In a square ABCD with side 14cm, 2 quadrants were made with centres A & B respectively and AB as radius. Find area of region I and II https://photos.app.goo.gl/5ibXDN5u6s0yo6KB3
I could do the following:
II + III = $\frac{1}{4}$ × $π$ × $14^2$ = $49π$ = $154cm^2$
II + IV = $\frac{1}{4}$ × $π$ × $14^2$ = $49π = 154 cm^2$
Area of square = $I + II + III + IV$ = $14^2 = 196 cm^2$
$(I + II + III +।V ) - ((II + III) + (II + IV)) = 196 - (154 + 154) = -112$ 
$\mapsto$ I - II = -112 
$\mapsto$ II - I = 112
Am I going on the correct path?
 A: Using help from the answer of Landuros, I figured out the answer.
$II + III$ = $\frac{1}{4}$ × $π$ × $14^2$ = $49π$ = $154cm^2$
$II + IV$ = $\frac{1}{4}$ × $π$ × $14^2$ = $49π = 154 cm^2$
Area of square = $I + II + III + IV$ = $14^2 = 196 cm^2$
$(I + II + III +।V ) - ((II + III) + (II + IV)) = 196 - (154 + 154) = -112$ 
$\mapsto$ I - II = -112 
$\mapsto$ II - I = 112
As seen from Landuros' answer, triangle EAB is equilateral.
Therefore, $\angle$ EBA = 60°.
Area of sector EBA = $\frac{60}{360} \times \pi 14^2$
                                   = $102.6 cm^2$
Area of $\triangle EAB$ = $ \frac{\sqrt 3}{4} \times 14^2$
                                           = $84.9 cm^2$
Area of each segment = $102.6 - 84.9$
                                        = $17.7$
Area of $II$ = $84.9 + 17.7 + 17.7$ 
                      = $120.3 cm^2$
We know that, $II - I = 112$
                          $\mapsto I = 8.3 cm^2$
A: HINT: The segment $II$ contains an equilateral triangle. The rest I think you can figure out. 
I think you can use the relation given in this question to solve your problem.
A: 
Start by constructing $\overline {AE}$ and $\overline {EB}$. $E$ is the intersection of the arcs created by $ABD$ and $BAC$.
Notice that $\overline{AE} = \overline{EB} = \overline{AB} = 14$ because they are all radii of the circle. The points $A$, $B$ and $E$ form an equilateral triangle $\triangle ABE$. Hence, we know $\angle EAB = \angle ABE = \angle BEA = \frac{\pi}{3}$rad. You will need to be confident with radians rather than degrees, as we are working with circular measure.
\begin{align}
A_{\text{segment}} & = \frac{1}{2}r^2(\theta-\sin \theta) \\
\text{Segment} ~ EB & = \frac {1}{2} \cdot 14^2 ( \frac{\pi}{3}-\sin {\frac{\pi}{3}})\\
& = 98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2})
\end{align}
Segment $EB$ is congruent to segment $AE$, so both are $98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2})$ cm$^2$.
The area of II can be found by the following formula:
\begin{align}
\text{II} & = \text{Quadrant $ABC$} - \text{Sector $EBC$} + \text{Segment $EB$} \\
& = \frac{1}{4}\pi r^2 - \frac{1}{2}r^2 \theta + \text{Segment $EB$}\\
& = \frac{1}{4} \pi \cdot 14^2 - \frac{1}{2} \cdot 14^2 \cdot \frac{\pi}{6} + 98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2}) \\
& = 49\pi - \frac{49}{3}\pi + 98\cdot(\frac{\pi}{3}-\frac{\sqrt3}{2}) \\
& = 98(\frac{\pi}{3}-\frac{\sqrt3}{2})+\frac{98}{3}\pi \\
& = 98(\frac{2}{3}\pi-\frac{\sqrt3}{2}) \\
& \approx 120.380
\end{align}
As you see, it's not a very nice answer, but it is an exact answer for the area of II.
As for the area of I:
\begin{align}
\text{I} & = ABCD - \text{Quadrant $ABD$} - \text{Quadrant $CBA$} + \text{II} \\
& = 14^2 - 49\pi - 49\pi + 98(\frac{\pi}{3}-\frac{\sqrt3}{2})+\frac{98}{3}\pi \\
& = 98(\frac{\pi}{3}-\frac{\sqrt3}{2})-\frac{196}{3}\pi + 196\\
& \approx 8.504
\end{align}
So to summarise, $\text{I} = 8.504 ~\text{cm}^2$ and $\text{II} = 120.380 ~\text{cm}^2$ which sound about right.
