Find the maximum possible area of a convex quadrilateral having sides of lengths 1, 4, 7 and 8.

The area of a cyclic quadrilateral is the maximum possible for any quadrilateral with the given side lengths (Mathworld).

8^2+1^2=4^2+7^2= 65; two right triangles with common hypotenuse sqrt65, which is therefore a diameter.

The figure must be cyclic, so its area is 4+14=18.

Posted by broll on 2016-07-03 08:38:38