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# Ok, so the circle is as drawn in the picture, and I have no

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Director
Joined: 31 Mar 2007
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Ok, so the circle is as drawn in the picture, and I have no [#permalink]

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09 Apr 2007, 03:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Ok, so the circle is as drawn in the picture, and I have no idea how to do it. Do you have to setup a system of equations based on a chord formula and something else? Any hints/suggestions on how to tackle this type of problem would be appreciated
Director
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09 Apr 2007, 03:30
pic
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hardgeomatry.jpg [ 17.04 KiB | Viewed 1046 times ]

Director
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09 Apr 2007, 06:54
Let's say centre of a circle is N.

We are told that <ORP =35
The diameter is a hypothenusis(always), so <OPR =90, hence <POR=180-90-35=55.
Because NO=NP, <OPN=55, too. Hence, <ONP=180-2*55=70.
Because NP and NQ are equal AND OR and PQ are parallel, <QNR=70, too.
Hence <PNQ=40.

minor arc = (40/360)*18Pi= 2pi

By now we are done
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Director
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09 Apr 2007, 07:29
Hi,

I am afraid if we can accept the solution given by botirvoy. There is no info given on Center of the circle.

I approach it this way.

Join OQ.

Angle OPR = 35. ( Given)

Angle PRO = Angle RPQ = 35 ( Corrosponding Angle of parllel lines)
Hence, Angle QOR = 35 = Angle QPR ( Angles inscribed in the same Arc)

Hence l(arc OP ) = 2 x Angle ORP = 70. ( Angle inscribed is half the Length of Intercepted Arc )

Similarly, l(arc QR ) = 2 x Angle QOR = 70.

Hence Sum of Minor Arc OP and Minor QR = 140.

Therefore, Minor Arc PQ = [ 180 - (Arc PO + Arc QR) ]
= 180 - 140 = 40.

Ratio of ARC PQ to Circle = 40/360 = 1/9.

Length of Circle is its circumference = 2 . pi . r = 2 . pi . 9

Therefore, Length of Arc PQ = 1/9 x 2 . pi . 9.

Cancelling 9, we get 2.pi as answer.

Hence A.
Director
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09 Apr 2007, 07:48
angle PRO is given = 35
Since OR = 2 * OC (C is the center of cirlce)
So, angle PCO = 2 * (Angle PRO) = 70
Line PQ || Line OR
So, angle PCO = angle CPQ = 70
So, angle PQC = 70 (Triangle PQC is issosles)

So, angle PCQ = 180 - 70 - 70 = 40

Arc Formula: S = r * theta
S = arc distance,
theta = radians (or in pi)

1 degree = 2*Pi/360
40 degree = 2*Pi * 40/360 = 2*Pi/9
SO, S = 9 * 2*Pi/9
Director
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09 Apr 2007, 09:12
asaf wrote:
So, angle PCO = angle CPQ = 70

asaf, how did you deduce that these angles are 70?
09 Apr 2007, 09:12
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