document.write( "Question 957148: a picture has a height that is 4/3 its width. it is to be enlarged so that the ratio of height to width remains the same, but the area is 192 in2. what are the dimensions of the enlargement? \n" ); document.write( "
Algebra.Com's Answer #585016 by macston(5194)\"\" \"About 
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W=width; L=height=(4/3)W; A=area =192 sq in
\n" ); document.write( "\"A=LW\" Substitute for L
\n" ); document.write( "\"A=%284%2F3%29%28W%29%28W%29\"
\n" ); document.write( "\"192in%5E2=%284%2F3%29W%5E2\" Multiply each side by 3/4.
\n" ); document.write( "\"144in%5E2=W%5E2\" Find the square root of each side.
\n" ); document.write( "\"12in=W\" ANSWER 1: The width will be 12 inches.
\n" ); document.write( "\"L=%284%2F3%29W=%284%2F3%2912in=16in\" ANSWER 2: The height will be 16 inches.
\n" ); document.write( "CHECK:
\n" ); document.write( "\"A=LW\"
\n" ); document.write( "\"192in%5E2=%2816in%29%2812in%29\"
\n" ); document.write( "\"192in%5E2=192in%5E2\"
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