The bases of a trapezoid have lengths 10 and 21, and the legs have lengths $\sqrt{34}$ and $3 \sqrt{5}$.  What is the area of the trapezoid?
1. **Define Variables and Set Up the Problem:**
   Let \( h \) be the height of the trapezoid. We draw a perpendicular from the top base to the bottom base, creating a rectangle with height \( h \) and width 10 (the length of the shorter base). This splits the trapezoid into a rectangle and two right triangles.

2. **Assign Horizontal Distances:**
   Let \( x \) be the horizontal distance from one end of the shorter base to the point where the perpendicular meets the longer base. The remaining horizontal distance on the longer base is \( 11 - x \) (since the difference between the bases is \( 21 - 10 = 11 \)).

3. **Use the Pythagorean Theorem:**
   For the right triangle with leg \( \sqrt{34} \):
   \[
   h^2 + x^2 = 34
   \]
   For the right triangle with leg \( 3\sqrt{5} \):
   \[
   h^2 + (11 - x)^2 = 45
   \]

4. **Subtract the Equations:**
   Subtract the first equation from the second:
   \[
   h^2 + (11 - x)^2 - (h^2 + x^2) = 45 - 34
   \]
   Simplify:
   \[
   (11 - x)^2 - x^2 = 11
   \]
   Expand and simplify:
   \[
   121 - 22x + x^2 - x^2 = 11
   \]
   \[
   121 - 22x = 11
   \]
   \[
   110 = 22x
   \]
   \[
   x = 5
   \]

5. **Find the Height \( h \):**
   Substitute \( x = 5 \) back into one of the original equations:
   \[
   h^2 + 5^2 = 34
   \]
   \[
   h^2 + 25 = 34
   \]
   \[
   h^2 = 9
   \]
   \[
   h = 3
   \]

6. **Calculate the Area of the Trapezoid:**
   The area \( A \) of a trapezoid is given by:
   \[
   A = \frac{1}{2} \times ( \text{sum of the bases} ) \times \text{height}
   \]
   Here, the sum of the bases is \( 10 + 21 = 31 \) and the height is \( 3 \):
   \[
   A = \frac{1}{2} \times 31 \times 3 = \frac{93}{2}
   \]

The final answer is \(\boxed{\frac{93}{2}}\)