Solid geometry in high school mathematics, solid geometry in high school mathematics

1.Point M is the midpoint of AB. Point e is the midpoint of ab1. So the straight line me is the median line of the triangle abb1, so me bb1, and because me is on the plane efm, then bb1 is parallel to the plane efm

2.Extend me to cross a1b1 to n, then n is the midpoint of a1b1. Take the midpoint h of b1c1.

If AH and NF are linked, AH is heavier than B1C1. Since bf:fc=1:

3, so point f is the midpoint of b1h. So NF is parallel to AH. So NF is perpendicular to B1C1.

And because Mn is perpendicular to the bottom A1B1C1, Mn is perpendicular to B1C1. So the straight line B1C1 is perpendicular to the plane ENF. So the straight line B1C1 is perpendicular to the straight line EF.

Because BC is parallel to B1C1, EF is perpendicular to BC

3.Lead the perpendicular line of the triangle A1B1D from A1. The vertical foot is g.

Connect ag, hg. Obviously, Ah is perpendicular to the plane B1C1C, so Ah is perpendicular to the straight line B1D on the plane, and A1G is perpendicular to B1D, so B1D is perpendicular to the plane A1GH. So B1D is perpendicular to GH.

Therefore, the angle a1gh is the dihedral angle sought.

Note that the triangle A1GH is a right-angled triangle and AH is perpendicular to GH. So the tangent is A1H divided by GH.

Suppose the prism length of the triangular prism is 4a, and the height is also 4aThen there is GH which is equal to half of DC1 and DC1 which is equal to half of CC1. So gh = a, and a1h is the height of the regular triangle, and the side length of the triangle is 4a, so a1h = 2 root number 3a

So the tangent is 2 roots and threes.

They are the midpoint of ab1 and ab, that is, em is the median line of the triangle abb1, so em is parallel to bb1

Again em belongs to the surface efm

SO,BB1 is parallel to the planar efm

2.As the midpoint m of AB, since BB1 is parallel to the plane EFM, and BB1 is perpendicular to BC, the surface EFM is perpendicular to BC

So EF vertical bc

3.Goo (b Finding dihedral corners has always been my weakness.

However, it can be easily solved using a spatial Cartesian coordinate system.

Let the center of the circle be O, connect OA, OB then there is OA=OB=R, then there is AB 2=OA 2+OB 2=2R 2, so the triangle AOB is a right triangle.

AOB = 90 degrees.

So v small cylinder = ( r 2 4 - r 2 2) * hv large cylinder = (3 r 2 4 + r 2 2) * hv small cylinder: v large cylinder = ( r 2 4 - r 2 2) * h: (3 r 2 4 + r 2 2) * h

r^2/4-r^2/2）：（3πr^2/4+r^2/2)

The volume ratio is the area ratio, so ab takes the area on the left as a fan minus the triangle = a quarter of a times *r 2 - half a times r 2, and the right side = *r 2 - the left, and the two can be compared.

The guide line diagram has been made.

If the point e is EF cd in F, and F is FM DB in M, then EMF is the dihedral angle E-DB-C

First of all, EF=BB1=1, DF=AB 2=1, DFM is similar to DBC, then FM DF=BC DB

df=1, so: fm=bc db

bc=1,dc=2,so:db=5;

So: limb loss fm = 5 5

In RT EFM, EF=1, FM= 晌imitation5 5 then: EM= 30 5

So,cos emf=fm em= 6 6 I hope it can help you, if you don't understand,Please hi me,I wish you progress in academic debate! Happy New Year's Day!

After E does EE1 bottom, there are: M is the midpoint of DB, E is the midpoint of D1C1, and E1 is the midpoint of DC. and ee1 vertical bottom surface.

So the dihedral pure stare is eme1

ee1=1, me=1 2, and ee1m is a right angle.

tan=2, sin=2cos, sin2+cos2=1, cos=root5 5

When E and F are the midpoints of Aa1 and CC1, respectively, the quadrilateral bfd1e is perpendicular to the plane B1D

Even EF, because E and F are the midpoints of Aa1 and CC1 respectively, so A1E=C1F, at this time A1EFC1 is rectangular, EF is parallel to A1C1, and A1C1 is perpendicular to B1D1.

Because BB1 is perpendicular to the bottom surface A1B1C1D1, it is perpendicular to A1C1, so A1C1 is perpendicular to the plane BB1D1D, so EF is perpendicular to the plane BB1D1D, so the plane BFD1E where EF is located is perpendicular to the flat code plane BB1D.

Take a look, I'll tell you if you don't understand.

Belch.

I'm also very, very afraid of solid geometry.

Our teachers often say that the three-dimensional geometry of the college entrance examination is to give points.

There are probably two ways to solve the problem.

1.Direct method. It's a very simple way to do it, as long as you think about how to solve it, of course, you have to understand the theorems very well.

Like what is the three-perpendicular line theorem to find the dihedral angle, the equal area method to find the distance from the point to the surface, the line angle by solving the triangle, etc.

As long as you think of the method, the process is relatively simple (mostly)2Vector method. In layman's terms, it is a violent solution. Forcibly create a spatial Cartesian coordinate system

Write out the coordinates of each point and find them in vectors.

This process is a bit troublesome, but you don't have to think about it, you can just get a question and build a department, of course, the process has to be written in detail, so as not to deduct points.

The above two are methods, as for which one to use, it depends on the difficulty of the question, generally you see the question, there is no idea, then don't waste time, hurry up to build a system, the first question is to prove a conclusion, it will not be too difficult, so try not to build a department, because it is slow above, finished.

I wish LZ an early victory.

High school mathematics cultivates three abilities: numeracy ability, logical thinking ability, and spatial imagination ability. Solid geometry cultivates the third ability.

Since junior high school is more exposed to plane geometry, the first confusion is that the drawings drawn on paper lack a sense of three-dimensionality, just like on an airplane, looking down, you will see a series of crashes, but nothing happened. The second is the lack of insight into the relationship between lines and surfaces in the graph, and the position relationship cannot be found, so that the quantitative relationship between them cannot be judged or considered. My suggestion:

Three-dimensional geometric diagram recognition is the key to learning, and when the figures are familiar, the interrelationships will be clear. You can do more geometry to play, from the observation of the real thing to the beauty of the real thing, you can also experience the interrelationship, and finally you can get rid of the real thing, and you can analyze and solve the problem with graphics.

If you read a lot and summarize the rules, you can do it, but at least the theorems and axioms must be remembered clearly, and they will be used to change everything.

Let's find a few geometric models to be visual.

The aab line and l intersect and the ab point is not the intersection point carrying the spike, then it is impossible to pull in parallel.

BAB straight line and L heteroplane then there is only one.

There are countless cab straight lines and l flat lines.

So choose D

Key points: 1. The ground of the regular triangular pyramid is a regular triangle, each side is a (you set), and each angle is 60°;

2. The line connecting any vertex and the opposite side of the regular triangle is not only the height of the regular triangle, but also the middle line, or the angular bisector; The length of the midline is [(root number 3)a 2];

3. The perpendicular line from the vertex intersects with the ground (i.e., the height of the regular triangular pyramid), and the intersection point (i.e., the perpendicular foot) is the midpoint of the regular triangle, which divides the midline of the regular triangle into two line segments of 1:2;

4. According to the above three articles, you can already know the length of the height and the bottom edge, make a diagram, sort out the relationship between the line and the surface, and find the oblique height and the length of the side edge according to the Pythagorean theorem - the oblique height, the height and the ground center line 1 3 line segments form a right-angled triangle; The side edge, height and the ground center line 2 3 line segments form a right triangle;

Judging by the way you look, you should be able to forget it.

The calculation method of the oblique height and the side edge length of the regular triangular platform has always been, but it is only a matter of quantitative relationship.

As for the distance between the midpoint of the bottom surface and the edges and vertices, after making the auxiliary line, you can use similar triangles.

The tangent is 1: If you connect cd, DC is perpendicular to AC, then the angle D1CD is 45°, and the dihedral angle D1-AC-D is the angle D1cd=45°

It's been a long time since I've done a high school math problem, and if you don't have a diagram, you can only rely on your imagination.

You turn this diagram sideways and take APQC as the ground, so that the height of the triangle ABC is a straight line perpendicular to AC through point B. If I'm not mistaken, the height should be 3 times the length of the side of the root number, (provided that it is a regular triangle) if you don't try to find the height on the side of the ac at point b, then it will be simple, the problem of the quadrangular pyramid is equal to (the area of apqc * the height of the side of ac) divided by 3

You may not have learned this method, but it must be the right thing to do.

v 6 doesn't have to take a special point.

AA1=CC1, PA=QC1, so PA1=QC So, the area of APQC = half of the area of AA1CCA.

Half of the area of AA1CCA h=v

The area of APQC H one-third = V 6

There are solutions, but it's really hard to say without a diagram.

The place where you need to use your brain for this problem is to calculate the area of ACQP S, the easiest way is to go to a special point to get it, and the safest way is to set PA QC1 X and then according to the ACQP area PQC1A ACC1A to get the ACQP area as 1 2 of the ACQ1A area

Then the volume is 1 3 s h

H is the distance from point B to surface ACQP, which can be obtained by crossing the perpendicular line of AC at point B.

This problem can also be solved by using the method of establishing a spatial Cartesian coordinate system.