Step-by-step explanation:
In this case, we will use Pythagoras theorem for all three cases.
Pythagoras theorem : Hypotenuse² = base² +perpendicular²
(1) a = 5, b = 12
[tex]c=\text{Hypotenuse}\\\\c=\sqrt{a^2+b^2} \\\\x=\sqrt{12^2+5^2} \\\\x=13[/tex]
(2) b = 30, c = 34
[tex]a=\sqrt{c^2-b^2} \\\\a=\sqrt{(34)^2-(30)^2} \\\\a=16[/tex]
(3) b = 15, c = 25
[tex]a^2=\sqrt{c^2-b^2} \\\\a=\sqrt{(25)^2-(15)^2} \\\\a=20[/tex]
Hence, this is the required solution.
