Suppose two circles $\Omega_1$ and $\Omega_2$ with centers $O_1$ and $O_2$ have radii $3$ and $4$, respectively. Suppose that points $A$ and $B$ lie on circles $\Omega_1$ and $\Omega_2$, respectively, such that segments $AB$ and $O_1O_2$ intersect and that $AB$ is tangent to $\Omega_1$ and $\Omega_2$. If $O_1O_2=25$, find the area of quadrilateral $O_1AO_2B$.
[asy]
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[/asy]

[i]Proposed by Deyuan Li and Andrew Milas[/i]
1. Draw a line connecting \( O_1 \) and \( O_2 \) and let its intersection with \( \overline{AB} \) be called \( C \). Note that \( \angle ABO_2 = \angle BAO_1 = 90^\circ \) because \( AB \) is tangent to both circles \( \Omega_1 \) and \( \Omega_2 \).

2. Since \( \angle BCO_2 = \angle ACO_1 \), by AA similarity, we have \( \triangle ACO_1 \sim \triangle BCO_2 \).

3. Given \( AO_1 = 3 \) and \( BO_2 = 4 \), from the similarity of triangles, we get:
   \[
   \frac{CO_2}{CO_1} = \frac{4}{3}
   \]

4. We are also given that \( O_1O_2 = O_1C + O_2C = 25 \). Let \( O_1C = x \) and \( O_2C = y \). Then:
   \[
   x + y = 25
   \]
   and
   \[
   \frac{y}{x} = \frac{4}{3} \implies y = \frac{4}{3}x
   \]

5. Substitute \( y = \frac{4}{3}x \) into \( x + y = 25 \):
   \[
   x + \frac{4}{3}x = 25 \implies \frac{7}{3}x = 25 \implies x = \frac{75}{7}
   \]
   \[
   y = \frac{4}{3} \cdot \frac{75}{7} = \frac{100}{7}
   \]

6. Using the Pythagorean theorem, calculate \( AC \) and \( BC \):
   \[
   AC = \sqrt{\left(\frac{75}{7}\right)^2 - 3^2} = \sqrt{\frac{5625}{49} - 9} = \sqrt{\frac{5625 - 441}{49}} = \sqrt{\frac{5184}{49}} = \frac{72}{7}
   \]
   \[
   BC = \sqrt{\left(\frac{100}{7}\right)^2 - 4^2} = \sqrt{\frac{10000}{49} - 16} = \sqrt{\frac{10000 - 784}{49}} = \sqrt{\frac{9216}{49}} = \frac{96}{7}
   \]

7. Therefore, \( AB = AC + BC = \frac{72}{7} + \frac{96}{7} = \frac{168}{7} = 24 \).

8. The area of quadrilateral \( O_1AO_2B \) can be found by summing the areas of \( \triangle O_2BA \) and \( \triangle O_1AB \):
   \[
   [O_1AO_2B] = [\triangle O_2BA] + [\triangle O_1AB] = \frac{1}{2} \cdot 4 \cdot 24 + \frac{1}{2} \cdot 3 \cdot 24 = 48 + 36 = 84
   \]

The final answer is \(\boxed{84}\)