Answer:
[tex]x=6\sqrt{3}[/tex]
Step-by-step explanation:
see the attached figure with letters to better understand the problem
step 1
In the right triangle ABC
Applying the Pythagoras Theorem
Find 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 2
In the right triangle ABD
Applying the Pythagoras Theorem
Find 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 A
step 3
In the right triangle BCD
Applying the Pythagoras Theorem
Find 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 B
step 4
equate 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]
