Question

determine whether the segment lengths form a triangle. if so, is the triangle acute, right, or obtuse?

1. 16, 30, and 34
2. 4, √67, and 9
3. √5, 5, and 5.5

Explanation:

Step1: Check Triangle Inequality

For sides \(a = 16\), \(b = 31\), \(c = 34\), check \(a + b>c\), \(a + c>b\), \(b + c>a\).
\(16 + 31 = 47>34\), \(16 + 34 = 50>31\), \(31 + 34 = 65>16\). So triangle exists.

Step2: Classify by Pythagorean Theorem

Compute \(a^{2}+b^{2}\) and \(c^{2}\).
\(16^{2}=256\), \(31^{2}=961\), \(34^{2}=1156\).
\(a^{2}+b^{2}=256 + 961 = 1217\). Since \(1217>1156\) (\(a^{2}+b^{2}>c^{2}\)), triangle is acute.

Step1 (Problem 5): Check Triangle Inequality

For \(a = 4\), \(b=\sqrt{67}\approx8.19\), \(c = 9\).
\(4+\sqrt{67}\approx12.19>9\), \(4 + 9 = 13>\sqrt{67}\), \(\sqrt{67}+9\approx17.19>4\). Triangle exists.

Step2 (Problem 5): Classify by Pythagorean Theorem

\(4^{2}=16\), \((\sqrt{67})^{2}=67\), \(9^{2}=81\).
\(a^{2}+b^{2}=16 + 67 = 83>81\) (\(a^{2}+b^{2}>c^{2}\)), so acute.

Step1 (Problem 6): Check Triangle Inequality

For \(a=\sqrt{5}\approx2.24\), \(b = 5\), \(c = 5.5\).
\(\sqrt{5}+5\approx7.24>5.5\), \(\sqrt{5}+5.5\approx7.74>5\), \(5 + 5.5 = 10.5>\sqrt{5}\). Triangle exists.

Step2 (Problem 6): Classify by Pythagorean Theorem

\((\sqrt{5})^{2}=5\), \(5^{2}=25\), \(5.5^{2}=30.25\).
\(a^{2}+b^{2}=5 + 25 = 30<30.25\) (\(a^{2}+b^{2}

Answer:

• Problem 4: Forms an acute triangle.
• Problem 5: Forms an acute triangle.
• Problem 6: Forms an obtuse triangle.