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)|EF||FE|sin(60) + (1/2)eg*sin(180-60)

    or

  eg = 6

|EG| = |EF|^2 + |FG|^2 - 2|EF||FG|cos(60)

     = e^2 + g^2 - 2eg*cos(180-60)

    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 2007-10-21 14:02:43