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.
The equation EFGH times 4 equals EFGH implies that multiplying EFGH by 4 results in the same value, which can only be true if EFGH is equal to 0. Therefore, the solution is EFGH = 0.
To determine the series of transformations that maps quadrilateral EFGH onto its image, we need the coordinates of the vertices of EFGH and its image. Typically, transformations can include translations, rotations, reflections, and dilations. For example, if EFGH is translated 3 units right and 2 units up, the new coordinates of its vertices would be calculated by adding (3, 2) to each vertex's coordinates. If further transformations are needed, such as a rotation of 90 degrees counterclockwise around the origin, the new coordinates can be calculated using the rotation matrix. Please provide the coordinates for precise calculations.
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.
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 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 of proportionality between the sides of abcd and the corresponding sides of efgh.
4
The question cannot be answered without information about the relative sizes of the two polygons.
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 ?"
12
It is the scale factor times the length of ad.
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.
It is k times the length of Ad where k is the constant of proportionality between the two shapes.
Without the diagram ; cannot do!!!! 'E' 5th letter of the alphabet 'F' 6th letter of the alphabet.