Lecture: Solving Geometry Problems with the Pythagorean Theorem

Introduction

• Overview of the Pythagorean Theorem
• Application in solving geometry problems

Pythagorean Theorem Formula

• For a right triangle with sides a, b, and hypotenuse c:
  • (c^2 = a^2 + b^2)
• Hypotenuse (c) is the longest side
• (a) and (b) are the legs of the triangle

Example Problems

Example 1: Finding the Hypotenuse

• Given triangle sides: (a = 5), (b = 12)
• Calculate hypotenuse (x)
• Use formula: (x^2 = 12^2 + 5^2)
  • (12^2 = 144)
  • (5^2 = 25)
  • (144 + 25 = 169)
  • (x = \sqrt{169} = 13)

Example 2: Finding a Leg

• Given hypotenuse (c = 10), leg (b = 5), find (a = y)
• Use formula: (10^2 = y^2 + 5^2)
  • (10^2 = 100)
  • (5^2 = 25)
  • (100 - 25 = 75)
  • (y = \sqrt{75})
  • Simplify (\sqrt{75} = 5 \sqrt{3})

Word Problems

Problem 1: Area of a Square

• Diagonal of square = 12 inches
• Use Pythagorean theorem for right triangle formed by diagonal:
  • (12^2 = x^2 + x^2)
  • (144 = 2x^2)
  • (x^2 = 72)
  • Area = (x^2 = 72)

Problem 2: Perimeter of a Rhombus

• Rhombus with diagonals (BE = 7), (CE = 24)
• Diagonals bisect each other at right angles
• Use Pythagorean theorem on triangle formed:
  • (s^2 = 24^2 + 7^2)
  • (s = 25)
  • Perimeter = (4 \times s = 100)

Problem 3: Area of an Isosceles Trapezoid

• Bases (B_1 = 12) and (B_2 = 20)
• Find height (h)
• Use additional points and the theorem:
  • (AD = 20), (EF = 12)
  • (x + 12 + x = 20); (2x = 8); (x = 4)
  • Height (h) from right triangle:
    • (5^2 = 4^2 + h^2)
    • (h = 3)
• Calculate area:
  • Formula: (\text{Area} = \frac{1}{2}(B_1 + B_2) \times h)
  • (\text{Area} = \frac{1}{2}(12 + 20) \times 3 = 48)