Of course. The vertices of a triangle, and any three vertices of any other polygon, are non-collinear. In the case of a triangle and a quadrilateral, if you had three collinear vertices, then you couldn't have the polygon.
It might depend on how you draw it, but I don't think it is possible. Draw the figure, and look at the points where lines meet. Count how many of those points are "odd", meaning that an odd number of lines meets there. Each of these points has to be either the starting point, or the end point, and you can only have one starting point and one end point.
No. To prove this, you can actually construct the plane.
draw two angles in three common points
Its pretty much 11 rectangles in one.
Of course. The vertices of a triangle, and any three vertices of any other polygon, are non-collinear. In the case of a triangle and a quadrilateral, if you had three collinear vertices, then you couldn't have the polygon.
To connect seven noncollinear points, draw six line segments connecting each point to every other point. This will create a network of lines that connect all seven points without crossing or overlapping.
point B lies in plane U
as in a 7 sided figure? a heptagon
a nonagon
No.
There are infinitely many spatial figures. I is not possible to draw or describe them.
Yes.
It might depend on how you draw it, but I don't think it is possible. Draw the figure, and look at the points where lines meet. Count how many of those points are "odd", meaning that an odd number of lines meets there. Each of these points has to be either the starting point, or the end point, and you can only have one starting point and one end point.
Yes, draw a 2 x 7 rectangle.
Take the 4 corners (vertices) of any 4-sided figure (quadrilateral). Draw the quadrilateral and its diagonals.
A cube would qualify...