As shown in the figure, a, B, C and D are the four vertices of rectangle ABCD, ab = 25cm and ad = 8cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until point B, and point Q moves to point d at the speed of 2cm / S. (1) P and Q, from the start to the second, PQ ‖ ad? (2) Question: P, Q two points, from the start to a few seconds, quadrilateral pbcq area of 84 square centimeters?

As shown in the figure, a, B, C and D are the four vertices of rectangle ABCD, ab = 25cm and ad = 8cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until point B, and point Q moves to point d at the speed of 2cm / S. (1) P and Q, from the start to the second, PQ ‖ ad? (2) Question: P, Q two points, from the start to a few seconds, quadrilateral pbcq area of 84 square centimeters?

(1) Let P and Q point, from the start to the X second, PQ ‖ ad, ∵ quadrilateral ABCD is parallelogram, ∵ ab ‖ CD, namely AP ‖ DQ, ∵ PQ ‖ ad, ∵ quadrilateral apqd is parallelogram, ∵ AP = DQ, ∵ 3x = 25-2x, x = 5, a: P and Q point, from the start to the 5 second, PQ ‖ ad. (2) let P and Q point, from the start to the a second, the area of quadrilateral PBC Q According to the trapezoidal area formula: 12 (25-3a + 2a) · 8 = 84, the solution is: a = 4, answer: P, Q two points, from the start to the fourth second, the area of quadrilateral pbcq is 84 square centimeters

Fold the square ABCD so that the vertex a coincides with the point m on the edge of CD. The crease intersects ad at e and BC at F. after the edge AB is folded, it intersects BC at g (as shown in the figure) (1) If the side length of a square is 2, M is the midpoint of the edge of CD. Find the length of EM. (2) if M is the midpoint of the edge of CD, prove de: DM: EM = 3:4:5; (3) if M is any point on the edge of CD, let ab = 2A, ask whether the perimeter of △ CMG is related to the position of point m? If it is relevant, please express the perimeter of △ CMG with the algebraic expression of length x with DM; if it is not relevant, please explain the reason

It is proved that: (1) if De is x, then DM = 1, EM = EA = 2-x, in RT △ DEM, ∠ d = 90 °, de2 + DM2 = em2x2 + 12 = (2-x) 2x = 34, EM = 54

Fold the square ABCD with side length 2A so that vertex a falls on the edge of CD, that is, point M. the crease intersects ad at e, BC at F, and edge AB intersects BC at point G after folding
(1) It is proved that: when the point m is located at the midpoint of CD, the ratio of three sides of △ DEM is calculated; when the point m moves on the edge of CD, whether the circumference of △ CMG has changed is explained

(1) When m is located in the middle of CD, DM = a, ed + Em = 2A, let em = x, ed = 2a-x. according to the Pythagorean theorem, we can calculate x = 5 / 4A, that is em

25. As shown in the figure, fold the square ABCD so that the vertex a coincides with the point m on the edge of CD, the broken line intersects ad at e, BC at F, and the edge AB after folding intersects BC at point G
(2) If M is any point of CD, let AB = 2A. Is the perimeter of △ CMG related to the position of point m? If so, let AB = 2A,
Please express the perimeter of △ CMG with the algebraic expression of length x (DM = x) with DM; if it is irrelevant, please explain the reason
Picture:

Let de = y
Triangle DEM is similar to triangle MCG
DM = x, de = y, EA = em, EM = 2a-de, and a Pythagorean theorem can be used. In this way, y can be represented by X and a, and GC can be represented by a and X
MG is represented by a, X
Finally, the perimeter is 2A

On the three vertices of an equilateral triangle with side length R, there are equal charges, all of which are Q
How much work does the electric field force do when the charge of 1C is moved from infinity to the center point

The point charge potential formula φ = KQ / R (take infinity as the zero potential surface),
Where k is the constant of electrostatic force, k = 9 × 10 ^ 9nm ^ 2 / C ^ 2, q is the electric quantity of the field source charge (the charge forming the electric field), and R is the distance between the studied point and the field source charge
Then the distance from the vertex of equilateral triangle to the center point is L = √ 3R / 3,
The total potential of three Q at the central point is φ = 3kq / L = 3 √ 3kq / R
When the charge of Q1 = 1 C is moved from infinity to the center point, the work done by the electric field force is w = uq1 = (0 - φ) Q1 = - 3 √ 3kq / R

When a positive charge q is placed at vertex a of equilateral triangle ABC, the field strength at C is e. after another positive charge q is placed at B, what is the field strength at C?

The root sign 3 is obtained by using the parallelogram rule of cosine theorem

The area of parallelogram ABCD is 48 square centimeters, AC = 3aE, BC = 4fC

14cm2, let sabcd = SS △ def = s △ dec-s △ DFC + s △ EFC △ Dec have the same height as △ DAC, and AE = 2 / 3aC, so s △ Dec = 2 / 3S △ DAC = 1 / 3S, and △ DFC has the same height as parallelogram ABCD, FC = 1 / 4bc, so s △ DFC = 1 / 8s

A trapezoid, if you increase the upper bottom by 2.4cm, will become a parallelogram, if you reduce the lower bottom by 5.8cm, will become a triangle
If you add 2.4cm to the upper part of the trapezoid, it will become a parallelogram. If you reduce 5.8cm to the lower part, it will become a triangle. At this time, the area of the trapezoid will be reduced by 21.46cm. What is the area of the original trapezoid?

If you reduce the bottom by 5.8 cm, it will become a triangle
It is concluded that the base is 5.8 cm
Because the area of the triangle is 21.46 square centimeters, the height of the triangle, that is, the height of the trapezoid, is: height = 2 times the area of the bottom
H=2×21.46÷5.8=7.4
So s ladder = (upper bottom + lower bottom) × height △ 2 upper bottom = lower bottom - 2.4
= （5.8-2.4+5.8）×7.4÷2
=34.04

If the upper and bottom of a trapezoid increase by 4cm, it becomes a parallelogram. If the upper and bottom decrease by 3cm, it becomes a triangle. At this time, the area is reduced by 7 or 5cm. What's the original trapezoid area in square centimeter? I'm poor in the process,
The height is trapezoidal,
So the trapezoid height = 7.5 × 2 △ 3 = 5cm, I can't understand 7.5x2 △ 3

You'd better draw the picture below. It's much easier if you look at it again
Then, if the top and bottom are reduced by 3 cm, they will become a triangle, and the area will be reduced by 7 or 5 cm
What is reduced is the area of a triangle. The bottom of the triangle is the top of the trapezoid, and the height is the height of the trapezoid,
The area of the triangle is s = base × height △ 2
Then height = s × 2 △ bottom = 7.5 × 2 △ 3 = 5cm
This can be understood. No, it is recommended to draw the figure below

A trapezoid has an upper base of 6cm. If its upper base is extended by 4cm, it will become a parallelogram. The area of the original trapezoid is ()
A trapezoid has an upper base of 6cm. If its upper base is extended by 4cm, it will become a parallelogram. The area of the trapezoid is increased by 10 cubic centimeters. The area of the original trapezoid is ()

(10+6)*(10*2/4)/2=40