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: Recall triangle - inequality for acute isosceles triangle

Let the length of the congruent sides be \(x\) and the longest side \(c = 12\) cm. For an acute - angled triangle, the following inequality must hold based on the Pythagorean theorem: \(x^{2}+x^{2}>c^{2}\) (if the triangle is acute, the sum of the squares of the two shorter sides is greater than the square of the longest side). Substituting \(c = 12\), we get \(2x^{2}>144\), so \(x^{2}>72\).

Step2: Solve for \(x\)

Taking the square - root of both sides of the inequality \(x^{2}>72\), we have \(x>\sqrt{72}\). Since \(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\approx6\times1.414 = 8.484\). Rounding to the nearest tenth, \(x\approx8.5\) cm.

Answer:

8.5 cm