document.write( "Question 1182434: The length of rectangular solids is four times the width and the height is twice the width.
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Algebra.Com's Answer #812444 by ikleyn(52903)\"\" \"About 
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document.write( "Let W be the width. \r\n" );
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document.write( "Then the length is 4W and the height is 2W.\r\n" );
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document.write( "The total surface area is  2*(LW + LH + WH) = 2*( (4W)*W + (4W)*(2W) + W*(2W) ) = 2*(4W^2 + 8W^2 + 2W^2) = 28W^2.\r\n" );
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document.write( "Thus we have this equation to determine W\r\n" );
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document.write( "    28W^2 = 198 in^2.\r\n" );
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document.write( "It implies\r\n" );
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document.write( "    W^2 = 198/28 = \"49%2F14\" = \"7%2F2\".\r\n" );
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document.write( "    W            = \"sqrt%287%2F2%29\".\r\n" );
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document.write( "Thus the dimensions are:  the length is  L = \"4%2Asqrt%287%2F2%29\";  the width  W = \"sqrt%287%2F2%29\";  the height  H = \"2%2Asqrt%287%2F2%29\".\r\n" );
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document.write( "The 3D diagonal of the solid is  D^2 = L^2 + W^2 + H^2 = \"21%2A%287%2F2%29\" = \"147%2F2\";   D = \"sqrt%28147%2F2%29\" = \"7%2Asqrt%283%2F2%29\"  in^2.    ANSWER\r\n" );
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document.write( "The volume is  LWH = \"8%2A%287%2F2%29%2Asqrt%287%2F2%29\" = \"28%2Asqrt%287%2F2%29\"  in^3.      ANSWER\r\n" );
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