<s>[INST] What is the area shaded in this figure? I was sent this problem by a friend, they say it is a year  6 question. Using trigonometry I got the shaded area as  $4 - \frac{\pi}{3}$.
Now Year six, I doubt if they do trigonometry. Is there another approach out there?

 [/INST] 
Label the vertices as shown in the diagram.
$\tan\alpha=\frac{1}{2}\implies \tan 2\alpha=\frac{4}{3}\implies 2\alpha=\tan^{-1}(\frac{4}{3})$
$\therefore$ Area of circular sector$ BAG=8\tan^{-1}(\frac{4}{3})$
Area of $\triangle AGC=16\cos\alpha\sin\alpha=8\sin2\alpha=\frac{32}{5}$
Thus, area of curved figure $BEG=\frac{48}{5}-8\tan^{-1}(\frac{4}{3})$
Therefore, Area of shaded part$=\frac{1}{2}$ (Area of $BEFC$)$-\frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $\frac{32}{5}-4\pi+8\tan^{-1}(\frac{4}{3})=\frac{32}{5}-4\tan^{-1}(\frac{3}{4})\approx1.252$
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