Lecture on Pythagorean Theorem

Formula

• For a right triangle with legs a and b and hypotenuse c:
  • Formula: ( c^2 = a^2 + b^2 )
  • c: Hypotenuse (longest side)
  • a and b: Legs of the triangle

Example Problems

Example 1: Find the Hypotenuse

• Given:
  • ( a = 12 )
  • ( b = 5 )
• Using the formula:
  • ( x^2 = 12^2 + 5^2 )
  • ( x^2 = 144 + 25 )
  • ( x^2 = 169 )
  • ( x = \sqrt{169} = 13 )

Example 2: Find a Leg

• Given:
  • Hypotenuse, ( c = 10 )
  • One leg, ( b = 5 )
• Using the formula:
  • ( 10^2 = y^2 + 5^2 )
  • ( 100 = y^2 + 25 )
  • ( 100 - 25 = y^2 )
  • ( y^2 = 75 )
  • ( y = \sqrt{75} )
  • Simplify: ( y = 5\sqrt{3} )

Word Problems

Example 3: Area of a Square with Diagonal

• Given:
  • Diagonal = 12 inches
• Steps:
  • Draw the square and diagonal: forms two right triangles
  • Using the formula:
    • ( c^2 = a^2 + b^2 )
    • ( 12^2 = x^2 + x^2 )
    • ( 144 = 2x^2 )
    • Divide by 2: ( x^2 = 72 )
    • Area, ( A = x^2 = 72 ) square inches
    • Simplify for exact value: ( x = 6\sqrt{2} )
  • Note: Area remains 72 square inches

Example 4: Perimeter of a Rhombus

• Given:
  • Diagonals: BE = 7, CE = 24
• Steps:
  • Properties: Diagonals bisect at right angles
  • Calculate sides using triangles:
    • ( s^2 = 7^2 + 24^2 )
    • ( s^2 = 49 + 576 )
    • ( s^2 = 625 )
    • ( s = \sqrt{625} = 25 )
  • Perimeter of rhombus:
    • ( P = 4s )
    • ( P = 4 \times 25 = 100 ) units

Example 5: Area of an Isosceles Trapezoid

• Given:
  • Bases: B1 = 12, B2 = 20
  • Side lengths = 5 each
• Steps:
  • Draw right triangles and solve for height (H)
  • Break into smaller segments:
    • Solve for X (extra segments):
      • ( 2X + 12 = 20 )
      • ( 2X = 8 )
      • ( X = 4 )
    • Solve for H (height):
      • ( 5^2 = 4^2 + H^2 )
      • ( 25 = 16 + H^2 )
      • ( H^2 = 9 )
      • ( H = \sqrt{9} = 3 )
  • Calculate area:
    • ( A = \frac{1}{2}(B1 + B2) \times H )
    • ( A = \frac{1}{2}(12 + 20) \times 3 )
    • ( A = \frac{1}{2} \times 32 \times 3 = 48 ) square units