Ariel completed the work below to show that a triangle with side lengths 9, 15, and 12 do not form a right triangle.92+152=122 81+225=144 306≠144 Is Ariel's work correct? Why or why not? NOTE: On the sample work, this is a multiple choice question, but on the test, you will be asked to explain your answer in complete sentences. Think about WHY she is right or wrong when completing this question, so you are prepared for the test.(1 point)ResponsesNo, when setting up the Pythagorean Theorem, the longest side needs to be by itself.No, when setting up the Pythagorean Theorem, the longest side needs to be by itself.No, she squared the numbers instead of multiplying each of them by 2.No, she squared the numbers instead of multiplying each of them by 2.Yes, Ariel's work is correct and this is not a right triangle.Yes, Ariel's work is correct and this is not a right triangle.Yes, Ariel's work is correct and this is a right triangle.

Answer 1

Ariel's work is not correct for determining whether the triangle with side lengths 9, 15, and 12 forms a right triangle. The Pythagorean Theorem states that for a triangle to be a right triangle, the square of the length of the longest side (the hypotenuse) should be equal to the sum of the squares of the lengths of the other two sides.

In this case:

1. Identify the longest side. The side lengths are 9, 15, and 12. Here, 15 is the longest side.

2. Apply the Pythagorean Theorem, which states: \[ c^2 = a^2 + b^2 \] where \(c\) is the longest side. Thus: \[ 15^2 = 9^2 + 12^2 \]

3. Calculate the squares: \[ 15^2 = 225 \] \[ 9^2 = 81 \] \[ 12^2 = 144 \]

4. Add the squares of the other two sides: \[ 81 + 144 = 225 \]

5. Compare: \[ 225 = 225 \]

Since both sides of the equation are equal, this means that the triangle with sides 9, 15, and 12 does form a right triangle.

Therefore, the correct response is, "Yes, Ariel's work is correct and this is a right triangle." Ariel's conclusion is incorrect because she did not correctly identify and separate the longest side in her calculations.