jmc

First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[ABC]\cdot \left(1-{\left(\frac{11}{h}\right)}^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[ABC]\cdot {\left(\frac{h-5}{h}\right)}^2$. These areas are equal, so $1-{\left(\frac{11}{h}\right)}^2={\left(\frac{h-5}{h}\right)}^2$. Simplifying yields $10h=146$ so $h=\boxed{14.6}$.