Objętość stożka wynosi 96 pi ,a średnica podstawy 12.Oblicz pole przekroju osiowego
skoro d =12, to r = 6
V=1/3Pp * h
96π = 1/3πr² *h /:π
96 = 1/3 * 6² *h
96 = 1/3 *36h
96= 12h /:12
h= 8
Pole:
P= 1/2 *a*h
P=1/2 * 12 * 8 = 6 * 8 = 48
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skoro d =12, to r = 6
V=1/3Pp * h
96π = 1/3πr² *h /:π
96 = 1/3 * 6² *h
96 = 1/3 *36h
96= 12h /:12
h= 8
Pole:
P= 1/2 *a*h
P=1/2 * 12 * 8 = 6 * 8 = 48