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Geometry’s Most Elegant Theorem

Geometry’s Most Elegant Theorem. Pythagorean Theorem Lesson 9.4. Theorem 69: The square of the measure of the hypotenuse of a right triangle is equal to the sum of the squares of the measures of the legs. (Pythagorean theorem). Theorem 69…. Theorem 69…. a 2 +b 2 =c 2. A. c. b. B. C.

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Geometry’s Most Elegant Theorem

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  1. Geometry’s Most Elegant Theorem Pythagorean Theorem Lesson 9.4

  2. Theorem 69:The square of the measure of the hypotenuse of a right triangle is equal to the sum of the squares of the measures of the legs. (Pythagorean theorem)

  3. Theorem 69…

  4. Theorem 69… a2+b2=c2 A c b B C a Draw a right triangle with squares on each side.

  5. Proof of Theorem 69 1. Given 2. From a point outside a line only one  can be drawn to the line. 3. A segment drawn from a vertex of a triangle to the opposite side is an altitude. 4. In a right triangle with an altitude drawn to the hypotenuse. (leg)2=(adjacent leg)(hypotenuse) ACB is a right angle Draw CD to AB CD is an altitude a2 = (c-x)c A c x D C-x b C B a

  6. Proof of Theorem 69 continued… Distributive property Same as 4 Addition property Algebra a2 = c2-cx b2= xc a2+b2= c2 - cx + cx a2 + b2 = c2

  7. Theorem 70:If the square of the measure of one side of a triangle equals the sum of the square of the measure of the other two sides, then the angle opposite the longest side is a right angle. If a2 + b2 = c2, then ∆ABC is a right ∆ and C is the right angle.

  8. If we increase c while keeping the a and b the same length, C becomes larger. If c is the length of the longest side of a triangle, a2+b2>c2, then the ∆ is acute a2+b2=c2, then the ∆ is right a2+b2<c2, then the ∆ is obtuse

  9. Find the perimeter of a rhombus with diagonals of 6 and 10. Remember that the diagonals of a rhombus are perpendicular bisectors of each other. 32 + 52 = x2 9 + 25 = x2 34 = x2 Since all sides of a rhombus are congruent, the perimeter is .

  10. Find the altitude of an isosceles trapezoid whose sides have lengths of 10, 30, 10, and 20. Draw in an altitude. Right ΔADE is congruent toright ΔBCF, and DE = FC =½ (30 – 20) = 5. A 20 B 10 10 x D 5 E 30 20 F 5 C In ΔADE, x2 + 52 = 102 x2 + 25 = 100 x2 = 75 x = Altitude =

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