# OP Malhotra Class 9 ICSE Maths Solutions Chapter 13

## OP Malhotra Class 9 ICSE Maths Solutions Chapter 13 Circle

Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 13, Circle. Here students can easily find step by step solutions of all the problems for Circle, Exercise 13a and 13b Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 13 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

Circle Exercise 13(a) Solution :

Question no – (1)

Solution : In given figure,

OA = 13 cm and OM ⊥ AB

Let, AB = 2x

OM ⊥ AB

M is the midpoint of AB

AM = 2x/2 = x

In OAM,

= OA2 = OM2 + AM2

= (b)2 = 122 + x2

= x2 = 132 – 121

= 169 – 144

= 25

x = 5

Length of chord,

AB = 2 × 5

= 10 cm

Therefore, the length of a chord will be 10 cm.

Question no – (2)

Solution : From the given circle,

CA = 6 cm, cm = 3 cm

let, Length AB = x cm

AMC = 1/2 AB = x cm

CA2 = CM2 + AM2

= 62 = 32 + (x/2)2

=  (x/2)2 = 62 – 32

= x2/4 = 36 – 9

= x2 = 7 × 8

= x2 = 108

= x = √6 × 6 × 3

= 6√3 cm

Therefore, the Length of AB will be 6√3 cm.

Question no – (3)

Solution :

From figure CO is the diameter of circle,

Here, EA = EB = 4 cm

CE = 3 cm

Let, ‘r’ radius of the circle

∴ OC = r and OE = (r – 3)

After join OB, in OBC

= OB2 = BE2 + OE2

= r2 = (4)2 + 9r – 3)2

= r2 = 16 + r2 – 6r + 9

= 6r = 25

= r = 25/6

= 4 1/6 cm

Therefore, the radius of the circle will be 4 1/6 cm.

Question no – (5)

Solution :

Here, the chords are in the opposite sides of center

Draw OL on AB and OM ⊥ CD

In △OCM

= OC2 + OM2 + CM2

= 52 = OM2 + 9

= OM2 = 25 – 9

= 16

= ON = 4

Now, In OLA

= OA2 = OL2 + AL2

= 562 = OL2 + 16

= OL2 = 25 – 16

= 9

= OL = 3 cm

Thus, LM = OL = OM = 3 + 4 = 7 cm

Question no – (7)

Solution :

From given figure, two circles are concentric with center ‘O’ chard AB interest the smallest circle at C and D.

OB = 17 cm, OC = 10 cm

OL ⊥ AB,

In OMB

= OB2 = OM2 + MB2

= (17)2 = (9)2 + MB2

= 289 = 81 + MB2

= MB2 = 289 – 81

= 208

= MB = 208 √208

= 14.42 cm

Question no – (8)

Solution :

From given figure, Two equal centre ‘O’ intersecting each other at A and B.

OO, AB, OA, OB, OA, OB, are joined

AB = 10 cm

OO, 6 cm

Now, MB = 10/2 = 5 cm

and, OM = MO = 6/2 = 3 cm

In △OAM

OA2 = OM2 + AM2

= 32 + 52

= 9 + 25

= 34

= 0A = √34

= 5.83 cm

Therefore, the radius of each circle is 5.83 cm.

Question no – (9)

Solution :

The center of circle is ‘O’

AB is the chord, OM ⊥ AB

AB = 8 cm, cm = 1 cm = OA, = x  (Let)

∴ OC = OA = x

OM = OC = – CM

= (x – 1)

In right OAM,

OA2 = AM2 + OM2

= x2 = 42 + (x – 1)2

= 16 + x2 – 2x + 1

= 2x = 17

= x = 17/2

= 8.5 cm

Hence, the x will be 8.5 cm.

Question no – (10)

Solution :

Here center of circle is ‘O’

AB is chord and CD is perpendicular of AB

AB = 2 cm , CD = 4 cm

OA = 2cm, CD = 4cm

= OA = OD = r

= OC = 4 – r

= AC = 1/2 AB

= 1/2 × 2 = 1 cm

In right OAC

OA2 = 0C2 + AC2

= r2 = (4 – r) + 12

= 16 + r2 – 8r + 1

= r = 17/8

= 2 1/8 cm

Thus, the radius of the circle will be 2 1/8 cm.

Question no – (11)

Solution :

According to the given figure,

AB is chord of circle OD AB and AC is join.

To prove CA = 20D

If OD ⊥ AB

O is the midpoint of BC

In △ ABC, DO || AC

So, AB/BD = BC/BO = AC/DO

AB/1/2AB = CA/OD = 2/1

= CA = 2 OD …(Proved)

Question no – (12)

Solution :

In the figure OMNP is a square, center of circle is ‘O’ is drawn on which interest the square to point X and Y

Then join OX and OY

Prove △OXM ≅ △OYP

and, NX = NY

Now, we have to proof :  In △OXM and △OYP

= OX = NY

Side OM = OP

△OXM ≅ △OYP

MX = PY

But, MN = PN

MN – MX = PN – PY

= NX = NY …(Proved)

Question no – (15)

Solution :

AB and CD are here two parallel chords of the circle and line ‘L’ is the perpendicular bisector of AB

AL || L and ∠ALM = 90

The prove ‘L” is the perpendicular bisector of CD.

Proof : If line ‘L’ is the perpendicular Defector  of chord AB If AB || CD and L is the Perpendicular to ‘AB’ and also perpendicular to CD

OM ⊥ CD

So, M is the midpoint of CD

Hence, ‘L’ Bisects CD. …(Proved)

Question no – (17)

Solution :

In fig two circle with center ‘O’ and interest each other at ‘P’

APB || OO is drawn

To prove : AB = 200

Now, from ‘O’ draw OM ⊥ AP

and from ‘O’ ON ⊥ PB

Proof : If OM ⊥ AP

M is the midpoint of AP

AM = MP

= MP = 1/2  = AP

Simplify = PN = NB

= PN = 1/2 PB

MP + PN = 1/2 AP + 1/2 PB

= M.N = 1/2 (AP + PB)

= OO’ = 1/2 AB

= AB = 200 …(Proved)

Question no – (18)

Solution :

From fig circle with center ‘O’ AB and CD are two equal chards interest at point ‘P’

To prove, AP = PD and BP = CP

Now, Draw OM ⊥ AB and, ON ⊥ CD.

Proof : From OMP and ONP,

OP = OP and OM = ON

OMP ≅ ONP

So, PM = PN

if OM ⊥ AB and ON ⊥ CD, and, M,N

are midpoint of AB and CD

AM = MP and CN = NP

AB = CD

AB – PB =CD – PC

A= AB – PB = CD – PC

= AP = PD

Simplify BM – PM = CN – PN

BP = CP …(Proved)

Question no – (19)

Solution :

With center ‘O’ of circle chard AB = CD,

OE ⊥ CD at A and OF AB at ‘G

To prove, EH = GF

Now, to proof : If OE ⊥ CD and

OF ⊥ AB

G, H are the midpoints of AB and CD

∴ OG = OH

But, OF = OE

OF – OG = OE = OH

= GF = EH

= EH = GE …(Proved)

Question no – (21)

Solution :

Here, in figure the center of circle is ‘O’ Two equal chards interest each other

OM ⊥ AD and ON ⊥ CD

Then MN are Joined

To prove : ∠OMN  = ∠ONM

Now, Proof : Chard AB = CD

OM = ON

Now, In △OMN

if OM = ON

∠OMN = ∠ONM …(Proved)

Circle Exercise 13(b) Solution :

Question no – (1)

Solution :

(i) Given, mQR

Here, 4x – 2 + 6x + 6 + 7x – 18 = 360

= 17x – 14 = 360

= 17x = 360 + 14

= x = 374/17

= 22

Now, MQR = 7x – 18

= (7 × 22 – 18)

= 136°

Therefore, the MQR is = 136°

(ii) Here, AD is the diameter of the circle with center ‘O’

m ∠AOD = 180°

∴ 10y = 180°

= y = 180°/10 = 18°

Now, m ∠BOC = 6y

= 6 × 18

= 108°

Thus, the m ∠BOC is = 108°

(iii) In figure AC = BC

∴ ∠AOC = ∠BOC

8Y – 8 = 6Y

= 8Y – 6Y = 8

= Y = 8/2

= 4

∴ m ∠AOC = 8Y – 8

= 8 × 4 – 8

= 32 – 8

= 24

Hence, the m ∠AOC is = 24°

(iv) In the fig, OA = OB

With center A and B are equal CD = EF

= 45 – 6x = 9n

= 6x + 9x = 45

= x = 45/15

= 3

Now, m ∠EBF = 9x = (9 × 3) = 27°

if m ∠CAD = 45 + 6x

and, 45 + 6x = 9x

= 3x = 45

= x = 45/3

= 15

m (∠EBF)

= 3x = 3 × 15

= 135°

Thus, m ∠EBF is = 135°

(vi) In the given figure,

Chord = AB = CD

∴ 4y + y = y + 68

= 4y = 68

= y 68/4 = 17°

MAB = 4y + y

= 5y

= 5 × 17

= 85°

Hence, m AB = 85°

Question no – (2)

Solution :

Here, ABC is inscribed in a circle and ∠P = ∠Q

∴ QR = PR

If equal chords inscribed equal angles at the center,

Therefore, MPR = MQR

Question no – (3)

Solution :

According to the given figure,

AB = CD

AC and BD are joined

We have to prove : AC = DB

If AB = DC

So, MAB = MDC

MAD – AB = MDC – MAB

= MAC = MDB

= AC = DB …(Proved)

Question no – (4)

Solution :

As per the given figure,

AC = DB

We have to prove :  AB = DC

= AC = DB

So, AC = DB

∴ AC + DB = AD + DB

= CD = AB

= CD = AB

= AB = CD …(Proved)

Question no – (5)

Solution :

According to the given circle,

AB and AC are the chards.

X and Y are the midpoint of are AB and, are AC.

XY is joined which meet

AB is P and AC in Q

To prove : AP = AQ

Join AX, AY and OY Proof : if AB = AC

So, arc  A × B are A

But, X and Y are the midpoint of AB and AC

A × = ×B, SA = YC

So, ∠XAY = ∠XBA and ∠YAC = ∠YCA

= ∠XAB = or ∠XAP = ∠YAQ

Now, in XAP and YAQ

= AY = AY

∠XAP = ∠YAQ

So, △XAP  ≅  △YAQ

Therefore, AB = AQ …(Prove)

Question no – (6)

Solution :

In a circle, with center ‘O’ chord AB = BC = CD = DF AB and BE  are joined.

Now, Join AO, BO, CO, DO and EO

Proof : If AB = BC = CD = DE

∴ AB + BC + CD – AD

Simplify BC + CD + DE = BE

∴ AB + BC + CD = BC + DE

So, Control ∠AOD and  ∠BOE

Now, In AOB and BOE

= OA = OB

= OD = DE

△AOD = BOE

△AOD = BOE Question no – (7)

Solution :

In a circle are APB = are A and D

AC and BD are Joined.

The prove AC = then

Proof : If are APB = are CQD

Equal are subtends equal angles at the circumference. But These are alternative

angles

= AC || BD …(Proved)

Question no – (9)

Solution :

A regular hexagon ABCDEF inscribed in a circle with center ‘O’ join AO, BO, CO, Do, EO and FO.

The prove,

Angle each of hexagon = 60°

Proof : If AB = BC = CD = DE = EF = FA

∴ AB = BC = CD = DE = EF = FA

∴ Each will be subtends angle at the canter,

= 360/6

= 60°

Next Chapter Solution :

Updated: June 20, 2023 — 5:44 am