In triangle $ABC,$ $AB = 10,$ $BC = 24,$ and $AC = 26.$ Find the length of the shortest altitude in this triangle.
+23246 Since 102 + 242 = 262, this is a right triangle with angle(B) the right angle.
The shortest altitude will be the one from angle(B).
Call the point where this altitude hits side(AC) "X".
AX/AB = AB/AC ---> AX/10 = 10/26 ---> AX = 50/13.
Triangle(ABX is a right triangle: (50/13)2 + (BX)2 = 102 ---> (BX)2 = 14400/169 ---> BX = 120/13.
