Question 1111551
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<pre>
Draw the altitude to the base from the vertex.


Then you will get a right angled triangle with the hypotenuse of 16 cm long.


The length of one leg is  {{{(32-20)/2}}} = 6 cm.


Then the altitude is {{{sqrt(16^2 - 6^2)}}} = {{{sqrt(256-36)}}} = {{{sqrt(220)}}} = {{{2*sqrt(55)}}}.


The area of the trapezoid is  {{{2*sqrt(55)*((32+20)/2)}}} = {{{52*sqrt(55)}}} cm^2.
</pre>

Solved.


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<U>Be aware</U>:  &nbsp;The solution and the answer by the @josgarithmetic are &nbsp;&nbsp;<U>W R O N G</U>.