t = √s2 – r2 = √102 - 62 = √100 – 36 = √64 = 8 cm
Ditanya: (1) luas permukaan…? (2) volume…?
Jawab:
(1) Luas Permukaan Kerucut
Lp = πr (r + s)
Lp = 6 π (6 + 10)
Lp = 6 π (16)
Lp = 96 π cm2
(2) Volume Kerucut
V = 1/3 π r2t
t = √s2 – r2 = √102 - 62 = √100 – 36 = √64 = 8 cm
Ditanya: (1) luas permukaan…? (2) volume…?
Jawab:
(1) Luas Permukaan Kerucut
Lp = πr (r + s)
Lp = 6 π (6 + 10)
Lp = 6 π (16)
Lp = 96 π cm2
(2) Volume Kerucut
V = 1/3 π r2t
Editor: Muhammad Khusaini
Sumber: buku.kemdikbud.go.id