In triangle $ABC,$ $AB = 15,$ $BC = 9,$ and $AC = 10.$ Find the length of the shortest altitude in this triangle.
By Heron's Formula
[ ABC ] = sqrt [ 17 * 2 * 8 * 7] = 4sqrt 119
Shortest altitude is always drawn to the longest side
So
4sqrt 119 = (1/2) 15 * altitude
4sqrt 119 = 7.5 *altitude
altitude = (4 / 7.5)sqrt 119 ≈ 5.82
+1609 
