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Diketahui dua kerucut sebangun. Jika r 1 ​ = 7 cm dan V 1 ​ = 616 cm 3 , serta r 2 ​ = 21 cm , tentukan: a. V 2 ​

Diketahui dua kerucut sebangun. Jika  dan , serta , tentukan:

a. 

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 straight V subscript 2 equals 16.632 space cm cubed.

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Diketahui dan . Maka Karena kedua kerucut sebangun, maka Dengan demikian, volume kerucut kedua sama dengan Jadi, .

Diketahui r subscript 1 equals 7 space cm dan straight V subscript 1 equals 616 space cm cubed. Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight V subscript 1 end cell equals cell 1 third straight pi r subscript 1 squared t subscript 1 end cell row cell 616 space cm cubed end cell equals cell 1 third straight pi times open parentheses 7 space cm close parentheses squared times t subscript 1 end cell row cell 1.848 space cm cubed end cell equals cell straight pi times open parentheses 49 space cm squared close parentheses times t subscript 1 end cell row cell fraction numerator up diagonal strike 1.848 end strike to the power of 264 space cm to the power of horizontal strike 3 end exponent over denominator up diagonal strike 49 to the power of 7 straight pi space horizontal strike cm squared end strike end fraction end cell equals cell t subscript 1 end cell row cell fraction numerator 264 over denominator 7 straight pi end fraction space cm end cell equals cell t subscript 1 end cell end table

Karena kedua kerucut sebangun, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell r subscript 1 over r subscript 2 end cell equals cell t subscript 1 over t subscript 2 end cell row cell fraction numerator up diagonal strike 7 to the power of 1 space horizontal strike cm over denominator up diagonal strike 21 cubed space horizontal strike cm end fraction end cell equals cell fraction numerator begin display style fraction numerator 264 over denominator 7 straight pi end fraction end style space cm over denominator t subscript 2 end fraction end cell row cell t subscript 2 end cell equals cell fraction numerator 792 over denominator 7 straight pi end fraction space cm end cell end table

Dengan demikian, volume kerucut kedua sama dengan

table attributes columnalign right center left columnspacing 0px end attributes row cell straight V subscript 2 end cell equals cell 1 third straight pi r subscript 2 squared t subscript 2 end cell row blank equals cell 1 over down diagonal strike 3 to the power of 1 horizontal strike straight pi times open parentheses 21 space cm close parentheses squared times fraction numerator down diagonal strike 792 to the power of 264 over denominator 7 horizontal strike straight pi end fraction space cm end cell row blank equals cell open parentheses 21 space cm close parentheses times open parentheses down diagonal strike 21 cubed space cm close parentheses times 264 over down diagonal strike 7 to the power of 1 space cm end cell row blank equals cell 16.632 space cm cubed end cell end table

Jadi, straight V subscript 2 equals 16.632 space cm cubed.

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