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CLASE 208

CLASE 208. Grupo de teoremas de Pitágoras (Ejercicios). C. D. G. CDEF es un cuadrado, AC  DE = {G} AF = 8,0 cm y AC = 10 cm. F. E. B. A. En la figura, E y F son puntos de la hipotenusa AB del triángulo rectán -gulo ABC.

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CLASE 208

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  1. CLASE208 Grupo de teoremas de Pitágoras (Ejercicios)

  2. C D G CDEF es un cuadrado, AC  DE = {G} AF = 8,0 cm y AC = 10 cm. F E B A • En la figura, E y F son puntos de la hipotenusa AB del triángulo rectán -gulo ABC.

  3. Identifica cinco triángulos rectán -gulos. Fundamenta en cada caso. • Prueba que BCF = CDG y FCA  AEG. • Halla la longitud de los segmentos AB, BC y DG. • Halla el área del AEG. • Construye un rectángulo que tenga igual área que el cuadrado CDEF.

  4. C D G F E B A 1.a)

  5. C D G F E B A 1.a)

  6. C D AF = 8,0 cm. AC = 10 cm. G F A E B CF: T. de Pitágoras (AFC) . FB: T. de la altura (ABC) . AB: Por suma de segmentos. BC: T. de Pitágoras o de los catetos (ABC) . DG: Elementos homólogos en trián – gulos iguales (FBC = CDG) . 1.c) Estrategia

  7. C D AF = 8,0 cm. AC = 10 cm. G = (a – b) (a +b) a2 – b2 F E B CF = 6,0 cm A h CF = h FB = q q (T. de Pitágoras) h2 = AC2 – AF2 h2 = (102 – 82) cm2 = 4 ·9 cm2 h2 = 2 ·18 cm2 h = 2 ·3 cm = 6 cm

  8. C D AF = 8,0 cm. AC = 10 cm. CF = h G F E B CF = 6,0 cm h2 = AF q FB = 4,5 cm q 36 9 = 8 2 A AB = h FB = q q (T. de la altura) (6cm)2 = 8 cm q 36 cm2= 8 cm q cm cm = 4,5 cm = 8 cm + 4,5 cm = 12,5 cm

  9. C D AF = 8,0 cm. AC = 10 cm. G FB = q F E B CF = 6,0 cm CB2 = AB  FB FB = 4,5 cm CB2 = 12,5 cm  4,5 cm CB2 = 56,25 cm2 CB = 7,5 cm A 12,5 cm AB = h q (T. de los catetos)

  10. C D h G AAHIF = AF q q E F A B h2 = AF q ACDEF = h2 pero; q (T. de la altura) I H ACDEF = AAHIF Entonces,

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