The plot is shown in Figure 1 on the right.
The point O is the center of the (semi) circle.
The diameter is AB and the equilateral triangle is ABC.   

The points D and E are the intersection points of the   
semicircle with the lateral sides of the triangle ABC.

The problem asks to find the area of a curved shape CDE.


Let us draw the radii OD and OE (Figure 2).

The triangle OEB is isosceles triangle (the radii OB and OD 
are congruent). It has the angle B equal to 60°.
It implies that the triangle OEB is equilateral.

Due to similar reasons, the triangle ODA is equilateral, too.   

Thus the triangles OEB and ODA are similar to the triangle 
ABC with the similarity coefficient .
 
  

 

              Figure 1. 

 
  

 

              Figure 2. 

  
  
Therefore, the areas of the triangles OEB and ODA are  of the area of the triangle ABC.

It means that the area of the quadrilateral CDOE (which is a rhombus, by the way) is half of the area of the triangle ABC.

Then the area of the curved shape CDE is the area of the rhombus CDOE minus the area of the sector ODE.

In other words,  =  = .