 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Largest circle in quadrilateral (Posted on 2021-09-17) Given a quadrilateral with side lengths 2, 4, 5, and 6 units, what is the radius of the largest circle which can be drawn completely inside of it?

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) shortcut approach | Comment 2 of 8 | If we consider the isosceles triangle with sides (2+4), 5, 6, something approaching this configuration might be the optimal shape of the quadrilateral. The question is: would any outward buckling of the 2+4 side, to make the figure a true quadrilateral, also make for a smaller incircle?  I suspect so!

Thus, the solution would be an infinitesimal departure from this triangle, and the solution would asymptotically approach the triangle solution.

In this case, with the base=5, and the two base angles=theta,
cos(theta) = 2.5/6
The incenter is at the intersection of the base angle bisector and the altitude.

The radius of the incircle is
(5/2) tan(theta/2) = (5/2) sqrt [ (1-cos(theta)) / (1+cos(theta)) ]

= (5/2) sqrt(7/17) = 1.604

Edited on September 19, 2021, 2:54 pm
 Posted by Steven Lord on 2021-09-19 07:44:57 Please log in:

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