Pole powierzchni calkowitej stozka jest rowne 100, a promien ma dlugosc 5. Oblicz objetosc tego stozka.
P=100pi
r=5
P=pir^2+pirl
pi*5^2+pi*5*l=100pi
25pi+5pil=100pi
5pil=100pi-25pi
l=75pi/5pi
l=15
H - wysokość stożka
H^2=l^2-r^2
H^2=15^2-5^2
H^2=225-25
H^2=200
H^2=100*2
H=10√2
V=1/3Pp*H
V=1/3pir^2*H
V=1/3*5^2*10√2 pi=1/3 *25*10√2pi=1/3 *250√2pi [j^3]
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P=100pi
r=5
P=pir^2+pirl
pi*5^2+pi*5*l=100pi
25pi+5pil=100pi
5pil=100pi-25pi
l=75pi/5pi
l=15
H - wysokość stożka
H^2=l^2-r^2
H^2=15^2-5^2
H^2=225-25
H^2=200
H^2=100*2
H=10√2
V=1/3Pp*H
V=1/3pir^2*H
V=1/3*5^2*10√2 pi=1/3 *25*10√2pi=1/3 *250√2pi [j^3]