Question

ramon wants to make an acute triangle with three pieces of wood. so far, he has cut wood lengths of 7 inches and 3 inches. he still needs to cut the longest side. what length must the longest side be in order for the triangle to be acute?

• exactly \\(\sqrt{58}\\) inches
• greater than \\(\sqrt{58}\\) inches but less than 10 inches
• less than \\(\sqrt{58}\\) inches but greater than 7 inches
• not enough information given

Explanation:

Step1: Recall triangle inequality and acute triangle conditions

For a triangle with sides \(a\), \(b\), \(c\) (where \(c\) is the longest side), the triangle inequality states \(a + b>c\). For an acute triangle, the Pythagorean inequality \(a^{2}+b^{2}>c^{2}\) (since \(c\) is the longest side, we check the square of the longest side against the sum of squares of the other two). Here, \(a = 3\), \(b = 7\), so first, triangle inequality: \(3 + 7>c\Rightarrow c<10\). Then, for acute, \(3^{2}+7^{2}>c^{2}\Rightarrow9 + 49>c^{2}\Rightarrow58>c^{2}\Rightarrow c<\sqrt{58}\approx7.62\). Also, since \(c\) is the longest side, \(c>7\) (because \(7\) is longer than \(3\), so the longest side must be longer than \(7\) to be the longest). So \(7 < c<\sqrt{58}\)? Wait, no, wait: Wait, if \(c\) is the longest side, then \(c\geq7\) (since \(7\) is one of the sides). But for acute, \(a^{2}+b^{2}>c^{2}\). Wait, no, actually, if \(c\) is the longest side, then the angle opposite \(c\) is the largest angle. For the triangle to be acute, that largest angle must be acute, so \(a^{2}+b^{2}>c^{2}\). Also, triangle inequality: \(c < a + b=10\). And since \(c\) is the longest side, \(c>7\) (because if \(c\leq7\), then the longest side would be \(7\), not \(c\)). So combining: \(7 < c\) (to be longest) and \(c^{2}<3^{2}+7^{2}=58\Rightarrow c < \sqrt{58}\), and \(c < 10\) (from triangle inequality, but \(\sqrt{58}\approx7.62<10\), so the stricter upper bound is \(\sqrt{58}\)). Wait, but the options: let's re - check. Wait, the options are:

1. exactly \(\sqrt{58}\) inches: If \(c = \sqrt{58}\), then \(a^{2}+b^{2}=c^{2}\), which is a right triangle, not acute.

1. greater than \(\sqrt{58}\) but less than 10: If \(c>\sqrt{58}\), then \(a^{2}+b^{2}

1. less than \(\sqrt{58}\) but greater than 7: Let's see, \(c>7\) (so it's the longest side, since \(7>3\)) and \(c < \sqrt{58}\) (so \(a^{2}+b^{2}>c^{2}\), acute). And \(c < 10\) (from triangle inequality, but since \(\sqrt{58}\approx7.62<10\), this range is \(7 < c<\sqrt{58}\)).

1. not enough info: No, we have enough info.

Wait, maybe I made a mistake earlier. Let's re - derive:

Let the three sides be \(3\), \(7\), and \(x\) (where \(x\) is the longest side, so \(x\geq7\)).

Triangle inequality: \(3 + 7>x\Rightarrow x < 10\).

For the triangle to be acute, the largest angle (opposite \(x\)) must be acute. By the Law of Cosines, \(\cos C=\frac{3^{2}+7^{2}-x^{2}}{2\times3\times7}\). For angle \(C\) to be acute, \(\cos C>0\), so \(3^{2}+7^{2}-x^{2}>0\Rightarrow x^{2}<9 + 49 = 58\Rightarrow x<\sqrt{58}\approx7.62\).

Since \(x\) is the longest side, \(x>7\) (because if \(x\leq7\), then the longest side would be \(7\), not \(x\)).

So combining these two conditions: \(7 < x<\sqrt{58}\), which is "less than \(\sqrt{58}\) inches but greater than 7 inches".

Step2: Analyze the options

• Option 1: If \(x = \sqrt{58}\), then \(\cos C = 0\), so the triangle is right, not acute. Eliminate.
• Option 2: If \(x>\sqrt{58}\), then \(\cos C<0\), so the triangle is obtuse. Eliminate.
• Option 3: \(7 < x<\sqrt{58}\) satisfies both the triangle inequality (since \(\sqrt{58}<10\)) and the acute condition (\(x^{2}<58\)) and the longest side condition (\(x > 7\)).
• Option 4: We have enough information (we used triangle inequality and Law of Cosines for acute triangles) to determine the range. Eliminate.

Answer:

less than \(\sqrt{58}\) inches but greater than 7 inches (the third option)