Oblicz objętość stożka którego promień podstawy r= 12 cm, a pole powierzchni bocznejjest o 36pi cm2 wieksze od pola podstawy.
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PP = \pi r^2 = 12^2 \pi = 144 \pi
Pb = 144 \pi + 36 \pi = 180 \pi
\pi r l = 180 \pi r * l = 180
12 * l = 180
l = 15
Teraz z pitagorasa liczymy H:
12^2 + H^2 = 15^2
144 + H^2 = 225
H^2 = 81
H = 9
V = 1/3 PP * H
V = 1/3 144 \pi * 9
V = 432