Algebra.Com's Answer #852546 by ikleyn(53475)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
.
\n" );
document.write( "A square is divided into three equal areas by two parallel lines drawn from opposite vertices.
\n" );
document.write( "Determine the  of the square in cm square if the distance between the two lines is 1 cm
\n" );
document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " Notice that I corrected/edited your post, to make \r
\n" );
document.write( "\n" );
document.write( " the problem's formulation correct from a Math point of view.\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( "\r\n" );
document.write( "Make a plot following my description.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Let ABCD be our square with the side length 'x', so A, B, C and D are its vertices.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The lines divide our square into three equal areas, so the area of each part is  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Draw the line AE from A to the side BC, so the intersection point E with BC\r\n" );
document.write( "divides side BC in proportion BE:CE = 2:1. \r\n" );
document.write( "In other words, BE =  , XE =  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Draw the line CF from the opposite vertex C to the side AD, so the intersection point F with AD\r\n" );
document.write( "divides side AD in proportion DF:AF = 2:1. \r\n" );
document.write( "In other words, DF = , AF = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So, now we have two right-angled triangles ABE and CDF of the area  each,\r\n" );
document.write( "and parallelogram AECF, whose area is also  , since it is the remaining area.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So, now we have exactly the configuration described in the problem.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "We can easy find the lengths of intervals AE and CF as hypotenuses of triangles ABE and CFD\r\n" );
document.write( "\r\n" );
document.write( " AE = CF =  =  =  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now the area of the parallelogram AECF is, from one hand side, ,\r\n" );
document.write( "and from other hand side it is the product of its base AE by the height, which is 1 cm.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So, we can write this equation for the area of parallelogram AECF\r\n" );
document.write( "\r\n" );
document.write( " =  .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Cancel common factors, and you will get\r\n" );
document.write( "\r\n" );
document.write( " x =  cm.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Hence, the area of the square ABCD is  = 13 cm^2. <<<---=== ANSWER\r\n" );
document.write( " \r
\n" );
document.write( "\n" );
document.write( "At this point, the problem is solved completely.\r
\n" );
document.write( "
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( " \n" );
document.write( " |