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Circle A, center X. XB is the radius. There is a chord which [#permalink]
 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?
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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 TIME-CONSUMING APPROACH!
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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}=203. Area = \frac{\pi*BY^2}{4}=100\pi
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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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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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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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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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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 TIME-CONSUMING APPROACH!sorry. i dont see how you get XD=8.
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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 TIME-CONSUMING 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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