The perimeter of parallelogram ABCD is 75cm, the height is 14cm with BC as the bottom (as shown in the figure); the height is 16cm with CD as the bottom. Calculate the area of parallelogram ABCD

The perimeter of parallelogram ABCD is 75cm, the height is 14cm with BC as the bottom (as shown in the figure); the height is 16cm with CD as the bottom. Calculate the area of parallelogram ABCD

According to the area formula of parallelogram, 14 × BC = 16 × CD, that is, 14bc = 16CD, then BC: CD = 16:14 = 8:7, BC = 87cd, and 2 × (BC + CD) = 75, then BC + CD = 37.5 (CM), 87cd + CD = 37.5 (CM), CD = 17.5 (CM). Therefore, the area of parallelogram ABCD is: 16 × 17.5 = 280 (square cm); answer: the area of parallelogram ABCD is 280 square cm

In rectangular ABCD, ab = 3, BC = 4, after translating rectangular ABCD 2 cm along the direction of ray AC, point a is translated to point P, then CP=

AC = 5, AP = 2, then CP = 3

As shown in the figure, in rectangular ABCD, ad = 8 cm, CD = 4 cm,
(1) If P is a moving point on edge ad, when p is at what position, PA = PC?
(2) In (1), when p is at p ', there is p'a = p'c and Q is a moving point on the edge of ab. if AQ = 15 / 4, is QP' perpendicular to p'c? Why?

Let PD = x, ab = 8-x ∵ PA = PC ∵ 8-x = √ (x ^ 2 + 4 ^ 2) get x = 3. When PA = 5, PA = PC (2) AQ = 15 / 4 ∵ BQ = 1 / 4, QC ^ 2 = 8 ^ 2 + (1 / 4) ^ 2 = 64 and 1 / 16pc ^ 2 = 25, PQ ^ 2 = 5 ^ 2 + (15 / 4) ^ 2 = 25 + 225 / 16, PC ^ 2 + PQ ^ 2 = 64 and 1 / 16

As shown in the figure, how to translate a part of the parallelogram ABCD to make the moving part and the non moving part form a rectangle

As shown in the figure

As shown in the figure, in the rectangle ABCD, BC = 2, DC = 4, the semicircle o with ab as the diameter is tangent to DC at e, then the area of the shadow part is______ (the results are expressed by exact values)

If AB is taken as the diameter, OA = ob = 12CD = 2, the area of semicircle is 12 × π × 22 = 2 π, the area of rectangle ABCD is 2 × 4 = 8, so the area of shadow is 8-2 π

In rectangle ABCD, BC = 2, DC = 4, the semicircle o with diameter AB is tangent to DC at point E, and the area of the shadow part is calculated. (results reserved)

Because tangent to DC, let the midpoint of AB be o, then EO = Ao = Bo is the radius of the circle. You mean the shadow area is the semicircle that intersects the rectangle. The radius is known. According to the area formula s = 2, if you say the shadow is the area of the remaining part of the rectangle, then subtract the semicircle area from the area of the rectangle

In trapezoidal ABCD, AD / / BC, AB is equal to DC, AC and BD intersect at point O, ad is equal to 1, BC is equal to 2, then the ratio of the shadow area (triangle AOB, triangle doc is shadow) to trapezoidal area is 0

1:2

The area of the circle in the figure is exactly 12 times that of the rectangle ABCD. The length of AB is 6.28cm, and the area of the shadow is______ Square centimeter

3.14 × (6.28 △ 2) 2 × 1.5 = 3.14 × 9.8596 × 1.5 = 30.959144 × 1.5 ≈ 46.44 (square centimeter); answer: the area of shadow part is 46 square centimeter. So the answer is: 46.44

As shown in the figure, the area of each circle in rectangle ABCD is 9 π, so the area of rectangle ABCD is 9 π______ ．

Let the radius of the circle be r, the area of ∵ circle be 9 π, and ∵ π R2 = 9 π. The solution is: r = 3, the length of ∵ rectangle ABCD is BC = 6 × 3 = 18, ab = 2 × 3 = 6, and the area of ∵ rectangle ABCD is 18 × 6 = 108

As shown in the figure, the area of circle is 6cm, the area of rectangle ABCD is equal to the area of circle. Calculate the area of shadow part

No picture, no truth