Treść zadania

Najlepsze rozwiązanie

• 

  1 0

  ewka13 14.10.2011 (14:06)

  Zad 1.
  Korzystamy ze wzorów (nie będę ich powtarzać) :
  
  P_{c} = \Pi r(r + l)
  
  V = \frac {1} {3}\Pi r^{2} h
  
  a)
  r = 5cm
  l = 15cm
  
  P_{c} = 5\Pi (5 + 15)=100 \Pi cm^{2}
  
  obliczamy h :
  
  h^{2} = l^{2} - r^{2}
  
  h^{2} = 15^{2} - 5^{2}=200cm^{2}
  
  h = \sqrt {200} = \sqrt {100\cdot 2} = 10\sqrt {2}cm
  
  V=\frac {1} {3}\cdot \Pi \cdot 5^{2} \cdot 10\sqrt {2} = \frac {250} {3}\sqrt {2}\Pi cm^{3}
  
  b)
  h = 3\sqrt {3}cm
  
  2r = l
  
  obliczamy r :
  
  (2r)^{2} = r^{2} + h^{2}
  
  4r^{2} -r^{2} =(3\sqrt {3})^{2}
  
  3r^{2} =27 / : 3
  
  r^{2} = 9
  
  r = 3cm
  
  ponieważ :
  
  l = 2r = 2\cdot 3cm=6cm
  
  P_{c}=3\Pi (3 + 6)=27\Pi cm^{2}
  
  V = \frac {1} {3}\Pi \cdot 3^{2} \cdot 3\sqrt {3}=\frac {1} {3}\cdot 27\sqrt {3}\Pi=9\sqrt {3}\Pi cm^{3}
  
  c)
  r = 14cm
  h= 48cm
  
  obliczamy l :
  
  l^{2} = r^{2} + h^{2}
  
  l^{2} = 14^{2} + 48^{2}
  
  l^{2} =196 + 2304=2500
  
  l = \sqrt {2500} =50cm
  
  P_{c} = 14\Pi (14 + 50)=14\cdot 64\Pi = 896\Pi cm^{2}
  
  V = \frac {1} {3}\Pi \cdot 14^{2} \cdot 48=16\cdot 196\Pi = 3136\Pi cm^{3}
  
  d)
  h= 20cm
  
  r = 75%*h=0,75*20=15cm
  
  obliczamy l (jak w c) :
  
  l^{2} = 15^{2} + 20^{2}
  
  l^{2} = 225 + 400=625cm^{2}
  
  l = \sqrt {625} = 25cm
  
  P_{c} = 15\Pi (15 + 25) =15\cdot 40\Pi = 600\Pi cm^{2}
  
  V = \frac {1} {3}\Pi \cdot 15^{2} \cdot 20=\frac {1} {3} \cdot 4500\Pi=1500\Pi cm^{3}.

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