Question

the longest side of an acute isosceles triangle is 8 centimeters. rounded to the nearest tenth, what is the smallest possible length of one of the two congruent sides? 4.0 cm 4.1 cm 5.6 cm 5.7 cm

Explanation:

Step1: Recall the triangle - inequality theorem and properties of acute - isosceles triangles

For an isosceles triangle with sides \(a\), \(a\), and \(b\) (\(b\) is the base), and for an acute - angled triangle, the following condition must hold: \(a^{2}+a^{2}>b^{2}\) (by the Pythagorean - like inequality for acute triangles). Let the congruent sides be \(x\) and the longest side \(b = 8\). Then \(2x^{2}>64\), so \(x^{2}>32\), and \(x>\sqrt{32}\approx5.66\).

Step2: Check the given options

We have the options: \(4.0\mathrm{cm}\), \(4.1\mathrm{cm}\), \(5.6\mathrm{cm}\), \(5.7\mathrm{cm}\). Since \(x>\sqrt{32}\approx5.66\), the smallest value among the given options that satisfies this condition is \(5.7\mathrm{cm}\).

Answer:

\(5.7\mathrm{cm}\)