All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Chord Harmonics (Posted on 2005-01-11) Difficulty: 3 of 5
Four chords of a circle are arranged in such a way that they form a quadrilateral shape, the length of the chords are as follows.

6cm, 7cm, 8cm and 9cm.

what is the area of the circle that will intersect all of the points?

No Solution Yet Submitted by Juggler    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Still no proof, but... | Comment 5 of 9 |
(In reply to Still no proof, but... by nikki)

Augh! I feel like an idiot. I checked my excel equation again, and I had 6 as all the chord lengths. Argh! I thought I checked for that tooÖ that I didnít accidentally have two 7s and not the 8 or something. Sigh.

So I did the zeroing in thing and got R=5.3636369363 as my answer. If that 3636369 sequence repeats, then that would make my answer 5+3636369/9999999 = 5+404041/1111111

So the area equals 90.37922614.

I wish I had a more scientific way of getting that answer. My zeroing in method is not very scientific, I donít think.  I just don't know what to do with a sum of inverse-sines.


  Posted by nikki on 2005-01-12 18:24:13
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information