A high school solid geometry math problem, a high school math solid geometry problem

Solution: Due to PA=PB=PC

Thus the projection of p on the bottom surface is the outer center of the bottom surface.

In the base triangle, the cosine theorem gives bc=21, and the sine theorem gives bc sina=2r

The radius of the circumscribed circle r=7 3

From the Pythagorean theorem, we get po =pa r

po = 14 (7 3) = 49 po = 7 then the distance from p to the triangle abc is 7

From the cosine theorem: BC=(ab 2+AC 2-2*ab*ac*cos120°) 1 2) =(4+1+2) (1 2)=7 (1 2) then ao=(bc 2) cos30°=(7 3) (1 2) The perpendicular line of AC and AC intersect D through o, and the parallel line of AB intersects E with the extension line of AC through o, then do=(ao 2-(ac 2) 2) (1 2)=(7 3-1 4) (1 2)=(25 12) (1 2) deo=60° do eo=cos30° eo=do cos30°=(25 12) (1 2)*(2 3 (1 2))=5 3 de=eo 2=5 6 ae=de+ac 2=5 6+1 2=4 3 If the parallel line of the o as ac intersects with ab at f, then the quadrilateral faeo is a parallelogram, and the vector ao=vector af+vector ae=m*vector a+n*vector b |Vector af|=m*|Vector a|，|Vector ae|=n*|Vector b|∵|Vector af|=eo=5/3,|Vector a|=2,|Vector ae|=4/3,|Vector b|=1∴5/3=2m,4/3=n∴m +n = 5/6 +4/3 = 13/6

What is called "then the distance from P to the triangle abc"?? Is there such a ?? The question is wrong.

(1) Connect AC, take the midpoint of AC as E, connect NE, and connect ME.

Because PN = NC, AE = EC

So pa ne, and because pa plane abcd so ne plane abcd

So ne cd....1]

Because AM = MB, AE = EC

So me bc, and because of ab bc

So me ab, and because ab cd, so me cd....2]

Because [1] and [2].

So cd planar mne

So mc cd

2) Take the midpoint F of PD and connect AF and NF

Because pn = nc, pf = fd

So nf cd, and because of ab cd

So nf ab

And because am = 1 2 ab = 1 2 cd = nf, amnf is a parallelogram.

So af mn

Because APD is an isosceles right triangle, and F is the midpoint of PD, AF PD, and because AF MN is located

So mn pd, again because mn cd

So mn planar pcd

(1) Take the midpoint F of CD, then NF CD (because the mapping of PD on the surface ABCD is AD, so PD CD is the midline NF PD, and NF CD can be obtained); And because of MF CD, CD mnf, you can launch CD MN.

2) Take the midpoint H of PD and connect AH and NH, then PD AH, PD AM, so, PD surface AMNH, so Mn PD, and because CD mn, so Mn plane PCD

1.Certificate: Connecting AC, AN, BN, PA plane ABCD, RT PAC, N is the midpoint of PC, AN=1 2PC; and pa plane abcd, pa ad, pa ab, ad planar pab, bc ad, bc planar pab, bc pb, rt pbc, n is the pc midpoint, bn=1 2pc; an=bn, isosceles nab, m is the ab midpoint, mn ab, ab cd, mn cd.

2.Evidence: as the midpoint Q of PD, connecting AQ, NQ, PA plane ABCD, PA AD, RT pad, PD=45°, PA=AD, and Q is the midpoint of PD, AQ PD; In PCD, N and Q are the midpoints of PC and Pd, Nq = 1 2cd, and Ma =1 2cd, Ma =nq, and the quadrilateral mnqa is a parallelogram, Mn Aq, Mn Pd, and Mn cd, Mn planar PCD have been proved.

Straight line a parallel straight line l parallel straight line b, then straight line a and b may be parallel because -l- is a straight dihedral angle, so the projection of straight line b in the plane is a straight line l and because straight line a is not perpendicular to straight line l, so straight line a and b cannot be perpendicular so choose c

The median line theorem, the side length of the resulting cross-section quadrilateral is equal to half of the diagonal, so the perimeter is 20

Let the top and bottom of the triangular prism de=ef=fg be long as a, and the triangular pyramid height sh is 15, so according to the similarity (similarity theorem).

A: 12 = Sg: 15, calculated as Sg, so the triangular prism is high gh=, because the triangular prism side area is 120, so.

s=120=a multiplied by (multiplied by 3, the solution is a=4, or a=8, so.

The high GH = 10, or GH = 5 of the triangular prism

The side area ratio is equal to the square of the similarity ratio.

The similarity ratio is 4:12 or 8:12, so the area ratio is.

1:9 or 4:9

Submit the answer to question (1) first, please see below, click to enlarge:

Feel free to ask, and continue to ask, otherwise that's it.

The picture is not very good-looking, I hope it is inclusive, and the focus is on the idea.

(Note: Lines that are not needed will not be drawn in the diagram, making the graph concise and easy to see).

Because in the cube ABCD-A B C D midpoint Q, E are C d, the midpoint of CD, it is easy to know QE plane ABCD, and AC is in the plane ABCD, so QE AC, and because in the cube ABCD-A B C D the quadrilateral ABCD is a square, and AC is a diagonal, the points L and E are the midpoints of BC and CD respectively, it is easy to know AC Le, and QE and LE are tangentially intersected at the point E in the plane QEL, so AC plane QEL, QL are on the plane QEL, So there's ac ql.

Because in the cube ABCD-A B C D midpoint f, p, q, l are respectively the midpoint of AB, A D, C D, BC, it is easy to know that PQ is parallel and equal to FL, that is, the quadrilateral FPQL is a parallelogram, and because PL is the diagonal of the parallelogram FPQL, so the area of PFL = the area of PQL, then in the pyramid D-FPQL, it can be seen that the tetrahedral DPFL and the tetrahedral DPQL are equal in height and equal in volume, which is calculated in the square ABCD The area of the DFL is 3A 8, so the volume of the tetrahedral DPFL is the area of the DFL AA 1 3=A 8, so the volume of the tetrahedral DPQL is also A 8.

Construct a right triangle.

The second question is mainly to find the dihedral angle of l-pq-d, and use the dihedral angle to solve the distance from the point to the surface, omitting some processes that I think are relatively simple.

If the spatial imagination ability is insufficient, it is recommended to directly construct the coordinate system between spaces, calculate the normal vectors of the two surfaces, find the dihedral angle, and then use the dihedral angle to solve the distance from the point to the surface, and the distance formula is:

d = distance from the target point to the flute of the dihedral angle * the sinusoidal value of the dihedral angle.

Question 1: It is very easy to establish a spatial coordinate system, especially for this kind of regular cube, which lists the vectors of two straight lines respectively, and the easy to obtain the vector product is 0, so the two straight lines are perpendicular.

Question 2: It is more difficult to deal with it directly, so I suggest that you start with the cutting method and observe the whole figure for cutting, which is easier to handle.

Let the cross-section of point A intersect with PB at D and PC at E, then the perimeter of the cross-section of A is the perimeter of the triangle ADE, and the three sides of PAB, PBC, PCA are cut along the PA and formed into a plane figure, and the PA is changed to Pa and Pa after being cut', then the plane formed by the three sides is a three-sceles triangle PAB, PBC, PCA'In a fan-shaped figure, you can see that the circumference of the triangle ADE = AD+DE+DA', the minimum value of which is AA'The length of the line (the shortest straight line between two points), in the triangle paa'Medium, APA'=3*40°=120°, ap=2*root number 3, so aa'= 30, that is, the minimum value of the circumference of the cross-section of the crossing point a is 30

The minimum circumference is 6, and the regular triangular pyramid is vested.