ML Aggarwal Circle Exe-15.1 Class 9 ICSE Maths Solutions

ML Aggarwal Circle Exe-15.1 Class 9 ICSE Maths APC Understanding Solutions. Solutions of  Exe-15.1. This post is the Solutions of  ML Aggarwal Chapter 15 – Circle for ICSE Maths Class-9.  APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-15 Circle for ICSE Board Class-9. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Circle Exe-15.1 Class 9 ICSE Maths Solutions

Board	ICSE
Publications	Avichal Publishig Company (APC)
Subject	Maths
Class	9th
Chapter-15	Circle
Writer	ML Aggarwal
Book Name	Understanding
Topics	Solution of Exe-15.1 Questions
Edition	2021-2022

Exe-15.1 Solutions of ML Aggarwal for ICSE Class-9 Ch-15, Circle

Note:- Before viewing Solutions of Chapter -15 Circle Class-9 of ML Aggarwal Solutions .  Read the Chapter Carefully. Then solve all example given in Exercise-15.1, Exe-15.2, MCQs, Chapter Test.

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Circle Exercise-15.1

ML Aggarwal Class 9 ICSE Maths Solutions

Page 335

Question 1. Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.

Answer :

AB is chord of a circle with center O and OA is its radius OM ⊥ AB

Hence, OA = 13 cm, OM = 12 cm

Now from right angled triangle OAM,

OA² = OM² + AM²

⇒ 13² = 12² + AM²

⇒ AM² = 13² – 12²

⇒ AM² = 169 – 144

⇒ AM² = 25

⇒ AM² = 5²

⇒ AM = 5

OM perpendicular to AB

Hence, M is the midpoint of AB

AB = 2 AM

⇒ AB = 2 (5)

⇒ AB = 10 cm

Question 2. A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the centre of the circle.

Answer :

AB is the chord of the circle with centre O and radius OA

OM is perpendicular to AB

Question 3. A chord of length 8 cm is at a distance of 3 cm from the centre of the circle. Calculate the radius of the circle.

Answer :

AB is the chord of a circle with center O

And radius OA and OM ⊥ AB

Question 4. Calculate the length of the chord which is at a distance of 6 cm from the centre of a circle of diameter 20 cm.

Answer :

AB is the chord of the circle with centre O

And radius OA and OM ⊥ AB

Diameter of the circle = 20 cm

Radius = 20/2 = 10 cm

OA = 10 cm, OM = 6 cm

Now in right ∆OAM,

OA² = AM² + OM² 

⇒ (10)² = AM² + (6)²

⇒ AM² = 10² – 6²

⇒ AM² = 100 – 36 = 64 = (8)²

⇒ AM = 8 cm

OM ⊥ AB

M is the mid-point of AB.

⇒ AB = 2× AM = 2×8 = 16 cm.

Question 5. A chord of length 16 cm is at a distance of 6 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 8 cm from the centre.

 

Answer :

AB is a chord a circle with centre O and

OA is the radius of the circle and OM ⊥ AB

AB = 16 cm, OM = 6 cm

OM ⊥ AB

AM = ½ AB = ½ × 16 = 8 cm

Now in right ∆OAM

OA² = AM² + OM²

= (8)² + (6)²

⇒ 64 + 36 = 100 = (10)²

ON ⊥ CD and OC is the radius.

In right ∆ONC

OC² = ON² + NC² 

⇒ (10)² = (8)² + (NC)²

⇒ 100 = 64 + NC²

⇒ NC ² = 100 – 64 = 36 = (6)²

⇒ NC = 6

But ON ⊥ AB

N is the mid-point of CD

⇒ CD = 2 NC

= 2×6 = 12 cm

Question 6. In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords if they are on :

(i) the same side of the centre.
(ii) the opposite sides of the centre

Answer :

Two chords AB and CD of a Circle with centre O and radius OA or OC

OA = OC = 5 cm

AB = 8 cm

CD = 6 cm

OM and ON are perpendiculars from O to AB and CD respectively.

M and N are the Mid-points of AB and

CD respectively

In figure (i) chord are on the same side

And in figure (ii) chord are on the opposite

OA² = AM² + OM² 

⇒ (5)² = (4)² + OM²

⇒ AM = ½ AB

⇒ 25 = 16 + OM²

⇒ OM² = 25 – 16 = 9= (3)²

⇒ OM = 3 cm

Again in right ∆OCN,

OC² = CN² + ON²

⇒ (5)² = (3)² + ON²

⇒ (CN = ½ CD)

⇒ 25 = 9 + ON²

⇒ ON² = 25 – 9 = 16 = (4)²

⇒ ON = 4

In fig (i), distance MN = ON – OM

= 4 – 3 = 1cm.

In fig (ii)

MN = OM + ON

= 3 + 4

= 7 Cm

Question 7. (a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle, OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the:

(i) radius of the circle.
(ii) length of chord CD.

(b) In the figure (ii) given below, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.

Answer :

(a) AB = 24 cm, OM = 5cm, ON = 12cm

OM ⊥ AB

M is midpoint of AB

AM = 12 cm

(i) Radius of circle OA = √OM² + AM²

(ii) Again OC² = ON² + CN²

⇒ 13² = 12² + CN²

⇒ CN = √(13² – 12²)

= √(169 – 144)

= √25

⇒ CN = 5 cm

As ON ⊥ CD, N is mid-Point of CD

CD = 2CN

= 2×5

= 10 cm

(b) AB = 8 cm,

EC = 3 cm

Let radius OB = OC = r

OE = (r-3) cm

Now in right ∆OBE

OB² = BE² + OE²

⇒ r² = (4)² + (r – 3)²

⇒ r² = 16 + r² – 6r + 9

⇒ 6r = 25

⇒ r = 25/6

= 4 1/6 cm

Question 8. In the adjoining figure, AB and CD ate two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Answer :

In the figure, chords AB ∥ CD

O is the centre of the circle

Radius of the Circle = 15 cm

Length of AB = 24 cm and CD = 18 cm

Join OA and OC

AB = 24 cm and OM ⊥ AB

AM = MB = 24/2

= 12 cm

ON ⊥ CD

CN = ND = 18/2 = 9 cm

Similarly In right ∆ CNO

OC² = CN² + ON² (15)²

= (9)² + ON²

⇒ 225 = 81 + ON²

⇒ ON² = 225 – 81

= 144 = (12)²

⇒ ON = 12 cm

Now MN = OM + ON

= 9 + 12

= 21 cm

Question 9. AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.

Answer :

Question 10. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm, find the radius of the circle.

Answer :

AB = AC 12√5 and

BC = 24 cm

Join OB and OC and OA

Draw AD ⊥ BC which will pass through

OD bisect BC in D

BD = DC = 12 cm

In right ∆ABD

AB² = AD² + BD²

⇒ (12√5)² = AD² + BD²

⇒ (12√5)² = AD² + (12)²

⇒ 144×5 =AD² + 144

⇒ 720 – 144

= AD²

⇒ AD² = 576

⇒ AD = √576

= 24

Let radius of the circle = OA = OB = OC = r

OD = AD – AO = 24 – r

Now in right ∆OBD,

OB² = BD² + OD²

⇒ r² = (12)² + (24 – r)²

⇒ r² = 144 + 576 + r² – 48r

⇒ 48r = 720

⇒ 48r = 720

⇒ r = 720/48

= 48r

⇒ 48r = 720

⇒ r = 720/48

= 15cm.

Radius = 15 cm.

Question 11. An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.

Answer :

ABC is an equilateral triangle inscribed in a

Circle with centre O. Join OB and OC,

From A, Draw AD ⊥ BC which will pass

Through the centre O of the circle.

Each side of ∆ABC = 6 cm.

AD = √3/2 a= √3/2 × 6 = 3 √3 cm.

OD = AD – AO = 3√3 – r

Now in right ∆OBD

OB² = BD² + OD²

⇒ r² = (3)² + (3√3-r)²

⇒ r² = 9 + 27 + r² -6√3r

⇒ 6√3r = 36

⇒ R = 36/6√3

= 6/√3 × √3/√3

= 6√3/3

= 2√3 cm

Radius = 2√3 cm

Question 12. AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.

Answer :

AM = 18 cm and MB = 8 cm

AB = AM + MB = 18 + 8

= 26 cm

Radius of the circle = 26/2

= 13 cm

Let CD is the shortest chord drawn through M.

CD ⊥ AB

Join OC

OM = AM – AO

= 18 – 13

= 5 cm

OC = OA = 13 cm

Now in right ∆OMC

OC² = OM² + MC²

⇒ (13)² = (5)² + MC² (MC² 

= 13² – 5²

⇒ MC² = 169 – 25

= 144 = (12)²

⇒ MC = 12

M is Mid-Point of CD

CD = 2 ×MC

= 2×12

= 24 cm

Question 13. A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the area of the rectangle.

Answer :

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Circle Exercise-15.1

ML Aggarwal Class 9 ICSE Maths Solutions

Page 336

Question 14. The length of the common chord of two intersecting circles is 30 cm. If the radii of the two circles are 25 cm and 17 cm, find the distance between their centres.

Answer :

Let the circles with centres A and B intersect each other at C and D.

RA = 25 cm, RB = 17 cm, CD = 30 cm.

Let CD and AB intersect at O.

OC=OD = 30/2 = 15 cm.

AO = [RA²-OC²]^0.5 = [25²–15²]^0.5

= [625–225]^0.5 = 20 cm.

BO = [RB²-OC²]^0.5 = [17²–15²]^0.5

= [289–225]^0.5 = 8 cm.

Hence AB = 20+8 = 28 cm.

Question 15. The line joining the mid-points of two chords of a circle passes through its centre. Prove that the chords are parallel.

Answer :

AB and CD are the two chords of a circle with centre O.
L and M are the midpoints of AB and CD and O lies in the line joining ML
To prove: AB ∥ CD
Proof: AB and CD are two chords of a circle with centre O.
Line LOM bisects them at L and M
Then, OL ⊥ AB
And, OM ⊥ CD
∴ ∠ALM = ∠LMD = 90°
But they are alternate angles
∴ AB ∥ CD.

Question 16. If a diameter of a circle is perpendicular to one of two parallel chords of the circle, prove that it is perpendicular to the other and bisects it.

Answer :

Question 17. In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.

Answer :

An equilateral triangle ABC in which D, E and F are the mid- points of sides BC,

CA and AB respectively.

To prove :

The centroid and circumcentre  are coincident.

Draw medians, AD, BE and CF.

Proof : 

Let G be the centroid of  ΔABC  i. e., the point of intersection of AD, BE and CF. In triangles BEC and BFC, we have

∠B = ∠C = 60

BC = CE     [∵  AB = AC  ⇒ ½ AB = ½ AC ⇒ BF = CE]

∴   ΔBEC ≅ ΔBFC

⇒ BE = CF     ……(i)

Similarly,   ΔCAF ≅  ΔCAD

CF = AD        …….(ii)

From (i) & (ii)   AD = BE = CF

⇒  2/3 AD = 2/3, BE = 2/3 CF

CG = 2/3 CF

GA = 2/3 AD, GB = 2/3 BE

⇒ GA = GB = GC

⇒ G is equidistant from the vertices

⇒ G is the circumcentre of  ΔABC

Hence,

the centroid and circumcentre are coincident.

Question 18.

(a) In the figure (i) given below, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, show that CA = 2 OD.
(b) In the figure (ii) given below, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ ⊥ AC, Prove that PB = QC.

Answer :

(a) Since,  OD ⊥ AB  and the perpendicular drawn from the centre to a chord bisects the chord.

∴  D is the mid-point of AB.

Also, O being the centre , is the mid – point of  BC.

Thus, in   ΔABC, D and O are mid-point of AB and BC respectively. Therefore, OD ∥ AC

And   OD = ½ CA    [∵ Segment joining the mid- points of two sides of a triangle is half of the third side]

⇒  CA = 2OD.

Hence  proved.

(b) AB and AC are two equal chords of a circle with centre O.

OP⊥AB and OQ⊥AC.

To prove: PB=QC

Proof:

OP⊥AB

⇒AM=MB …. (perpendicular from centre bisects the chord)….(i)

Similarly, AN=NC….(ii)

But, AB=AC

⇒ 2AB = 2AC

⇒ MB = NC …(iii) ( From (i) and (ii) )

OP=OQ (Radii of the circle)

and OM=ON (Equal chords are equidistant from the centre)

⇒ OP − OM = OQ − ON

⇒ MP = NQ ….(iv) (From figure)

In ΔMPB and ΔNQC, we have

∠PMB =  ∠QNC (Each =90∘)

MB = NC ( From (iii) )

MP = NQ ( From (iv) )

∴ ΔPMB ≅ ΔQNC (SAS)

⇒ PB = QC (CPCT)

Question 19.

(a) In the figure (i) given below, a line l intersects two concentric circles at the points A, B, C and D. Prove that AB = CD.
(b) In the figure (it) given below, chords AB and CD of a circle with centre O intersect at E. If OE bisects ∠AED, Prove that AB = CD.

Answer :

(a) Since, perpendicular from the centre to a chord bisects the chord, therefore,

AM = MD …(i)

and BM = MC …(ii)

(i) and (ii) gives, AM – BM = MD – MC

⇒ AB = CD.

(b)

Draw OM⊥AB and ON⊥CD.

In △OEM and △OEN,

∠OME = ∠ONE=90o

∠OEM = ∠OEN

OE = OE

Hence,
by AAS congruence,

△OEM ≅ △OEN

So, OM=ON [By cpct ]
Chords AB and CD are equidistant from the centre.
So,
Chord AB = Chord CD

Question 20.

(a) In the figure (i) given below, AD is a diameter of a circle with centre O.
If AB || CD, prove that AB = CD.
(b) In the figure (ii) given below, AB and CD are equal chords of a circle with centre O. If AB and CD meet at E (outside the circle) Prove that :
(i) AE = CE (ii) BE = DE.

Answer :

(a) We can join

in triangles ABO and DOC,

BO = OD (Radius)
Ang. BOA = Ang. DOC (Vertically opp angle)
AO= OC (Radius)

By SAS, ABO is congruent to DOC.

By CPCT, AB = CD

Hence proved

(b)

Two equal chords AB and CD intersecting at a point E.

To prove :

BE = DE and AE = CE

 Join OE. Draw OL ⊥ AB and OM ⊥ CD.

Proof :

AB = CD

⇒  OL = OM   [∵  Equal chords are equidistant from the centre ]

In triangles OLE and OME, we have

OL = OM

∠OLE = ∠OME    [Each equal to 90˚]

OE = OE      [Common]

So,

by SAS – criterion of congruences

△OLE  ≅  △OME

⇒    LE = ME             ……..(i)

Now,       AB = CD

⇒   ½ AB = ½ CD ⇒  BL = DM    …….(ii)

Subtracting (ii) from (i), we get

LE – BL = ME – DM

⇒  BE = DE

AB = CD  and  BE = DE

⇒   AB + BE = CD + DE

⇒   AE = CE

Hence,

BE = DE  and  AE = CE

Hence proved.