$AB<AC$

Probability Level 4

Triangle $ABC$ has $AB=10$ and $AC=14$ . A point $P$ is randomly chosen in the interior or on the boundary of triangle $ABC$ . What is the probability that $P$ is closer to $AB$ than to $AC$ ?

\frac{5}{12}

\frac{1}{4}

\frac{5}{7}

\frac{2}{3}

\frac{3}{7}

\frac{1}{3}

\frac{3}{4}

0 solutions

No explanations have been posted yet. Check back later!

A B < A C AB<AC A B < A C

0 solutions

0 pending reports

$AB<AC$