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Home / Geometry / Class Eight(8)-Circle-Exercise-10.1|Soltuion

Class Eight(8)-Circle-Exercise-10.1|Soltuion

Exercise-1: prove that, if two chords of a circle bisect each other , their point intersection will be the centre of the circle.

General enunciation: We have to prove that, if two chords of a circle bisect each other, their point of intersection will be the centre of the circle.

Screenshot_4

Particular enunciation: Let in the circle ABCD the chords AB and CD bisect each other at E.

We have to prove that the point E is the centre of the circle ABCD.

Construction: Let the centre of the circle is O rather than E. We join O,E.

Proof: O be the centre of the circle. E be the midpoint of the chords AB and CD respectively.

∴OE⊥AB and OE⊥CD

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

So ∠OEC = 1 right angle

AB and CD are two intersector straight lines.

Both ∠OEA and ∠OEC are not equal to 1 right angle.

∴ No point other than E can be the centre of the circle.

∴E be the centre of the circle. [proved]

Exercise-2: Prove that the line joining the midpoints of two parallel chords passes through the centre and is perpendicular to the two chords.

General enunciation: we have to prove that the line joining the midpoints of two parallel chords passes through the centre and is perpendicular to the two chords.

Screenshot_5

Particular enunciation: Let in the circle of centre O, AB and CD are two parallel chords. E and F are the midpoints of AB and CD respectively. The line segment EF is the joining  line of the midpoints E and F. It is required to prove that EF passes through the centre and is perpendicular to the chords.

Construction: We join O and F.

Proof: O be the centre of the circle and E and F be the midpoints of the chords AB and CD.

∴OE⊥AB and OF⊥CD

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

So ∠OEB = 1 right angle

And ∠OFD= 1 right angle

∴∠OEB=∠OFD.

The two angles ∠OEB and ∠OFD are corresponding and equal.

∴OF and OE will be the transversals    [AB∥CD]

i.e. lie on the same straight line   [O be the centre of the circle]

therefore EF passes through the centre (proved)

Exercise -3: Two chords AB and AC of  a circle make circle make equal angles with the radius through A. Prove that AB=AC.

General Enunciation: Two chords AB and AC of  a circle make circle make equal angles with the radius through A. We have to prove that AB=AC.

Particular enunciation: Let O is the centre of the circle ABC. The two chords AB and AC has made equal angles ∠BAO and ∠CAO with the radius OA through A.

It is required to prove that AB=AC.

Screenshot_6

 

Construction: from the centre O we draw the perpendicular OM and ON respectively to AB and AC.

Proof: O is the centre and OM⊥AB

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

∴AM= ½ AB  Again ON⊥AC

∴AN= ½ AC

In ∆AOM and ∆AON

∠AMO=∠ANO      [1 right angle]

∠MAO=∠NAO  [supposition]

And AO=AO      [common side]

So, ∆AOM≌∆AON  [ASA theorem]

∴AM=AN

Or, ½ AB= ½ AC    [from 1]

Or, AB=AC  [proved]

Exercise-4: In the figure , O is the centre of circle and the chord AB = chord AC . Prove that ∠BAO=∠CAO.

General Enunciation: In the figure , O is the centre of circle and the chord AB = chord AC . We have to prove that ∠BAO=∠CAO.

Particular enunciation: Given that O is the centre of the circle and the chord AB=AC. We join O and A.

Screenshot_1

We have to prove that ∠BAO=∠CAO

Construction: We join O, B and O, C

Proof: In ∆AOB and ∆AOC

AB=AC    [given]

OB=OC   [radius of same circle]

OA=OA   [common side]

So ∆AOB≌∆AOC [SSS theorem]

∴∠BAO=∠CAO.  [proved]

Exercise-5: A circle passes through the vertices of a right – angled triangle. Show that the centre lies on the midpoint of the hypotenuse.

General enunciation: A circle passes through the vertices of a right-angled triangle. It is required to show the centre lies on the midpoint of the hypotenuse.

Screenshot_2

Particular enunciation: Let ABC is a right – angled triangle. Its ∠ABC= 1 right angle and AC is the hypotenuse, The circle passes through the vertices A,B and C.

Construction: From O we draw the perpendicular OD to AB and the perpendicular OE to BC. Again we join O, B.

Proof: O is the centre of the circle and OD⊥AB

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

 

∴AD=BD  ——————————–(1)

In ∆AOD and ∆BOD

AD=BD   [from 1]

OD=OD    [common side]

And ∠ADO=∠BDO= 1 right angle  [OD⊥AB]

So ∆AOD≌∆BOD  [SAS theorem]

∴AO=BO

Similarly from ∆BOE and ∆COE it is proved that CO=BO

So AO=BO=CO ——————————-(2)

Therefore O is the centre of the circle.

Hypotenuse AC=AO+CO

=CO+CO   [from 2]

=2CO

Or, 2CO=AC

Or, CO= ½ AC

Therefore O lies on the midpoint of the hypotenuse AC. (proved)

Exercise-6: A chord AB of one of two concentric circles intersect the other at C and D. Prove that AC=BD.

General enunciation: A chord AB of one of two concentric circles intersect the other at C and D. We have to prove that AC=BD.

Screenshot_3

Particular enunciation: Let O be the two circles ABE and CDF.  The chord AB of the circle ABE intersects the circle CDF at C and D.

It is required to prove that AC=BD.

Construction: From O we draw the perpendicular OP on AB or CD.

Proof: O is the centre and OP⊥CD, OP⊥AB.

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

∴CP=PD and AP=BP     ————————- (1)

Again, AP=AC+CP

And  BP=PD+BD

Since AP=BP  [from 1]

So AC+CP=PD+BD

Or, AC+CP=CP+BD    [CP=PD]

Or,  AC+CP=BD+CP

Or, AC=BD.   (proved)

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