In 2010, how to solve the eighth problem of Xicheng junior high school and second mode mathematics

If you can't finish this multiple-choice question in 2 minutes, don't do it.

You can analyze it. Since it is the maximum distance、。

Just look for the biggest one. But option D doesn't look right.

It's not the same as those three. So choose A

a The hypotenuse midline is equal to half of the hypotenuse. The sum of the two sides is greater than the third side. Therefore, the connection can pass through the midpoint of the hypotenuse.

You give me some bounty points first, and I'll tell you how to do it! Really, not kidding you.

That's too general! What about the topic?

2013 junior high school Xicheng mathematics model 1 model problem detailed explanation to some specific urgency.

That's a good calculation. CDD' is a fan with C as the center and Cd as the radius, and Caa' is a fan with C as the center and Ca as the radius.

Total area = S sector caa' + s abc + s cd'a' = 3+ *2*2 4 = 3+

Shaded part area = total area - sabcd-s sector cdd' = 3+ -3- *3*3*

Do r with respect to the symmetry point r on the x-axis"(2,-2), the problem becomes from r"(2,-2) The minimum distance to the triangle area enclosed by the three straight lines.

Obviously, the distance to the point (1,2) is the smallest, and the long root number 17 is obtained by the Pythagorean theorem

Answer: Q point is on the x-axis, the Qr should be the smallest, according to the shortest vertical line between the two points, that is, the Q point coordinates are (0, 2);

The point p is in the area or boundary enclosed by the line y=-x+3, the line y=4 and the line x=1, and the coordinates of the point p can be found as (1,2) according to the principle of equality of the inner and opposite angles of the triangle.

The minimum value of qp+qr is 3

This problem should be done as y=min(x -1,1-x).

With (x -1) - (1-x).

Get 2x -2 so that it is equal to zero.

The solution is x= 1

When x -1 y=1-x is small.

When -1 x 1 y=x -1 is small.

When x -1 y=1-x is small.

In summary, the image should be an image of A.

I just took a general look, it should be a positive similarity or a four-point contour, and then convert the angles, and the specific answer is here.

Because of the rotation, PBC is all equal to AP'b

So ap'=pc=1 p'b=pb= 2, so p'BP is an isosceles triangle.

Because p'bp=90°

So p'p= √￣2* √2=2

Because of the AP'=1 , ap=√￣5

So ap'p is a right-angled triangle.

So ap'p=90°

Because isosceles right triangle ap'p

So bp'p=45°

So ap'b=135°

So bpc=135°

The second question is the same as above.

Because of the rotation, PBC is all equal to AP'b

So ap'=pc= 2 p'b=pb= 4, so p'BP is an isosceles triangle.

Because p'bp=120°

So p'p=4√￣3

Because of the AP'=2, ap=2√￣13

So ap'p is a right-angled triangle.

So ap'p=90°

Because of the isosceles triangle AP'p

So bp'p=30°

So ap'b=120°

So bpc=120°

bp=4 cp=2 is known

So bc = 4 +2 -2 * 4 * 2 * cos120° = 28 so bc = 2 7

Xiao Ming's idea is that the known conditions are relatively scattered, and the scattered known conditions can be concentrated together through rotational transformation, so he rotates BPC 90° counterclockwise around point B to obtain bp a (as shown in Figure 2), and then connects PP

1) The power of the BPC in Figure 2 is ;

2) As shown in Figure 3, if there is a point p in the regular hexagon abcdef, and p a= , pb=4, pc=2, then the degree of bpc is , and the side length of the regular hexagon abcdef is

You have your own picture (1).

After rotation pp'=2, angular bp'p = 45° while in the triangle ap'ap2=ap in p'2+pp'2, so angular ap'p=90°, so angular bpc=135°

2) In the same way, rotate the triangle BPC by 120° to get the angle BPC = 120°, BC = 2 root number 7