1967 AHSME Problems/Problem 40

Problem

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:

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

Solution

$\fbox{D}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 40
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