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Ask the math problem of compulsory two! Please ask questions! Please! Ask! La!
The first problem is the problem of spatial imagination. This question examines your familiarity with geometric features. A trapezoid can be translated to form a geometry with six faces, and the four sides are equal (because of the translation effect). The trapezoid naturally becomes two layers. Obviously, this geometry can only be a quadrangular prism.

The second question examines the normal cross section of the geometry. The cross section of cylinder is quadrangle (rectangle), the cross section of frustum is quadrangle (trapezoid), and the cross section of pyramid is triangle.

The third question examines the calculation of geometric volume. First, you draw a picture. It's not difficult to draw. So if every edge is 1, you can know that the total volume is 1, and then you will find that all the other four geometries are the same except the tetrahedron BDA 1C 1. The volume is1*1* 0.5 *1*1/3 =1/6, four of which are 2/3 of the total * *, then the remaining tetrahedron BDA1c/kloc. The landlord can ask if he doesn't understand.

The fourth question mainly investigates the five-core problem of geometric figures. You can refer to this:

Center of gravity: the intersection of three midlines, the three midlines of a triangle intersect at a point, and the distance from the point to the vertex is twice the distance from the midpoint of the opposite side; The ratio of the center of gravity to the centerline is1:2; Vertical center: the intersection of three heights of a triangle; Inner heart: the intersection of the bisectors of three internal angles is the abbreviation of the center of the inscribed circle of a triangle; Outer center: the intersection of three perpendicular lines, which is the abbreviation of the center of the triangle circumscribed circle; Paracenter: the intersection of an inner bisector and two other outer bisectors. ((* * There are three. ) is the abbreviation of triangle tangent circle center. If and only if the triangle is a regular triangle and the four centers are one, it is called the center of the regular triangle. All these will help you solve the problem in the future. You draw this problem, and everything will be solved.

The fifth question is a bit difficult, but it is not difficult to do the third. Tell you a rule, the tetrahedron in the third question is two regular tetrahedrons, and the edge connected by each vertex is 60 degrees. Then it is required to be a straight line with different planes, so it is only necessary to pair the parallel edges corresponding to one edge in a pair. One tetrahedron can help find twelve pairs, so two tetrahedrons can find twelve pairs, so the answer is 24 pairs.

Sixth, we can use the folding coincidence property of the middle vertical line. If the projections of AC and BD in the C option are on the same straight line, it can be explained that the intersection of the surface composed of AC and BD and the β surface is perpendicular to EF. So EF can be perpendicular to BD. Even if the included angle between AC and A and β in item D is equal, BD and EF are not necessarily perpendicular, for example, the included angle is 45 degrees. It makes no difference whether those two straight lines are vertical or not. . .

You got it?

You can ask again if you don't understand anything. Remember to adopt it and give me extra points.