The two diagonals of a
cyclic quadrilateral EFGH are EG and FH while
the respective lengths of its adjacent sides EF and FG are 2 and 5.
It is known that Angle EFG = 60
^{o} and the area of the quadrilateral is 4√3.
Determine the sum of the lengths GH + HE.
Let e = GH and g = HE.
Area = 4*sqrt(3)
= (1/2)EFFEsin(60) + (1/2)eg*sin(18060)
or
eg = 6
EG = EF^2 + FG^2  2EFFGcos(60)
= e^2 + g^2  2eg*cos(18060)
or
19 = e^2 + g^2 + eg
or
25 = 19 + eg = (e+g)^2
or
GH+HE = 5
One of the two possible quadrilaterals is an
Isosceles Trapezoid.

Posted by Bractals
on 20071021 14:02:43 