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 triangle conditions

For an isosceles triangle with congruent sides \(a\) and base \(b = 12\) (since it's the longest side), the triangle inequality gives \(2a>b\), so \(2a > 12\) or \(a>6\). For it to be acute, the square of the longest side must be less than the sum of the squares of the other two sides. So \(b^{2}<2a^{2}\), substituting \(b = 12\), we get \(12^{2}<2a^{2}\), i.e., \(144 < 2a^{2}\), then \(a^{2}>72\), so \(a>\sqrt{72}\approx8.485\). But we need the smallest possible \(a\) (rounded to nearest tenth) that satisfies both. Wait, wait, maybe I mixed up. Wait, in an acute isosceles triangle, if the base is the longest side, then the vertex angle is acute. The condition for acute: \(b^{2}<2a^{2}\). But also, if the equal sides were the longest, but here the longest side is 12, so it's the base. So we have \(2a>12\) (triangle inequality) and \(12^{2}72\implies a > \sqrt{72}\approx8.485\). Rounded to nearest tenth, the smallest possible is \(8.5\)? Wait, no, wait \(\sqrt{72}\approx8.485\), so to be greater than that, the smallest possible (rounded to tenth) is \(8.5\)? Wait, but let's check the options. Wait, maybe I made a mistake. Wait, maybe the longest side is one of the equal sides? Wait, the problem says "the longest side of an acute isosceles triangle is 12 centimeters". So if the equal sides are the longest, then \(a = 12\), and the base \(b\) must satisfy \(2\times12>b\) and \(b^{2}<2\times12^{2}\). But the problem says "the longest side" is 12, so if equal sides are 12, then base \(b < 12\), but we need the smallest possible length of one of the two congruent sides. Wait, maybe the longest side is the base. So let's re-express. Let the two equal sides be \(a\), base \(b = 12\) (longest side). Then for acute: \(b^{2}72\implies a>\sqrt{72}\approx8.485\). So the smallest \(a\) (rounded to tenth) is \(8.5\)? Wait, but \(\sqrt{72}\approx8.485\), so to be greater than that, the smallest possible (rounded to nearest tenth) is \(8.5\)? Wait, no, wait, if we round \(8.485\) to nearest tenth, it's \(8.5\)? Wait, \(8.485\) is closer to \(8.5\) than \(8.4\). Wait, but let's check the options. The options are 6.0, 6.1, 8.4, 8.5. So from the acute condition, \(a>\sqrt{72}\approx8.485\), so the smallest possible (rounded to tenth) is \(8.5\). Wait, but let's check triangle inequality: \(2a>12\implies a > 6\), which is satisfied by 8.4 and 8.5. But the acute condition is stricter. So \(a\) must be greater than \(\sqrt{72}\approx8.485\), so the smallest possible (rounded to nearest tenth) is \(8.5\).

Step2: Evaluate the options

• Option 6.0: \(6.0\) is less than \(\sqrt{72}\), so invalid (not acute).
• Option 6.1: Also less than \(\sqrt{72}\), invalid.
• Option 8.4: \(8.4^{2}=70.56\), \(2\times8.4^{2}=141.12\), and \(12^{2}=144\). But \(144>141.12\), so \(12^{2}>2\times8.4^{2}\), which would make the triangle obtuse (since if \(c^{2}>a^{2}+b^{2}\), it's obtuse). So 8.4 is invalid.
• Option 8.5: \(8.5^{2}=72.25\), \(2\times8.5^{2}=144.5\). \(12^{2}=144\), and \(144<144.5\), so it satisfies the acute condition. Also, \(2\times8.5 = 17>12\), so triangle inequality holds. And it's the smallest among the options that satisfies the acute condition (since 8.4 is too small as shown).

Answer:

D. 8.5 cm