Answer:
[tex]\huge\boxed{x=2\sqrt{286}}[/tex]
Step-by-step explanation:
Use the Pythagorean theorem:
[tex]leg^2+leg^2=hypotenuse^2[/tex]
We have:
[tex]leg=x\\leg=15\\hypotenuse=37[/tex]
Substitute and solve for x:
[tex]x^2+15^2=37^2\\\\x^2+225=1269\qquad|\text{subtract 225 from both sides}\\\\x^2+225-225=1269-225\\\\x^2=1144\to x=\sqrt{1144}\\\\x=\sqrt{4\cdot286}\\\\x=\sqrt4\cdot\sqrt{286}\\\\x=2\sqrt{286}[/tex]
