The trapezoid has a top side of $32$ and two equal legs of $10$. The bottom base is shorter. Drop perpendiculars from the ends of the shorter base to form right triangles.

The difference in the bases is $32 - 12 = 20$, so each right triangle has a horizontal leg of:

$\frac{32 - 12}{2} = \frac{20}{2} = 10$

Wait — let me reconsider. The base is $32$ and each leg is $10$. Drop altitudes from the top vertices to the base. Each horizontal segment is:

$\frac{32 - 12}{2} = 10$

But that would make the horizontal leg equal to the hypotenuse, which is impossible. Let me re-read: the top is $32$ and the bottom is shorter. Actually, the base (bottom) of the trapezoid can be found:

$\text{Half the difference} = \frac{32 - \text{bottom}}{2}$

Using the Pythagorean theorem with leg $= 10$ and the altitude $h$:

$h^2 + 6^2 = 10^2$

$h^2 = 100 - 36 = 64$

$h = 8$