Circle In A Semicircle In A Quarter Circle

Relevant wiki: Length and Area - Composite Figures

Let the radius of the semicircle be $r$

$\therefore AF=DF=FG=r$

$2r=14\sqrt{2}$

$r=7\sqrt{2}$

The area of the green region

$= \frac{\pi r^2}{2} - \pi(\frac{r}{2})^{2}$

$= \pi(\frac{r^2}{4})$

$= (\frac{22}{7}) (\frac{98}{4})$

$=77$

Any point on a circle connected to the extremities of a diameter of said circle forms a 90 $^{\circ}$ angle: which means EGD is 90 $^{\circ}$ ; which makes AEGD a square, having AG and ED as its equal diagonals. Which makes ED $14\sqrt{2}$ . So AD = AE = 14, using Pythagoras. In the semicircle, EF = FD = FG = $\frac{14\sqrt{2}}{2}$ . So the area of the full green semicircle, including the small circle is half of $\pi\left(\frac{14\sqrt{2}}{2}\right)^{\!\!2} = 49\pi$ . The area of the small circle is $\pi\left(\frac{14\sqrt{2}}{4}\right)^{\!\!2} = \frac{49}{2}\pi$ . So the Green area is $\frac{49}{2}\pi = \frac{49}{2}\frac{22}{7} = 77cm^{2}$