The purple area is the sum of the circle sector and the triangle:
![A = A_s+A_t](https://img.qammunity.org/2020/formulas/mathematics/high-school/ybnt54jl5khjw92n6sj5yib4jltft8k3qw.png)
Let's compute them one at the time:
The sector is identified by an angle of 260°, because it is the remainder of a 100° angle.
We can build this simple proportion
![\text{total area}/\text{total angle}=\text{sector area}/\text{sector angle}](https://img.qammunity.org/2020/formulas/mathematics/high-school/t7pvz8k01gm0haaxuaty7h8njpuz7rpg04.png)
The area of the circle is
, so we have
![\pi 8.35^2/ 360= A_s/ 260](https://img.qammunity.org/2020/formulas/mathematics/high-school/1v6dz6fhz7he7rcqqxujsfyg437auuvrr8.png)
Solving for the sector area, we have
![A_s = (\pi\cdot8.35^2\cdot260)/(360) = (8.35^2\cdot 13\pi)/(18)](https://img.qammunity.org/2020/formulas/mathematics/high-school/97b420n86mg67gazuhqmh6asyn0u5grk96.png)
The triangle is an isosceles triangle, because two of the sides are radii. This means that the height is also a bisector, so we can cut the triangle in two 90-50-40 triangles.
Using the law of sines, we can deduce that the height is
![8.35\sin(40)](https://img.qammunity.org/2020/formulas/mathematics/high-school/tb7sxujaiaahpwhjuzlfi1i9gojfsdozil.png)
And half the base is
![8.35\sin(50)](https://img.qammunity.org/2020/formulas/mathematics/high-school/3824s50pgllhbrv3h8kvttwcnysgskbtfo.png)
So, the area of the triangle is
![A_t = 8.35^2\sin(40)\sin(50)](https://img.qammunity.org/2020/formulas/mathematics/high-school/yp8i7gd6tyogjksa4a1xjhtu2ue4bw89ha.png)
So, the purple area is
![A = A_s+A_t =\\(8.35^2\cdot 13\pi)/(18)+8.35^2\sin(40)\sin(50) = 8.35^2\left((13)/(18)\pi+\sin(40)\sin(50)\right)\approx 192.5](https://img.qammunity.org/2020/formulas/mathematics/high-school/9k77mvhsh2m6gih3css9bc1z2uiuwy4bj5.png)