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

$\fbox{D}$

1967 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 40
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1967 AHSME Problems/Problem 40

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