A midsegmnet parallel to the the base will create two similar triangles, as shown in the figure. We know that in similar triangles the ratios of the lengths of their corresponding sides are equal; therefore, we can establish a ratio between the longer and the shorter sides of our triangles and solve for x as follows:
[tex] \frac{11+121}{11} = \frac{10+5x+10}{10} [/tex]
[tex] \frac{132}{11} = \frac{5x+20}{10} [/tex]
[tex](10)(132)=(11)(5x+20)[/tex]
[tex]1320=55x+220[/tex]
[tex]55x=1100[/tex]
[tex]x= \frac{1100}{55} [/tex]
[tex]x=20[/tex]
The answer is x=20.
