Final answer:
The smallest possible perimeter of a triangle with sides that are squares of distinct positive integers is obtained using sides of lengths 32, 42, and 52, totaling a perimeter of 50. The options provided in the question do not match this correct solution. The provided options are irrelevant or incorrect, and none of them reflect the correct solution to the problem.
Step-by-step explanation:
The student is tasked with finding the smallest possible perimeter of a triangle where the lengths of the sides are squares of distinct positive integers. To solve this, we look for the smallest square numbers that can form a triangle. The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
If we start with the smallest squares, we have 12=1, 22=4, and 32=9. However, 1+4 is not greater than 9, so these cannot form a triangle. The next smallest combination that satisfies the triangle inequality is 12=1, 32=9, and 42=16. The sum of these sides gives us a perimeter of 1+9+16=26, but this is not one of the options provided.
The correct answer is found by using 32=9, 42=16, and 52=25. These lengths satisfy the triangle inequality because the sum of any two side lengths is greater than the third. Therefore, the smallest possible perimeter is 9+16+25=50.
The provided options are irrelevant or incorrect, and none of them reflect the correct solution to the problem.