A trapezoid's diagonals are perpendicular, its height is 4 units, and one of its diagonals is 5 units. What is the area of the trapezoid?
Since the diagonals of the trapezoid are perpendicular, its area can be easily calculated as half the product of its diagonals. The length of one diagonal is given, 5 units. The position of the other diagonal is not determined, but we know that it moves between two parallel lines 4 units apart, so its length is constant. Therefore, the area of the trapezoid is uniquely determined.

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Let the parallel sides of the trapezoid be denoted by $e$ and $f$, and the 5-unit diagonal by $AC$. Our task is to determine the length of the segment between $e$ and $f$ that is perpendicular to $AC$. Draw a perpendicular from $C$ to $f$ and to $AC$, intersecting $f$ at $C_1$ and $AC$ at $C_2$. $CC_1$ is the height of the trapezoid, which is 4 units, and by the Pythagorean theorem, the leg $AC_1$ of the right triangle $ACC_1$ is 3 units. The right triangles $ACC_2$ and $AC_1C$ are similar, so

$$
AC_1 : C_1C = AC : CC_2
$$

from which

$$
CC_2 = \frac{C_1C \cdot AC}{AC_1} = \frac{20}{3}
$$

Thus, the other diagonal of the trapezoid is 20/3 units long, and its area, as stated at the beginning of the solution, is $\frac{1}{2} \cdot 5 \cdot \frac{20}{3} = \frac{50}{3}$ square units.