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:
\(\sqrt{60^2 - 36^2} = 48\)
\(\sqrt{(x - 36)^2 + 48^2} = \sqrt{x^2 - 60^2}\)
\((x - 36)(x - 36) - x^2 = - 60^2 - 48^2\)
\(x^2 - x^2 - 72x + 1296 = -5904\)
\(-72x = -4608\)
\(x = 64\)
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