Question

If a right triangle has a leg A that measures 7 units and a hypotenuse, C, that measures 9 units. A. Is it a Pythagorean triple? B. Show your steps AND explain how you know. (3 points)

Answer 1

Make it shorter

Answer 2

To determine if the right triangle with leg \( A = 7 \) units and hypotenuse \( C = 9 \) units is a Pythagorean triple, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (\( C^2 \)) is equal to the sum of the squares of the lengths of the other two legs (\( A^2 + B^2 \)), where \( B \) is the other leg.

**Step 1: Calculate the square of the hypotenuse.**
\[
C^2 = 9^2 = 81
\]

**Step 2: Calculate the square of leg \( A \).**
\[
A^2 = 7^2 = 49
\]

**Step 3: Set up the equation to find leg \( B \).**
According to the Pythagorean theorem, we need to find whether there exists a leg \( B \) such that:
\[
C^2 = A^2 + B^2
\]
which means:
\[
81 = 49 + B^2
\]

**Step 4: Solve for \( B^2 \).**
Subtract \( 49 \) from both sides:
\[
B^2 = 81 - 49 = 32
\]

**Step 5: Determine the value of leg \( B \).**
Take the square root of both sides to find \( B \):
\[
B = \sqrt{32} = 4\sqrt{2} \approx 5.66
\]

**Conclusion:**
A. The triangle does not consist of whole-number side lengths, so it is not a Pythagorean triple.

To confirm if it is a Pythagorean triple, we recall that a Pythagorean triple consists of three positive integers \( (a, b, c) \) such that \( a^2 + b^2 = c^2 \). In this case, we found \( B = \sqrt{32} \) is not an integer.

B. **Explanation:**
The values we found (7, 5.66, 9) are not all integers, as Pythagorean triples must consist exclusively of whole numbers. The leg \( B \)rounded to 5.66 demonstrates that it cannot form a Pythagorean triple with the leg \( A \) and hypotenuse \( C \). Therefore, the three lengths do not satisfy the conditions of being a Pythagorean triple.