Patrick from Woodbridge School offered the following insight to this problem, as did Chip from King's Ely School.

The difference arises because there is a difference in the line gradient between the trapezium and the triangle. The sloping line of the trapezium has a gradient of $\frac{2}{5} = 0.4$ , whereas the hypotenuse of the triangle has a gradient of $\frac{3}{8} = 0.375$.This means that the lines made when you join the triangle and trapezium together are not straight, so there is a small amount of space, as shown on my diagram. This space adds up to the extra $1^2$cm.

An anonymous solver also wrote: The area of the original square was $64$ square units and the
area of the rectangle was $65$ square units. What appears to be a
diagonal of the rectangle is not in fact a straight line and the
extra unit of area comes from the long thin parallelogram in the
middle of the rectangle.
You can confirm this by making a careful scale drawing for
yourself on graph paper.
The long thin parallelogram occurs because the slopes of the trapezia are not the same as explained by Patrick and Chip: the slopes of the triangles. For each trapezium the slope is in the ratio of 2 up to 5 across and for each triangle the slope is in the ratio of 3 up to 8 across. |

Finally, Susie from DEECD Victoria offered an example of other numbers that work (Do they have the same properties as the numbers used for the lengths in the problem?). She said:

Another set of dimensions are $4,7,11$ in the corresponding
places of $3,5,8$.

The areas are $121$ square units compared to $126$ square
units.