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.
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b. complete the inequality so that it describes possible lengths for x in an obtuse triangle.
type your answers in the boxes.
<x<

Explanation:

Part a: Acute Triangle

Step1: Triangle Inequality

For any triangle, the sum of two sides must be greater than the third side.

• \(5 + 7>x\), so \(x < 12\)
• \(5+x>7\), so \(x>2\)
• \(7 + x>5\) (always true for \(x>0\))

Step2: Acute - Triangle Condition

By the Pythagorean theorem for acute triangles \(a^{2}+b^{2}>c^{2}\) (where \(c\) is the longest side).
Case1: If \(x\) is the longest side (\(x\geq7\)), then \(5^{2}+7^{2}>x^{2}\), \(25 + 49>x^{2}\), \(x^{2}<74\), \(x<\sqrt{74}\approx8.6\)
Case2: If \(7\) is the longest side (\(x < 7\)), then \(5^{2}+x^{2}>7^{2}\), \(x^{2}>49 - 25=24\), \(x>\sqrt{24}\approx4.9\)

Combining with \(2 < x<12\), we get \(\sqrt{24}

Part b: Obtuse Triangle

By the Pythagorean theorem for obtuse triangles \(a^{2}+b^{2}Case1: If \(x\) is the longest side (\(x\geq7\)), then \(5^{2}+7^{2}74\), \(x>\sqrt{74}\approx8.6\)
Case2: If \(7\) is the longest side (\(x < 7\)), then \(5^{2}+x^{2}<7^{2}\), \(x^{2}<24\), \(x<\sqrt{24}\approx4.9\)
Combining with \(2 < x<12\), we get \(2

Answer:

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