AHSME 1967 Problem 40

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: $\textbf{(A)}\ 159 \textbf{(B)}\ 131 \textbf{(C)}\ 95 \textbf{(D)}\ 79 \textbf{(E)}\ 50$