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Which classification best represents a triangle with side lengths 10 in., 12 in., and 15 in.?

A. Acute, because [tex]$10^2 + 12^2 \ \textgreater \ 15^2$[/tex]
B. Acute, because [tex]$12^2 + 15^2 \ \textgreater \ 10^2$[/tex]
C. Obtuse, because [tex]$10^2 + 12^2 \ \textgreater \ 15^2$[/tex]
D. Obtuse, because [tex]$12^2 + 15^2 \ \textgreater \ 10^2$[/tex]



Answer :

GinnyAnswer
To classify the type of triangle given its side lengths of 10 inches, 12 inches, and 15 inches, we can use the relationship between the squares of the sides. We start by calculating the squares of each side:

[tex]\[ 10^2 = 100 \][/tex]
[tex]\[ 12^2 = 144 \][/tex]
[tex]\[ 15^2 = 225 \][/tex]

Next, we compare the sum of the squares of the two shorter sides with the square of the longest side. There are several steps you can follow to classify the triangle:

1. Calculate sum of squares of the shorter sides:
[tex]\[ 10^2 + 12^2 = 100 + 144 = 244 \][/tex]

2. Compare this sum (244) with the square of the longest side (225):

- Check if \(10^2 + 12^2 > 15^2\)
[tex]\[ 100 + 144 = 244 > 225 \][/tex]
- Since \(244 > 225\), this indicates that the triangle is acute, because the sum of the squares of the two shorter sides is greater than the square of the longest side.

3. Verification for other inequalities:
[tex]\[ 12^2 + 15^2 = 144 + 225 = 369 > 100 \][/tex]
[tex]\[ 15^2 + 10^2 = 225 + 100 = 325 > 144 \][/tex]

All comparisons show the same result, but the critical comparison \(10^2 + 12^2 > 15^2\) is sufficient.

Thus, the classification based on the side lengths 10 in., 12 in., and 15 in. is acute, because [tex]\(10^2 + 12^2 > 15^2\)[/tex].

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