$\angle \mathrm{A}_{1} \mathrm{OA}_{2}=\frac{2 \pi}{9}=40^{\circ}$
$\cos \frac{2 \pi}{9}=\frac{x^{2}+x^{2}-4}{2 x^{2}}$
$\left.x^{2} \cos 40^{\circ}\right)=x^{2}-2$
$x^{2}\left(1-\cos 40^{\circ}\right)=2$
$x=\frac{1}{\sin 20^{\circ}}$ .........(i)
Now $\cos \frac{8 \pi}{9}=\frac{x^{2}+x^{2}-\left(A_{1} A_{5}\right)^{2}}{2 x^{2}}$ (in $\left.\Delta A_{1} O A_{5}\right)$
$\begin{array}{l}
\left(A_{1} A_{5}\right)^{2}=2 x^{2}\left(1-\cos \frac{8 \pi}{9}\right) \\
=2 x^{2}\left(1-\cos 160^{\circ}\right) \\
=4 x^{2} \sin ^{2} 80^{\circ}
\end{array}$
$\mathrm{A}_{1} \mathrm{~A}_{5}=2 \mathrm{x} \sin 80^{\circ}$ .........(ii)
Similarly in $\triangle \mathrm{A}_{2} \mathrm{OA}_{4}$
$\mathrm{~A}_{2} \mathrm{~A}_{4}=2 \mathrm{x} \sin 40^{\circ}$ .........(iii)
(ii) $-$ (iii)
$\begin{aligned} \mathrm{A}_{1} \mathrm{~A}_{5}-\mathrm{A}_{2} \mathrm{~A}_{4} &=2 \mathrm{x}\left(\sin 80^{\circ}-\sin 40^{\circ}\right) \\ &=2(\text { using }(\mathrm{i})) \end{aligned}$