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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)

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Solution (some cheating) | Comment 1 of 9

Note that opposite angles of an inscribed quadrilateral sum to 180.

Using the law of cosines with one of the diagonals

c^2 = 6^2 + 7^2 -2*6*7*cos(A)

c^2 = 8^2 + 9^2 -2*8*9*cos(180-A)

Allows us to find the length of the diagonal (10.349) as well as the opposite angles (105.26 and 74.74)

I figued I only need to then find the circumradius, so I looked for a formula on mathworld and found too much.

R = 1/4*sqrt(((6*8+7*9)(6*9+7*8)(6*7+8*9)/((15-6)(15-7)(15-8)(15-9))
R = 1/4*sqrt(703/6)

A = pi*R^2 = pi*1/16*703/6 = 703/96*pi = 23.0056212

Ok, so that was a lot of cheating.  Cool formula, though.


  Posted by Jer on 2005-01-11 19:47:55
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