Kunci Jawaban Matematika Kelas 9 Halaman 280-283 Esai, Latihan 5.1 Menghitung Luas Permukaan dan Volume Tabung

a. r = 4 cm dan t = 10 cm
LP = 2 π r (r + t)
= 2 × π × 4 (4 + 10) cm²
= 112 π cm²
= 112 × ²²/₇ cm²
= 352 cm²
V = π r² t
= π × 4 × 4 × 10 cm³
= 160 π cm³
= 160 × 3,14 cm³
= 502,4 cm³

b. r = 7 cm dan t = 6 cm
LP = 2 π r (r + t)
= 2 × π × 7 (7 + 6) cm²
= 182 π
= 182 × ²²/₇ cm²
= 572 cm²
V = π r² t
= π × 7 × 7 × 6 cm³
= 294 π
= 294 × ²²/₇ cm³
= 924 cm³

 Baca Juga: Kunci Jawaban Matematika Kelas 9 Halaman 280-283 Latihan 5.1 Tabung, Lengkap Terbaru 2022

c. r = 4 cm dan t = 12 cm
LP = 2 π r (r + t)
= 2 × π × 4 (4 + 12) cm²
= 128 π
= 128 × 3,14 cm²
= 401,92 cm²
V = π r² t
= π × 4 × 4 × 12 cm³
= 192 π cm³
= 192 × 3,14 cm³
= 602,88 cm³

d. d = 2 m dan t = 8 m
LP = 2 π r (r + t)
= 2 × π × 1 (1 + 8) m²
= 18 π cm²
= 18 × 3,14 m²
= 56,25 m²
V = π r² t
= π × 1 × 1 × 8 m³
= 8 π cm³
= 25,12 m³

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e. d = 4 m dan t = 10 m
LP = 2 π r (r + t)
= 2 × π × 2 (2 + 10) m²
= 48 π m²
= 48 × 3,14 m²
= 150,72 m²
V = π r² t
= π × 2 × 2 × 10 m³
= 40 π m³
= 125,6 m³

f. d = 7 dm dan t = 20 dm
LP = 2 π r (r + t)
= 2 × π × ⁷/₂ (⁷/₂ + 20) dm²
= 7 π × ⁴⁷/₂ dm²
= ³²⁹/₂ π
= ³²⁹/₂ × ²²/₇ dm²
= 517 dm²
V = π r² t
= π × ⁷/₂ × ⁷/₂ × 20 dm³
= 245 π dm³
= 245 × ²²/₇ dm³
= 770 dm³

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