Pythagorean Theorem Lecture Notes

Introduction

• Focus on the Pythagorean theorem and its applications in geometry.

The Pythagorean Theorem Formula

• For a right triangle:
  • Hypotenuse (C): longest side
  • Legs (A and B): shorter sides
• Formula: C² = A² + B²

Example 1: Finding the Hypotenuse

1. Given a right triangle with sides 5 and 12.
2. Assign values:
   • A = 12
   • B = 5
   • C = x (hypotenuse)
3. Calculation:
   • x² = 12² + 5²
   • x² = 144 + 25 = 169
   • x = √169 = 13

Example 2: Finding a Leg

1. Given a hypotenuse of 10 and one leg of 5.
2. Assign values:
   • C = 10
   • A = y
   • B = 5
3. Calculation:
   • 10² = y² + 5²
   • 100 = y² + 25
   • y² = 100 - 25 = 75
   • y = √75 = 5√3 (simplified)

Example 3: Area of a Square with Diagonal

1. Diagonal = 12 inches.
2. All sides equal (X):
3. Using Pythagorean theorem:
   • 12² = X² + X²
   • 144 = 2X²
   • X² = 72
   • X = √72 = 6√2 (simplified)
4. Area of square: A = X² = 72.

Example 4: Perimeter of a Rhombus

1. Diagonals bisect each other.
2. Given B/E = 7 and C/E = 24.
3. Using Pythagorean theorem:
   • C² = A² + B² where C = s (side of rhombus)
   • s² = 24² + 7²
   • s² = 576 + 49 = 625
   • s = 25
4. Perimeter = 4s = 4 * 25 = 100 units.

Example 5: Area of an Isosceles Trapezoid

1. Formula: A = 1/2 (B1 + B2) * H
   • B1 = 12, B2 = 20
2. Need to calculate height (H).
3. Use congruency of sides:
   • AD = 20, BC = 12, and BE = 7 (congruent segments).
4. Set up equation for X:
   • X + 12 + X = 20
   • 2X + 12 = 20
   • 2X = 8
   • X = 4
5. Use the right triangle:
   • C² = A² + B², where C = 5, A = 4, B = H.
   • 5² = 4² + H².
   • 25 = 16 + H²
   • H² = 9
   • H = 3.
6. Area calculation:
   • A = 1/2 (12 + 20) * 3
   • A = 1/2 (32) * 3 = 16 * 3 = 48.

Summary

• Pythagorean theorem is a fundamental concept for solving problems in geometry, including right triangles, squares, rhombuses, and trapezoids.