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.

Explanation:

Step1: Define variables & acute condition

Let the length of each congruent side be $x$. For an acute triangle with longest side 30, use the acute triangle inequality: the square of the longest side is less than the sum of the squares of the other two sides.
$$30^2 < x^2 + x^2$$

Step2: Simplify the inequality

Calculate and rearrange to solve for $x$.
$$900 < 2x^2$$
$$x^2 > 450$$
$$x > \sqrt{450} = 15\sqrt{2} \approx 21.213$$

Step3: Calculate minimum perimeter

Perimeter $P = 2x + 30$. Substitute the smallest $x$ value.
$$P > 2\times15\sqrt{2} + 30 \approx 42.426 + 30 = 72.426$$

Answer:

72.4 in.