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The following is a list of centroids of various two-dimensional and three-dimensional objects. The centroid of an object in -dimensional space is the intersection of all hyperplanes that divide into two parts of equal moment about the hyperplane.
For a semicircle with a diameter of a + b, the length of its radius is the arithmetic mean of a and b (since the radius is half of the diameter). The geometric mean can be found by dividing the diameter into two segments of lengths a and b, and then connecting their common endpoint to the semicircle with a segment perpendicular to the diameter ...
Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the shape must lie on this line . Divide the shape into two other rectangles, as shown in fig 3. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids.
The "vertex centroid" comes from considering the polygon as being empty but having equal masses at its vertices. The "side centroid" comes from considering the sides to have constant mass per unit length. The usual centre, called just the centroid (centre of area) comes from considering the surface of the polygon as having constant density ...
Regular polygons; Description Figure Second moment of area Comment A filled regular (equiliteral) triangle with a side length of a = = [6] The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.
If is held constant, and the radius is allowed to vary, then we have = As the central angle approaches π, the area of the segment is converging to the area of a semicircle, π R 2 2 {\displaystyle {\tfrac {\pi R^{2}}{2}}} , so a good approximation is a delta offset from the latter area:
If and are the radii of the small semicircles of the arbelos, the radius of an Archimedean circle is equal to = + This radius is thus = +.. The Archimedean circle with center (as in the figure at right) is tangent to the tangents from the centers of the small semicircles to the other small semicircle.
Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length. Using the fact that one part of one chord times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2r − x)x = (y / 2) 2. Solving for r, we find the required result.