Notice that the area of the shaded region is the area of the big circle minus the area of the triangle, then with the area of the small circle added.
\begin{gathered}
A_\text{region}=A_\text{big circle}-A_\text{triangle}+A_\text{small circle}
\end{gathered}
The radius of the big circle is r=8 mm. Now, we can use the formula for the area of a circle.
The area of the circle is about 201.06 square millimeters. Now we will find the area of the triangle and the area of the small circle. Let's analyze an equilateral triangle with three heights, the inscribed circle, and the circumscribed circle.
Notice that the heights of the equilateral triangle are also its medians. From the Centroid Theorem, the centroid C is two-thirds of the distance from each vertex to the midpoint of the opposite side. This tells us that AC is two times smaller than BC.
AC=1/2* BC=1/2* 8=4 mm
Therefore, the radius of the small circle is 4 mm. Now we can use the formula for the area of a circle.
The area of the small circle is about 50.27 square millimeters. Let's find the area of the triangle. To do this, first, let's find the height of the triangle, h=AB.
