Mysterious Quadrilateral

By the law of cosines on triangle ABC, we get that $AC = 3\sqrt{2} - 2 = CD = DA$ , so ACD is equilateral.

Let the perpendicular to BC at C hit AB at E; then BCE is a 30-60-90 triangle, and $CE = \sqrt{2}$ . Also, $AE = 4 - 2\sqrt{2}$ , and $\angle CEA = 120$ .

We can then use the law of cosines and sines on ACE to get the sine and cosine of $\angle ACE$ . Since we have $\angle BCE = 90, \angle ACD = 60$ , we have $\angle BCD = \angle ACE + 150$ . Then we use the cosine addition formula to compute the cosine of $\angle BCD$ , and by the law of cosines on triangle BCD we find that $BD = \sqrt{22}$ .

The answer is then just $3\sqrt{2} - 2 + \sqrt{22} \approx 6.933$ .

Let the diagonal AC be denoted by P. Then, in triangle ABC, by the Law of Cosines, P^2 = 4^2 + [sqrt(6)]^2 - 2 4 sqrt(6) cos(30(, or P = 2.242640687. Define < BAC = t. then, by the Law of sines, sqrt(6)/sin(t) = P/sin(30). Substituting, t = 33.10104895. Define < CAD = a. Then, in triangle ACD, by the Law of sines, P/sin(180 - 2a) = (3sqrt(2) - 2)/sin(a), or P/sin(2a) = 2.242640687/sin(a), or P/(2sin(a)cos(a) = P/sin(a). or cos(a) = .5, and a = 60. Then t + a = 93.10104895. Defining the diagonal BD as Q, in triangle BAD, we have by the Law of cosines, Q^2 = 4^2 + (3sqrt(2) - 2)^2 - 4 (3(sqrt(2) - 2)cos(93.10104895. Then Q = 4. 690415761, and P + Q = 6.932. Ed Gray