[tex]|SM|= \frac{1}{2} |SU|\\\\\Rightarrow\ \ \ \frac{1}{4} |SU|=x+15\ \ \ and\ \ \ \frac{3}{4} |SU|=4x-45\\\\|SU|=4(x+15)\ \ \ and\ \ \ |SU|= \frac{4}{3} (4x-45)\\\\4(x+15)=\frac{4}{3} (4x-45)\ /\cdot \frac{3}{4} \\\\3(x+15)=4x-45\\3x+45=4x-45\\3x-4x=-45-45\\-x=-90\\x=90\\\\|SU|=4(x+15)=4\cdot(90+15)=4\cdot105=420[/tex]