 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Where is the center? (Posted on 2019-03-15) Consider the cyclic quadrilateral with sides 1, 4, 8, 7 in that order. What is its circumdiameter?

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 3
The area of a cyclic quadrilateral can be calculated using A^2 = (s-a)*(s-b)*(s-c)*(s-d) where s is the semiperimeter.

An alternative area formula involves the circumradius, A^2 = (a*b+c*d)*(a*c+b*d)*(a*d+b*c)/(16r^2).

From the values give, s = (1+4+7+8)/2 = 10.
Then A^2 = (10-1)*(10-4)*(10-8)*(10-7) = 324
And A^2 = (1*4+8*7)*(1*8+4*7)*(1*7+4*8)/(16r^2) = 5265/r^2

Equate the two area results to yield 324 = 5265/r^2, which implies the circumradius is sqrt(65)/2, which means the circumdiameter sought is sqrt(65)

Edited to fix error 5625->5265

Edited on March 17, 2019, 12:55 pm
 Posted by Brian Smith on 2019-03-15 12:42:01 Please log in:

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