Determine the Difference

1.Connect AB to create sector

1. Area of sector ABB'= 26/360 pi r^2, = 3.62

2. Notice triangle EAB

3. Drop perpendicular from A to EB, call this point M

4. angle(B'EB)=1/2 angle(B'AB), angle(B'EB)=13

5. Length of AM= sin(13)=x/4, = 3.44

6. Combine areas of triangle AEB and sector B'AB, = 7.06

7. Area of Yellow area= $\frac{pi*r^2}{2}$ - 7.06 ,= 18.06

8. Diff of areas= 18.06-7.06, approximately 11.

We can find the area of both by easily: $\frac{\pi \cdot 16}{2} \approx 25.1327$

If $BAB' = 26^{\circ}$ , $FAE = 64^{\circ}$ . This means $EAB = 154^{\circ}$ . This gives us an isosceles triangle, and we can find $\overline{EB}$ by using the law of sines: $\overline{EB} = \frac{4 \cdot sin(154)}{sin(13)} \approx 7.7950$ . Now, you can use Heron's formula to find the area of the isosceles to be $\approx 3.5070$ . To find the rest of the red area, we simply find the area of the portion of the circle: $\frac{16 \pi 26}{360} \approx 3.6303$ . We add the two areas together to get a total red area of $\approx 7.1373$ . Now, the yellow area equal the half of the circle minus the red, so to find the difference between the two is simple: $25.1327 - \left ( 2 \cdot 7.1373 \right ) \approx \boxed{10.86}$ .

Join AB to form a sector.

Area of sector BAB' = $\pi*r^{2}*\frac{26}{360}$ = 3.63

Area of $\triangle$ EAB = $\frac{1}{2}*r^{2}*sin(180-26)$ = 3.507

Total red area = 3.63 + 3.507 = 7.137

Area of semicircle EBB' = $8\pi$ = 25.133

Yellow area = 25.133 - 7.137 = 17.996

Yellow area - Red area = 10.86

$\angle \ EAB=180-26=154^o=\dfrac \pi {180}154=2.6878^c. \\ Yellow\ segment \ area =\dfrac{4^2} 2*\{2.6878 - Sin(2.6878)\}=17.99548 cm^2. \\ Yellow\ +\ Red\ area=area\ of\ semicircle=8\pi.\\ \therefore\ Red\ area\ =8\pi- 17.99548.\\ Diffrance\ in\ area\ =\ 17.99548 -(8\pi\ - 17.99548)=2*17.99548 - 8\pi=10.8592 cm^2.$