+1234 In triangle $ABC,$ $AB = 15,$ $BC = 10,$ and $AC = 12.$ Find the length of the shortest altitude in this triangle.
+129771 The shortest altitude is drawn to the longest side
Let tha altitude we are looking for be CM
Let AM = x and BM = 15 - x
We can solve this (equating squares of the altitude CM )
12^2 - x^2 = 10^2 - (15 - x)^2
144 - x^2 = 100 - (225 - 30x + x^2)
30x = 369
x = 269 / 30
The altitude is sqrt [ 12^2 - (269/30)^2 ] ≈ 7.97
