In \(\triangle ABC\), the \(A\)-median, \(B\)-median, \(C\)-median have lengths \(13,14,15\), respectively. Find area of \(\triangle ABC\).
Solution
Answer: \(112\)
The triangle with side lengths \(13,14,15\) has semiperimeter \(21\) and area
\[\sqrt{21(21-13)(21-14)(21-15)}=84.\]
Let the area of \(ABC\) be \(S\). Reflect \(B\) across the midpoint of \(AC\) to obtain \(D\), so \(ABCD\) is a parallelogram. The centroid lies two-thirds of the way along each median. Comparing the resulting areas gives
\[2S=6\left(\frac23\right)^2\cdot84,\qquad S=\frac43\cdot84=112.\]