So the shadow area =1/2 *10 * 20-π (10/2)&; # 178; - 1.79= 100-25π- 1.79= 19.7 1
In this way, with auxiliary lines, the figures on both sides are approximately regarded as triangles, so the heights of the two triangles are basically 10/20? Therefore, the small triangle should account for 1/3 of the total area. It is considered that the base of the triangle is the radius of the circle, equal and the base is the same.