2. (1) Solution: Move the two shadow parts inward to connect them with the outermost shadow part.
Then the area of the shadow part is less than half of the great circle times 1/4 the minimum circle area, that is, the area of the shadow part is:
1/2*3. 14*6*6- 1/4*3. 14*2*2=53.38.
(2) Solution: If the side length of a square is 20, then the diameter of the small circle is 20 and the radius is 10.
The square area is 20*20=400, and the circular area is 3.14 *10 *10 = 314.
The diagonal line connecting the upper left corner and the lower right corner, the blank part in the middle is divided into two halves, and the area solution refers to the solution of the first question.
For: 1 14*2=228。 The sum of the blank areas in the upper right corner and the lower left corner is equal to the square area minus half of the circular area, that is, (400-3 14)/2=43, so the blank area * * * is 228+43=27 1 and the shadow area is 400-27 1 = 65438.
3. Solution: the diameter of the great circle is 20, the radius is 10, the diameter of the small circle is 10, and the radius is 5. Four tangents (tangents, that is, points next to the small circle and the big circle) connecting the four small circles and the big circle form a square. The diagonal of this square is the diameter of the great circle, which is 20. A square can be divided into four isosceles right triangles with equal area, with an area of 200.
Then, like the shadow of the first question, the areas of the four parts around the square are10 *10 * 3.14-200 =14, and one area is11. The shadow area can be divided into three parts: the part where two small circles intersect, the part where two small circles intersect with the big circle, and the rest of the small circle. Move the shadow by I, and we can get a square and a figure like the shadow in the first question, so the shadow area is 5*5/2*4+28.5=78.5.
4 Solution: Let the diplomatic points of three circles be A, B and C, and the internal intersections be D, E and F, and connect the AB intersection, then connect the two internal intersections to make them parallel to AB, and then make the shadow part form a semicircle with the same radius as the small circle, which is 5CM, so the area of the shadow part is 1/2.
5. Solution: The triangle ABC is an isosceles right triangle, 12= 1/2*AB*BC, AB 2 = 24. The area of the small sector is 45/360 * ab * ab * 3.13 = 45/360 * 24 * 3.14 = 9.42, so the area of the blank part below is the area of the triangle minus the area of the sector:12-9.42 = 2. The area of shaded part is the area of semicircle minus the area of blank part.
Namely:1/2 * (1/2bc) * (1/2bc) * 3.14-2.58 =1/8 * 3.14 *.
6. Solution: The shadow area is equal to the area of the big triangle minus the area of the small triangle.
Let the radius of the big circle be a and the radius of the small circle be b, then 1/2 * A- 1/2 * B * B = 25, so a 2-b 2 = 50.
The area of the ring is the area of the big circle minus the area of the small circle, 3.14 * a-3.14 * b * b = 3.14 * (A2-B2) = 3.14 * 50 =/kloc.