Area of shaded portion inside a circle.

Question

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

Answer 1

$\textbf{Hint}$: 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 $I+II+III+IV$ = $98\pi-196=98(\pi-2)\;sq.units$.

Answer 2

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: $$x = 2\pi 7^2-14^2 = 2 \cdot 7^2 \cdot (\pi-2)$$