Draw a circle with the vertex a 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

Draw a circle with the vertex a 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

From the square, the side is root 24 cm,
The shadow area is 1 / 4 of the circle area
Shadow area = 1 / 4 π R ^ 2 = 1 / 4 × π ×√ 24 ^ 2 = 6 π = 18.84 square centimeter

The side length of the square is 8 cm. Take the vertex ABCD of the square as the center of the circle and draw arcs with the radius of 3 cm respectively to calculate the area of the shadow part

Shadow area = square area - area of four small arcs,
The area of each small arc is exactly 1 / 4 of a circle, so the sum of the areas of the four small arcs is exactly the area of a circle = π times the square of 3
That is, the shadow area = 8.8-π.3.3 = 64-28.26 = 35.74

If the side length of the square ABCD is greater than 4cm, draw a line along a 45 degree angle from 2cm away from the four vertices to divide the square into five parts, then the area of the shadow part in the middle The area of the middle quadrilateral

Let the side length of the square ABCD be a and a > 4;
The shadow in the middle is a square with A-2 √ 2,
So its area s = (A-2 √ 2) ^ 2

As shown in the figure, ABCD is a square with side length A. draw semicircles with AB, BC, CD and DA as diameters respectively, and calculate the area of shadow part surrounded by these four semicircle arcs

π（A
2）2×1
2×4-A2
= Pi
2A2-A2
=（π
2-1）A2；
So the answer is: (π
2-1）A2

The square ABCD is the largest square in a quarter circle with a radius of 10 cm. What is the area of the shadow part? I need a formula!

The shadow area should be 25 π - 50

In the parallelogram ABCD, AE = EF = FB. Ag = 2CG, the area of triangle GEF is 6 square centimeters, and what is the area of parallelogram?

Let the height of △ ABC with BC as the base edge is h, and the height of △ EFG with FG as its base is h; because AC: Ag = AB: AF = 3:2, so FG ∨ BC, so △ ABC ∷ AFG, FG: BC = 2:3, that is BC = 32fg, because AE = EF = FB, so h: H = 13, that is h = 3h, because the area of triangle EFG = FG × h ﹤ 2 = 6 (square centimeter) ᛿

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______ .

The diameter of AE is 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 the triangle ABC, ab = AC, angle a equals 120 degrees, point D is the midpoint of BC, and DF is perpendicular to F. what is the ratio of AB to AF

If ad = x, then AB = 2x, AF = x / 4, the ratio is 4 / 1. Draw the icon to show the degree of the angle, and you can see that there are three angles in the right triangle, 90, 30, 60

As shown in the figure, D is known to be a point on the extension line of BC side of triangle ABC, DF is perpendicular to AB and F, AC is at e, angle a is equal to 35 ° and angle D is equal to 42 ° to find the degree of angle ACD

∵DF⊥AB
∴∠AFE=90°
∵∠A=35°
∴∠AEF=55°
∴∠DEC=55°
∵∠D=42°
∴∠ACD=180°-55°-42°=83°

As shown in the figure, in the right triangle ABC, the angle BAC is equal to 90 degrees, ab = AC, point D is any point on BC, De is perpendicular to point E, DF is perpendicular to point F Point m is the midpoint of BC Try to determine the shape of the triangle MEF and explain your reasons

Connecting am
Judging the congruence of △ BFM and △ AEM
BM=AM
∠B=∠MAE=45°
BF=FD=AE
∴△BFM≌△AEM(SAS)
∴FM=ME
∠BMF=∠AME
The △ FME is an isosceles right triangle