document.write( "Question 1030634: Benjamin visited an art show and saw a painting he really wanted to purchase but it was very expensive! In the gift shop, he found that he could by post cards of the paintings at a much more affordable price. The post card was 4 inches by 6 inches and it said that it was 2/7 the size of the original. What are the dimensions of the original painting? \n" ); document.write( "
Algebra.Com's Answer #645474 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Size is in reference to area, not distances.\r
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\n" ); document.write( "\n" ); document.write( "4 by 6 is area \"4%2A6=24%2Ainch%5E2\".\r
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\n" ); document.write( "\n" ); document.write( "\"4%2A6=%282%2F7%29A\" using original area A.\r
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\n" ); document.write( "\n" ); document.write( "\"A=%284%2A6%29%287%2F2%29\"
\n" ); document.write( "\"A=2%2A6%2A7\"
\n" ); document.write( "\"A=84\"\r
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\n" ); document.write( "\n" ); document.write( "Some linear factor k will change 4 by 6 into the area 84.
\n" ); document.write( "\"4k%2A6k=84\"
\n" ); document.write( "\"4%2A6%2Ak%5E2=84\"
\n" ); document.write( "\"k%5E2=84%2F%284%2A6%29\"
\n" ); document.write( "\"k%5E2=%287%2A2%2A6%29%2F%284%2A6%29\"
\n" ); document.write( "\"k%5E2=%287%2A2%29%2F4\"
\n" ); document.write( "\"k%5E2=7%2F2\"
\n" ); document.write( "\"k=sqrt%287%2F2%29\"
\n" ); document.write( "\"k=sqrt%287%29sqrt%282%29%2F2\"
\n" ); document.write( "\"k=%281%2F2%29sqrt%2814%29\"\r
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\n" ); document.write( "\n" ); document.write( "Original dimensions were \"highlight%284%281%2F2%29sqrt%2814%29=2sqrt%2814%29%29\"
\n" ); document.write( "and
\n" ); document.write( "\"highlight%286%281%2F2%29sqrt%2814%29=3sqrt%2814%29%29\".
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