My solution for this is using the cosine theorem to get the side i am missing and then using Heron's formula to find the area of the triangle and then doubling it to get the wanted area. So i have two questions:
1) When i make a $\Delta BDA$ and use the cosine theorem i get the equation $x^2-x*10\sqrt(3)+64=0$ so i get two solutions ($x=5\sqrt(3)-\sqrt(11)$ and $x=5\sqrt(3)+\sqrt(11)$) how would i choose the side?
2) Could someone give me a hint or a solution on how i would solve this problem without cosine theorem? Edit: Perhaps we see that since angle is $60$ we use it to find $AC=10$ and then substract the triangle $\Delta ACD$ which has sides $6,6,10$ ?