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L = 2πl = 2π*12cm = 24πcm
¼L = ¼*24πcm = 6πcm
2πr=6πcm, stąd 2r = 6cm, r = 3cm
πr² = π(3cm)² = 9πcm²
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h² + r² = l²
h² = l² - r²
h² = (12cm)² - (3cm)²= 144cm² - 9cm² = 135 cm²
h = √135 cm = √(9*15)cm = 3 √15 cm
V stożka
V = ⅓πr²h = ⅓*9πcm²*3 √15 cm = 9√15πcm³
Odp. Objętość stożka wynosi 9√15πcm³.