Question

which classification best represents a triangle with side lengths 6 cm, 10 cm, and 12 cm?
acute, because $6^{2}+10^{2}<12^{2}$
acute, because $6 + 10>12$
obtuse, because $6^{2}+10^{2}<12^{2}$
obtuse, because $6 + 10>12$

Explanation:

Step1: Calculate the sum of squares of two shorter sides and the square of the longest side

The two shorter sides are \(a = 6\) and \(b = 10\), and the longest side is \(c=12\).
Calculate \(a^{2}+b^{2}\): \(6^{2}+10^{2}=36 + 100=136\).
Calculate \(c^{2}\): \(12^{2}=144\).
So, \(6^{2}+10^{2}<12^{2}\).

Step2: Determine the type of triangle

For a triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side):
If \(a^{2}+b^{2}=c^{2}\), the triangle is right - angled.
If \(a^{2}+b^{2}>c^{2}\), the triangle is acute - angled.
If \(a^{2}+b^{2}Since \(6^{2}+10^{2}<12^{2}\), the triangle is obtuse.
The condition \(6 + 10>12\) is just the triangle - inequality (to check if a triangle with these side lengths can exist), not for classifying the triangle as acute or obtuse.

Answer:

obtuse, because \(6^{2}+10^{2}<12^{2}\)