No. A line can be contained by many, many planes, Picture this, A rectangle with corners - going clockwise - A, B, C and D is the screen of your computer. This is a plane figure. 1 inch away from it a line runs from A1 to C1. The line is parallel to the plane. Now, take a sheet of paper with corners E, F, G and H, and place corner E at corner A of the screen, and place corner F at corner C of the screen. The Line AI is now 'contained' in the plane EFGH. and EFGH is perpendicular to ABCD.
One possible combination is ABCD v EFGH ABEF v CDGH ABGH v CDEF This has been worked with people staying in pairs. So that at different times AB are partnered with CD, EF and GH and the same applies to the other three pairs. There are many more combinations where constant pairings are not important.
Suppose the top face of the pyramid is ABCD with the square EFGH directly below it.Suppose AC and BD meet at P, the apex of the pyramid.Make a cut with a plane through P which is parallel to AB and goes through EF.Make a cut with a plane through P which is parallel to BC and goes through FG.Make a cut with a plane through P which is parallel to CD and goes through GH.Make a cut with a plane through P which is parallel to DA and goes through HE.The result will be the square-based pyramid PEFGH.
If your definition of parallel lines is that they never meet, then the answer is yes. If, however, the definition is that they remain equidistant from one another at all points, then, in my opinion, the answer is no. It is difficult to explain the second without recourse to diagrams which are very difficult to manage on this site. So consider a cube with vertices ABCD forming the top face and EFGH (in corresponding order) forming the bottom face. Now AB is parallel to the bottom plane - EFGH. And AB is clearly parallel to EF and any line parallel to EF. But is AB parallel lines such as FG? True, they will never meet but the distance between them increases as you move away from BF - ie they are not the same distance apart. Incidentally, Euclid's parallel postulate was phrased in a very different way from the one most mathematicians come across it. That version is a much later equivalent statement.
Each coordinate of the midpoint of a line segment is the average of the correspondingcoordinates of the end points.'x' of the midpoint = average of 'x' of the endpoints'y' of the midpoint = average of 'y' of the endpointsFor segment GF, you only need the coordinates of 'G' and 'F'.G . . . (b, 0)F . . . (3b, 2b)'x' of the midpoint = 1/2 (b + 3b) = 1/2 (4b) = 2b'y' of the midpoint = 1/2 (0 + 2b) = 1/2 (2b) = bCoordinates of the midpoint . . . (2b, b)
4
It is k times the perimeter of eh where k is the constant ratio of the sides of abcd to the corresponding sides of efgh.
It is k times the perimeter of EFGH where k is the constant ratio of the sides of ABCD to the corresponding sides of EFGH.
12
It is k times the perimeter of abcd where k is the constant ratio of the sides of efgh to the corresponding sides of abcd.
The question cannot be answered without information about the relative sizes of the two polygons.
It is the scale factor times the length of ad.
It is k times the perimeter of efgh, where k is the constant of proportionality between the sides of abcd and the corresponding sides of efgh.
It is k times the length of Ad where k is the constant of proportionality between the two shapes.
Wonderful! If you had told us something about polygon efgh, and mentioned some small tidbit of information regarding the ratio of similarity, we might have had a fighting chance. The question is a lot like asking: "Bob is older than Jim. How old is Bob ?"
2.50
touch "abcd efgh" touch 'abcd efgh' touch abcd\ efgh are three possibilities, given that you use a Linux shell. Otherwise, it may depend on the specifics of the software (e.g. libreoffice, emacs, firefox...), usually you can do it staghtforwardly when saving a file.