BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Friday , July 28 2017
Home / Geometry / Solution of Exercise- 10.2| Circle |Class- eight (8)

# Solution of Exercise- 10.2| Circle |Class- eight (8)

Exercise-1: If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

General enunciation: if two equal chords of a circle intersect within the circle, it is required to prove that the segments of the one chord.

Particular enunciation: Let in the circle ABCD with centre O, AB and CD are two equal chords. They intersect each other at E. it is required to prove that, AE=CE and BE=DE.

Construction: We draw the perpendicular OP and OQ from the centre O to the chords AB and CD respectively.

Proof: O be the centre and OP⊥AB

∴AP=BP

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

i.e. BP= ½ AB

again O be the centre and OQ⊥CD

∴CQ=DQ

i.e. DQ= ½ CD

now AB=CD   [supposition]

or, ½ AB =  ½ CD

∴BP=DQ  ———————— (1)

The distances of the chords AB and CD from the centre O are OP and OQ respectively and AB=CD

∴OP=OQ  [equal chords of a circle are equidistant from the centre]

Now in the right angled ∆OPE and OQE

Hypotenuse OE= hypotenuse OE    [common side]

And OP=OQ

So ∆OPE≌∆OQE

∴ PE=QE ————————– (2)

BP+PE=DQ+QE  [from 1 and]

Or, BE= DE ————————(3)

AB= CD

Or, AB-BE = CD-BE       [from 3]

Or,AB-BE=CD-DE

Or,AE=CE ————————–(4)

Therefore AE=CE and BE=DE  [proved]   [from 3 and 4]

Exercise-2: prove that the bisecting points of equal chords lie on a circle

General enunciation: To prove that the bisecting points of equal chords lie on a circle.

Particular enunciation: let O be the centre of the circle ABCD. AB, CD and EF are three chords and equal to each other. M, P and N are the bisecting points of AB, CD and EF respectively.

We have to prove that M, N and P lie on a circle.

Construction: We join O, M; O,P and O,N.

Proof: O be the centre of the circle and M, P and N are the midpoints of AB,CD and EF respectively.

The line segment joining the centre of a circle to midpoint of a chord other than diameter is perpendicular to the chord.

∴OM⊥AB, OP⊥CD and ON⊥EF.

The distance of AB, CD and EF from the centre O be OM, OP and ON respectively.

Again, AB= CD=EF

∴OM=OP=ON

If with centre at O and radius equal to OM or ON or OP a circle is drawn, the circle passes through the points M, P and N. Therefore, M,N and P lie on the circle.   [proved]

Exercise-3: Shaw that equal chords drawn from the end poins on opposite sides of a diameter are parallel.

General enunciation: We have to show that equal chords drawn from the end points on opposite sides of a diameter are parallel.

Particular enunciation: Let in the circle of centre O, AD is a diameter and AB and CD are two equal chords opposite the diameter.

We have to show that AB∥CD

Construction: We drawn the perpendicular OM and ON from the centre O to AB and CD respectively.

Proof: In ∆OAM and ∆ODN

OM=ON  [equal chords of a circle are equidistant from the centre]

And AM=DN  [the midpoints of equal chords are M and N]

So ∆OAM≌∆ODN  [SSS theorem]

∴∠OAM=∠ODN

The angles ∠OAM and ∠ODN produced by the transversal AD of AB and CD are equal and these are alternate angles.

Therefore AB∥CD . (proved)

Exercise-4: Show that parallel chords drawn from the end points of a diameter are equal.

General enunciation: To show that parallel chords drawn from the end points of a diameter are equal.

Particular enunciation: Let in the circle with centre O, AD is a diameter . AB and CD are two parallel chords opposite the diameter AD.

We have to prove that AB=CD.

Construction: We join O,B and O,C.

Proof: In ∆OAB and ∆OCD

And included ∠AOB= included ∠COD   [vertically opposite angle]

So ∆AOB≌∆OCD  [SAS theorem]

∴AB=CD [proved]

Exercise-5: Prove that between the two chords the larger one is nearer to the centre than the smaller one.

General enunciation: We have to show that between the two chords the larger one is nearer to the centre than the smaller one.

Particular enunciation: Let ABCD is a circle with centre O. AB and CD are two chords and AB>CD. OE and OF are respectively perpendiculars from the centre O to the chords AB and CD.

We have to prove that OE< OF.

Proof: O be the centre and OE⊥AB

The perpendicular from the centre of a circle to a chord bisects the chord.

∴AE= BE= ½ AB

Again OF⊥CD

∴CD=DF= ½ CD ——————————-(1)

Now, ∆OAE and ∆OCF are right angled.  [OE⊥AB and OF⊥CD]

∴OA2=OE2+AE2  [Pythagoras theorem]

And OC=OF+CF

But OA=OC  [radius of same circle]

Or, OA2=OC2

Or, OE2+AE2=OF2+CF2

Or,AE2-CF2=OF2-OE2 —————————(2)

AB>CD   [supposition]

Or, ½ AB> ½ CD  [multiplying by ½ ]

Or, AE>CF  [from 1]

Or, AE2>CF2  [by squaring]

Or,AE2-CF2>0

OF2-OE2>0  [from (2) and( 3)]

OF2>OE2

Or, OF>OE

∴OE<OF   (proved)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...