Answer:
Step-by-step explanation:
[tex]\text{Use the Pythagorean theorem:}\\\\leg^2+leg^2=hypotenuse^2\\\\\text{We have}\\\\leg=a=3b\\leg=b\\hypotenuse=2\sqrt{10}\\\\\text{Substitute}\\\\(3b)^2+b^2=(2\sqrt{10})\qquad\text{use}\ (ab)^n=a^nb^n\\\\3^2b^2+b^2=2^2(\sqrt{10})^2\qquad\text{use}\ (\sqrt{a})^2=a\ \text{for}\ a\geq0\\\\9b^2+b^2=4(10)\qquad\text{combine like terms}\\\\10b^2=40\qquad\text{divide both sides by 10}\\\\b^2=4\to b=\sqrt4\\\\b=2[/tex]
