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PΔ=a²√3/4
√3=a²√3/4 /·4
4√3=a²√3 /:√3
a²=4
a=√4=2 --->bok Δ w przekroju osiowym tego stozka
zatem stozek ma
h=a√3/2=2√3/2=√3
r=1/2a=1/2·2=1
l=a=2
to objetosc stozka wynosi:
V=1/3πr²·h=1/3π·1²·√3=1/3·π·√3=√3/3 π [j³]
pole calkowite bryly
Pc=πr²+πrl=1²π+1·2π=π+2π=3π [j²]