Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$, $PB=6$, and $PC=10$.  To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$
[asy] draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); label("$A$",(0,10),N); label("$B$",(-9.5,-5.2),N); label("$C$",(9.5,-5.2),N);  dot((-3,0)); label("$P$",(-3,-2),N); draw((-3,0)--(0,10)); draw((-3,0)--(-8.66,-5)); draw((-3,0)--(8.66,-5));  dot((-9,7.5)); label("$P'$",(-9.2,7.5),N); draw((-9,7.5)--(0,10)); draw((-9,7.5)--(-8.66,-5)); draw((-9,7.5)--(-3,0));  [/asy]
Notice that $6^2+8^2=10^2.$ That makes us want to construct a right triangle.
Rotate $\triangle APC$ $60^{\circ}$ about A. Note that $\triangle PAC \cong \triangle P'AB$, so 
\[\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.\]
Therefore, $\triangle APP'$ is equilateral, so $P'P=8$, which means $\angle P'PB = 90^{\circ}.$
Let $\angle BP'P = \alpha .$ Notice that $\cos\alpha = \frac{8}{10}=\frac{4}{5},$ and $\sin\alpha = \frac{3}{5}.$
Applying the Law of Cosines to $\triangle APC$ (remembering $\angle APC = \angle AP'B$):
\begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}
We want to find the area of $\triangle ABC$, which is \[AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{(D) 79}.\]
~pfalcon