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As shown in the figure, ∵ s shadow = S1 + S2 + S3 + S4,
The area of the four semicircles is (S1 + S2 + S5) + (S2 + S6 + S3) + (S3 + S7 + S4) + (S1 + S8 + S4) = (S1 + S2 + S3 + S4) + (S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8),
The area of the square is s Square = S1 + S2 + S3 + S4 + S5 + S6 + S7 + S8,
/ / s shadow = the area of four semicircles - the area of a square,
=2×π×（a
2）2-a2
=1
2π a2−a2．
So the answer is: 1
2π a2−a2．

The side length of the square is a, and the diameter of each side is taken as the diameter. Draw a semicircle inside the square and calculate the area of the shadow part in the figure Why is the shadow area in the graph equal to the area of two circles minus the area of the square I can't think of it

If you think about it, half circles are drawn on the four sides of a square. The four semicircles are two circles. The shadow part is exactly where the four semicircles overlap. Since the overlap is the area that the two circles overlap, isn't subtracting the square shape the more shadow area?

As shown in the figure, the side length of the square is a. take each side as the diameter, make a semicircle in the square, and calculate the area of the shadow part in the figure

S shadow = 1
2π×（a
2）2=1
8πa2．

As shown in the figure, the side length of square ABCD is 4, Mn ∥ BC intersects AB respectively, CD is at points m and N, and any two points P and Q are taken at Mn, then the area of shadow part in the graph is______ .

According to the meaning of the title, the area of the shadow part is half of the area of the square. Because the side length of the square is 4, the area of the square is 16, so the area of the shadow part is 8
So the answer is 8

As shown in Fig. 4-7, the side length of square ABCD is 4cm. Arc BD is drawn with point a as the center and ab as the radius. Semicircles are drawn with BC and CD as diameters to calculate the shadow area If I could paint, I would have done it.

Connect BP and make the vertical line of BC through point E, and the vertical foot is f
It's easy to know △ bef ∷ BDC
So be / BD = EF / DC, where DC = be = 2cm, BD = √ (2? 2 + 2?) = 2 √ 2
So EF = √ 2
S△BEC=BC*EF/2=2*√2/2=√2
S △ BEC = s △ BPE + s △ BPC = be * pN / 2 + BC * PM / 2, where BC = be = 2
So there is be * pN / 2 + BC * PM / 2 = √ 2
So PM + PN = √ 2

As shown in the figure, the side length of square ABCD is 4, Mn ∥ BC intersects AB respectively, CD is at points m and N, and any two points P and Q are taken at Mn, then the area of shadow part in the graph is______ .

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If O is a point in the square ABCD and the distance from each vertex is equal to 3, then o is the intersection point of the diagonal
So the length of the diagonal is: 3 * 2 = 6
Area of square ABCD = diagonal * diagonal / 2 = 36 / 2 = 18

As shown in the figure, given that the side length of square ABCD and square CEFG are 8 cm and 6 cm respectively, then the area of shadow part is______ Square centimeter

（8×8+6×6）-（8+6）×8÷2-6×6÷2-（8-6）×8÷2，
=（64+36）-14×8÷2-18-2×8÷2，
=100-56-18-8，
=44-18-8，
=26-8，
=18 (square centimeter);
Answer: the area of shadow part is 18 square centimeter
So the answer is: 18

As shown in the figure, the side length of the square is 8cm. Take the four vertices of the square as the center of the circle, and take 4 as the radius, draw four circles?

Subtract the area of a circle from the area of a square
8*8-3.14*4*4=64-50.24=13.76
I don't know if the shadow part you said is the middle part of the square? If it is, that's it

As shown in the figure, the side length of square ABCD is 8 cm, and M is the midpoint of ad side

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