Each coordinate of the midpoint of a line segment is the average of the corresponding
coordinates of the end points.
'x' of the midpoint = average of 'x' of the endpoints
'y' of the midpoint = average of 'y' of the endpoints
For 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) = b
Coordinates of the midpoint . . . (2b, b)
To find the perimeter of polygon efgh, you need the ratio of similarity between polygons abcd and efgh, as well as the perimeter of polygon abcd. Once you have the perimeter of abcd, multiply it by the ratio to obtain the perimeter of efgh. If the ratio is not provided, it cannot be determined.
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.
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.
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.
EF
The identity transformation.
To find the perimeter of polygon efgh, you need the ratio of similarity between polygons abcd and efgh, as well as the perimeter of polygon abcd. Once you have the perimeter of abcd, multiply it by the ratio to obtain the perimeter of efgh. If the ratio is not provided, it cannot be determined.
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.
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.
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.
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Vertices: E (-2, -1); F (-4, 3); G (1, 5); and H (3, 1).Plot these points in a coordinate system and connect them. The quadrilateral EFGH is formed.The slope of the line where the side EF lies is -2.[(3 - -1)/(-4 - -2)] = 4/-2 = -2The slope of the line where the side HG lies is -2.[(5 - 1)/(1 - 3)] = 4/-2 = -2Thus the opposite sides EF and HG of the quadrilateral EFGH are parallel.The slope of the line where the side EH lies is 2/5.[(1 - -1)/(3 - -2)] = 2/5The slope of the line where the side FG lies is 2/5.[(5 - 3)/(1 - -4)] = 2/5Thus the opposite sides EH and FG of the quadrilateral EFGH are parallel.Since any of the slopes is not a negative reciprocal of the others, then the two adjacent sides are not perpendicular. And if we look at the diagram, we clearly see that this quadrilateral can be a parallelogram.The length of the side EF is square root of 20.Square root of [(5 - 1)^2 + (1 - 3)^2] = Square root of [4^2 + (-2)^2] = Square root of 20.The length of the side HG is square root of 20.Square root of [(3 - -1)^2 + (-4 - -2)^2] = Square root of [4^2 + (-2)^2] = Square root of 20.Thus, the opposite sides EF and HG of the quadrilateral EFGH are congruent.Two other parallel sides, EH and FG, must be congruent also, since they intersect two other parallel lines (their length is equal to square root of 29).So we verified that the quadrilateral EFGH is a parallelogram.
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.
false
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.
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