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Volume = 1/3 πr² t
= 1/3 x 3,14 x 7² x 24
= 1/3 x 3.14 x 49 x 24
= 3.14 x 49 x 8
Volume = 1230,88 cm³
e). Diketahui:
t = 3 cm
s = 4 cm
Ditanya: luas permukaan dan volume kerucut ?
Dijawab:
s² = r² + t²
4² = r² + 3²
16 = r² + 9
r² = 16 – 9
r² = 7
r = √7
Baca Juga: KUNCI JAWABAN Matematika Kelas 9 Halaman 212 213 214 215 Latihan 4.1 Kekongruenan Bangun Datar
L = πr (r + s)
= 3.14 x √7 (√7 + 4)
= 3.14√7 (√7 + 4)
= 3.14 x 7 + 12.56√7
= 21.98 + 33.22
L = 55,2 cm²
Volume = 1/3 πr² t
= 1/3 x 3,14 x √7² x 3
= 1/3 x 3.14 x 7 x 3
= 3.14 x 7
Volume = 21,98 cm³
f). Diketahui:
