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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?
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22 Mar 2008, 01:16
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az780
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!
22 Mar 2008, 01:34
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Another way:
1. ABY is geometrically similar to DBA
2. \(\frac{AB}{2}=\frac{BY}{AB}\) -->\(BY=\frac{AB^2}{2}=\frac{6^2+2^2}{2}=20\)
3. \(Area = \frac{\pi*BY^2}{4}=100\pi\)
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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