Question

The longest side of an acute triangle measures 30 inches. The two remaining sides are congruent, but their length is unknown.

What is the smallest possible perimeter of the triangle, rounded to the nearest tenth?

41.0 in.
51.2 in.
72.4 in.
81.2 in.

Answer 1

Let the equal sides each be x. For an acute triangle, the square of the longest side must be less than the sum of the squares of the other two sides:
30^2 < x^2 + x^2 = 2x^2
So x > 15√2. The perimeter is P = 30 + 2x, which is minimized as x approaches 15√2 from above. The infimum is
P = 30 + 2(15√2) = 30(1 + √2) ≈ 72.4264,
which rounded to the nearest tenth is 72.4 in.

Answer: 72.4 in.