c4Question 11: The figure shows two concentric circles and AD is the chord of larger circle. Prove that AB = CD .

Answer:

If you drop a perpendicular from O to the chord, it would bisect the chord (Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.). Let us say that it intersects BC and AD at E .

Therefore BE = CE and AE = DE

If you subtract these two, we get

AE - BE = DE - CE \Rightarrow AB = CD 

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c5Question 12: A straight line is drawn cutting two equal circles and passing through the midpoint M of the line joining their centers O \ and \  O' .  Prove that the chords AB \ and \ CD , which are intercepted by two circles are equal.

Answer:c12.jpg

First draw perpendiculars OP and O'P'

Now consider \triangle OPM and \triangle O'P'M

\angle OMP = \angle P'MO' (opposite angles)

\angle OPM = \angle O'P'M = 90^o

OM = O'M (M is the mid point of OO' – Given)

Therefore  \triangle OPM \cong \triangle O'P'M

Therefore OP = O'P'

Therefore AB = CD (Theorem 8 (Converse of Theorem 7) : Chords of circle which are equidistant from the center of the circle are equal in length.)

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c6Question 13: M \ and \ N are mid points of two equal chords AB \ and \ CD respectively of a circle with center O . Prove that:

(i) \angle BMN = \angle DNM

(ii) \angle AMN = \angle CNM

c13Answer:

First drop perpendiculars OM and ON on AB and CD respectively.

OM will bisect AB and ON will bisect CD . (Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)

Therefore BM = \frac{1}{2} AC and DN = \frac{1}{2}CD

Given AB = CD . Therefore BM = DN

Also OM^2 = OB^2-MB^2 = OD^2-DN^2 = ON^2

\Rightarrow OM = ON

Therefore in \triangle OMN, \angle OMN = \angle ONM  (since angles opposite equal sides of a triangle are equal)

We know \angle OMB = \angle ONC

(i) Therefore \angle OMB - \angle OMN = \angle ONC - \angle ONM

\Rightarrow \angle BOM = \angle DNM

(ii) We know \angle AMO = \angle CNO = 90^o

Therefore \angle AMO + \angle OMN = \angle CNO + \angle ONM

\Rightarrow \angle AMN = \angle CNM

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c7.jpgQuestion 14: In the following figure; P and Q are the points of intersection of two circles with centers O   and O' . If straight lines APB and CQD are parallel to OO' ; prove that

(i) OO'=\frac{1}{2} AB

(ii) AB = CD

c14Answer:

MP = \frac{1}{2} AP \ and \  NP = \frac{1}{2} PB

M'Q = \frac{1}{2} CQ \ and \  N'Q = \frac{1}{2} DQ

OO'=MN = MP+NP = \frac{1}{2} (AP+BP) = \frac{1}{2} AB  … … … … (i)

OO'=M'N' = M'Q+N'Q = \frac{1}{2} (CQ+DQ) = \frac{1}{2} CD … … … … (ii)

Therefore from (i) and (ii) we get AB = CD . Hence proved.

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Question 15: Two equal chords AB and CD of a circle with center O , intersect each other at point P inside the circle. Prove that:

(i)  AP = CP

(ii) BP = DP

Answer:

Draw ON \perp CD and OM \perp AB

(Theorem 5: A perpendicular to a chord, from the center of the circle, bisects the chord.)

MB = AM  \Rightarrow  AM = \frac{1}{2} BA  c15

CN = ND   \Rightarrow  CN = \frac{1}{2} CD 

\Rightarrow MB = CN = ND

In \triangle OMB  \ \ OM^2 = OB^2 -MB^2

In \triangle OND  \ \ ON^2 = OD^2 -ND^2

OM^2 = OB^2-ND^2

ON^2 = OD^2-ND^2

OB = OD (radius of the circle)

\Rightarrow OM = OM

Consider \triangle OPM and \triangle OPN

OM = ON

\angle OMP = \angle ONP = 90^o

OP is common

Therefore  \triangle OPM \cong \triangle OPN

\Rightarrow PM = PN

PM+MB = PN+ND \Rightarrow PB = DP

AB = CD (Given)

Since BP = DP

Therefore AB - BP = CD - DB \Rightarrow AP = CP

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c9Question 16: In the following figure, OABC is a square. A circle is drawn with O as the center which meets OC at P and OA at Q . Prove that:

(i) \triangle OPA \cong \triangle OQC

(ii) \triangle BPC \cong \triangle BQA

c16Answer:

(i) In \triangle OPA \ and \  \triangle OQC

OP=OQ (radius)

\angle AOP = \angle QOC = 90^o

OA = OC (side of square)

Therefore \triangle OPA \cong \triangle OQC (S.A.S postulate)

(ii) OP = OQ (radius)

OC = OA (side of a square)

Therefore OC - OP = OA - OQ

CP = QA

Consider \triangle BPC \ and \  \triangle BQA

BC = BA (side of a square)

\angle PCB = \angle QAB = 90^o

CP = QA 

Therefore \triangle BPC \cong \triangle BQA

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Question 17: The length of common chord of two intersecting circles is 30 \ cm . If the diameter of the circles be 50 \ cm and 34 \ cm , calculate the distance between their center.

c17Answer:

OA = 17 \ cm and O'A = 25 \ cm (given)

In \triangle OAD \ \ OA^2=AD^2+OD^2

\Rightarrow OD^2 = OA^2-AD^2

OD^2 = 17^2 -15^2 = 289-225 = 64

\Rightarrow OD = 8 \ cm

Similarly in \triangle O'DA \ \ O'D^2 = O'A^2 - AD^2 = 25^2-15^2 = 400

\Rightarrow O'D = 20 \ cm

Therefore OO' = 8 + 20 = 28 \ cm

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Question 18: The line joining the mid points of two chords if a circle passes through the center. Prove that the chords are parallel.

Answer:c18.jpg

AL = LB (given)

OL bisects AB (given)

Therefore OL \perp AB (if line drawn from the center bisects the chord, then is is perpendicular to the chord)

Similarly OM \perp CD

Therefore \angle DML = \angle BLP = 90^o (alternate angles)

Therefore AB \parallel CD

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c10.jpgQuestion 19: In the following figure, the line ABCD \perp PQ ; where P and Q are the centers of the circles. Show that:

(i) AB = CD

(ii) AC = BD

Answer:

QO \perp  AD \therefore AO = OD … … … … (i)

PO \perp BC \therefore BO = CO … … … … (ii)

(i) – (ii) we get

AO - BO = AD - CO \Rightarrow AB = CD . Hence proved.

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Question 20: AB and CD are two equal chords of a circle with center O which intersect each other at right angle at point P . If OM \perp AB and ON \perp CD , show that OMPN is a square.

c19Answer: 

AB = CD and AB \perp CD (given)

OM \perp AB \Rightarrow AM = MB

ON \perp CD \Rightarrow CN = ND

\therefore all \angle \ in \ OMPN \ are 90^o 

BM = \frac{1}{2} AB = \frac{1}{2} CD = CN

Since AB = CD and OM = ON and because all angles are 90^o we can say

OM = NP, ON = MP \Rightarrow NP = MP 

MB = NC

MP + PB= NP+PC \Rightarrow PB = PC

\therefore CN = BM

CP+PN=PB+PM \Rightarrow PN = PM

Therefore OMPN is  square.

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c11Question 21: In the given figure, O is the center of the circle. AB and CD are two chords of circle. OM \perp AB and ON \perp CD . AB = 24 \ cm, OM = 5 \ cm , ON = 12 \ cm . Find the 

(i) radius of the circle

(ii) length of chord CD [2014]

c20.jpgAnswer:

AB = 24 \ cm , OM = 5 \ cm , ON =12 \ cm

OM \perp AB

\Rightarrow AM = BM

Therefore AM = MB = 12 \ cm

(i) Consider \triangle AOM

AO^2 = AM^2 +OM^2 = 12^2+5^2 = 169

\therefore AO = 13 \ cm = Radius of the circle.

(ii) Now consider \triangle CNO

CO^2 = CN^2+ON^2

\Rightarrow CN^2 = CO^2-ON^2 = 13^2 - 12^2 = 169 - 144 = 25

\therefore CN= 5 \ cm

 

 

 

 

 

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