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: Check for triangle inequality

For any set of three numbers \(a,b,c\) (\(c\) is the largest side), \(a + b>c\) must hold. All the given sets satisfy this.

Step2: Use the obtuse - triangle condition

If \(a,b,c\) are the sides of a triangle (\(c\) is the largest side), the triangle is obtuse if \(a^{2}+b^{2}

For the set \(4,7,8\)

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

For the set \(3,4,5\)

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

For the set \(2,2,3\)

\(a = 2\), \(b = 2\), \(c = 3\)
\(a^{2}+b^{2}=2^{2}+2^{2}=4 + 4=8\)
\(c^{2}=3^{2}=9\)
Since \(8<9\), it is an obtuse triangle.

For the set \(6,8,9\)

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

For the set \(3,5,6\)

\(a = 3\), \(b = 5\), \(c = 6\)
\(a^{2}+b^{2}=3^{2}+5^{2}=9 + 25=34\)
\(c^{2}=6^{2}=36\)
Since \(34<36\), it is an obtuse triangle.

Answer:

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