document.write( "Question 269130: A 4-inch by 6-inch picture is enlarged for framing by tripling its dimensions. A 2-inch-wide border is then placed around each side of the enlarged picture, as shown. Thin metal framing is sold only in increments of one foot. What is the minimum number of linear feet of framing that must be purchased to go around the perimeter of the border? \n" ); document.write( "

Algebra.Com's Answer #197268 by mananth(16946) $\"\"$ $\"About$

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A 4-inch by 6-inch picture is enlarged for framing by tripling its dimensions. A 2-inch-wide border is then placed around each side of the enlarged picture, as shown. Thin metal framing is sold only in increments of one foot. What is the minimum number of linear feet of framing that must be purchased to go around the perimeter of the border?\r
\n" ); document.write( "\n" ); document.write( "width 4 inch tripled so width = 12
\n" ); document.write( "length 6 inch tripled = 18\r
\n" ); document.write( "\n" ); document.write( "border width 2inch\r
\n" ); document.write( "\n" ); document.write( "total length = 18+4= 22 inch
\n" ); document.write( "total width = 12 +4 =16 inch\r
\n" ); document.write( "\n" ); document.write( "perimeter = 2L + 2W\r
\n" ); document.write( "\n" ); document.write( "2*22+2*16\r
\n" ); document.write( "\n" ); document.write( "44+32= 76\r
\n" ); document.write( "\n" ); document.write( "so 7 running feet will be required.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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