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.

Answer 1

To determine if the triangle is a Pythagorean triple, we need to check whether the lengths of the sides satisfy the Pythagorean theorem. The Pythagorean theorem states that for a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \):

\[
a^2 + b^2 = c^2
\]

In this case, we know:

- One leg \( A = 7 \) units
- Hypotenuse \( C = 9 \) units

We need to find the length of the other leg, which we will call \( B \).

### Step 1: Apply the Pythagorean theorem

According to the Pythagorean theorem:

\[
A^2 + B^2 = C^2
\]

Substituting the known values:

\[
7^2 + B^2 = 9^2
\]

Calculating \( 7^2 \) and \( 9^2 \):

\[
49 + B^2 = 81
\]

### Step 2: Solve for \( B^2 \)

Now, subtract \( 49 \) from both sides:

\[
B^2 = 81 - 49
\]
\[
B^2 = 32
\]

### Step 3: Calculate \( B \)

To find \( B \), we take the square root:

\[
B = \sqrt{32} = 4\sqrt{2}
\]

### Step 4: Check if the sides form a Pythagorean triple

A Pythagorean triple consists of three integers. Here, we have the sides as:

- \( A = 7 \) (integer)
- \( B = 4\sqrt{2} \) (not an integer)
- \( C = 9 \) (integer)

Since \( B \) is not an integer, the sides \( 7, 4\sqrt{2}, 9 \) do not form a Pythagorean triple.

### Conclusion

**A. Is it a Pythagorean triple?**
No, it is not a Pythagorean triple.

**Explanation:** For a set of three lengths to form a Pythagorean triple, all three lengths must be integers. In this case, while \( A \) and \( C \) are integers, \( B \) is not, as it equals \( 4\sqrt{2} \), which is an irrational number. Therefore, the triangle does not satisfy the criteria for being a Pythagorean triple.