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gmat prep PS

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Balvinder
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gmat prep PS [#permalink] New post   13 Oct 2007, 16:57
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please explain



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Re: gmat prep PS [#permalink] New post   13 Oct 2007, 17:16
R is 35 -> 35 * 2 = 70, Thus: OP is 70 (you should memorize that the measure of minor arc = angel *2 )

Because PQ is parallel to OR -> angle QPR is also 35. and the minor arc QR is 35 * 2 = 70

Now, the minor arc of PQ is 180 - 70 - 70 = 40

In case you do not how we got 40:

The entire circle is 360 which mean above OR is 180 and below OR is 180. Then, we only need to focus on above OR, which we have OP= 70, QR = 70, and now we need to find: PQ which is = 180 - 70 - 70 = 40

NOW:
We need to find the length of minor arc PQ which is: (x / 360) 2pi r

In our case: x is 40 and r is : diameter(which is given) / 2 => 18 /2 = 9

Thus: (40/360)2*pi*9 = 2pi

The correct answer should be 2pi.
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GMATBLACKBELT
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Re: gmat prep PS [#permalink] New post   13 Oct 2007, 17:51
Balvinder wrote:
please explain



The angle is inscribed so the central angle is 70.

Now we just take 140/360 = 7/18.

Heres where the mistake can happen.

7/18*9pi=7/2pi, this is not correct.


should be 7/18*18pi---> 7pi (which is the length of the arcs minus QP)

Now we just take half the circumfrence: 9pi-7pi---> 2pi


Even if u couldnt get this, u should realise that 7/2pi is not the answer b/c what u just calculated wouldn't account for QP. It doesn't make sense.
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Re: gmat prep PS [#permalink] New post   17 Oct 2007, 18:04
I don't understand why the angle goes from 35 degrees to 70 .

The way that I would think about this is as follows:

The total circumference is 18pi. I can exclude half of the circle (9pi), and 2 minor arcs PO and QR.

The minor arc PO is 35/360 * 18pi. As the two lines are parallel, angle POR and PRO are equivalent, so the minor arc QR is 35/360 * 18pi

Summing together

18pi - 9pi - 2(35/360*18pi) = 9pi - 70/360*18pi = 9pi- 7/36*18pi = 9pi - 126/36pi = (7/2)pi ...
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Re: gmat prep PS [#permalink] New post   17 Oct 2007, 19:55
Big-O wrote:
I don't understand why the angle goes from 35 degrees to 70 .

The way that I would think about this is as follows:

The total circumference is 18pi. I can exclude half of the circle (9pi), and 2 minor arcs PO and QR.

The minor arc PO is 35/360 * 18pi. As the two lines are parallel, angle POR and PRO are equivalent, so the minor arc QR is 35/360 * 18pi

Summing together

18pi - 9pi - 2(35/360*18pi) = 9pi - 70/360*18pi = 9pi- 7/36*18pi = 9pi - 126/36pi = (7/2)pi ...


The angle 35 goes to 70 b/c 35 is an inscribed angle. Inscribed angles, when doubled will equal the central angle.
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Re: gmat prep PS [#permalink] New post   17 Oct 2007, 20:17
Nevermind, I think I actually figured it out :)

You can make a right triangle out of OPR. Angle OPR is 90 degrees, and so you can figure out that Angle POR is 70 degrees.

What you ultimately have to do is find the angle measurement btwn P and Q, and for this i used C. I drew a triangle using PCQ. You know that PC is 9, and so is QC. You can figure out that Angle QPC is 70 degrees, and so that must mean since PC=QC, Angle PQC is 70 degrees as well. That leaves 180 -(70+70) = 40 degrees for Angle PCQ.
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Re: gmat prep PS [#permalink] New post   15 Nov 2007, 03:50
Assume 'C' is the point represent the center of the circle. Now, our objective is to find the angle PCQ.

1. Since angle-PRO is 35, angle-PCO is 70.
2. Since lines PQ and OR are parallel, angle-PCO = angle-CPQ = 70
3. Consider triangle-PQC;
Line-PC = Line-QC = radius
=> angle-CPQ = angle- CQP = 70
=> angle-PCQ = 40 degrees [180 - 70 - 70]

Lenght of minor arch = ratio of angles * circumefrence
= 40/360 * 2Pi*radius
= 1/9 * 2 * pi * 9
= 2Pi
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Re: gmat prep PS [#permalink] New post   15 Nov 2007, 06:59
Length of arc OP = 2*35 = 70
Length of arc QR = 3*35 = 70
Length of arc PQ = 180 - (70+70) = 40

Length of arc = 40/360 * 2 pi 9 = 2 pi



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Re: gmat prep PS [#permalink]
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