The topic of three dimensional geometry in the second year of high school, the master is advanced!

The triangle ABC is an isosceles right triangle. and AB is perpendicular to AC.

As the midline AD of the BC side, then AD bisects BC perpendicularly.

As the height of the side of the hg in the bottom triangle ehg is ef, then ef bisects hg perpendicularly.

Since D and E are the midpoints of BC and Hg, respectively, then DF BH AE.

Since AD, EF are both perpendicular to the plane BCGH, then AD EF.

Since DF is perpendicular to EF, ADFE is rectangular. then ad=ef=2 3

So bc = 2ad=4 3.

The regular triangular prism is: ABC-A'b'c'

The isosceles right triangle inside is def, and d, e, and f are on aa'bb'cc', respectively.

And if the angle e is a right angle, then there is:

Over e respectively as em perpendicular aa' in m, do en perpendicular cc' in n, and because the bottom side length of the regular triangular prism is 4, so there is:

em=en=4, let the length of the right-angled side be a, then the length of the hypotenuse is (a*2 under the root number) in the right-angled triangle efn.

fn=open root number (a 2-16).

The same can be obtained for DM = open root number (A 2-16).

Over d to dq perpendicular cc' to q

In the right triangle DFQ.

2a 2 = 16 + (fn+dm) 2 = 16 + 4a 2-64 solution yields: a = 2 6 under the root number

So the hypotenuse is 4 times longer than the root number 3

fn=dm=2, root number 2

That is, the three edges are respectively taken from the lower bottom surface of the distance of 0, 2 root number 2, 4 root number 2, can form a right triangle. (Of course, the first point does not have to be 0, but the vertical distance between the three points cannot be changed).

Let's be a year of high school...

I thought, you just look at the isosceles right triangle projection will be an equilateral triangle, and then do a prism according to that projection, so that you can put the right triangle in...

If all the edges of a regular quadrangular pyramid S-ABCD are 1, we can see that ASC and BSD are right-angled triangles, E is perpendicular to the section of the SC is two trapezoids, the area = 2(1-2x+1-x)X, the volume of the pentagonal pyramid formed by the cross-section of vertex C and SC = 2(2-3x)x(1-x) 3, the volume of two triangular pyramids on both sides of the pentagonal pyramid = 2(1-2x) 6, the function y = 2(2-3x)x(1-x) 3+ 2(1-2x) 6= 2(6x -6x +1) 6, (<0x 1 2), when 1 2 x < 1 and y = 2(1-x) 3, then the image of the function y=v(x) is roughly as follows: y= 2(6x -6x +1) 6 (<0x 1 2) and y= 2(1-x) 3 (1 2 x<1), all of which are subtractive functions in the defined domain, and the value ranges are: ( 2 6, 2 24], [2 24,0), and when x=1 2 the values of the two functions are equal to 2 24.

In the end, just calculate the rate of change and you can easily get the answer a.

If you don't understand anything, please ask.

I will get back to you as soon as possible.

I hope you will adopt o(o......

1) BC vertical AF and FD, so BC vertical plane AFD, so BC vertical EF, the same way AD vertical EF. The length of EF can be found from the right triangle CFD.

2) Take the midpoint G of FD and find the angle from the triangle CEG!

Proof that the midpoint of dc is e: me pq, ne ad and pd and ad intersect d, me and ne intersect e, so; Planar APD Planar MNE

Whereas, MN is within the planar MNE.

So; MN planar pad

To make an auxiliary line, let the midpoint of PD be E, connect AE, EM, and get the parallelogram AEMN (it is easy to prove that it is a parallelogram), so MN is parallel to AE, and AE is in the plane pad, so it is proven.

(1) 3 3, that is, the root number of 3 (2) 2 12*a 3, that is, the root number of 12 points 2 multiplied by a to the third power, when folded in half, it is just vertical, you can prove it yourself.

Hello girl stealing candy :

The first problem is on a circular surface (formed by cutting a ball on the equatorial plane), given a circumferential angle of 270 degrees, find the arc length = radius r*270

The second problem is that using the Pythagorean theorem twice, the circle O1 and the circle O2 intersect the sphere respectively, and on the same surface in a, b, then rt a01o, rtbo2o,

Connect A'c', take B'c'The focus is on the knowledge of N, connecting Mn, then Mn A'c'//ac。

Therefore, the angle formed by the straight line AC and BM on the opposite plane is BMN.

Let the side length of the cube be 2a, and connect bn, then bm=bn=5a, mn=2a under the root.

So cos bmn=(bm +mn -bn ) 2bm mn(5a +2a -5a ) 2 under 5a 2 under the pin, 2a1 under the root, 10 under the root

10 10 under the root

Isn't it simple?,Hunger Fight Connect A."c"Take it and write it as n.,Even the rotten Chang grind to connect cn,an,Hehe's a good ball now.,Brother high school three-dimensional fast geometry is a master.。