Difference between revisions of "1989 AHSME Problems/Problem 3"

(Created page with "Let <math>x</math> be the width of a rectangle so that the side of the square is <math>3x</math>. Since <math>2(3x)+2(x)=24</math> we have <math>x=3</math>. Thus the area of the ...")
 
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Let <math>x</math> be the width of a rectangle so that the side of the square is <math>3x</math>. Since <math>2(3x)+2(x)=24</math> we have <math>x=3</math>. Thus the area of the square is <math>(3\cdot3)^2=81</math>.
 
Let <math>x</math> be the width of a rectangle so that the side of the square is <math>3x</math>. Since <math>2(3x)+2(x)=24</math> we have <math>x=3</math>. Thus the area of the square is <math>(3\cdot3)^2=81</math>.
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Revision as of 12:47, 5 July 2013

Let $x$ be the width of a rectangle so that the side of the square is $3x$. Since $2(3x)+2(x)=24$ we have $x=3$. Thus the area of the square is $(3\cdot3)^2=81$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png