An isosceles triangle has at least two congruent sides. The perimeter of a certain isosceles triangle is at most 12 in. The length of each of the two congruent sides is 5 in. What are the possible lengths of the remaining side?
Let the length of the remaining side be $x$. Given that the perimeter of the triangle is at most 12 in., we have $5+5+x\leq 12$. Solving, we find $x\leq 2$.
Since the remaining side cannot be longer than 2 in., the possible lengths of the remaining side are $\boxed{0 \text{ in.}, 2 \text{ in.}}$.
Since the triangle is isosceles, it has two congruent sides. Let's call the length of each of the congruent sides "a" and the length of the remaining side "b".
According to the problem, each of the congruent sides is 5 in. So, we have a = 5 in.
The perimeter of a triangle is the sum of the lengths of all three sides. In this case, we know that the perimeter is at most 12 in. So we can set up an inequality:
a + a + b ≤ 12
Substituting the value of a as 5 in, we have:
5 + 5 + b ≤ 12
Simplifying, we get:
10 + b ≤ 12
Subtracting 10 from both sides:
b ≤ 2
So, the possible lengths of the remaining side, b, are 2 inches or any value less than or equal to 2 inches.