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.
52.2 in.
72.4 in.
81.2 in.

Explanation:

Step1: Use the Pythagorean theorem for acute triangles

For an acute triangle with sides \(a\), \(a\), and \(c = 30\) (where \(c\) is the longest side), by the property of acute triangles \(a^{2}+a^{2}>c^{2}\). Substitute \(c = 30\) into the inequality: \(2a^{2}>30^{2}\), so \(2a^{2}>900\), then \(a^{2}>450\), and \(a>\sqrt{450}\approx21.21\).

Step2: Calculate the perimeter

The perimeter \(P=2a + c\). Since \(a>\sqrt{450}\approx21.21\), when \(a = 21.2\) (rounded to the nearest tenth), \(P=2\times21.2+30\).

$$P = 42.4+30=72.4$$

Answer:

72.4 in.