The diagonals of a trapezoid are 20 cm and 15 cm, and its height is 12 cm. Find its face.

Let's denote the bases of the trapezoid as b_1 and b_2 where b_1 is the longer base. We can use the Pythagorean theorem to relate the diagonals and bases of the trapezoid.

Let's consider the right triangle formed by half the longer base, half the shorter base, and the diagonal of length 20 cm:

According to Pythagorean theorem:

(b_1/2)^2 + (b_2/2)^2 = (20/2)^2
(b_1/2)^2 + (b_2/2)^2 = 10^2

Similarly, for the right triangle formed by half the longer base, half the shorter base, and the diagonal of length 15 cm:

(b_1/2)^2 + (b_2/2)^2 = (15/2)^2
(b_1/2)^2 + (b_2/2)^2 = 7.5^2

Solving the system of equations, we get:
b_2 = 6 cm
b_1 = 19 cm

The area of a trapezoid is given by the formula:
A = \frac{1}{2} \times h \times (b_1 + b_2)
A = \frac{1}{2} \times 12 \times (19 + 6)
A = 126 cm^2

\boxed{126} cm² is the area of the trapezoid.