Let $F$ be the equation $y=x$, let $G$ be the equation $y=125 x$, and let $H$ be the equation $y=\frac{1}{x^{2}}$.
Find $h$ so that the area of the region enclosed by the graphs of these equations is equal to that of an isosceles triangle of base 11 and height $h$.
The displayed graph of the region is not drawn to scale.
(Express numbers in exact form. Use symbolic notation and fractions where needed.)

Step 1 ：The area of the isosceles triangle is given by \(\frac{1}{2} \times 11 \times h\).

Step 2 ：The area between the curves $F$ and $G$ is given by the integral \(\int_{a}^{b} (G(x) - F(x)) dx\), where $a$ and $b$ are the x-coordinates where the curves intersect.

Step 3 ：Setting $F$ and $G$ equal to each other gives $x = 125x$, which has solutions $x = 0$ and $x = 1$. So, $a = 0$ and $b = 1$.

Step 4 ：Substituting $F(x) = x$ and $G(x) = 125x$ into the integral gives \(\int_{0}^{1} (125x - x) dx = \int_{0}^{1} 124x dx\).

Step 5 ：Evaluating this integral gives \(\frac{124}{2} x^2 \Big|_0^1 = 62\).

Step 6 ：The area between the curves $F$ and $H$ is given by the integral \(\int_{a}^{b} (F(x) - H(x)) dx\), where $a$ and $b$ are the x-coordinates where the curves intersect.

Step 7 ：Setting $F$ and $H$ equal to each other gives $x = \frac{1}{x^2}$, which has solutions $x = -1, 0, 1$. So, $a = 0$ and $b = 1$.

Step 8 ：Substituting $F(x) = x$ and $H(x) = \frac{1}{x^2}$ into the integral gives \(\int_{0}^{1} (x - \frac{1}{x^2}) dx\).

Step 9 ：Evaluating this integral gives \(\frac{1}{2} x^2 + \frac{1}{x} \Big|_0^1 = \frac{1}{2} + 1 = \frac{3}{2}\).

Step 10 ：The total area enclosed by the graphs of the equations is $62 + \frac{3}{2} = \frac{127}{2}$.

Step 11 ：Setting this equal to the area of the triangle gives \(\frac{1}{2} \times 11 \times h = \frac{127}{2}\).

Step 12 ：Solving for $h$ gives $h = \frac{127}{11} = \boxed{\frac{127}{11}}$.