Question

which sets of three of numbers represent the sides of an obtuse triangle? check all that apply. 4,7,8 3,4,5 2,2,3 6,8,9 3,5,6

Explanation:

Step1: Recall the obtuse - triangle condition

For a triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side), the triangle is obtuse if \(a^{2}+b^{2}

Step2: Check the set \(4,7,8\)

Let \(a = 4\), \(b=7\), \(c = 8\)
\(a^{2}+b^{2}=4^{2}+7^{2}=16 + 49=65\)
\(c^{2}=8^{2}=64\)
Since \(65>64\) (\(a^{2}+b^{2}>c^{2}\)), it is not an obtuse triangle

Step3: Check the set \(3,4,5\)

Let \(a = 3\), \(b = 4\), \(c=5\)
\(a^{2}+b^{2}=3^{2}+4^{2}=9 + 16=25\)
\(c^{2}=5^{2}=25\)
Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle

Step4: Check the set \(2,2,3\)

Let \(a = 2\), \(b = 2\), \(c = 3\)
\(a^{2}+b^{2}=2^{2}+2^{2}=4 + 4=8\)
\(c^{2}=3^{2}=9\)
Since \(8<9\) (\(a^{2}+b^{2}

Step5: Check the set \(6,8,9\)

Let \(a = 6\), \(b = 8\), \(c=9\)
\(a^{2}+b^{2}=6^{2}+8^{2}=36+64 = 100\)
\(c^{2}=9^{2}=81\)
Since \(100>81\) (\(a^{2}+b^{2}>c^{2}\)), it is not an obtuse triangle

Step6: Check the set \(3,5,6\)

Let \(a = 3\), \(b = 5\), \(c = 6\)
\(a^{2}+b^{2}=3^{2}+5^{2}=9 + 25=34\)
\(c^{2}=6^{2}=36\)
Since \(34<36\) (\(a^{2}+b^{2}

Answer:

\(2,2,3\) and \(3,5,6\)