Lesson A trapezoid divided in four triangles by its diagonals

A trapezoid divided in four triangles by its diagonals

Problem 1

Triangles AMD and BMC are similar (since their three angles are congruent, in pairs).

The similarity coefficient is  k = 5/3, from the greater triangle to the smaller.

Lat  a = AM,  b = BM,  c = CM  and  D = DM.


Then  b = ;  c = ,  due to similarity.


        For any triangle with the side p and q and the concluded angle , 
             the area of this triangle is  .


So, for the area of triangle AMD we have   =  = 315 cm^2.


Here M is the angle between sides a = AM and d = DM .


Next,  =  = ,

                  where N is the supplementary angle to M.


Since sin(N) = sin(M), we can re-group the formula above and to get

       = . = *area(AMD) =  =  =  25*35 = 875.



          This calculation re-tells us a well known fact that the areas 
    of similar triangles do relate as the square of the similarity coefficient.



Now let's consider the most interesting things - the areas of triangles adjacent to lateral sides.


       =  =  =  =  = 5*105 = 525,

       =  =  =  =  = 5*105 = 525.


Thus the total area of the trapezoid ABCD is 315 + 875 + 525 + 525 = 2240 cm^2.