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Using the cosine rule to find $\angle AOD$ we have $AC=CD=1$ and so $AD = \sqrt 2$ and hence

$$\cos \mathrm{\angle }AOD=\frac{O{A}^2+O{D}^2-A{D}^2}{2OA.OD}=\frac{1}{\sqrt{2}}.$$

Hence $\angle AOD = 45$ degrees.

The area of the triangle $AOD$ is

$$\frac{1}{2}OA\times OD\times \sin \mathrm{\angle }AOD=\frac{\sqrt{2}}{4}(2+\sqrt{2}).$$

To find the area of the minor segment $AD$ we subtract the area of triangle $AOD$ from the area of sector $AOD$ which gives

$$\frac{\pi }{8}(2+\sqrt{2})-\frac{\sqrt{2}}{4}(2+\sqrt{2}).$$

To get the total red shaded area we now add the area of triangle $ACD$ and multiply by $4$ which gives:

$$4[\frac{\pi }{8}(2+\sqrt{2})-\frac{\sqrt{2}}{4}(2+\sqrt{2})+\frac{1}{2}]=\frac{\pi }{2}(2+\sqrt{2})-2\sqrt{2}\approx 2.535\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}.$$