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