right triangle

5) The hypotenuse of the "12 by 16" triangle, using pythagoras' theorem, is $\sqrt{12^2 + 16^2} = \sqrt{400} = 20$

Then, the hypotenuse of the "12 by x" triangle, using the big triangle, is $\sqrt{(x+16)^2 - 20^2}$

But we also know that that side is $\sqrt{12^2 + x^2}$

So we can make an equation since the two are equivalent:
$\sqrt{(x + 16)^2 - 20^2} = \sqrt{12^2 + x^2}$

$(x + 16)^2 - 20^2 = 12^2 + x^2$

$(x + 16)(x + 16) - x^2 = 144 + 400$

$x^2 - x^2 + 32x + 256 = 544$

$32x = 288$

$x = 9$

6) This question is similar to the previous one so I won't explain in depth:
$\sqrt{60^2 - 36^2} = 48$

$\sqrt{(x - 36)^2 + 48^2} = \sqrt{x^2 - 60^2}$

$(x - 36)(x - 36) - x^2 = - 60^2 - 48^2$

$x^2 - x^2 - 72x + 1296 = -5904$

$-72x = -4608$

$x = 64$