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 and acute condition

For an isosceles triangle with congruent sides \(a\) and base \(b = 12\) (longest side), by triangle inequality \(2a>b\). For it to be acute, using the Pythagorean inequality: \(a^{2}+a^{2}>b^{2}\) (since the largest angle is opposite the longest side, so we check the angle opposite the base is acute). So \(2a^{2}>b^{2}\).

Step2: Solve the inequality for \(a\)

Substitute \(b = 12\) into \(2a^{2}>12^{2}\). Divide both sides by 2: \(a^{2}>72\). Take square root: \(a>\sqrt{72}\approx8.485\). But also from triangle inequality \(2a > 12\Rightarrow a>6\). The smallest \(a\) satisfying the acute condition is just above \(\sqrt{72}\), but wait, maybe I mixed up. Wait, if the longest side is 12, maybe the congruent sides are not the base? Wait, no—if the triangle is acute and isosceles, the longest side must be the base (otherwise, if the congruent sides were longer, the base would be shorter, but we are told the longest side is 12). Wait, no, correction: if the congruent sides are longer than 12, then 12 would not be the longest side. So the longest side is the base \(b = 12\), and the congruent sides are \(a\), with \(a\) as legs (in the right triangle case, \(a^{2}+a^{2}=b^{2}\Rightarrow a=\frac{b}{\sqrt{2}}\)). For acute, the angle opposite the base (the vertex angle) must be acute, so \(a^{2}+a^{2}>b^{2}\) (since in a triangle, \(c^{2}144\Rightarrow a^{2}>72\Rightarrow a > \sqrt{72}\approx8.485\). But the options are 6.0, 6.1, 8.4, 8.5. Wait, maybe I had the sides reversed. Wait, maybe the longest side is one of the congruent sides? No, because it's isosceles—if two sides are 12, then the base would be less than 24, but we need the longest side to be 12, so the two congruent sides must be \(\leq12\), but no—wait, no, the longest side is 12, so the other two sides (congruent) are \(a\), with \(a\leq12\), but for it to be acute, if the two congruent sides are \(a\) and the base is 12, then the angle opposite the base (between the two \(a\) sides) is acute, so \(a^{2}+a^{2}>12^{2}\) (as the angle opposite 12 is acute). Wait, no—angle opposite 12 is between the two \(a\) sides? No, angle opposite a side is at the vertex opposite. So side 12 is opposite the angle between the two \(a\) sides? No, side 12 is opposite the angle that is between the two \(a\) sides? Wait, no: in triangle \(ABC\), with \(AB = AC = a\), \(BC = 12\). Then angle at \(A\) is between \(AB\) and \(AC\), opposite \(BC = 12\). So to have angle at \(A\) acute, \(BC^{2}72\Rightarrow a>\sqrt{72}\approx8.485\). So the smallest \(a\) (rounded to nearest tenth) is 8.5? Wait, but \(\sqrt{72}\approx8.485\), so rounded to nearest tenth is 8.5. Wait, but let's check the options. The options are 6.0, 6.1, 8.4, 8.5. So 8.5 is the answer. Wait, but maybe I made a mistake in which side is which. Let's re-express: if the triangle is isosceles with two sides \(a\) and one side 12, and 12 is the longest side, then \(a\leq12\), and the triangle inequality: \(a + a>12\Rightarrow a > 6\). For acute, the largest angle is opposite 12, so by the law of cosines, \(\cos\theta=\frac{a^{2}+a^{2}-12^{2}}{2a\cdot a}\). For \(\theta\) acute, \(\cos\theta>0\), so \(2a^{2}-144>0\Rightarrow a^{2}>72\Rightarrow a>\sqrt{72}\approx8.485\). So the smallest \(a\) (rounded to nearest tenth) is 8.5, since 8.4 is less than \(\sqrt{72}\) (8.485), and 8.5 is more. So 8.5 cm.

Answer:

8.5 cm (the option: 8.5 cm)