It's summing up

$\begin{aligned} \displaystyle \sum_{r=1}^5 {4\cos^2{\left( \frac{r\pi}{11}\right)}} & = 2 \sum_{r=1}^5 {\left[ \cos{\left( \frac{2r\pi}{11}\right)}+1\right]} = 2 \sum_{r=1}^5 {\cos{\left( \frac{2r\pi}{11}\right)}} + 10 \end{aligned}$

Now we note that $\displaystyle \sum_{r=0}^{10} {\cos{\left( \frac{2r\pi}{11}\right)}}$ is the real part of the $11^{th}$ roots or unity, $z^{11} = 1$ . And because of symmetry, we have:

$\displaystyle \sum_{r=0}^{10} {\cos{\left( \frac{2r\pi}{11}\right)}} = 0\quad \Rightarrow \sum_{\color{#D61F06} {r=1}}^{10} {\cos{\left( \frac{2r\pi}{11}\right)}} = -1 \\ \Rightarrow \displaystyle \sum_{\color{#D61F06}{r=1}}^{\color{#D61F06}{5}} {\cos{\left( \frac{2r\pi}{11}\right)}} = - \frac{1}{2}$

Therefore,

$\displaystyle \sum_{r=1}^5 {4\cos^2{\left( \frac{r\pi}{11}\right)}} = 2 \sum_{r=1}^5 {\cos{\left( \frac{2r\pi}{11}\right)}} + 10 = 2\left(-\frac{1}{2} \right) + 10 = \boxed{9}$