Senior 2 Mathematics Geometry , urgently needed, thank you

The plane of A, B1, and D1 does not intersect with the bottom ABCD, as long as the intersection point A

Some of the questions are correct.

Idea: L is the intersection line between the vertex and the bottom ABCD, so the plane of A, B1, and D1 should be extended to make the intersection line with ABCD.

Method:1Make another cube on the left and name it a b3 c3 d-a1 b2 c2 d, connecting a c3 and c3 d.

then the composition l is the intersection of the two planes, i.e., a c3Because the square a b1 d1 c3So a c3 is parallel to b1d1.

2.The distance is a. Because the distance from L to D1 is the length of AA1 (the perpendicular distance of the two faces of ABCD-A1B1C1D1).

Because d= bad=90°, ce ae, and ad=cd=a

So, the quadrilateral ADCE is a square.

So, ae=ad=a

It is known that ab = 2a

Therefore, the point e is the midpoint of ab.

Again, ce ab

Therefore, ACB is an isosceles triangle.

Whereas, in a square ADCE, cae = 45°

So, b= cae=45°

Therefore, ACB is an isosceles right triangle.

In the figure, cross the dot d'as the perpendicular line of ac, the perpendicular foot is f; The perpendicular line of the point F as AC intersects AB at the point G and connects with D'g

Because d'f⊥ac，gf⊥ac

So, d'fg is the planar angle of the dihedral angle AC.

So, d'fg=90°

So, d'f⊥fg

Again, FG AC, BC AC

So, fg bc

So, d'f⊥bc

While, BC AC

So, BC face AD'c

So, BC D'c

So, d'Ca is the planar angle of the dihedral angle bc.

Known in isosceles right-angled triangle d'ac, d'ca=45°

So, the dihedral angle bc is 45°

As shown in the figure, the same point D is the perpendicular line of AC, the perpendicular foot is E, and the perpendicular line of Ac is made in the surface ABC, and the intersection of AB is at the point F. Pass the dot d'As the perpendicular line of the bottom ABC, the vertical foot is O

Because d'e⊥ac，ef⊥ac

So, d'EF is the planar angle of the dihedral angle AC.

So, d'ef=60°

Again, d'ac is an isosceles right triangle, d'e⊥ac

Therefore, point e is the midpoint of AC.

While, BC AC, EF AC

So, EF BC

, EF is the median line of the isosceles right triangle ACB.

Therefore, the point f is the midpoint of ab.

Because of the face d'EF face ABC, and D'o Facet ABC

So, point o on the surface d'EF on the intersection line with the surface ABC.

Well, in RT D'OE, D'eo=60°，d'e=(√2/2)cd=(√2/2)a

So, d'o=d'e*sin∠d'eo=(√2/2)a*(√3/2)=(√6/4)a

The bottom surface abc is an isosceles right triangle with an area of s abc = (1 2) ac 2 = (1 2) * (2a) 2 = a 2

So, triangular pyramid d'Volume of -abc v=(1 3)*s abc*d'o

1/3)*a^2*(√6/4)a

√6/12)a^3

Semicircle area s=1 2(r)=a

r = 2a

The circumference of the bottom surface of the cone.

l=πr=2πr

The base radius r= can be obtained

It's hard to write and type by yourself.

It's a very annoying process, and it's a drawing process.

Let me give the landlord some hints for understanding and changing the subject.

First of all, you should prove that the angle between the surface EAC and the surface A1B1C1D1 is also 45 degrees.

2 Then it is proved that d1b and the face a1b1c1d1 are also 45 degrees.

3 Then, on the face dd1b1b, as the midpoint of ac, connect the auxiliary wires, and you should be able to prove that de is equal to half of dd1, and dd1 is equal to d1b1, and ac is equal to ab of root number 2, then the combined calculation is that d1b is equal to 2a, and the area of face eac is equal to the square of root number 2 times a.

Question 2: I forgot the specific concept, but I don't think the two lines are parallel or intersecting.

Then the distance should be perpendicular to both lines at the same time.

The key to proof is to prove that aa1 is perpendicular to the face abcd and face a1b1c1d1 respectively, and then according to the length of some line segments calculated in question 1, then the result should be the root number 2 multiplied by a

3 ca is right, cut the large triangular pyramid from the vertex to the bottom midpoint, and it becomes a triangular pyramid and a quadrangular pyramid.

B is right, a round table is cut horizontally parallel to the bottom surface.

C No, it can't be done anyway.

D is right, and B is the same.

The number 4 a is fine, and the bottom of the prism 4 edges and 4 top 4 add up to 12

Putting this cube into a spatial Cartesian coordinate system, we may assume that the edge length of the cube is a(a>0), and the equation for the three lines in the problem is:

l1 (ab): x=0, z=a;

l2 (cc1): x=y=a;

l3 (a1d1): x=z=0.

Suppose the distance from one point p(x, y, z) to three lines in space can be expressed as:

d1 = [x 2+(z-a) 2] under the root number

d2=under the root number [(x-a) 2+(y-a) 2]d3=under the root number [x 2+z 2].

Equidistant requirements d1=d2=d3.

You can first use d1=d3 to get z=a2.

It can then be solved using d1=d2 or d2=d3.

x=(y-a)^2/(2a)+3a/8

That is, the trajectory of the three straight lines equidistant to the question is a parabola on the plane z=a2, so the number of points that meet the requirements of the question is infinite.

I haven't solved an equation for a long time, and I don't know if it's right.

But anyway, that's the way of thinking.

It was originally given to you yesterday, but I wrote it twice, and it was almost done, but the computer was stuck, and it was gone, and I had something to do at the time, so I'll help you today.

Proof that: (1) The parallel lines of the point O as BC intersect AB and Cd are m and n respectively, because O is the intersection point of the diagonal line of the rectangle, so on = 1 2BC, and because EF = 1 2BC, then.

on//=ef

Therefore, the quadrilateral onfe is a parallelogram, so oe nf, and it is known that nf is on the plane cdf, so oe plane cde

2)on=1/2bc,bc=√3cd；So on= 3 2cd, cde is an equilateral triangle, oe is its centerline, then.

OE = 3 2cd, by above.

oe=on, onfe is a parallelogram, can be obtained.

onfe is diamond-shaped, fo and ne are diagonal, so there is fo ne;

The projection of FO on the planar ABCD is on, and the ON cd is known, so it is by the projective theorem.

Fo cd, fo ne, cd and ne are all on the plane cde, hence the perpendicular theorem.

fo planar cde

If the slope of the line on the axis of symmetry is 1, you can find it in a simple way

From x-y-2=0, x=y+2, y=x-2, substituting the curve f(x,y) to obtain f(y+2,x-2)=0

This is the equation for the symmetry curve. i.e.: Answer C

General solution: Let (x0, y0) be any point on the curve, then there is f(x0, y0)=0 and let (x,y) and (x0,y0) about x-y-2=0 symmetry (y0-y) (x0-x)=-1 (vertical, slope product is equal to -1) (x0+x)-(y0+y)-4=0 (the midpoint is on the axis of symmetry) to solve the above two equations:

x0=y+2, y0=x-2, substitute f(x,y)=0 to obtain: f(y+2,x-2)=0, this is the equation found.

Answer: Select C.

The reason is as follows: Let any point (x,y) on f(x,y) be symmetrical with respect to the straight line x-y-2=0 as (x',y'), then:

1. The midpoint of these two points is on the straight line x-y-2=0;

2. The straight line and the symmetrical line x-y-2=0 through these two points are perpendicular.

Example of two systems of equations, solved: x=y'+2;y=x'-2。i.e. (y'+2，y=x'-2) on the curve f(x,y).

So f(y'+2，y=x'-2)=0, i.e.: f(y+2,x-2)=0.