Pb = πrl = 120π
l = 2r
z warunków zadania
wstawiamy drugie do pierwszego
πr*2r = 120π
r2 = 60
r = 2√15
l = 2r = 4√15
z Pitagorasa
h = √l2 − r2 = √(4√15)2 − (2√15)2 = √240 − 60 = √180 = 6√5
| 1 | 1 | |||
V = | πr2*h = | π*60*6√5 = 120π√5 | ||
| 3 | 3 |