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The equation of the tangent is of the form $y-(9-a^2)=m(x-a).$ Substituting $y=9-x^2,$ we get $$9-x^2-(9-a^2)=m(x-a),$$or $x^2+mx-ma-a^2=0.$ Since we have a tangent, $x=a$ should be a double root of this quadratic. In other words, the quadratic is identical to $(x-a{)}^2=x^2-2ax+a^2,$ so $m=-2a.$

The equation of the tangent is then $$y-(9-a^2)=(-2a)(x-a).$$When $x=0,$ $$y-(9-a^2)=2a^2,$$so $y=a^2+9,$ which is the height of the triangle.

When $y=0,$ $$-(9-a^2)=(-2a)(x-a),$$so $x=\frac{a^2+9}{2a},$ which is the base of the triangle. Hence, $$\frac{1}{2}\cdot (a^2+9)\cdot \frac{a^2+9}{2a}=25.$$Expanding, we get $a^4+18a^2-100a+81=0.$

Since $a$ is rational, by the Rational Root Theorem, $a$ must be an integer divisor of 81. Furthermore, $a$ must lie in the range $0\le a\le 3.$ Checking, we find that $a=\boxed{1}$ is the only solution.