As shown in the figure, in diamond ABCD, point E is on BC, and AE = ad, ∠ ead = 2 ∠ BAE, find the degree of ∠ BAE

As shown in the figure, in diamond ABCD, point E is on BC, and AE = ad, ∠ ead = 2 ∠ BAE, find the degree of ∠ BAE

In diamond ABCD, ab = ad, ∵ AE = ad, ∵ AB = AE, let ∠ BAE = x, then ∠ ead = 2x, ∵ Abe = 12 (180 ° - x), ∥ ad ∥ BC, ∵ bad + ∥ Abe = 180 °, ∥ x + 2x + 12 (180 ° - x) = 180 ° and the solution is x = 36 °, i.e. ∠ BAE = 36 °

When the image of a function is known to pass through points (1,1) and (2, - 1), the expression of the function is obtained and the range of X that makes the value of the function positive is obtained
This linear equation is y = ax + B, which brings (1,1) and (2, - 1) into the equation
1=a+b
-1 = 2A + B a = - 2 b = 3, so the analytic formula is y = - 2x + 3
When the function value is positive, Y > 0, so - 2x + 3 > 0 x

This linear equation is y = ax + B, which brings (1,1) and (2, - 1) into the equation
1=a+b
-The solution of 1 = 2A + B is a = - 2, B = 3, so the analytic formula is y = - 2x + 3
When the function value is positive, Y > 0, so the solution of - 2x + 3 > 0 is X

Let the midpoint of the edge CD of the square ABCD be e, and f be the midpoint of Ce (graph)

It is proved that: as shown in the figure, if the bisector ah of ∠ BAF intersects with the extension of DC at h, then ∠ 1 = ∠ 2 = ∠ 3, so FA = FH. Let the side length of the square be a, in RT △ ADF, af2 = ad2 + df2 = A2 + (3A4) 2 = 2516a2, so AF = 54a = FH. Thus ch = fh-fc = 54a-a4 = a, so RT △ ABG ≌ RT △ hCG (AAS), G

The image of a given function passes through a (- 2, - 3) B (1,3) two points 1
2. Try to judge whether the point P (- 1,1) is on the image of this linear function?
There is no graph

1. Let the expression of this first-order function be y = KX + B,
Substituting a (- 2, - 3) and B (1,3), we get
{-2k+b=-3
k+b=3
The solution is {k = 2
b=1
The expression of this linear function is y = 2x + 1
2. When x = - 1, y = 2 × (- 1) + 1 = - 2 + 1 = - 1 ≠ 1
The point P (- 1,1) is not on the graph of this linear function

In the square ABCD, e is the midpoint of DC and F is the midpoint of EC. It is proved that ∠ DAE = 1 / 2 ∠ BAF

Tan ∠ BAF = 1 / (3 / 4) = 4 / 3, Tan ∠ DAE = 1 / 2, Tan ∠ BAF = 2tan ∠ DAE / (1-tan ^ 2 ∠ DAE) = Tan (2 ∠ DAE), so ∠ BAF = 2 ∠ DAE, that is ∠ DAE = 1 / 2 ∠ BAF

It is known that the straight line is equal to KX + B and the straight line y is equal to - 3x + 5, and through the point (2,9), the function expression is obtained

Parallel, then k = - 3
The function is y = - 3 (X-2) + 9
That is y = - 3x + 15

It is known that the line y = KX + B is parallel to the line y = 3x-1 and passes through (0,12). The analytic expression of the function of this line is______ ．

If the ∵ line y = KX + B is parallel to the line y = 3x-1, and ∵ k = 3, and ∵ passes through (0, 12), then the analytic expression of the function of the ∵ line is y = 3x + 12

The gate of a factory is shown in the figure, in which the quadrilateral ABCD is a rectangle, and the upper part is a semicircle with ab as the diameter, in which ad = 2.3m, ab = 2m
Ad = 2.3 meters, ab = 2 meters, there is a truck full of goods, 5 meters, 1.6 meters wide, ask if this car can pass through the factory gate? Explain the reason

x^2+y^2=1 (y>0)
Substituting x = 0.8 into the solution, we get | y | = 0.6
Because 2.3 + 0.6 = 2.9 > 2.5
So this car can go through the factory gate

Mathematical problem: the gate of a factory is shown in the figure, in which the quadrilateral ABCD is a rectangle, and the upper part takes AB as the diameter
The gate of a factory, in which the quadrilateral ABCD is rectangular, the upper part is a semicircle with ab as the diameter, in which ad = 2.3m, ab = 2m, there is a truck full of goods, 2.5m high and 1.6m wide, can this car pass through the gate? And explain the reason

The height of the truck is 2.5m, ad = 2.3m, so it is 0.2m higher. Therefore, we only need to consider whether the top of the truck with height of 0.2m and width of 1.6m can pass through the semicircle with diameter of 2m
Make the center of the lower side of the top of the truck coincide with the center of the circle, and set it as the coordinate point. The radius of the circle is 1 m, and the half width of the truck is 0.8 M. according to the Pythagorean theorem, the half chord length at 0.8 m is 0.6 M
0.6m is larger than 0.2m, so the truck can pass
You can draw pictures to illustrate

As shown in the figure, the quadrilateral ABCD and aefd are parallelograms

It is proved that ∵ quadrilateral ABCD and aefd are parallelograms, ∵ ad ∥ BC, ad = BC, ad ∥ EF, ad = EF, ∵ BC ∥ EF, BC = EF, ∵ quadrilateral BCFE is also parallelograms