document.write( "Question 1198554: The area of a base of a cone is two-thirds the area of a base of a cylinder and their volumes are equal. Find the ratio of their altitudes. \n" ); document.write( "
Algebra.Com's Answer #832341 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "VL = volume of cylinder
\n" ); document.write( "VC = volume of cone\r
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\n" ); document.write( "\n" ); document.write( "VC = VL since we're told these particular cone and cylinder have equal volume.\r
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\n" ); document.write( "\n" ); document.write( "HL = height of the cylinder
\n" ); document.write( "HC = height of the cone\r
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\n" ); document.write( "\n" ); document.write( "AL = area of the cylinder circular base
\n" ); document.write( "AC = area of the cone circular base
\n" ); document.write( "AC = (2/3)*AL\r
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\n" ); document.write( "\n" ); document.write( "VL = (area of base)*(height)
\n" ); document.write( "VL = (AL)*(HL)\r
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\n" ); document.write( "\n" ); document.write( "VC = (1/3)*(area of base)*(height)
\n" ); document.write( "VC = (1/3)*(AC)*(HC)
\n" ); document.write( "VC = (1/3)*((2/3)*AL)*(HC)
\n" ); document.write( "VL = (2/9)*AL*HC
\n" ); document.write( "(AL)*(HL) = (2/9)*AL*HC
\n" ); document.write( "HL = (2/9)*HC
\n" ); document.write( "HL/HC = 2/9
\n" ); document.write( "HC/HL = 9/2\r
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\n" ); document.write( "\n" ); document.write( "The ratio of the cone's height (HC) over the cylinder's height (HL) is 9/2.
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