To determine the classification of the triangle with side lengths 10 inches, 12 inches, and 15 inches, we will use the relationship between the squares of the side lengths which is derived from the Pythagorean theorem.
Step 1: Calculate the squares of the side lengths:
- [tex]\( a = 10 \)[/tex] inches, so [tex]\( a^2 = 100 \)[/tex].
- [tex]\( b = 12 \)[/tex] inches, so [tex]\( b^2 = 144 \)[/tex].
- [tex]\( c = 15 \)[/tex] inches, so [tex]\( c^2 = 225 \)[/tex].
Step 2: Analyze the results:
- [tex]\( 10^2 = 100 \)[/tex]
- [tex]\( 12^2 = 144 \)[/tex]
- [tex]\( 15^2 = 225 \)[/tex]
Step 3: Compare the sum of the squares of two sides with the square of the third side:
- Check [tex]\( 10^2 + 12^2 > 15^2 \)[/tex]:
[tex]\( 100 + 144 > 225 \rightarrow 244 > 225 \)[/tex]
This is true, which suggests we may have an acute or right triangle, but further inspection is needed.
- Repeat this process for the other combinations:
[tex]\( 12^2 + 15^2 > 10^2 \)[/tex]:
[tex]\( 144 + 225 > 100 \rightarrow 369 > 100 \)[/tex]
This is true.
[tex]\( 10^2 + 15^2 > 12^2 \)[/tex]:
[tex]\( 100 + 225 > 144 \rightarrow 325 > 144 \)[/tex]
This is true.
Since all the sums [tex]\( a^2 + b^2 \)[/tex], [tex]\( a^2 + c^2 \)[/tex], and [tex]\( b^2 + c^2 \)[/tex] are greater than the third side squared individually, the triangle is either acute, right, or obtuse. To further classify it:
Step 4: Determine the specific type:
- Check if it is a right triangle:
[tex]\( a^2 + b^2 = c^2 \)[/tex]:
[tex]\( 100 + 144 = 225 \rightarrow 244 \neq 225 \)[/tex]
This is not a right triangle.
Step 5: For acute triangles, the sum of the squares of any two sides should be greater than the square of the third side.
- We already established that [tex]\( 10^2 + 12^2 > 15^2 \)[/tex].
Conclusion:
Since all the side conditions fulfill the inequalities and since we determined it is not a right triangle, our final check indicates that this is an acute triangle.
Thus, the correct classification best representing the triangle with side lengths 10 in., 12 in., and 15 in. is:
- acute, because [tex]\( 10^2 + 12^2 > 15^2 \)[/tex]
