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Math problems in the sixth grade math circle
Firstly, it is proved that the area of (1) in the graph is equal to the area of (2).

As an auxiliary line, the great circle is divided into 1/4 sectors, and the area of the great circle is π r 2? =? 225π, so the area of the big sector is 56.25π. The big sector contains two small semicircles, and the radius of the small semicircle is 7.5, so the sum of the areas of the two small semicircles is equal to 56.25π. Since the area of the big sector is equal to the sum of the areas of the two small semicircles, the areas of (1) and (2) are equal.

Secondly, calculate the area of (1).

As an auxiliary line, draw a square inside the big fan, with a side length of 7.5 and an area of 56.25. And because a square consists of two overlapping small sectors, the area of the small sectors is1/4× π r 2? = 14.0625π, and the sum of the areas of the two small sectors is 28. 125π, so the area of the (1) part is (28. 125π-56.25).

So far, we have calculated the shadow area of a large sector. So the total shadow area is 225π-8? ×? (28. 125π- 56.25)? =? 450