If we obtain either a pentagon (parallelogram truncated at a vertex) or a trapezium depending on whether or respectively.įinally, if are less than 1 and are greater than 1, we obtain a hexagon.įor details on calculating the areas of such polygons refer to, especially the method applying the area cosine principle that relates an area of a figure to its projection. Next, if we obtain the following parallelogram. Similarly if all are less than 1, the oppositely similar triangle on the red vertices would be chosen. Firstly, if are all greater than 1 we choose the following triangle. Note that for the plane to intersect the cube at all we require to be on the different side of the plane from the origin, or in other words. A red vertex has a conflicting constraint with its neighbouring two blue vertices, so either a red point or one or more blue points in this area can be chosen. In this figure a vertex for the cross-sectional polygon is chosen if the constraint associated with it is satisfied. The type of polygon obtained depends on which vertices of the figure below are selected, as determined by the values of. To sum up, all of the possible cross sections of a cube where the plane is not parallel to an edge can be described by the intersection of two oppositely similar triangles with corresponding sides parallel. Side lengths of the triangles and distances between corresponding parallel sides may be found by Pythagoras’ theorem and are shown below for one pair of corresponding sides (the remaining lengths can be found by cyclically permuting ).
The centre of similarity of the two triangles is the intersection of two lines joining corresponding sides – this can be found to be the point, which is the intersection of the unit cube’s diagonal from the origin (to ) and the plane. The following diagram shows the coordinates of the vertices of the two triangles, which in this case intersect in a hexagon. Hence the two triangles are oppositely similar with a centre of similarity. Note that the triangles have parallel corresponding sides, being bounded by the pairs of parallel faces of the cube. This can be considered the intersection of the two regionsĮach of which is an acute-angled triangle in the same plane (acute because one can show that the sum of the squares of any two sides is strictly greater than the square of the third side). The cross section satisfies and the inequalities, and. In other words, we may assume the plane has equation and intercepts at and, where are positive. If the plane is not parallel to a face, we may set up a coordinate system where a unit cube is placed in the first octant aligned with the coordinate axes and the normal to the plane has positive x, y and z coordinates. That rectangle becomes a square if the plane is parallel to a face.
Firstly, if the plane is parallel to an edge (any of four parallel edges), the cross section can be seen to be a line or rectangle with the longer dimension of length at most times the other. Let us be systematic in determining properties of the cross sections above.
#CROSS SECTIONAL AREA OF RECTANGLE FORMULA HOW TO#
One can use to experiment interactively with cross sections given points on the edges or faces, while shows how to complete the cross section geometrically if one is given three points on the edges. Some are demonstrated in the video below too. The last five of these (the non-degenerate cases) are illustrated below and at.
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