As shown in the figure, a square ABCD with a side length a is connected inside the circle, and semicircles are made to the outside of the square with the diameter of each side of the square. Then, the area of the four crescent shaped by the four semicircles and the four arcs of the circumscribed circle of the square is______ .

As shown in the figure, a square ABCD with a side length a is connected inside the circle, and semicircles are made to the outside of the square with the diameter of each side of the square. Then, the area of the four crescent shaped by the four semicircles and the four arcs of the circumscribed circle of the square is______ .

According to the above analysis, the areas of the four crescent shapes are as follows:
4×1
2×π×(a
2)2+a2−1
2a2π，
=1
2a2π+a2-1
2a2π，
=a2．
So the answer is: A2

Draw four quarter circles in the shape with the radius of the four sides of the square, and calculate the area of the shadow part

Let the side length of a square be a, draw a square, point a on the top left, point B at the bottom left, mark C and D anticlockwise, then the area of an olive shape will be (1 / 2) π a ^ 2-A ^ 2 (the area of two quarter circles is more than the square). If the area is surrounded by four small arcs in the middle, let x, x + 2Y = the area of olive

As shown in the figure, the side length of the square ABCD is 4cm. Draw a circle in the square with B and D as the center and 4cm as the radius to calculate the area of the shadow part

One
4. Area of circle: 3.14 × 42 × 1
4 = 12.56 (square centimeter);
Area of square: 4 × 4 = 16 (square centimeter);
Shadow area: 12.56 × 2-16 = 9.12 (square centimeter)
A: the shadow area is 9.12 square centimeters

As shown in the figure, the side length of the square ABCD is 4cm. Draw a circle in the square with B and D as the center and 4cm as the radius to calculate the area of the shadow part

One
4. Area of circle: 3.14 × 42 × 1
4 = 12.56 (square centimeter);
Area of square: 4 × 4 = 16 (square centimeter);
Shadow area: 12.56 × 2-16 = 9.12 (square centimeter)
A: the shadow area is 9.12 square centimeters

Draw a circle with the vertex of the square ABCD as the center and the side length as the radius. The area of the square is known to be 24 square centimeters. Calculate the area of the shadow part Help

The area of a quarter circle
3.14*24/4=18.84
Remaining area
24-18.84=5.12

As shown in the figure, the side length of the square ABCD is 4cm. Draw a circle in the square with B and D as the center and 4cm as the radius to calculate the area of the shadow part

One
4. Area of circle: 3.14 × 42 × 1
4 = 12.56 (square centimeter);
Area of square: 4 × 4 = 16 (square centimeter);
Shadow area: 12.56 × 2-16 = 9.12 (square centimeter)
A: the shadow area is 9.12 square centimeters

Circle O is the inscribed circle of square ABCD and triangle EFG, and the side length of square is 2. Find the area of square GEF

The radius of the inscribed circle is 1, right? Then the radius from the center of the circle to the top of the regular triangle is also 1. Then the two vertices and the center of the triangle form an isosceles triangle. You can go over the center of the circle to the opposite side, and the three lines are in one, which is also the center line

As shown in the figure, △ ABC is inscribed in ⊙ o, ad ⊥ BC is at point D, ad = 2cm, ab = 4cm, AC = 3cm, then the diameter of ⊙ o is______ .

Make the diameter AE of ⊙ o, connect CE, as shown in the figure,
∵ AE is the diameter,
∴∠ACE=90°，
And ∵ e = ∠ B,
∴Rt△AEC∽Rt△ABD，
∴AE
AB=AC
AD，
Ad = 2cm, ab = 4cm, AC = 3cm,
∴AE=AB•AC
AD=3
2×4cm=6cm．
So the diameter of ⊙ o is 6cm
So the answer is: 6cm

As shown in the figure, in triangle ABC, AB is equal to AC, circle O with diameter AB intersects with D at edge BC, intersects with e at edge AC, and makes DF perpendicular to AC and F through point D If de equals five under the radical and ab equals five fifths, find the length of AE

In RT △ DGB and RT △ OGB, BD ⊥ DG ⊥ BD, ed ⊥ BD,

As shown in the figure, in the triangle ABC, AB equals AC, the circle O with ab diameter intersects AC at point E, intersects BC at point D, connects be, ad intersects point P. verification: 1. D is the midpoint of BC; 2. Triangle BEC is similar to triangle adc3. AB times CE equals 2dp multiplied by AD Mei A kind of Mei A kind of Mei A kind of

According to the secant theorem, AP * ad = AE * AC, ① AB = AC, ᙨ PBD = AC, ᙨ PBD =  CAD =  CAD, ⊙ PBD,  Pd / BD = BD / AD, ᚠ BD ^ 2 = ad * PD. From ①, AB * ce = AC * ce = AC ^ 2-ae * AC = AB ^ 2-AP * ad = ad ^ 2-AP * ad = ad ^ 2-ad = ad * PD, AB * ce = AC * ce = AC ^ 2-ae * AC = AB ^ 2-AP * ad = ad ^ 2 + BD ^ 2-AP * ad = 2A...It's a good idea