Find high school solid geometry example questions, high school solid geometry problems

The angle between EH and the planar pad is the largest, let AB 2, then AE 3, 3 Ah (6) 2 ah＝√2.

Set ap x and see . i.e. 2 (x +4) 2x, solution x 2.

PAC isosceles right angle. Note the PAC ABCD, as EQ has EQ 3 2

qo＝cf-cq/√2＝√2-1/（2√2）.tan∠qoe＝eq/qo＝√（2/3）

cos qoe (3 5), cosine value of dihedral e-af-c (3 5).

Note the EQ PACqoe is the planar angle of the dihedral angle e-af-c].

(1) Take the midpoint G of CD, connect FG, FG is parallel to the plane PCE, and find the distance H from G to the plane PCE;

2) ligation eg, in p-ceg, s ceg*pa=s pce*h;

3) Right angle CEG, the area can be found, PD CD, PDA=45 degrees, PA=AD=2, PCE can be found on three sides, it is an isosceles triangle, and the area can also be found.

4) Find h.

(1) Do PE DB and cross CD to E, then E is the midpoint of CD and CE=CC'/4 pe∥db

In doing em da'Submit AA'Yu m

then the quadrilateral da'me is the parallelogram a'm=de=ce=cc'4 then plane PEM and plane A'db pe∥db em∥da'

Then PM is parallel to plane A'db then the point m is aa'one near A'The quarter of the point is a'm=aa'/4(2)

It's actually very simple, use the volume method.

Set the coordinates, use the term to solve the problem, this kind of problem is universal.

When M is the midpoint of A'C', BC1 is parallel to the plane MB1A, take the midpoint of AC M', and connect M'B, M'C'.

It is easy to prove that the plane MB'A is parallel to the plane M'BC', and if the planes are parallel, then the line planes are parallel, so BC' is parallel to the plane MB'A.

Obviously, M is the midpoint of A1C1 and BC1 is parallel to the plane MB1A at this time. Connect A1B and pass Ab1 to O

In A1BC1, O is the midpoint of A1B and M is the midpoint of A1C1.

om //bc1

OM is inside the planar MB1A.

BC1 is outside planar MB1A.

So BC1 is parallel to the planar MB1A

Solution: (1) When M is the midpoint of A1C1, BC1 plane MB1A

m is the midpoint of a1c1, extending am and cc1, and let am and cc1 extension line intersect at the point n, then nc1=c1c=a

If NB1 is connected and the extension intersects with the CB extension at the point G, then BG=CB, NB1=B1G

In CGN, BC1 is the median line and BC1 GN

GN plane mab1, bc1 plane mab1, bc1 plane mab1

m is the midpoint of a1c1, connecting the diagonal intersection of the rectangle abb1a1 with mThe midpoint of the two sides is parallel to the third side, and BC1 is parallel to the line and parallel to the plane.

If you want the line to be parallel to the plane, you have to prove that the line is parallel to the line, and continue to find conditions!

The height is 2r obtained by tangent to the upper and lower sides. From the tangent or middle of the three sides of the side, we can get a figure of such an argument, and the radius of an inscribed circle in the regular triangle is r, and the side length of the regular triangle is (2 roots and 3), so the area is (3 roots and 3) *r 2, and then the volume is equal to (6 roots and 3) and the group let * r 3

The Cartesian coordinate system d(0,0,0) of the built-up space

d1(0,0,silver a)a1(a,0,a) p(a 2,0,a)c(0,2a,0) b(a,2a,0) e(a 2,2a,0)n(0,a,a 2).

e(a2,2a,0) n(0,a,a2) p(a2,0,a) three-edged envy angular pne area is root number 5a 2 4 surface pne normal (0,1,2).

Vector dp(a2,0,a).

The distance from D to the surface PNE is 2 root 5A 5

v = 1 3 * root number 5a 2 4 * 2 root number 5a 5 = a 3 6