Answer:
[tex]CD=(3-\sqrt{3})\ cm[/tex]
Step-by-step explanation:
step 1
Find the length side AB
In the right triangle ABD
[tex]sin(30\°)=\frac{AB}{AD}[/tex]
[tex]AB=sin(30\°)(AD)[/tex]
we have
[tex]AD=2\sqrt{3}\ cm[/tex]
[tex]sin(30\°)=\frac{1}{2}[/tex]
substitute
[tex]AB=\frac{1}{2}(2\sqrt{3})[/tex]
[tex]AB=\sqrt{3}\ cm[/tex]
step 2
Find the length side BD
In the right triangle ABD
[tex]cos(30\°)=\frac{BD}{AD}[/tex]
[tex]BD=cos(30\°)(AD)[/tex]
we have
[tex]AD=2\sqrt{3}\ cm[/tex]
[tex]cos(30\°)=\frac{\sqrt{3}}{2}[/tex]
substitute
[tex]BD=\frac{\sqrt{3}}{2}(2\sqrt{3})[/tex]
[tex]BD=3\ cm[/tex]
step 3
Find the length side BC
In the right triangle ABC
we know that
BC=AB -----> is an 45°-90°-45° triangle
therefore
[tex]BC=\sqrt{3}\ cm[/tex]
step 4
Find the length side CD
we know that
[tex]BD=BC+CD\\CD=BD-BC[/tex]
substitute the values
[tex]CD=(3-\sqrt{3})\ cm[/tex]