A **Pythagorean triplet** consists of three positive integers \( a \), \( b \), and \( c \) such that:

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

To find a triplet involving \( 48 \), we can assume \( 48 \) to be one of the legs \( a \) or \( b \), and we need to find the other leg and the hypotenuse \( c \).

Let’s consider \( 48 \) as one of the legs, say \( a = 48 \). We need to find \( b \) and \( c \) such that:

\[
48^2 + b^2 = c^2
\]

\[
2304 + b^2 = c^2
\]

\[
c^2 - b^2 = 2304
\]

This can be solved using factorization methods or trial and error to find a valid pair of \( b \) and \( c \).

Let me calculate the pair that satisfies this equation.The Pythagorean triplet that includes 48 is \( (48, 55, 73) \). This means:

\[
48^2 + 55^2 = 73^2
\]

To verify:

\[
48^2 = 2304, \quad 55^2 = 3025, \quad 73^2 = 5329
\]

And indeed:

\[
2304 + 3025 = 5329
\]

Let me know if you want further details or explanations.

Here are 5 related questions:
1. What are Pythagorean triplets, and how are they useful in geometry?
2. Can any number be part of a Pythagorean triplet?
3. How would you find all Pythagorean triplets for a given number?
4. How can you prove that certain numbers do not form a triplet?
5. What is the formula for generating Pythagorean triplets?

**Tip:** In any right triangle, the hypotenuse is always the longest side and is opposite the right angle.

Discover how to find the Pythagorean triplet that includes 48 as one of the legs. This solution explores the Pythagorean theorem and number theory to solve for the triplet (48, 55, 73). Ideal for students in Grades 9-10.

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