Lesson A square divided into three equal areas by two parallel lines

A square divided into three equal areas by two parallel lines

Problem 1

Make a plot following my description.


Let ABCD be our square with the side length 'x',  so A, B, C and D are its vertices.


The lines divide our square into three equal areas, so the area of each part is  .


Draw the line AE from A to the side BC, so the intersection point E with BC
divides side BC in proportion BE:CE = 2:1.  
In other words,  BE = ,  XE = .


Draw the line CF from the opposite vertex C to the side AD, so the intersection point F with AD
divides side AD in proportion DF:AF = 2:1.  
In other words,  DF = ,  AF = .


So, now we have two right-angled triangles ABE and CDF of the area  each,
and parallelogram AECF, whose area is also  , since it is the remaining area.


So, now we have exactly the configuration described in the problem.


We can easy find the lengths of intervals AE and CF as hypotenuses of triangles ABE and CFD

    AE = CF =  =  = .


Now the area of the parallelogram AECF  is, from one hand side, ,
and from other hand side it is the product of its base AE by the height, which is 1 cm.


So, we can write this equation for the area of parallelogram AECF

     = .


Cancel common factors, and you will get

    x =  cm.


Hence, the area of the square ABCD is   = 13 cm^2.    <<<---===  ANSWER