A geometry problem by Hana Wehbi

$Area \space of \space shaded \space region \space=\space Area \space of \space sector \space ADC \space-\space Area \space of\space minor \space sector \space in \space small \space circle$ $\space+\space Area \space of \space major \space sector\space in \space small \space circle$ $\space \space \space -\space \boxed{1}$

$Let \space \angle{ADC} \space=\space x°$

$Area \space of \space larger \space circle \space = \space \pi\space(2)^{2}\space =\space 4\pi$

$Area \space of \space smaller \space circle \space = \space \pi\space(1)^{2}\space =\space \pi$

$Area \space of \space sector \space ADC \space= \space \frac{x}{360}\times 4\pi$

$Area \space of \space minor \space sector \space in \space smaller \space circle \space= \space \frac{x}{360} \times \pi$

$Area \space of \space major \space sector \space in \space smaller \space circle \space= \space \frac{360-x}{360} \times \pi$

$Putting \space the \space values \space in \space eq \space 1$

$\rightarrow \space \frac{5}{12} \space\times \space4\pi \space= \space \frac{4\pi\space x}{360} \space-\space \frac{\pi\space x }{360} \space+\space \frac{(360-x)\space\pi}{360}$

$After \space solving \space, \space we \space will \space get \space x \space=\space 120°$

$\therefore \space \boxed{\angle{ADC} \space=\space 120°}$

The total area of the larger circle is $4\pi$ , thus the area of the shaded region must be $\frac{5}{12} of 4\pi = \frac{5}{3}\pi$ .

Suppose $\angle ADC =y$ ,

The area of the unshaded portion of the inner circle is thus $\frac{y}{360}$ of the total area of the

inner circle, or $\frac{y}{360}(\pi(1^2))= \frac{y}{360}\pi$ , since $\angle ADC is \frac{y}{360}$ of the largest possible central angle $360$ .

The area of the shaded portion of the inner circle is thus $\pi-\frac{y}{360}\pi= \frac{360-y}{360}\pi$ .

The total area of the outer ring is the difference of the areas of the outer and inner circles, or. $\pi(2^2)-\pi(1^2)=3\pi$ .

The shaded area in the outer ring will be $\frac{y}{360}$ of this total area, since $\angle ADC$ is $\frac{y}{360}$ of the largest possible central angle $360$ .

So the shaded area of the outer ring is $\frac{y}{360}(3\pi)$ = $\frac{3y}{360}\pi$ .

So the total shaded area (which must be equal to $\frac{5}{3}\pi$ , interms of $y$ is:

$\frac{3y}{360}\pi+\frac{360-y}{360}\pi= \frac{360+2x}{360}\pi$ .

Therefore $\frac{360+2y}{360}=\frac{5}{3}=\frac{600}{360}, \text {so} \ 360+2y=600 \text \ {or} \ y=120$ .

Thus $\angle ADC = 120$ .

Keep in mind all angles are in degrees.