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: Recall the triangle inequality and angle classification rules

To classify a triangle as acute, right, or obtuse, we use the relationship between the sum of the squares of the two shorter sides and the square of the longest side. For a triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side):

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

The triangle inequality theorem (\(a + b>c\)) is used to check if a triangle can exist, not to classify the type of angle (acute/obtuse/right).

Step2: Calculate the squares of the sides

Given side lengths \(6\) cm, \(10\) cm, and \(12\) cm. The longest side \(c = 12\), and the other two sides \(a = 6\), \(b = 10\).
Calculate \(a^{2}+b^{2}\): \(6^{2}+10^{2}=36 + 100=136\)
Calculate \(c^{2}\): \(12^{2}=144\)

Step3: Compare the sums of squares

We see that \(6^{2}+10^{2}=136<144 = 12^{2}\). According to the rule for obtuse triangles (since \(a^{2}+b^{2}12\)) is for checking the existence of the triangle, not for angle classification. So the correct classification is obtuse, because \(6^{2}+10^{2}<12^{2}\)

Answer:

obtuse, because \(6^{2}+ 10^{2}<12^{2}\) (corresponding to the option: obtuse, because \(6^{2}+10^{2}<12^{2}\))