smc

Given the equations $x^2+y^2=25$ and $(x-4{)}^2+9y^2=81$, we can substitute $y^2=25-x^2$ from the first equation and plug it in to the 2nd equation, giving us $(x-4{)}^2+9(25-x^2)=81$. After rearranging, $8x^2+8x-160=0$ or $x^2+x-20=0$. The solutions are $x=-5$ and $x=4$. This gives us the points $(-5,0),(4,3)$,and $(4,-3)$. The area of the triangle formed by these points is $27=\boxed{27}$