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Manager
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Circle A, center X. XB is the radius. There is a chord which
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22 Mar 2008, 00:36
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Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area? == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: PS (geometry)
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22 Mar 2008, 01:16
az780 wrote: Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area? My computer does not support drawing, pls get it by yoursefl 1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6 2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 > r = 8+2 = 10 > Area of cirle = 100pi EXPECT A LESS TIMECONSUMING APPROACH!
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Re: PS (geometry)
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22 Mar 2008, 01:34
Another way: 1. ABY is geometrically similar to DBA 2. \(\frac{AB}{2}=\frac{BY}{AB}\) >\(BY=\frac{AB^2}{2}=\frac{6^2+2^2}{2}=20\) 3. \(Area = \frac{\pi*BY^2}{4}=100\pi\)
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Re: PS (geometry)
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13 May 2008, 13:09
I think I need some serious work in geometry.
Could someone please explain the following... 2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 > r = 8+2 = 10 >
or walker's method? Thanks



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Re: PS (geometry)
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13 May 2008, 17:25
walker wrote: Another way:
1. ABY is geometrically similar to DBA
2. \(\frac{AB}{2}=\frac{BY}{AB}\) >\(BY=\frac{AB^2}{2}=\frac{6^2+2^2}{2}=20\)
3. \(Area = \frac{\pi*BY^2}{4}=100\pi\) the first answer is straight forward Pythagorean theorem. i like the walkers answer, but don’t really get it. walker could you please shed some light!



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Re: PS (geometry)
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13 May 2008, 19:40
the first step is the most important in my approach: the triangles DBA and ABY are right triangles, because angles ADB and YAB are right angles. the triangles DBA and ABY have the same angle ABY=ABD and, therefore, third angles in triangles are equal: AYB=DAB So, 1. ABY is geometrically similar to DBA Next step is easer, because I only use the formula for similar triangles: \(\frac{BY}{AB}=\frac{AY}{AD}=\frac{AB}{BD}\)
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Re: PS (geometry)
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14 May 2008, 09:47
Thanks Walker.
If anyone is wondering how we know that angle BAY is right, look up Thales' theorem. Until I looked at walker's method, I had no idea this theorem existed. I really don't remember this from high school. Did I simply forget it, or is it not taught in US public schools?
Also, if anyone is confused by the first (simple) method, the trick is to recognize that XB = XA = XD + DB



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Re: PS (geometry)
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14 May 2008, 20:34
sondenso wrote: az780 wrote: Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area? My computer does not support drawing, pls get it by yoursefl 1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6 2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 > r = 8+2 = 10 > Area of cirle = 100pi EXPECT A LESS TIMECONSUMING APPROACH!sorry. i dont see how you get XD=8.



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Re: PS (geometry)
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15 May 2008, 07:41
seongbae wrote: sondenso wrote: az780 wrote: Circle A, center X. XB is the radius. There is a chord which intersects XB. D is the point of intersection between XB and AC. BD = 2, AC = 12, XDA = 90 degrees. What is the cirle's area? My computer does not support drawing, pls get it by yoursefl 1. XDA = 90 degr. so XB perpendicular with AC, and so D is mid point of AC, that is AD=DC =6 2 Triangle XDA: (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 ==> XD=8 > r = 8+2 = 10 > Area of cirle = 100pi EXPECT A LESS TIMECONSUMING APPROACH!sorry. i dont see how you get XD=8. (XA)^2 = (XD +2)^2 = (XD)^2 + (DA)^2 = (XD)^2 + 36 take one thing at a time. (XA)^2 = (XD +2)^2 (Because XA and (XD + 2) are both equal to the radius of the circle) (XD)^2 + (DA)^2 (Imagine a triangle ADX, with angle ADX a right angle. This is the pythagorean theorem. Leg AD squared plus leg XD squared equals hypotenuse AX squared. This equals XD squared plus 36. So AX^2 equals both (XD +2)^2 and (XD)^2 + 36, and since (XD +2)^2 = (XD)^2 + 36, we can solve for XD (XD)^2 + 2 XD + 2 XD + 4 = (XD)^2 + 36 4 XD = 32 XD = 8



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Re: Circle A, center X. XB is the radius. There is a chord which
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Re: Circle A, center X. XB is the radius. There is a chord which
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