Question

find an integer, x, such that 2, 4, and x represent the lengths of the sides of an obtuse triangle. a. 2 b. 3 c. 4 d. 6 please select the best answer from the choices provided a b c d

Explanation:

Step1: Recall triangle inequality and obtuse triangle condition

For a triangle with sides \(a\), \(b\), \(c\) (where \(c\) is the longest side), the triangle inequality states \(a + b>c\), \(a + c>b\), \(b + c>a\). For an obtuse triangle, if \(c\) is the longest side, then \(a^{2}+b^{2}

First, find the possible range of \(x\) using triangle inequality:
\(\vert2 - 4\vert

Step2: Check each option

• Option A: \(x = 2\). Check triangle inequality: \(2+2 = 4\), which does not satisfy \(2 + 2>4\) (it's equal, so it's a degenerate triangle, not a valid triangle). So A is out.
• Option B: \(x = 3\). Check if it's obtuse. Let's see which side is the longest, \(4\) is the longest. So check \(2^{2}+3^{2}\) vs \(4^{2}\). \(2^{2}+3^{2}=4 + 9=13\), \(4^{2}=16\). Since \(13<16\), by the obtuse triangle condition (\(a^{2}+c^{2}
• Option C: \(x = 4\). Check if it's obtuse. The triangle is isoceles with sides \(2\), \(4\), \(4\). Check the obtuse condition. Let the longest side be \(4\). \(2^{2}+4^{2}=4 + 16 = 20\), \(4^{2}=16\). \(20>16\), so it's acute (since \(a^{2}+b^{2}>c^{2}\) for the longest side \(c = 4\)).
• Option D: \(x = 6\). Check triangle inequality: \(2+4=6\), which is degenerate (not a valid triangle). So D is out.

Answer:

B. 3