Lecture Notes: Pythagorean Theorem and Related Geometric Problems

Introduction to the Pythagorean Theorem

• Formula: In a right triangle with legs a and b, and hypotenuse c, the Pythagorean theorem states: [ c^2 = a^2 + b^2 ]
• Definitions:
  • c: Hypotenuse, the longest side of the triangle.
  • a, b: Legs of the triangle.

Example Problems

Example 1: Finding the Hypotenuse

• Given:
  • One leg a = 12
  • Other leg b = 5
• Problem: Calculate hypotenuse x.
• Solution:
  1. Use formula: [ x^2 = 12^2 + 5^2 ]
  2. Calculate: [ 12^2 = 144 ] and [ 5^2 = 25 ]
  3. Sum: [ 144 + 25 = 169 ]
  4. Find ( x ): ( x = \sqrt{169} = 13 )

Example 2: Finding a Triangle Leg

• Given:
  • Hypotenuse c = 10
  • One leg a = 5
• Problem: Calculate other leg y.
• Solution:
  1. Use formula: [ 10^2 = y^2 + 5^2 ]
  2. Calculate: [ 10^2 = 100 ] and [ 5^2 = 25 ]
  3. Solve for y: [ 100 - 25 = 75 ]
  4. Find ( y ): ( y = \sqrt{75} = 5\sqrt{3} )

Word Problems

Area of a Square

• Problem: Find the area of a square with diagonal 12 inches.
• Solution:
  1. Use theorem for diagonal: [ 12^2 = x^2 + x^2 ]
  2. Calculate: [ 12^2 = 144 ]
  3. Simplify: [ 2x^2 = 144 ] → [ x^2 = 72 ]
  4. Area: ( x^2 = 72 )

Perimeter of a Rhombus

• Given:
  • Diagonals ( BE = 7 ) and ( CE = 24 )
• Problem: Calculate perimeter.
• Solution:
  1. Use theorem on each triangle: [ s^2 = 24^2 + 7^2 ]
  2. Calculate: [ 24^2 = 576 ] and [ 7^2 = 49 ]
  3. Sum: [ 576 + 49 = 625 ]
  4. Find ( s ): ( s = 25 )
  5. Perimeter: ( 4s = 100 )

Area of an Isosceles Trapezoid

• Formula: ( \text{Area} = \frac{1}{2} (B_1 + B_2) , \times , H )
• Given:
  • Bases ( B_1 = 12 ) and ( B_2 = 20 )
  • Side lengths = 5 (isosceles)
• Problem: Calculate area.
• Solution:
  1. Solve for height H using right triangle.
  2. ( H = 3 ) from calculations: [ 5^2 = 4^2 + H^2 ]
  3. Area: ( \frac{1}{2} (12 + 20) , \times , 3 = 48 )