It is given that, $z_i\mathit{=}\left(x_i\mathit{-}\overline{x}\right)\left(y_i\mathit{-}\overline{y}\right)$

And $\sum _{i=1}^{10}{z}_i=c$

Rewrite the above equation.

$\sum _{i=1}^{10}\left\{\left(x_i-\overline{x}\right)\left(y_i-\overline{y}\right)\right\}=c$

$\Rightarrow \sum _{i=1}^{10}\left\{x_i{y}_i-x_i\overline{y}-\overline{x}{y}_i+\overline{x}·\overline{y}\right\}=c$

$\Rightarrow \underset{i=1}{\overset{10}{ \sum }}{x}_i{y}_i-\sum _{i=1}^{10}{x}_i\overline{y}-\sum _{i=1}^{10}\overline{x}{y}_i+\sum _{i=1}^{10}\overline{x}·\overline{y}=c$

$\Rightarrow \sum _{i=1}^{10}{x}_i{y}_i-10\overline{x}·\overline{y}-10\overline{x}·\overline{y}+10\overline{x}·\overline{y}=c$

$\Rightarrow \sum _{i=1}^{10}{x}_i{y}_i-10\overline{x}·\overline{y}=c$

Now,

${\sigma }_{\left(x_1-y_1\right)}=\sqrt{\frac{1}{10}\sum {\left(x_i-y_i\right)}^2-\left(\overline{x}-\overline{y}\right){}^2}$

$=\sqrt{\left(\frac{1}{10}\sum x_i^2+\frac{1}{10}\sum y_i^2-\frac{1}{5}\sum x_i{y}_i-{\overline{x}}^2-{\overline{y}}^2+2\overline{xy}\right)}$

$=\sqrt{\left(\frac{1}{10}\sum x_i^2-{\overline{x}}^2\right)+\left(\frac{1}{10}\sum y_i^2-{\overline{y}}^2\right)-\frac{1}{5}\left(\sum x_i{y}_i-10\overline{xy}\right)}$

$=\sqrt{a^2+b^2-\frac{c}{5}}$