smc

Clearly, the other side of $R_1$ has length $\frac{12}{2}=6$. Now call the sides of $R_2$ a and b, with $b>a$. We know that $\frac{b}{a}=\frac{3}{1}=3$, because the two rectangles are similar. We also know that $a^2+b^2=15^2=225$. But $b=3a$, so substituting gives $$a^2+(3a{)}^2=225$$ $$10a^2=225$$ $$a^2=\frac{45}{2}$$ $$a=\frac{3\sqrt{10}}{2}$$ $$b=\frac{9\sqrt{10}}{2}$$ $$[R_2]=ab=\boxed{\frac{135}{2}}$$