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: Recall the cosine - law for an acute - angled triangle

For an acute - angled triangle with sides \(a\), \(b\), \(c\) (\(c\) is the longest side), \(a^{2}+b^{2}>c^{2}\). Let the longest side \(c = 30\) inches and the other two congruent sides be \(x\). Then \(2x^{2}>30^{2}\), so \(x^{2}>450\), and \(x>\sqrt{450}\approx21.213\).

Step2: Consider the triangle - inequality theorem

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. So \(2x>30\), which gives \(x > 15\). The smallest value of \(x\) that satisfies \(x^{2}>450\) is \(x\approx21.213\).

Step3: Calculate the perimeter

The perimeter \(P\) of the triangle is \(P=2x + 30\). Substituting the smallest value of \(x\approx21.213\) into the perimeter formula, we get \(P=2\times21.213+30=42.426 + 30=72.426\approx72.4\) inches.

Answer:

72.4 in.