This is just an old geometry problem which I think is good

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This is just an old geometry problem which I think is good [#permalink]

02 Jul 2004, 15:24

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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?
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Paul

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02 Jul 2004, 15:42

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
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02 Jul 2004, 15:44

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

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12 Jul 2004, 05:38

Sorry guys, just remembered about this one :oops:

. I don't have OA but I got 100pi too
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Paul

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16 Jul 2004, 15:16

Same as ashkg. Got 100pi as well. Good explanation ashkg!
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Davefor MBA

[#permalink] 16 Jul 2004, 15:16

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This is just an old geometry problem which I think is good

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This is just an old geometry problem which I think is good

This is just an old geometry problem which I think is good

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