Materi : Bangun Ruang Sisi Lengkung
t²/t¹ = r²/r¹
x/( x + 4 ) = 4/10
10x = 4( x + 4 )
10x = 4x + 16
10x - 4x = 16
6x = 16
x = 16/6 cm
x + 4 = 40/6 cm
V = ⅓πr²t - ⅓πr²t
V = ⅓.π.10².40/6 - ⅓.π.4².16/6
V = 4.000/18π - 256/18π
V = 3.744/18π
V = 208 × 3,14
V = 653,12 cm³
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
x/x + 4 = 4/10
4(x + 4) = 10(x)
4(x) + 4(4) + 10x
4x + 16 = 10x
(10 - 4)x = 16
x1 = 16/6
x2 = 40/6
MAKA :
V = 1/3(3,14(100(40/6) - (1/3(3,14(16(16/6)
= 1/3(3,14(100(40)/6) - (1/3(3,14(16(16)/6)
= 1/3(3,14(4.000/6) - (1/3(3,14(256/6)
= 1/3(3,14(666,6) - (1/3(3,14(42,6)
= 1/3(2.093,124) - 1/3(133,764)
= 2.0932,124/3 - 133,764/3
= 697,708 - 44,588
= 653,12 CM³
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Verified answer
Materi : Bangun Ruang Sisi Lengkung
Tinggi Kerucut yang terpotong
t²/t¹ = r²/r¹
x/( x + 4 ) = 4/10
10x = 4( x + 4 )
10x = 4x + 16
10x - 4x = 16
6x = 16
x = 16/6 cm
x + 4 = 40/6 cm
Volume Kerucut Terpancung
V = ⅓πr²t - ⅓πr²t
V = ⅓.π.10².40/6 - ⅓.π.4².16/6
V = 4.000/18π - 256/18π
V = 3.744/18π
V = 208 × 3,14
V = 653,12 cm³
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
x/x + 4 = 4/10
4(x + 4) = 10(x)
4(x) + 4(4) + 10x
4x + 16 = 10x
10x = 4x + 16
(10 - 4)x = 16
6x = 16
x1 = 16/6
x2 = 40/6
MAKA :
V = 1/3(3,14(100(40/6) - (1/3(3,14(16(16/6)
= 1/3(3,14(100(40)/6) - (1/3(3,14(16(16)/6)
= 1/3(3,14(4.000/6) - (1/3(3,14(256/6)
= 1/3(3,14(666,6) - (1/3(3,14(42,6)
= 1/3(2.093,124) - 1/3(133,764)
= 2.0932,124/3 - 133,764/3
= 697,708 - 44,588
= 653,12 CM³