Question 1195835
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Let "a" be the shorter base of the trapezoid;

let "b" be the longer base of the trapezoid.


The triangles of the areas 9 and 25 square units are similar, 
with the similarity coefficient of {{{3/5}}} (smaller to larger).


Let h be the height of the smaller triangle drawn to its base "a";

Let H be the height of the larger triangle drawn to its base "b".


Then from the similarity of triangles,  b = {{{(5/3)a}}};  H = {{{(5/3)h}}}.


We have then  {{{(a+b)/2}}} = {{{(a + (5/3)a)/2}}} = {{{(8/6)a}}} = {{{(4/3)a}}};

              h + H = {{{h + (5/3)h}}} = {{{(8/3)h}}}.


The area of the trapezoid is  

    Area = {{{((a+b)/2)*(h+H)}}} = {{{(4/3)a*(8/3)h}}} = {{{(32/9)*ah}}};

    but  {{{(ah)/2)}}} = the area of the upper triangle = 9;  hence  ah = 2*9;  therefore

    Area = {{{(32/9)*(2*9)}}} = 32*2 = 64 square units.


<U>ANSWER</U>.  The area of the trapezoid is 64 square units.
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Solved.