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GMAT Prep 1 - Circle

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by shovan85 » Tue Dec 14, 2010 10:49 am
pradeepspanchal wrote:Image

Please help.
Follow the image attached below.

Let X be the center.

By the property of circle angel subtended at Center will be twice that of the angle subtended by two point at the circle.

So angle OXP = 70 as angle ORP = 35

As RO is parallel to QP angle QXR = 70 as angle QOR = 35

Thus remaining angle PXQ = 180 - 70 -70 = 40

Angle 360 represents the permieter of circle = pie * diameter = 18pie
Angle 40 represents the arc PQ of circle = 18pie * (40/360) = 2pie

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by pradeepspanchal » Tue Dec 14, 2010 11:11 am
Shovan85, Thanks for the solution .

However, I am not able to get how you get angle QOR = 35 . Is it symmetry of the circle or any property of circle .
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by rainribbon » Tue Dec 14, 2010 11:25 am
pradeepspanchal wrote:Shovan85, Thanks for the solution .

However, I am not able to get how you get angle QOR = 35 . Is it symmetry of the circle or any property of circle .
angle OXP = 70
since, RO is parallel to QP, angle XPQ=70 and angle PQX=70 (PX and QX are both radius of the circle and hence the corresponding angles, XPQ and PQX are equal)

Therefore, PXQ = 40
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by shovan85 » Tue Dec 14, 2010 11:31 am
rainribbon wrote:
pradeepspanchal wrote:Shovan85, Thanks for the solution .

However, I am not able to get how you get angle QOR = 35 . Is it symmetry of the circle or any property of circle .
angle OXP = 70
since, RO is parallel to QP, angle XPQ=70 and angle PQX=70 (PX and QX are both radius of the circle and hence the corresponding angles, XPQ and PQX are equal)

Therefore, PXQ = 40
Thanks rainribbon!! Great explanation. Pradeep hope you doubt is clear. Sorry to make this ambiguous by stating the other 35. This is the basic property of parallelism.
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by N:Dure » Tue Dec 14, 2010 6:48 pm
Shovan's way is definetly faster but what I did is get the lenght of arcs PO and QR, then subtract them from 1/2 the circle's circumference. The result is also 2 TT.

1/2 circumference - lenght of both arcs
TT 9 - ( 2 * 70/360 2 TT 9)

TT 9 - TT 7 = 2 TT
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by pradeepspanchal » Tue Dec 14, 2010 8:12 pm
Thanks Shoven, rainribbon, N:Dure.

@Shoven : please elaborate a bit more on the property you talking about.
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by shovan85 » Tue Dec 14, 2010 11:05 pm
pradeepspanchal wrote:
@Shoven : please elaborate a bit more on the property you talking about.
A central angle equals twice an inscribed angle if they are subtended by the same chord (if both angles are on the same side of the chord). The image below illustrates it, COB = 2 ∠CDB = 2∠CAB.

See the Image...
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by pradeepspanchal » Wed Dec 15, 2010 9:04 pm
@Shoven : Many thanks for the explanation , but I have trouble in understanding text in red


As RO is parallel to QP angleQXR = 70 as angle QOR = 35


Image

how you get angle QOR = 35 , based on the parallel lines in a circle ? is this a property of circle with two parallel lines . Please highlight the concept I am missing.
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by shovan85 » Thu Dec 16, 2010 1:45 am
pradeepspanchal wrote:@Shoven : Many thanks for the explanation , but I have trouble in understanding text in red


As RO is parallel to QP angleQXR = 70 as angle QOR = 35


Image

how you get angle QOR = 35 , based on the parallel lines in a circle ? is this a property of circle with two parallel lines . Please highlight the concept I am missing.
Please let me know if you understood how QXR is 70. If so then the 70 is happening in Center. Then follow the formula what I have given just last but one post.
The QOR ought to be half of QXR :)
Last edited by shovan85 on Thu Dec 16, 2010 2:54 am, edited 1 time in total.
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by pradeepspanchal » Thu Dec 16, 2010 2:48 am
As RO is parallel to QP angleQXR = 70 as angle QOR = 35

I got the property you pointed, but I am not able get how you get angle QXR = 70 or angle QOR = 35.

what I concluded it may because circle is symmetrical . Please suggest.
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by shovan85 » Thu Dec 16, 2010 5:37 am
pradeepspanchal wrote:As RO is parallel to QP angleQXR = 70 as angle QOR = 35

I got the property you pointed, but I am not able get how you get angle QXR = 70 or angle QOR = 35.

what I concluded it may because circle is symmetrical . Please suggest.
Sorry!! There is not cut short formula like this. I thought it is obvious so stated like that. Sorry for the ambiguous post :)
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by fskilnik@GMATH » Thu Dec 16, 2010 5:46 am
Hi there,

Have a look at this pictures (taken from my "Express Class #3" online course material)!

(The reason for presenting BOTH the 2nd and 3rd diagrams -- as if they were different scenarios -- is to show the students that the property of inscribed angles being half the central angles corresponding to the same arc does NOT depend on the relative position of the center of the circle to these other "elements".)

Best Regards,
Fabio.
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by beat_gmat_09 » Thu Dec 16, 2010 5:57 am
Hi Guys,

I think triangles formed using parallel lines have equal angles in this case. Please have a look at attached image, such two triangles are formed by the parallel lines and transversal's in the circle.
What Shovan interpreted angle QOR=35 is correct, i think so. let me know your thoughts.

Image
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