Question

.
The value of x is:


12.
6.
$12\sqrt{2$
None of the choices are correct.

Answer 1

$\bf \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad \begin{cases} c=hypotenuse=x\\ a=adjacent=12\\ b=opposite=12\\ \end{cases} \\\\\\ x=\sqrt{12^2+12^2}\implies x=\sqrt{288}\implies x=\sqrt{12^2\cdot 2} \\\\\\ x=12\sqrt{2}$