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Circle A, center X. XB is the radius. There is a chord which [#permalink]
21 Mar 2008, 23:36

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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?

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! _________________

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:

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

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

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

(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