Let $x$ be the length of the base of the right triangle, and let $y$ be the length of the height. Using the Pythagorean Theorem, we have $x^2+y^2=(16-x-y)^2$. Simplifying, we have $x^2+y^2=256-32(x+y)+x^2+2xy+y^2$. Rearranging and simplifying again, we have $32(x+y)=256+2xy$. Since we want to maximize the area of the right triangle, we want to maximize $xy$. To do this, we will maximize $x$ and $y$. To start, we can write $y$ in terms of $x$ using the equation we derived earlier: $y=\dfrac{256+2xy-32x}{32}$ which simplifies to $y=\dfrac{8+x}{2}$. Substituting this into the equation for $xy$, we have $xy=x\dfrac{8+x}{2}$. Expanding and simplifying, we have $xy=\dfrac{x^2+8x}{2}$. Taking the derivative of this expression with respect to $x$, we have $\dfrac{d(xy)}{dx}=\dfrac{d}{dx}(\dfrac{x^2+8x}{2})=\dfrac{2x+8}{2}=x+4$. Setting this equal to $0$ to find critical points, we have $x=-4$. Since negative lengths are non-existent in this context, we can conclude that there are no critical points. Since we are looking for the dimensions that maximize the area of the triangle, we know that $xy$ must be at its maximum value when $x$ and $y$ are at their maximum values. Since $x$ and $y$ cannot be longer than $16$ meters, we conclude that $x=16$ and $y=16$. Therefore, the dimensions that maximize the area of the triangle are a base length of $\boxed{16}$ meters and a height of $\boxed{16}$ meters.

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