Pythagorean Theorem Lecture Notes

Overview

• The Pythagorean theorem is used to solve problems related to geometry, particularly involving right triangles.

Pythagorean Theorem Formula

• For a right triangle:
  • C = hypotenuse (longest side)
  • A and B = legs of the right triangle
• Formula: C² = A² + B²

Example 1: Finding the Hypotenuse

• Given: A = 12, B = 5
• Find: C (hypotenuse)
• Calculation:
  • C² = 12² + 5²
  • C² = 144 + 25 = 169
  • C = √169 = 13

Example 2: Finding a Leg

• Given: C = 10, B = 5
• Find: A (leg)
• Calculation:
  • C² = A² + B²
  • 10² = A² + 5²
  • 100 = A² + 25
  • A² = 100 - 25 = 75
  • A = √75 = 5√3 (simplified)

Area of a Square with Diagonal

• Given: Diagonal = 12 inches
• Area Formula: Area = X² (where X = side length)
• Calculation:
  • C² = A² + B² (right triangle)
  • 12² = X² + X²
  • 144 = 2X²
  • X² = 72; Area = 72
  • X = √72 = 6√2 (simplified)

Example 3: Perimeter of a Rhombus

• Given: B/E = 7, C/E = 24
• Calculation:
  • Diagonals of a rhombus bisect each other at 90 degrees.
  • Create right triangles:
    • B/E = 7, C/E = 24
  • Use Pythagorean theorem:
    • C² = A² + B²
    • S² = 24² + 7² = 576 + 49 = 625
    • S = √625 = 25
  • Perimeter = 4S = 4 * 25 = 100 units.*

Example 4: Area of an Isosceles Trapezoid

• Formula: Area = 1/2 (B₁ + B₂) * H
  • B₁ = 12, B₂ = 20
  • Find height (H).
• Setup right triangles:
  • Sides are equal (5 units each).
  • Let: X = length of the segments from the base to the top.
  • Equation: 2X + 12 = 20
  • Solve for X: X = 4.
• Calculation for height (H) using right triangle:
  • C² = A² + B²
  • 5² = 4² + H²
  • 25 = 16 + H²
  • H² = 9; H = 3.
• Area Calculation:
  • Area = 1/2(12 + 20) * 3
  • Area = 1/2(32) * 3 = 16 * 3 = 48.

Summary

• Key formula: C² = A² + B² for right triangles.
• Applications include finding lengths, areas, and perimeters in geometric shapes.