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

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GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
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Kudos [?]: 217 [0], given: 0

This is just an old geometry problem which I think is good [#permalink]  02 Jul 2004, 14:24
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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

Senior Manager
Joined: 05 Feb 2004
Posts: 290
Location: USA
Followers: 1

Kudos [?]: 3 [1] , given: 0

1
KUDOS
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
Senior Manager
Joined: 21 Mar 2004
Posts: 444
Location: Cary,NC
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Kudos [?]: 20 [0], given: 0

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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ash
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I'm crossing the bridge.........

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 217 [0], given: 0

Sorry guys, just remembered about this one . I don't have OA but I got 100pi too
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Best Regards,

Paul

Intern
Joined: 07 Jun 2004
Posts: 29
Location: Sunnyvale
Followers: 0

Kudos [?]: 0 [0], given: 0

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

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