In triangle $ABC,$ $AB = 15,$ $BC = 10,$ and $AC = 12.$ Find the length of the shortest altitude in this triangle.
Find the area using Heron's Formula
A = sqrt [ 18.5 (3.5) (6.5) (8.5) ] ≈ 59.8
The shortest altitude is drawn to the longest side
59.8 = (1/2) (15) (altitude)
59.8 / [ 15/2 ] = altitude ≈ 7.97
+436 
