Question

a triangle has sides of length 5, 7, and x.
a. complete the inequality so that it describes possible lengths for x in an acute triangle.
type your answers in the boxes.
<x<
b. complete the inequality so that it describes possible lengths for x in an obtuse triangle.
type your answers in the boxes.
<x<

Explanation:

Step1: Determine the range of \(x\) for a triangle

By the triangle - inequality theorem, \(|5 - 7|\lt x\lt5 + 7\), so \(2\lt x\lt12\).

Step2: Use the Pythagorean theorem for acute and obtuse triangles

For an acute triangle, if \(c\) is the longest side, then \(a^{2}+b^{2}>c^{2}\).
Case 1: If \(x\) is the longest side (\(x\geqslant7\)), then \(5^{2}+7^{2}>x^{2}\), \(25 + 49>x^{2}\), \(x^{2}<74\), \(x<\sqrt{74}\approx8.6\).
Case 2: If \(7\) is the longest side (\(x<7\)), then \(5^{2}+x^{2}>7^{2}\), \(x^{2}>49 - 25=24\), \(x>\sqrt{24}\approx4.9\).
For an obtuse triangle, if \(c\) is the longest side, then \(a^{2}+b^{2}Case 1: If \(x\) is the longest side (\(x\geqslant7\)), then \(5^{2}+7^{2}74\), \(x>\sqrt{74}\approx8.6\).
Case 2: If \(7\) is the longest side (\(x<7\)), then \(5^{2}+x^{2}<7^{2}\), \(x^{2}<49 - 25 = 24\), \(x<\sqrt{24}\approx4.9\), but also \(x>2\) (from triangle - inequality).

Answer:

a. \( \sqrt{24}b. \(2