Overview
This lecture covers the Pythagorean Theorem, its formula, and applying it to solve a variety of geometry problems involving right triangles, squares, rhombuses, and trapezoids.
Pythagorean Theorem Basics
- The Pythagorean theorem formula is ( c^2 = a^2 + b^2 ) for a right triangle.
- ( c ) is the hypotenuse (longest side), ( a ) and ( b ) are the legs.
Example Problems with Right Triangles
- For sides 5 and 12, hypotenuse ( x = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 ).
- If hypotenuse is 10, one leg is 5, then ( y^2 = 10^2 - 5^2 = 75 ) so ( y = 5\sqrt{3} ).
Applying the Theorem to Squares
- A square with diagonal 12: diagonal forms a right triangle with sides ( x ).
- ( 12^2 = x^2 + x^2 = 2x^2 \Rightarrow x^2 = 72 ).
- Area of the square is ( x^2 = 72 ) square units.
- Side length ( x ) simplifies to ( 6\sqrt{2} ).
Rhombus Problem
- Diagonals of a rhombus bisect at 90° and lengths BE = 7, CE = 24.
- Each side ( s = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 ).
- Perimeter of the rhombus is ( 4 \times 25 = 100 ) units.
Isosceles Trapezoid Area
- Area formula: ( \text{Area} = \frac{1}{2}(b_1 + b_2)h ), where ( b_1 = 12, b_2 = 20 ).
- Non-parallel sides are each 5; height needs to be calculated.
- Split shape to get right triangles; solve for segment ( x ): ( 2x + 12 = 20 ) so ( x = 4 ).
- Height ( h = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3 ).
- Area ( = \frac{1}{2}(12 + 20) \times 3 = 48 ).
Key Terms & Definitions
- Hypotenuse — The longest side of a right triangle, opposite the right angle.
- Legs — The two sides of a right triangle that form the right angle.
- Rhombus — A quadrilateral with all sides equal and diagonals that bisect at 90°.
- Isosceles Trapezoid — A trapezoid with non-parallel sides of equal length.
Action Items / Next Steps
- Practice similar problems using the Pythagorean theorem.
- Review area formulas for quadrilaterals.
- Complete any assigned homework on triangle and quadrilateral properties.