Tentukan nilai x pada tripel pythagoras berikut A.16,24,x B.32,x,56 C.x,27,35

A).
[tex]x = \sqrt{ {16}^{2} + {24}^{2} } \\ x = \sqrt{256 + 576} \\ x = \sqrt{832} = 28.84[/tex]
B).
[tex]x = \sqrt{ {56}^{2} - {32}^{2} } \\ x = \sqrt{3136 - 1024} \\ x = \sqrt{2122} = 46.06[/tex]
C).
[tex]x = \sqrt{ {35}^{2} - {27}^{2} } \\ x = \sqrt{1225 - 729} \\ x = \sqrt{496} = 22.27[/tex]
Semoga bermanfaat.....
-Prisco