The value of $\sum\limits_{r = 1}^9 {{{\sin }^2}\dfrac{{\pi r}}{{18}}} \;is\;equal\;to\;$ 

$\begin{array}{l} \sum_{r=1}^{9} \sin ^{2} \frac{r \pi}{18} \\ =\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\ldots+\sin ^{2} \frac{8 \pi}{18}+\sin ^{2} \frac{9 \pi}{18} \\ \text { As } \frac{\pi}{18}+\frac{8 \pi}{18}=\frac{\pi}{2} \end{array}$

Similarly,

$\frac{2 \pi}{18}+\frac{7 \pi}{18}=\frac{\pi}{2}$

So,

$\begin{array}{l} \sum_{r=1}^{9} \sin ^{2} \frac{r \pi}{18} \\ =\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{8 \pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2} \frac{7 \pi}{18} \\ +\sin ^{2} \frac{3 \pi}{18}+\sin ^{2} \frac{6 \pi}{18}+\sin ^{2} \frac{4 \pi}{18}+\sin ^{2} \frac{5 \pi}{18} \\ +\sin ^{2} \frac{9 \pi}{18} \\ =\left(\sin ^{2} \frac{\pi}{18}+\cos ^{2} \frac{\pi}{18}\right)+\left(\sin ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{2 \pi}{18}\right) \\ +\left(\sin ^{2} \frac{3 \pi}{18}+\cos ^{2} \frac{3 \pi}{18}\right)+\left(\sin ^{2} \frac{4 \pi}{18}+\cos ^{2} \frac{4 \pi}{18}\right) \\ +\sin ^{2} \frac{\pi}{2} \\ =1+1+1+1+1=5 \end{array}$