Question

the longest side of an acute isosceles triangle is 12 centimeters. rounded to the nearest tenth, what is the smallest possible length of one of the two congruent sides?
6.0 cm
6.1 cm
8.4 cm
8.5 cm

Explanation:

Step1: Use the property of an acute isosceles triangle

For an acute isosceles triangle with the longest side \(c = 12\) and the congruent sides \(a=b\). By the Pythagorean - like condition for an acute triangle \(a^{2}+a^{2}>c^{2}\) (using the fact that if \(a = b\), and the triangle is acute, the sum of the squares of the two equal sides is greater than the square of the longest side).

Step2: Substitute \(c = 12\) into the inequality

We have \(2a^{2}>12^{2}\), which simplifies to \(2a^{2}>144\). Then \(a^{2}>72\), and \(a>\sqrt{72}\).

Step3: Calculate \(\sqrt{72}\)

\(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\approx6\times1.414 = 8.484\)

Answer:

\(8.5\mathrm{cm}\)