Here is your answer
B. [tex]x=\sqrt{90}[/tex]
Using Pythagoras ' theorem we know that
Hypotenuse^2= base^2 + perpendicular^2
So here,
[tex]{x}^{2}= {9}^{2}+{3}^{2}[/tex]
[tex]{x}^{2}= 81+9[/tex]
[tex]{x}^{2}= 90[/tex]
[tex]x= \sqrt{90}[/tex]
HOPE IT IS USEFUL
