ Equal chords subtend equal angles at the centre

∴ mPR = mQR

Question 3.

Given AB = CD. Prove that AC = DB.

Solution:

In the given figure,

AB = CD

AC and BD are joined

To prove : AC = DB

∵ AB = DC

∴ m\(\overparen{A B}\) = m\(\overparen{D C}\)

Subtracting m\(\overparen{A D}\) from both sides

∴ m\(\overparen{A B}\) – \(\overparen{A D}\) = m\(\overparen{D C}\) – m\(\overparen{A D}\)

⇒ m\(\overparen{A C}\) = m\(\overparen{D B}\)

⇒ AC = DB

Hence proved.

Question 4.

Given AC = BD. Prove that AB = CD.

Solution:

In the figure, AC = DB

To prove : AB = DC

∵ AC = DB

∴ \(\frac { 1 }{ 2 }\)

Adding m\(\overparen{A D}\) to both sides

\(\overparen{\mathrm{AC}}+\overparen{D B}=\overparen{A D}+\overparen{D B}\)

\(\overparen{C D}=\overparen{A B}\)

CD = AB ⇒ CD = AB

Hence AB = CD

Question 5.

In figure, X, Y are the middle points of the arcs AB, AC. Prove that AP = AQ.

Solution:

Given : in the circle,

AB and AC are two arcs.

X and Y are the midpoints of arc AB and arc AC.

XY is joined which meet AB in P and AC in Q

To prove : AP = AQ

Construction : Join AX, AY, BX and BY

Proof: ∵ AB = AC

∴ arc AXB = arc AYC

But X and Y are the midpoints of \(\overparen{A B}\) and \(\overparen{A C}\)

∴ AX = XB and AY = YC

∴ ∠XAY = ∠XBA and ∠YAC = ∠YCA

∴ ∠XAB or ∠XAP = ∠YAQ

Now in ∆XAP and ∆YAQ

AY = AY (common)

∠XAP = ∠YAQ (proved)

∠AXP = ∠AYQ (proved)

∴ ∆XAP ≅ ∆YAQ (AAS axiom)

∴ AP = AQ (c.p.c.t.)

Hence proved.

Question 6.

Circle O with chords AB = BC = CD = DE. Prove that AD = BE.

Solution:

Given : In a circle with centre O, chord AB = BC = CD = DE

AD and BE are joined

To prove : AD = BE

Construction : Join AO, BO, CO, DO and EO

Proof:

∵ AB = BC = CD = DE(given)

∴ \(\overparen{A B}+\overparen{B C}+\overparen{C D}=\overparen{A D}\)

Similarly \(\overparen{B C}+\overparen{C D}+\overparen{D E}=\overparen{B E}\)

\(\overparen{A B}+\overparen{B C}+\overparen{C D}=\overparen{B C}+\overparen{C D}+\overparen{D E}\)

⇒ \(\overparen{\mathrm{AD}}=\overparen{\mathrm{BE}}\)

∴ Central ∠AOD = central ∠BOE Now in ∆AOD and ∆BOE,

OA = OB

OD = OE (radii of the same circle)

∠AOD = ∠BOE (proved)

∴ ∆AOD = ∆BOE (SAS axiom)

∴ AD = BE (c.p.c.t.)

Hence proved.

Question 7.

In figure, APB and CQD are two congruent arcs. Prove that AC || BD.

Solution:

Given : In a circle, arc APB = arc COD

AC and BD are joined

To prove : AC = BD

Construction : Join AD

Proof: ∵ Arc APB = arc CQD (given)

∴ ∠ADB = ∠CAD

{Equal arcs subtends equal angles at the circumference}

But these are alternate angles

∴ AC || BD

Hence proved.

Question 8.

In figure, ABC is equilateral, P and S are midpoints of arcs AB and AC.

Prove that PQ = QR = RS.

Solution:

In the figure, equilateral ∆ABC is inscribed in a circle

P and S are midpoints of arcs AB and AC respectively

To prove : PQ = QR = RS

Construction : Join AP, BP, AS and CS

Proof: ∵ P is the midpoint of arc AB and S

is the midpoint of arc AC

∴ AP = PB and AS = SC

∠PAB = ∠PBA and ∠CAS = ∠SAC

But AB = AC (Side of equilateral triangle)

∴ ∠PAB = ∠PBA = ∠CAS = ∠SAC

Now in ∆APQ and ∆ASR,

AP = AS

∠PAQ = ∠SAR (∵ ∠PAB = ∠SAC)

and ∠APQ = ∠ASR (∵ AP = AS)

∴ ∆PAQ ≅ ∆ASR (AAS axiom)

∴ PQ = RS (c.p.c.t) … (i)

Now ∵ ∠PAQ = ∠SAR

Adding ∠QAR to both sides

∴ ∠PAQ + ∠QAR = ∠QAR + ∠SAR

⇒ ∠PAR = ∠SAQ

Now in ∆PAR and ∆SAQ,

AP = AS (proved)

∠PAR = ∠SAQ (proved)

∠APQ = ∠ASR (proved)

∴ ∆PAR ≅ ∆SAQ (AAS axiom)

∴ PR = QS … (ii)

Subtracting (i) and from (ii)

PR – PQ = QS – RS

QR = QR

∴ PQ = QR = RS

Hence proved.

Question 9.

Each side of a regular hexagon, inscribed in a circle subtends an angle of 60° at the centre and is equal to the radius of the circle. Prove it.

Solution:

Given : A regular hexagon ABCDEF inscribed in a circle with centre O

Join AO, BO, CO, DO, EO and FO

To prove : Each angle of hexagon = 60°

Proof: ∵ AB = BC = CD = DE = EF = FA

(Sides of regular hexagon)

∴ \(\overparen{\mathrm{AB}}=\overparen{\mathrm{BC}}=\overparen{\mathrm{CD}}=\overparen{\mathrm{DE}}=\overparen{\mathrm{EF}}=\overparen{\mathrm{FA}}\)

∴ Each arc will subtends angle at the centre

= \(\frac { 360° }{ 6 }\) = 60°

Hence proved.