Question

pythagorean theorem converse assignment
given the side lengths below, first determine whether they form a triangle. if they do, determine whether the triangle is acute, right, or obtuse.
3, 16, 17
acute triangle
right triangle
obtuse triangle
not a triangle

Explanation:

Step1: Check triangle inequality

First, check if the sides \(3\), \(16\), \(17\) can form a triangle. The sum of any two sides must be greater than the third side.

• \(3 + 16 = 19>17\)
• \(3 + 17 = 20>16\)
• \(16 + 17 = 33>3\)

So, they can form a triangle.

Step2: Apply Pythagorean theorem converse

Let \(a = 3\), \(b = 16\), \(c = 17\) (where \(c\) is the longest side).
Calculate \(a^{2}+b^{2}\) and \(c^{2}\):

• \(a^{2}+b^{2}=3^{2}+16^{2}=9 + 256 = 265\)
• \(c^{2}=17^{2}=289\)

Since \(a^{2}+b^{2}=265<289 = c^{2}\), by the converse of the Pythagorean theorem and the triangle type rule (if \(a^{2}+b^{2}eq 289=17^2\). Wait, but wait, maybe I miscalculated. Wait, \(3, 16, 17\): Wait, \(3 + 16 = 19>17\), so triangle exists. Now, check the type. The formula for triangle type: for sides \(a\leq b\leq c\),

• If \(a^{2}+b^{2}=c^{2}\): right triangle.
• If \(a^{2}+b^{2}>c^{2}\): acute triangle.
• If \(a^{2}+b^{2}

So \(a = 3\), \(b = 16\), \(c = 17\). \(3^{2}+16^{2}=9 + 256 = 265\), \(17^{2}=289\). Since \(265<289\), so \(a^{2}+b^{2}

Wait, but the options include "Obtuse Triangle". So the answer should be Obtuse Triangle. Wait, but let me check again. Wait, maybe I made a mistake in the side lengths. Wait, the problem says \(3, 16, 17\). Yes. So the steps are:

1. Check triangle inequality: holds.
2. Check Pythagorean theorem converse: \(3^2 + 16^2

eq 17^2\), so not right.

1. Check \(a^2 + b^2\) vs \(c^2\): \(265 < 289\), so obtuse.

Answer:

Obtuse Triangle