High school space geometry, dihedral problems

Pass C as the AD perpendicular, and then make the PD perpendicular line and then COSA seek.

Hehe, the upstairs is amazing! First of all, I would like to say that these should be directly asked to the teacher!

Add one upstairs: the coordinate method! I don't know if you've heard of it!

It is very effective, and everything is complicated, for example, this problem can make a perpendicular line at point C, and then you can establish a right-angled left system! Then use vectors and formulas to calculate! I don't remember the specific formula too much!

It's been a few years since I graduated! This method is used when your traditional method is inconclusive, and it should be computationally intensive!

Using the coordinate method, the distance of the PA is set to 1, and then the coordinates of the ABCDE point are calculated, and the other problems are solved by yourself.

If the visual diagram looks like this, set the height to h

Then the original trapezoidal a1b1 = 2ab

abc=45°

Get ab = root number 2

h then a1b1 = 2 root number 2

HA1B1 is the height of the original trapezoid.

then the area of the original trapezoid is.

Root number 2 2 root number 2 = 4

There are also no closed understanding files can be asked.

Hope is defeated.

Visualize the area of the graph.

Be. The area of the actual plot. of the calendar.

Fold. The reason is that the length of the filial beam on the y-axis is regarded as the height, and the length becomes the original 1 2, and there is an angle of 45 ° in Qiaoluoyun, so it is (2) 4

Fold. The area of the original trapezoidal is: 2 [(2) 4]=4

The ground floor has helped you. In fact, you have not applied the following knowledge flexibly: 1. The sufficient condition for the perpendicularity of two planes is that one plane is perpendicular to the straight line of the intersecting line perpendicular to the other plane, and 2 the line is perpendicular to the surface of the line, and the line is perpendicular to any straight line in the plane.

When the straight line is perpendicular to the two straight lines in the face, the line surface is pushed out perpendicular. The key to this problem is that you don't know how to make auxiliary lines, which means that you are not proficient in the three-perpendicular line theorem and the typical graphs of the three-perpendicular line theorem. Usually, you should think about the simplest and typical diagram of the three perpendicular lines, and then do a few more questions to consolidate it.

The root of the quarter is three to one, and the interval is closed.

The projection of EF on BC is equal to AD*SINA, which is shown in the figure below

In the cube ABCD-A1B1C1D1, the dihedral angle B1-MB-C1 is 60 degrees because the size of the dihedral angle A-MB-C1 is 120 degrees.

Because C1B1 is perpendicular to the plane A1B1Ba, the perpendicular line B1E of the MB is crossed by the point B1E, which connects C1E. So the angle c1-e-b1 is the planar angle of the dihedral angle b1-mb-c1. That is, the angle b1ec1 = 60 degrees.

In the RT triangle C1B1E, C1B1 = 1 so B1E is equal to the root number of thirds.

From the similarity of triangles, b1e mb1=bb1 bm is given and b1m=x is given

b1e*bm=mb1*bb1

Simplifying to give x is equal to the root number of two.

So the point m is the point whose distance from b1 is two-part of the root number.

Dihedral angle a-mb-c'The size is 120°, i.e. dihedral angle b'-mb-c'The size is 60°

Make b'n mb in n, even c'n

b'c'Face MBB'

c'nb'It's dihedral angle B'-mb-c'The plane angle is 60° b'c'=1

b'n=√3/3

Set MB'=x, then.

mb=√(x²+1)

mb·b'n=2s△bb'm=bb'·mb'

3(x²+1)]/3=x

x²+1=3x²

x= 2 2, take a positive value.

When MB'= 2 2 at dihedral a-mb-c'The size is 120 °

Solution: The body diagonal of the cuboid is the selling of the outside ball only straight in the middle of the diameter, let the cuboid ab=a, bc=b, cc1=c, the radius of the outside ball r, then.

2r) = a + b + c, the external solanum sphere surface area s = 4 r = a + b + c.