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Circle and Chord

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Junior | Next Rank: 30 Posts
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by keizer Soze » Wed Apr 29, 2009 7:34 pm
Any ideas how to solve this? I couldn´t, don´t know if it´s right.

Circle X has chord AB that intersects radius XC in point D and angle XDB is 90°. If AB=12 and DC=2. What is the circumference of the circle?


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by dumb.doofus » Wed Apr 29, 2009 8:15 pm
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by keizer Soze » Wed Apr 29, 2009 8:20 pm
Thanx doofus!
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by 4seasoncentre » Wed Apr 29, 2009 8:24 pm
kinda hard to explain without a diagram, but.....

We know that XC is the radius. Since the chord intersects at a 90 degree angle, we know that AD=6.

DC=2
Therefore XD = XC -2
note XC is also the radius of the circle, so XD = R - 2
Note that XA is also the radius.

Using Pythagoras:

XD^2 + 6^2 = XA ^2
(r-2)^2 + 36 = r^2
r^2 - 4r + 4 + 36 = r^2

work that out and you get r = 10
Circumference of a circle is defined by
2πr
=2π(10)
=20π

DAMN IT, someone beat me to it!
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by PAB2706 » Wed Apr 29, 2009 10:02 pm
why is the answer different when i take XB=XD+2
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by dumb.doofus » Thu Apr 30, 2009 12:33 am
PAB2706 wrote:why is the answer different when i take XB=XD+2
It can't be :-) both are equal.. recheck the steps what you are doing.. or post the steps here..
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