Two circles, centered at A and B have radius 1 and 2, respectively, with AB=6. The common internal tangent intersects the smaller circle at

Let's approach this step by step:

1. Draw the diagram with circles centered at A and B and connect the centers A and B to form the line segment AB of length 6.

2. Draw the tangent line that is common to both circles and intersects AB at a point, let's call it E. The point where the tangent line touches the smaller circle is C, and the point where it touches the larger circle is D.

3. You have two right triangles: ACE and BDE. The hypotenuses AC and BD are the radii of the circles, which are 1 and 2 respectively.

4. You can calculate the lengths of CE and DE using the Pythagorean theorem:
   - In triangle ACE: \(CE^2 = AC^2 - AE^2 = 1^2 - 4^2 = -15.\)
   - In triangle BDE: \(DE^2 = BD^2 - BE^2 = 2^2 - 4^2 = 0.\)

   Since CE squared is negative, it's not a real number, and DE squared is 0, which implies DE is 0. However, this doesn't make sense geometrically, so there seems to be an issue with the problem statement or the setup.

The situation described doesn't seem to align with the properties of the given circles and their common internal tangent. Double-check the information provided and the diagram to make sure everything is accurate and coherent.

I have created a diagram, and I added a few points for convenience. 

Notice that \(\overline{\rm BC} \perp \overleftrightarrow{\rm CD} \text{ and } \overline{\rm AD} \perp \overleftrightarrow{\rm CD}\) because of the Tangent Perpendicular to Radius Theroem. Since both segments are perpendicular to the same line, \(\overline{\rm AD} \parallel \overline{\rm BC}\). Quadrilateral ACBD has a pair of parallel sides and a pair of non-parallel sides, so this quadrilateral is classified as a trapezoid. The area formula for a trapezoid is \(A_{\text{trapezoid}} = \frac{1}{2}\left(b_1 + b_2\right)h \) where b1 and b2 are the lengths of each parallel side and h is the perpendicular distance between b1 and b2 of the trapezoid. I have constructed the diagram such that \(\overline{\rm CE} \parallel \overline{\rm AB}\). Now, pay close attention to \(\triangle {\rm EDC}\). This is a right triangle, so we can use Pythagorean's Theorem to find \(\rm CD\), which is the height of the trapezoid.

\({\rm CD}^2 + {\rm ED}^2 = {\rm EC}^2 \\ {\rm CD}^2 + (2 + 1)^2 = 6^2 \\ {\rm CD}^2 = 36 - 9 \\ {\rm CD}^2 = 27 \\ {\rm CD} = 3\sqrt{3} \text{ or } {\rm CD} = -3\sqrt{3}\)

Since we are dealing with lengths, we should reject \({\rm CD} = -3\sqrt{3}\). Now, we have all the ingredients to find the area of the trapezoid ACBD.

\(A_{\text{trapezoid}} = \frac{1}{2}\left(b_1 + b_2\right)h \\ A_{\text{trapezoid}} = \frac{1}{2}(2 + 1)*3\sqrt{3} \\ A_{\text{trapezoid}} = \frac{9\sqrt{3}}{2} \approx 7.7942 \)