jmc

We set $a$ as the length of the side of the first triangle and $b$ as the length of the side of the second. We know that the sum of the perimeter is $45$, so $3a+3b=45\to a+b=15$.

We also know that the area of the second is $16$ times the area of the first, so $b^2=16a^2$. Solving and taking the positive root, we get that $b=4a$. Thus, $a+4a=15\to a=3$. Therefore, the side of the larger triangle $b=4\cdot 3=12$.

The area of an equilateral triangle with side length $s$ is $\frac{s^2\cdot \sqrt{3}}{4}$, so the desired area is $\frac{12^2\cdot \sqrt{3}}{4}=\boxed{36\sqrt{3}}$.