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? 5.7 cm 4.1 cm 5.6 cm 4.0 cm

Explanation:

Step1: Recall the property for acute isosceles triangle

For an acute isosceles triangle with sides \(a\), \(a\), and \(b\) (\(b\) is the longest side), by the Pythagorean - like inequality for acute triangles \(a^{2}+a^{2}>b^{2}\). Here \(b = 8\) cm.

Step2: Solve the inequality for \(a\)

We have \(2a^{2}>64\), then \(a^{2}>32\), and \(a>\sqrt{32}\).

Step3: Calculate the value of \(\sqrt{32}\)

\(\sqrt{32}=\sqrt{16\times2}=4\sqrt{2}\approx4\times1.414 = 5.656\).

Step4: Round to the nearest tenth

Rounding \(5.656\) to the nearest tenth gives \(5.7\).

Answer:

A. \(5.7\) cm