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Ariel completed the work below to show that a triangle with side lengths of 9, 15, and 12 does not form a right triangle.

[tex]\[
\begin{array}{l}
9^2 + 15^2 \neq 12^2 \\
81 + 225 \neq 144 \\
306 \neq 144
\end{array}
\][/tex]

Is Ariel's answer correct?

A. No, Ariel should have added [tex]\(12^2\)[/tex] and [tex]\(15^2\)[/tex] and compared that to [tex]\(9^2\)[/tex].
B. No, Ariel should have subtracted [tex]\(12^2\)[/tex] from [tex]\(9^2\)[/tex] and compared that to [tex]\(15^2\)[/tex].
C. No, Ariel should have added [tex]\(9^2\)[/tex] and [tex]\(12^2\)[/tex] and compared that to [tex]\(15^2\)[/tex].
D. Yes, Ariel's work is correct.


Sagot :

To determine the validity of Ariel's work, let’s carefully review the steps and check all possible right triangle configurations. Here are the side lengths provided: 9, 15, and 12.

### Calculation Steps:
1. Calculate the squares of each side:
- [tex]\( 9^2 = 81 \)[/tex]
- [tex]\( 15^2 = 225 \)[/tex]
- [tex]\( 12^2 = 144 \)[/tex]

2. Check the Pythagorean Theorem with all possible combinations:

#### First Combination
Check if [tex]\( 9^2 + 15^2 = 12^2 \)[/tex]:
[tex]\[ 81 + 225 = 306 \][/tex]
[tex]\[ 306 \neq 144 \][/tex]

Hence, [tex]\( 9^2 + 15^2 \neq 12^2 \)[/tex].

#### Second Combination
Check if [tex]\( 9^2 + 12^2 = 15^2 \)[/tex]:
[tex]\[ 81 + 144 = 225 \][/tex]
[tex]\[ 225 = 225 \][/tex]

Hence, [tex]\( 9^2 + 12^2 = 15^2 \)[/tex].

#### Third Combination
Check if [tex]\( 12^2 + 15^2 = 9^2 \)[/tex]:
[tex]\[ 144 + 225 = 369 \][/tex]
[tex]\[ 369 \neq 81 \][/tex]

Hence, [tex]\( 12^2 + 15^2 \neq 9^2 \)[/tex].

### Conclusion:
- Ariel's statement that [tex]\( 9^2 + 15^2 = 12^2 \)[/tex] is incorrect because [tex]\( 306 \neq 144 \)[/tex].
- The correct approach should use the combination where [tex]\( 9^2 + 12^2 = 15^2 \)[/tex], which holds true since [tex]\( 225 = 225 \)[/tex].

Based on this information, the correct answer is:
No, Ariel should have added [tex]\( 9^2 \)[/tex] and [tex]\( 12^2 \)[/tex] and compared that to [tex]\( 15^2 \)[/tex].