Final answer:
A parallelogram with diagonals of different lengths can only be a rhombus because such diagonals must be perpendicular to allow congruent halves, resulting in four right-angled triangles with equal sides within the parallelogram.
Step-by-step explanation:
To prove that a parallelogram with diagonals of lengths 2.3cm and 3.2cm is a rhombus, we must show that all four sides of the parallelogram have equal length. A rhombus is a special type of parallelogram where not only opposite sides are parallel, but all sides are of equal length as well. One of the properties of a parallelogram is that its diagonals bisect each other. In a parallelogram that is also a rhombus, the diagonals not only bisect each other, but also perpendicular to one another.
If we assume this parallelogram is not a rhombus, meaning the sides are not equal, the diagonals would still have to bisect each other at some angle. However, the fact that the diagonals are given different lengths implies that the only way for each half of the diagonals to be congruent (since they bisect each other) is if they are perpendicular, as it would form four right-angled triangles with the sides of the parallelogram acting as the hypotenuse for these triangles.
Therefore, the only way for a parallelogram with two distinct diagonal lengths to exist is if it's a rhombus, proving that our parallelogram with diagonals of 2.3cm and 3.2cm is indeed a rhombus, with the diagonals intersecting each other at right angles, thereby creating four right-angled triangles within the shape.