perhatikan gambar disamping. Luas permukaan benda tersebut.. cm2 a. 292,02 b. 244,92 c. 235,5 d. 103,62

1/2×4×phi×r^2 +2×phi×r×t+phi×r×s
=phi×r (2×r+2×t+s)
=3,14×3 (2×3+2×10+5)
=9,42(6+20+5)
=9,42× 31
=292,02 (A)

Luas permukaan setengah bola
D = 6 cm
r = D/2 = 6/2 = 3 cm

L = 2.π.r²
L = 2 x 3,14 x 3²
L = 6,28 x 9
L = 56,52 cm²

Luas selimut tabung
r = 3 cm
t = 10 cm

L = 2.π.r.t
L = 2 x 3,14 x 3 x 10
L = 31,4 x 6
L = 188,4 cm²

Luas selimut kerucut
r = 3 cm
t = 4 cm
s² = r² + t²
s² = 3² + 4²
s² = 9 + 16
s² = 25
s = √25
s = 5 cm

L = π.r.s
L = 3,14 x 3 x 5
L = 3,14 x 15
L = 47,1 cm²

Luas permukaan seluruhnya
L = 56,52 cm² + 188,4 cm² + 47,1 cm²
L = 292,02 cm²

Jawaban A