Consider the isosceles trapezoid with base lengths actually 5 and 3. The longer base overlaps the shorter one by 1 unit on either end. If the height of the trapezoid is h, the angle a diagonal makes with the longer base is
.. α = arctan(h/4)
The angle the slant end of the trapezoid makes with the longer base is
.. β = arctan(h/1)
The angles are related by
.. tan(β) = tan(2α) = 2×tan(α)/(1 -tan(α)^2)
.. h = 2*(h/4)/(1 -(h/4)^2)
.. 1 -(h/4)^2 = 1/2 . . . . . . . . . . . multiply by the denominator and divide by h
.. 16 -h^2 = 8 . . . . . . . . . . . . . . multiply by 16
.. h = √8
The length of the slant side (s) will be
.. s = √(1^2 +h^2) = √(1 +*) = 3
so the perimeter of the trapezoid will be
.. 5 + 3 + 2*3 = 14
The perimeter of the trapezoid of interest is twice that, so its side lengths will be twice these.
The lengths of the sides of the trapezoid with perimert 28 are 6 and 6.
The base lengths are 10 and 6.