<PRE><B>Question 1152198</B>
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Let E and F be the midpoints of bases AB and CD respectively.  Then AE=x and DF=y; and segment EF is a diameter of the inscribed circle, so its length is 2r, where r is the radius of the inscribed circle.&lt;br&gt;
Let H be on DF such that AH is perpendicular to DF; then the length of AH is also 2r.&lt;br&gt;
Let G be the point of tangency of the inscribed circle with side AD of the trapezoid.&lt;br&gt;
By the theorem about the lengths of tangents to a circle, AG=AE=x and DG=DF=y.&lt;br&gt;
Finally, the length of DH is y-x.&lt;br&gt;
Now look at right triangle ADH.  It has legs of length (y-x) and 2r and hypotenuse of length (x+y).&lt;br&gt;
Use of the Pythagorean Theorem with those lengths will lead you to the desired result.&lt;br&gt;
I leave the calculations to you....&lt;br&gt;
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