The lengths of the parallel sides of a trapezium are x cm and y cm and the area of the trapezium is $$\frac{1}{2}(x^{2} - y^{2}) cm^{2}$$. What is the distance between the parallel sides (in cm)?

The lengths of the parallel sides of a trapezium are x cm and y cm.

Area of the trapezium is $$\frac{1}{2}(x^{2} - y^{2}) cm^{2}$$.

Area of the trapezium = $$\frac{1}{2}\times\ sum\ of\ the\ lengths\ of\ the\ parallel\ sides\times\ perpendicular\ distance\ between\ parallel\ sides$$

$$\frac{1}{2}(x^{2} - y^{2}) = \frac{1}{2}\times (x+y) \times height$$
$$(x-y)(x+y)=(x+y)\times height$$

height = (x-y)

Distance between the parallel sides = (x-y) cm