+1427 In triangle $ABC,$ $AB = 15,$ $BC = 9,$ and $AC = 10.$ Find the length of the shortest altitude in this triangle.
+129849 The shortest altitude will be drawn to the longest side
Using Heron's formula, the semi-perimeter = [ 9 + 10 + 15 ] / 2 = 17
The area = sqrt [ 17 * (17 - 9) * (17 -10) * (17 -15) ] = sqrt [ 17 * 8 * 7 * 2 ] =
sqrt [ 1904]
So area = (1/2) (AB) (altitude drawn to AB)
sqrt [ 1904] = (1/2)(15) ( altitude drawn to AB)
sqrt [ 1904] / 7.5 = altitude drawn to AB ≈ 5.82
