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02 Jul 2004, 15:24
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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This is just an old geometry problem which I think is good practice for all. There has not been that many lately and this will, I believe, sharpen your skills.
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 circle'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 circle's area?
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02 Jul 2004, 15:42
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r^2 = 6^2 + (r-2)^2
r = 10
A = 100pi
I should have realized that its a 3-4-5 triplet. Did all the calculations and then got it. Could have saved some time .Bummed
- ash
r = 10
A = 100pi
I should have realized that its a 3-4-5 triplet. Did all the calculations and then got it. Could have saved some time .Bummed
- ash
02 Jul 2004, 15:44
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100PI....!!
let XC = a = XB = XA = radius
AD= DC = 12/2 = 6
XC^2 = XD^2 + DC^2
a^2 = (a-2)^2 + 6^2
Solving a = 10 = radius.
Area = pi * 10^2
let XC = a = XB = XA = radius
AD= DC = 12/2 = 6
XC^2 = XD^2 + DC^2
a^2 = (a-2)^2 + 6^2
Solving a = 10 = radius.
Area = pi * 10^2
. I don't have OA but I got 100pi too 
