Treść zadania

Rozwiązania

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  0 1

  banpioszy 16.2.2013 (23:51)

  Zad 1
  r = 5 cm
  h = 8,2 cm
  s = √[5² + 8,2²] = √92,24 = 1/5√2306
  .............
  s – tworząca, bo litera „el” mało widoczna
  ...................................
  Pp = πr² = π×5² = 25π
  Pb = πrs = π×5×1/5√2306 = π√2306
  .............
  V= ⅓Pph = ⅓×25π×8,2 = 205π/3
  ..................
  Pc = 25π + π√2306
  =======================
  Zad 2
  r = 6 dm = 60 cm
  Pp = π×60² = 3600π
  .........
  V = ⅓×3600πh ....ale :
  ⅓×3600πh = 80π
  1200πh = 80π
  ................
  h = 80/1200 = 1/15 (cm)
  s = √[60² + (1/15)²] = 1/15√810001
  Pb = π×60×1/15√810001 = 4π√810001
  ................
  Pc = 3600π + 4π√810001
  .=======================
  Zad 3
  s = 13
  πr² + 13πr = 90π
  r² + 13r – 90 = 0
  Δ = 169 + 360 = 529
  √Δ=23
  r = [ - 13 – 23]/2 = (- 18) < 0 .. nie może być !
  r = [- 13 + 23]/2 = 5
  h = √[13² – 5²] = 12
  ...........
  V = ⅓π×5²×12 = 100π
  ======================
  Zad 4
  r = 2√3
  Pp = πr² = π×(2√3)² = 12π
  h = √6
  Pb = 2πrh = 2π×2√3×√6 = 4π√18 = 12π√2
  Pc = 2×12π + π12√2 = 12π(2 + √2)
  ..........................
  V = 12π×√6 = 12π√6
  ======================
  Zad 5 (BŁĄD W TREŚCI ZADANIA)
  r = 25
  Pp = π×25² = 625π
  Pb = 2π×25× h = 50hπ
  Pc = 2×625π + 50hπ = 1250π + 50hπ
  ...................
  1250 + 50h = 42,2
  50h = 42,2 – 1250 < 0 ......???
  ..........................
  coś tu nie tak :
  r = 25 cm, więc pole jednej tylko podstawy mierzy 625π... w zadaniu Pc = 42,2 π ? Pc < Pp ....???

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