The ratio of the top to the bottom of a trapezoid is 3:5. The top is 1 meter shorter than the bottom, and the bottom is 25% higher. The area of the trapezoid is ()

How to turn a rectangle 44 meters long and 28 meters wide into an equal small square without remaining? The maximum length of each side is (4) meters, and the total length is (77)?

Given the three vertices a (1,2), B (3,4) and C (2,6) of the parallelogram ABCD, the point direction equation of the line where the four edges are located is solved

The vector AB = (2,2), AB: (x-1) / 2 - (Y-2) / 2 = 0, DC: (X-2) / 2 - (y-6) / 2 = 0
Vector BC = (- 1,2), BC: (X-2) / (- 1) - (y-6) / 2 = 0, ad: (x-1) / (- 1) - (Y-2) / 2 = 0

Given the vertex a (- 2,2), B (- 1,4) C (4,5) of ladder ABCD and ab = 1 / 2dc, the coordinates of vertex D are obtained

Let D (x, y), then AB = (1,2),
DC=(4-x,5-y),
∵ DC=2AB,
∴ 4-x=2,
5-y=4,
∴ x=2,
y=1,
D (2,1)

As shown in the figure, in the rectangular trapezoid ABCD, ad is parallel to BC, vertex D.C moves on am and BN respectively (point d does not coincide with point a, point C does not coincide with B), e is the midpoint on the edge of AB, and De is always vertical during the movement
Verification: AD + BC = CD
De and CE bisect angle ADC and BCD respectively

Make the median line MH through the midpoint h of DC, AD + BC = 2mh
And because h is a right angle △ Dec, the midpoint eh = 1 / 2dc
So CD = AD + BC
EH = HC, so ∠ HEC = ∠ HCE
EH / / BC so ∠ HEC = ∠ ECB
Therefore, ECB = HCE
EC split BCD equally
Another same reason can be proved

If the four vertices of a trapezoid are on the same circle, then the trapezoid must be isosceles. Do you agree? Why?

Agree
If the two bottoms of a trapezoid are parallel and the two arcs between them are equal, then the two bottom angles on both sides of the lower bottom are equal to the sum of one of the two arcs and the inferior arcs opposite the upper bottom, that is, the two bottom angles are equal. Therefore, a circular inscribed trapezoid must be an isosceles trapezoid

The edge length of rectangular paper ABCD is ab = 4, ad = 2. Fold the rectangular paper along EF so that point a and point C coincide. After folding, color it on one side (as shown in the figure), the area of the colored part is ()
A. 8B. 112C. 4D. 52

In RT △ GFC, there are fc2-cg2 = fg2, fc2-22 = (4-fc) 2, the solution is FC = 2.5, and the area of 〈 shadow part is ab · ad-12fc · ad = 112, so B

If a △ 0.2 = B × 0.2 = C △ 1.02 = D × 1.02 (ABCD is greater than 0), the largest number in ABCD is（
)

B because 5A = B / 5 = 100C / 102 = 102d / 100

At the bottom of the pyramid p-abcd, ABCD is a rectangle PA perpendicular to the plane ABCD PA = ad = 4 AB = 2 in BD
At the bottom of the pyramid p-abcd, ABCD is a rectangle, PA is perpendicular to the plane ABCD, PA = ad = 4, ab = 2, and M is the midpoint of PD
(1) It is proved that the plane ABM is perpendicular to the plane PCD
(2) Finding the angle between the line PC and the plane ABM
Find the distance from point C to plane ABM

(1) It is proved that ∵ ABCD is a rectangle ⊥ CD ⊥ ad ∩ PA ⊥ plane ABCD ⊥ PA ∩ ad = a ∩ CD ⊥ plane pad ⊥ CD ⊥ am ∩ PA = ad, M is the midpoint of PD ⊥ am ∩ PD = D ∩ am ⊥ plane PCD ∵ am in plane ABM ∩ plane ABM ⊥ plane PCD (2) takes the midpoint n of PC, connects Mn, BN ∵ m is the midpoint of PD ∩ Mn / / CD / / AB ∩ a

ABCD is a rectangle, ab = 2, BC = 1, O is the midpoint of ab. take a point randomly in the rectangle ABCD, and the probability that the distance from the point to o is greater than 1 is ()
A. π4B. 1−π4C. π8D. 1−π8

As shown in the figure, the area of the rectangle is 2, with o as the center and 1 as the radius, and the area of the part (semicircle) inside the rectangle is π 2. Therefore, the probability that the distance between the point and O is greater than 1 is p = 2 − π 22 = 1 - π 4, so B is selected

Two rectangles, 2cm in length and 1cm in width, are placed on the straight line L (as shown in Figure 1), CE = 2cm. Rotate rectangle ABCD clockwise around point C and rectangle efgh anticlockwise around point e at the same angle
(1) When D and H coincide, connect Ag (as shown in Figure 2) and calculate the distance from D to Ag;
2) When α = 45 ° as shown in Figure 3, it is proved that mhnd is a square

（1）∠ADG=120°
If D is used as DM ⊥ AG, then
DM=AD/2=1/2
(2)∠CNE=∠H=∠D=90°
The quadrilateral mhnd is a rectangle
∠NCE=∠NEC=45°
∴CN=NE
∴DN=HN
The quadrilateral mhnd is a square

The ratio of the top to the bottom of a trapezoid is 3:5. The top is 1 meter shorter than the bottom, and the bottom is 25% higher. The area of the trapezoid is ()