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How do you find the area of orange shaded region given the inner diameter of the green circle is $14\sqrt{2}$ units?

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closed as off-topic by José Carlos Santos, Travis, Jyrki Lahtonen, Peter Foreman, Shailesh Mar 10 at 0:07

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  • $\begingroup$ First of all, can you specify how the region is constructed? Second, even if you do know it (and so you should edit it into the question), what have you tried towards solving the problem? $\endgroup$ – Parcly Taxel Mar 9 at 19:06
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The inner diameter of the green circle is the diagonal of the square formed by the outer points of the orange region. This diagonal is $14\sqrt{2}$, so the sides of the square have length $14$. This is also the diameter of the smaller half-circles whose intersections form the orange region. If we add the area of $4$ of these half-circles, we get an area that is by $x$ greater than the area of the square, where $x$ is the desired area of the orange region. Therefore: $$x = 2\pi 7^2-14^2 = 2 \cdot 7^2 \cdot (\pi-2)$$

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$\textbf{Hint}$: Diameter of the orange circle will be $(7\sqrt{2})^2+(7\sqrt{2})^2=196$ or $14$ units.

For the elaborate answer:

$\textbf{Area of sector ABCD}$=$\dfrac{\theta}{2\pi}\times\pi r^2=\dfrac{49\pi} {2}\;sq.units$

$\textbf{Area of $\triangle$ ABD}$ = $49\;sq.units$.

$\textbf{Area of the region DBC}$ = $\dfrac{49(\pi-2)} {2}\;sq.units$.

Removing the left and right semicircle of the orange circle of radius $7 \;units$ from the inner green circle gives the area of part $I$ and $II$ along with two parts each of area $\dfrac{49(\pi-2)} {2}\;sq.units$.

Similarly removing the upper and lower semicircle of the orange circle of radius $7 \;units$ from the inner green circle gives the area of part $II$ and $IV$ along with two parts each of area $\dfrac{49(\pi-2)} {2}\;sq.units$.

$\begin{align}\textbf{Area of the four parts $\big($I+II+III+IV$\big)$} &=2\times \big(\pi((7\sqrt{2})^2-7^2)-49(\pi-2)\big)\\&=196\;sq.units\end{align}$.

Thus, $\textbf{Area of orange shaded region}$ = Area of inner green circle - Area of the four parts $I+II+III+IV$ = $98\pi-196=98(\pi-2)\;sq.units$.

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