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Title: Find The Radius Post by ThudanBlunder on Jan 18th, 2009, 3:37pm Given that the chords AC = CD = 3 and BD = 7, find the radius of the circle below. |
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Title: Re: Find The Radius Post by Grimbal on Jan 18th, 2009, 3:53pm I find it easier to [hide]switch one 3 and the 7 line to make the 7 line horizontal.[/hide] My result: [hide] ----> 9 <---- [/hide]. |
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Title: Re: Find The Radius Post by ThudanBlunder on Jan 18th, 2009, 4:00pm The resulting diameter? Anyway, a neat trick, Grimbal. I never thought of that. ::) |
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Title: Re: Find The Radius Post by codpro880 on Jan 18th, 2009, 8:05pm [hide]4.5[/hide] ;D |
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Title: Re: Find The Radius Post by Grimbal on Jan 19th, 2009, 12:49am on 01/18/09 at 16:00:12, ThudanBlunder wrote:
Err... yes, that would be the diameter. :-[ |
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Title: Re: Find The Radius Post by Barukh on Jan 19th, 2009, 2:51am on 01/18/09 at 16:00:12, ThudanBlunder wrote:
It's really a very nice idea! Grimbal is ingenious, as always :D |
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Title: Re: Find The Radius Post by Immanuel_Bonfils on Jan 19th, 2009, 3:58pm With Grimbal & Ptolemy it becomes cool. |
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Title: Re: Find The Radius Post by chronodekar on Jun 24th, 2009, 2:19am If you switch the 3 and 7 lines, you get a trapezoid. Two of the lines are parallel, we know the lengths of the side-lines (pun not intended). But how does that give us the length of the bottom line? ??? I know that if you connect a point on a circle with the ends of a diameter, you get a right-angled triangle. Is there some similar law of geometry I'm missing here? -chronodekar |
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Title: Re: Find The Radius Post by Ronno on Jun 24th, 2009, 5:59am By Ptolemy's theorem, 3*3+7d=d^2-3^2 |
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Title: Re: Find The Radius Post by towr on Jun 24th, 2009, 6:48am For easy reference: http://en.wikipedia.org/wiki/Ptolemy's_theorem (http://en.wikipedia.org/wiki/Ptolemy's_theorem) http://mathworld.wolfram.com/PtolemysTheorem.html |
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Title: Re: Find The Radius Post by chronodekar on Jun 24th, 2009, 9:30am I'll brush up on my geometry. Thanks Ronno and towr. -chronodekar |
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Title: Re: Find The Radius Post by 1337b4k4 on Jun 24th, 2009, 10:18am Or if you don't remember much about Ptolemy like me, [hide]extend BD to meet AC at point E. AOC, CDE and ABE are all similar, and the result quickly follows.[/hide] |
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Title: Re: Find The Radius Post by ecoist on Jun 24th, 2009, 5:42pm I must be having a senior moment. I don't know Ptolemy and I don't know where 1337b4k4 goes after the similarities. Here's another solution. [hide]Extend AC and BD to meet at E. Since AE and BC are orthogonal and angles ABC and CBE are equal, triangles ABC and BCE are congruent. Hence AE=6 and BE=d, where d is the diameter of the circle. Further, AD is orthogonal to BE, whence (AD)^2=6^2 -(d-7)^2=d^2-7^2, which simplifies to d^2-7d-18=0. Hence d=9.[/hide] |
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Title: Re: Find The Radius Post by 1337b4k4 on Jun 26th, 2009, 12:45pm on 06/24/09 at 17:42:44, ecoist wrote:
[hide] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifAOD is twice http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifABD, thus http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifAOC is the same as http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifABD, AOC and ABE are similar (two angles in common). Consequently, CE has length 3. http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifACO, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifOCD, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifDCE add to 180, thus http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifDCE is the same as http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/angle.gifAOC. As CDE is also isosceles, it is also similar to AOC and ABD. side lengths of similar triangles are proportional, thus r/3 = 3/(2r-7) so r = 4.5[/hide] |
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Title: Re: Find The Radius Post by ecoist on Jun 26th, 2009, 6:57pm Danke sehr, 1337b4k4. Trivial details are often left out, but some of these details are not so trivial to me. Nice to know that there are those willing provide such details. I figured out your solution shortly after my post but I still struggled with your third similarity, triangle CED. Fortunately, that problem disappeared when I paid attention to the fact that the sum of angles CED and CDE equals angle ACD. Nice work, 1337b4k4, much better than my tedious solution. |
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