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 acute triangle inequality

For an acute triangle with sides \(a\), \(b\), \(c\) (where \(c\) is the longest side), \(a^{2}+b^{2}>c^{2}\). Here, \(c = 30\), and \(a=b=x\) (since the two remaining sides are congruent). So \(x^{2}+x^{2}>30^{2}\), which simplifies to \(2x^{2}>900\), then \(x^{2}>450\), so \(x>\sqrt{450}\approx21.21\). Also, from triangle inequality (sum of two sides > third side), \(x + x>30\), so \(2x>30\) or \(x > 15\) (but the acute condition is stricter here).

Step2: Find minimum perimeter

The perimeter \(P=x + x+30=2x + 30\). To minimize \(P\), we minimize \(x\). The minimum \(x\) satisfying \(x>\sqrt{450}\approx21.21\). So minimum \(x\approx21.21\) (we can take \(x=\sqrt{450}\) for the limit, but since \(x\) must be greater, we use the smallest \(x\) that makes it acute). Then \(P = 2\times21.21+30\approx42.42 + 30=72.42\approx72.4\) (wait, wait, no: wait, \(\sqrt{450}\approx21.21\), so \(2x\approx42.42\), plus 30 is \(72.42\approx72.4\). Wait, but let's check again. Wait, the options: 72.4 is one of them. Wait, maybe I made a mistake earlier. Wait, the acute triangle: for a triangle to be acute, the square of the longest side must be less than the sum of the squares of the other two sides. So \(30^{2}450\), \(x>\sqrt{450}\approx21.21\). Then perimeter \(P = 2x+30\). So minimum \(x\) is just above \(\sqrt{450}\), so minimum perimeter is just above \(2\times\sqrt{450}+30\). Calculate \(2\times\sqrt{450}=\sqrt{1800}\approx42.426\), so \(42.426 + 30=72.426\approx72.4\).

Answer:

72.4 in.