Final answer:
After finding the hypotenuse (rounded to 10 units), we conclude that the area of the smallest square, based on the length of the side calculated from the hypotenuse, is not directly listed among the answer choices. An area of 49 square units would be expected, but if an error is assumed and we must select from the given options, the closest would be 4, possibly referring to a square with a side of 2 units.
Step-by-step explanation:
To find the area of the smallest square, we begin by understanding that the hypotenuse of a right triangle represents the diagonal of the square in question. In this scenario, we calculate the hypotenuse using the Pythagorean theorem:
\(\sqrt{(9 \text{ blocks})^2 + (5 \text{ blocks})^2} = 10.3 \text{ blocks}\)
Since this hypotenuse is the diagonal of the square, we use the property that the diagonal of a square with side length \(s\) is \(s\sqrt{2}\). Thus, if \(d = s\sqrt{2}\), we can solve for \(s\) as follows:
\(s = \frac{d}{\sqrt{2}} = \frac{10.3}{\sqrt{2}} = \frac{10.3}{1.414} \approx 7.28 \text{ blocks}\)
After rounding \(7.28\) to the nearest whole number, we get \(7\) blocks as the side length of the square. The area of the square is then \(s^2 = 7^2 = 49 \text{ block}^2\). However, since this value is not available in the answer choices, it suggests there may have been a rounding error or misinterpretation in the process. Assuming we need the smallest integral value of the area due to rounding the hypotenuse before calculating the side length, we round our hypotenuse \(10.3\) to \(10\) and proceed with:
\(s = \frac{10}{\sqrt{2}} \approx 7.07\)
Rounding \(7.07\) to the nearest whole number gives us \(7\) again. Therefore, the area of the square is:
\(s^2 = 7^2 = 49\)
Among the answer choices provided, the closest to \(49\) is \(\textbf{4},\) assuming that there has been an error in the question or choices given since none match the calculated area. It's possible that the square in question refers to one with a side length of \(2\), hence having an area of \(2^2 = 4\).