Question 1149735
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I will borrow the picture from the Edwin' solution to produce my own, much more simple.


<pre>
{{{drawing(400,700/3,-3,33,-3,18,
green(line(7.416198487,0,7.416198487,15.29705854),locate(7.47,7.6,h),

locate(7.7,1.3,E)), 
locate(2.4,8.7,17),locate(11,8.7,17), locate(19,8.7,27),
locate(7,0,x), locate(21,0,x),locate(7.4,16.6,A),
locate(0,0,B), locate(14.7,0,C), locate(29.7,0,D), 
triangle(0,0,7.416198487,15.29705854,2sqrt(55),0),
triangle(0,0,7.416198487,15.29705854,4sqrt(55),0) )}}}


The triangle ABC is isosceles, therefore, its altitude AE is the median, at the same time.


So, I introduce the variable y = x/2.


Then BE = EC = y;  CD = x = 2y;  ED = 3y;  BD = 4y.


Therefore,

    h^2 = 17^2 - y^2 = 27^2 - (3y)^2,   or

    17 - y^2 = 27^2 - 9y^2

    9y^2 - y^2 = 27^2 - 17^2

    8y^2       = (27-17)*(27+17) = 10*44 = 440

     y^2                                 = 440/8 = 55

     y                                   = {{{sqrt(55)}}}.


Then   h^2 = 17^2 - 55 = 234  and  h = {{{sqrt(234)}}} = {{{sqrt(9*26)}}} = {{{3*sqrt(26)}}}.


Now the area of the triangle ABD is 


     area = {{{(1/2)*BD*AE)}}} = {{{(1/2)*4y*h}}} = 2y*h = {{{2*sqrt(55)*3*sqrt(26)}}} = {{{6*sqrt(1430)}}} = 226.89  (approx.)
</pre>

Solved.