Answer:[tex]x=6\sqrt{3}[/tex]Step-by-step explanation:see the attached figure with letters to better understand the problemstep 1In the right triangle ABCApplying the Pythagoras TheoremFind out the length side AB[tex]AC^2=AB^2+BC^2[/tex][tex]AB^2=AC^2-BC^2[/tex]substitute the given values[tex]AB^2=12^2-x^2[/tex][tex]AB^2=144-x^2[/tex]step 2In the right triangle ABDApplying the Pythagoras TheoremFind out the length side BD[tex]AB^2=AD^2+BD^2[/tex][tex]BD^2=AB^2-AD^2[/tex]substitute the given values[tex]BD^2=144-x^2-3^2[/tex][tex]BD^2=144-x^2-9[/tex][tex]BD^2=135-x^2[/tex] -----> equation Astep 3In the right triangle BCDApplying the Pythagoras TheoremFind out the length side BD[tex]BC^2=DC^2+BD^2[/tex][tex]BD^2=BC^2-DC^2[/tex]substitute the given values[tex]BD^2=x^2-9^2[/tex][tex]BD^2=x^2-81[/tex] -----> equation Bstep 4equate equation A and equation B[tex]BD^2=135-x^2[/tex] -----> equation A[tex]BD^2=x^2-81[/tex] -----> equation B[tex]x^2-81=135-x^2[/tex][tex]x^2+x^2=135+81[/tex][tex]2x^2=216[/tex][tex]x^2=108[/tex][tex]x=\sqrt{108}[/tex]Simplify[tex]x=6\sqrt{3}\ units[/tex]