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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.

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.

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.

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.

We have to prove that ∠BAO=∠CAO

Construction: We join O, B and O, C

Proof: In ∆AOB and ∆AOC

AB=AC    [given]

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.

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]

In ∆AOD and ∆BOD

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.

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)

## 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 ...