Answer:
[tex]CD=3\sqrt{3}\ cm,\\ \\A_{ABC}=18\sqrt{3}\ cm^2[/tex]
Step-by-step explanation:
Triangle ABC is right triangle. Since m∠ACB = 90° and m∠ACD = 60°, you get m∠BCD = 90°-60°=30°.
Triangle BCD is rigth triangle, then the leg that is opposite to the angle of 30° is half of the hypotenuse. Thus,
[tex]BD=\dfrac{1}{2}BC=\dfrac{1}{2}\cdot 6=6\ cm.[/tex]
By the Pythagorean theorem,
[tex]CD^2=BC^2-BD^2=6^2-3^2=36-9=27,\\ \\CD=3\sqrt{3}\ cm.[/tex]
Then
[tex]CD^2=AD\cdot BD,\\ \\27=3\cdot AD,\\ \\AD=9\cm.[/tex]
Hypotenuse AB of the triangle ABC is equal to
[tex]AB=AD+BD=9+3=12\ cm.[/tex]
The area of the triangle ABC is
[tex]A_{ABC}=\dfrac{1}{2}\cdot AB\cdot CD=\dfrac{1}{2}\cdot 12\cdot 3\sqrt{3}=18\sqrt{3}\ cm^2.[/tex]
