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Pythagorean Theorem

Pythagorean Theorem. By: SAMUEL M. GIER. DRILL. SIMPLIFY THE FOLLOWING EXPRESSION. 1. 4. + 2. 5 . 3. REVIEW ABOUT RIGHT TRIANGLES. A. LEGS. &. The perpendicular side. HYPOTENUSE. B. C.

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Pythagorean Theorem

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  1. Pythagorean Theorem By: SAMUEL M. GIER

  2. DRILL • SIMPLIFY THE FOLLOWING EXPRESSION. 1. 4. + 2. 5. 3.

  3. REVIEW ABOUT RIGHT TRIANGLES A LEGS & The perpendicular side HYPOTENUSE B C The side opposite the right angle

  4. PYTHAGOREAN THEOREM • For any right triangle, the square of the length of the HYPOTENUSE is equal to the SUM of the squares of the length of the LEGS.

  5. ILLUSTRATION A HYPOTENUSE (AC)² = (AB)² +(BC)² LEG B C LEG

  6. ILLUSTRATIVE EXAMPLE Find AB, if AC = 15 and BC = 12. Solution 1: (AC)² = (AB)² +(BC)² (15)² = (AB)² +(12)² (225) = (AB)² + 144 225 - 144 = (AB)² 81 = (AB)² AB = AB = 9 A HYPOTENUSE LEG B C LEG (Squaring both sides)

  7. ILLUSTRATIVE EXAMPLE Find BC, if AC = 20 and AB = 12. Solution 1: (AC)² = (AB)² +(BC)² (20)² = (12)² +(BC)² 400 = 144 +(BC)² 400 - 144 = (BC)² 256 = (BC)² BC = BC = 16 A HYPOTENUSE LEG B C LEG (Squaring both sides)

  8. ILLUSTRATIVE EXAMPLE Find BC, if AC = 20 and AB = 12. Solution 2: (BC)² = (AC)² - (AB)² (BC)² = (20)² - (12)² (BC)² = 400 - 144 (BC)² = 256 BC = BC = 16 A HYPOTENUSE LEG B C LEG (Squaring both sides)

  9. EXERCISES 1. • Find the value of k. Solution 1: Write an equation for k (k)² = (15)² +(20)² k² = 225 +400 k² = 625 k = k = 25 15 20 k (Squaring both sides) Answer : 25

  10. EXERCISES 2. • Find the value of k. Solution 1: Write an equation for k (14)² = (k)² +(6)² 196 = k² +36 196-36 = k² 160= k² =k k = 4 k 6 14 Answer: 4

  11. EXERCISES 2. • Find the value of k. Solution 2: Write an equation for k (k)² = (14)² -(6)² k² = 196 - 36 k² = 160 k = k = 4 k 6 14 Answer: 4

  12. EXERCISES 3. • Find the value of k. Solution 1: Write an equation for k (k)² = (13)² -(5)² k² = 169 - 25 k² = 144 k = k= 12 5 h 13 Answer: 12

  13. SPECIAL RIGHT TRIANGLES • 30 – 60 – 90 TRIANGLE • 45 – 45 – 90 TRIANGLE (Isosceles Right Triangle) 30º 45º 60º 45º

  14. SPECIAL RIGHT TRIANGLES • THEOREM “In a 45 -45 -90 triangle, the length of the hypotenuse is equal to the SQUARE ROOT of the LEG. • 45 – 45 – 90 TRIANGLE (Isosceles Right Triangle) 45º 45º

  15. SPECIAL RIGHT TRIANGLES • Hence, the legs of an Isosceles Right Triangle are equal. So, either equation is true. • THEOREM “In a 45 -45 -90 triangle, the length of the hypotenuse is equal to the SQUARE ROOT of the LEG. c = a c a a c = b 45º c b 45º b

  16. Examples • 1. Find the value of c. • 2. Find the value of z 45º z 7 c Solution: c = a c = 5 45º 5 SOLUTION: z = a =(7 )( ) = 7 = 7 ( 2) Z = 14

  17. SPECIAL RIGHT TRIANGLES • 30 – 60 – 90 TRIANGLE • THEOREM “ In a 30- 60- 90 triangle, the side opposite the 30º is equal to ONE – HALFthe length of the HYPOTENUSE.” THUS, Shorter leg = ½ (hypotenuse) Or a = ½ c 30º LONGER LEG HYPOTENUSE (c) ( b) 60º SHORTER LEG (a)

  18. SPECIAL RIGHT TRIANGLES • 30 – 60 – 90 TRIANGLE • THEOREM “ In a 30- 60- 90 triangle, the side opposite the 60º is equal to the length of the SHORTER LEG times the square of 3.” THUS, longer leg = (shorter leg) x Or b = a 30º HYPOTENUSE LONGER LEG (c) ( b) 60º SHORTER LEG (a)

  19. Summary 1. Shorter leg = ½ (hypotenuse) Or a = ½ c 2. Hypotenuse =( 2 ) x ( shorter leg) Or C = 2a 3. longer leg = (shorter leg) x Or b = a • 30 – 60 – 90 TRIANGLE 30º LONGER LEG HYPOTENUSE (c) ( b) 60º SHORTER LEG (a)

  20. In the figure, find the value of c and b. SOLUTION: 1. TO FIND c. ( side Opposite 90º) c= 2a = 2( 5) C = 10 2. TO FIND b. (side Opposite 60º) b = a = 5 30º c b 60º a =5

  21. In the figure, find the value of x and y. SOLUTION: 1. TO FIND y. ( side Opposite 30º) y = ½ ( 20) y = 10 2. TO FIND x. (side Opposite 60º) x = y = 10 30º 20 x 60º y

  22. QUIZ • 2. Find the value of a and b. • 1. Find the value of x. 30º 45º x 16 b 7 60º a

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