Final answer:
The width of the rectangle must be an irrational number since the product of a rational and an irrational number is irrational. Any irrational width would suffice, except those that involve division by sqrt(2). However, from the measurement context, the width is 1.25 cm to three significant figures.
Step-by-step explanation:
The student’s question pertains to the area of a rectangle and the dimensions that can lead to an irrational area. Given that the length of the rectangle is 3sqrt(2) units, and the area must be irrational, we can deduce that the measure of the width must also be an irrational number. This is because the product of a rational and an irrational number is irrational.
Therefore, any irrational number could be the width, as long as it is not the result of a division by sqrt(2), which would rationalize the area.
For example, the width could be sqrt(5), sqrt(7), or any irrational number that does not include sqrt(2) as a divisor. However, based on the additional information provided, it seems the width is a measured value rather than an exact irrational number.
Therefore, the measured width of the rectangle is 1.25 cm and is reported to three significant figures based on the estimation between tick marks.