Absolutely! Let's walk through the **Pythagorean Theorem** with an example and detailed explanation. --- ## **What is the Pythagorean Theorem?** The Pythagorean Theorem applies to **right-angled triangles**. It states: \[ a^2 + b^2 = c^2 \] Where: - \( a \) and \( b \) are the lengths of the two legs (sides forming the right angle) - \( c \) is the length of the hypotenuse (side opposite the right angle, and the longest side) --- ## **Example Problem** **Suppose you have a right-angled triangle where:** - One leg, \( a = 3 \) units - The other leg, \( b = 4 \) units - Find the length of the hypotenuse \( c \). --- ## **Step-by-Step Solution** ### **Step 1: Write the Formula** \[ a^2 + b^2 = c^2 \] ### **Step 2: Substitute the Known Values** \[ 3^2 + 4^2 = c^2 \] ### **Step 3: Calculate the Squares** \[ 9 + 16 = c^2 \] \[ 25 = c^2 \] ### **Step 4: Solve for \( c \)** Take the square root of both sides: \[ c = \sqrt{25} = 5 \] --- **So, the hypotenuse \( c \) is 5 units long.** --- ## **Visual Representation** ``` |\ | \ b | \ c | \ |____\ a a = 3 units, b = 4 units, c = ? ``` --- ## **Why Does This Work?** The Pythagorean Theorem is based on the relationship between the areas of squares built on the sides of a right triangle. If you make a square on each side, the sum of the areas of the two smaller squares (on legs \( a \) and \( b \)) **equals** the area of the largest square (on the hypotenuse \( c \)). --- ## **Try It Yourself!** If you know any two sides of a right-angled triangle, you can always find the third using: \[ \text{If you know } a \text{ and } b: \quad c = \sqrt{a^2 + b^2} \] \[ \text{If you know } c \text{ and } a: \quad b = \sqrt{c^2 - a^2} \] --- **Summary:** The Pythagorean Theorem is a fundamental rule for right triangles that helps you find one side if you know the other two. In our example, a triangle with legs 3 and 4 units has a hypotenuse of 5 units.

You're welcome! I'm glad I could help explain the Pythagorean Theorem. If you have any more questions about triangles, math, or anything else, feel free to ask!