Question

practice classifying triangles with known side lengths. 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 inequality

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

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

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

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

Let \(a = 3\), \(b = 4\), \(c = 5\). Then \(a^{2}+b^{2}=3^{2}+4^{2}=9 + 16=25\) and \(c^{2}=5^{2}=25\). This is a right - triangle (\(a^{2}+b^{2}=c^{2}\)), not an obtuse - triangle.

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

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

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

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

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

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

Answer:

C. \(2,2,3\), E. \(3,5,6\)