[tex]\triangle BCA\ \text{and}\ \triangle DCB\ \text{are similar (AAA). Therefore the sides are in proportion:}\\\\\dfrac{BC}{CD}=\dfrac{AC}{BC}\\\\\text{We have}\\\\BC=m\\CD=7\\AC=7+11=18\\\\\text{Substitute:}\\\\\dfrac{m}{7}=\dfrac{18}{m}\qquad\text{cross multiply}\\\\m^2=(7)(18)\\\\m^2=126\to\boxed{m=\sqrt{126}}\to\boxed{A.}[/tex]
