Math Solid Geometry Problems, math problems about solid geometry

1) Connect A1B

Because the straight triangular prism ABC-A1B1C1

So the face abb1a1 is perpendicular to the bottom surface a1b1c1, and a1b1 is the intersection of the two sides.

Because abc-a1b1c1 is a straight triangular prism and ab=cc1=a, the quadrilateral abb1a1 is square.

So ab1 is perpendicular to a1b

Because ab1 is perpendicular to bc1

So a1b is a projection of BC1 inside the face abb1a1.

So a1c1 is perpendicular to the face abb1a1

So a1c1 is perpendicular to a1b1

Because AC is parallel to A1C1 and AB is parallel to A1B1, AC is perpendicular to AB

2)v=ac*s(abb1a1)

Under the root number (b 2-a 2) * a 2

ab=cc1

So abb1a1 is a square.

So ab1 is vertical a1b

And because ab1 is vertical bc1

So ab1 is in the vertical plane a1bc1

So ab1 is vertical a1c1

ac//a1c1

So ab1 vertical ac

Because AA1 vertical plane ABC

So aa1 vertical ac

So the ac vertical plane aa1b1

So AC vertical ab

Which tetrahedral volume to find?

1) a; (2) 45°

Tips: (1) Make so plane abc in o, in plane abc as oe ac, of ab, sab sac 60°, ase asf, o point is on the bisector of bac, ae sacos60°=a, ao==a;

2) In RT SAO, cos SAO==, SAO=45°

(1) Connect AO and extend the intersection of BC at point D.

Because the projection of point P in the plane is O, the po surface ABC and BC belong to the surface ABC

So BC Po

And because it is a regular triangle, it is BC AD

And PO, AO intersection PAO at O point, so BC PAO and PA belong to the surface PAO

So pa bc

2) It is best to establish a spatial Cartesian coordinate system.

The angle between the normal vector of PA and the surface PBC is used to solve the problem.

(8) Connect the line B8C, BC8, let the intersection point D must be the midpoint of the line B8C, and E is the midpoint of AC, then in the triangle Ab8C, ed is the median line, and Ab8 is the side opposite the median line, so Ab8 CE, and because CE is in the plane BEC8, so AB8 surface BEC8 (8) is simple by volume method by (8)...

Take the midpoint E on BC and connect EP and EA

BC is perpendicular to EA, PO, so BC is perpendicular to surface EAP, so perpendicular to PA

2.It doesn't seem to be a special horn.

The high line of 90 degrees PBC ABC is the root of the half, and the number 3 and the side PA form a missing triangle of the right angle isosceles triangle, and the hole bihedral angle P-BC-A is the right angle.

The short answer is as follows: AF BC in the face ABC is in F, and EF is connected.

It is easy to know that AC 2FC, combined with BC 2AC, has BC 4FC so EF DC, EF surface ABC, EF BC, then has BC surface AFE, BC AE.

In addition, it is okay to set AC to 2 and other values, which means 2 units, so that there is BC 4 and so on, but if you want to set AC to 2, you may wish to set AC 2T, so that BC=4T, which is clearer.

No, if you fill in the blanks, you can, but you can't do that, because that's the only way to do it.

1) After S as the bottom projection H, connecting Ah, then Ah is the bisector of the angle BAC, and then the angle SAH is 45 degrees, S

a=a, then the projective long comma is the root of the two-part number two annihilation sells a

2) 45 degrees of finger teasing.