Question

Przekatna podstawy graniastosłupa prawidłowego czworokątnego ma długosc 6√2 cm. wysokosc bryly wynosi 10 cm.Oblicz v i Pc bryły

Answer 1

p = 6 p(2) cm

h = 10 cm

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p = a p(2)

zatem

a = 6 cm

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V = Pp *h = a^2 * h = ( 6 cm)^2 * 10 cm = 36*10 cm^3 = 360 cm^3

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Pc = 2 Pp + Pb = 2 a^2 + 4*a*h = 2 * ( 6 cm)^2 + 4* 6 cm* 10 cm =

  = 2* 36 cm^2 + 240 cm^2 = (72 + 240 ) cm^2 = 312 cm^2

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Answer 2

d=a√2

6√2=a√2|:√2

6=a

Pc=2Pp+Pb

Pp=a² Pp=6²=36

Pb=4·a·b

Pb=4·6·10

Pb=240

Pc=2·36+240=312 (cm²)

V=Pp·H

V=36·10=360(cm³)