To solve for the altitude of the trapezoid, we can break this down step by step.

Information provided:

• Upper base $= 6$
• Lower base $= 16$
• Diagonal $= 12$
• Trapezoid has equal nonparallel sides (isosceles trapezoid).

Step 1: Calculate the difference between the upper and lower base.

The difference between the upper and lower bases is: $16 - 6 = 10$ This difference is split into two equal parts because the trapezoid is isosceles. So, each part is: $\frac{10}{2} = 5$ This means that the horizontal distance from the base of the right triangle to the point where the altitude intersects the lower base is 5 units (as shown in the diagram).

Step 2: Use the Pythagorean theorem.

In the right triangle, we know:

• The horizontal leg (adjacent to the right angle) is 5 (calculated earlier).
• The hypotenuse is the diagonal $= 12$.

Let $h$ represent the altitude (which is the vertical leg of the right triangle).

Applying the Pythagorean theorem: $h^2 + 5^2 = 12^2$ $h^2 + 25 = 144$ $h^2 = 144 - 25 = 119$ $h = \sqrt{119} \approx 10.91$

Step 3: Approximate the altitude.

The correct answer must be approximately $10.91$. However, the question asks for the altitude of the trapezoid, not just the right triangle.

There seems to be a misalignment between the question's multiple-choice answers and this calculation. I will reconsider the steps or check whether additional clarifications are required.