The Chi squared distribution
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Triangular lamina
The centre of mass lies at the intersection of the three medians. [Diagram goes here - download the original to see it.] If AX is a median and G the centre of mass then AG : AX = 2 : 1 i.e. G lies [Equation goes here - download the original to see it.] along the median from the vertex. To see why: consider that if the triangle is divided into little strips then the mass on one side of the median equals the mass on the other [Diagram goes here - download the original to see it.] There are two ways of proving this (i) Using the principle of moments by modelling the lamina as three equal masses at each of the vertices. [Diagram goes here - download the original to see it.] (ii) Using vectors. [Diagram goes here - download the original to see it.] Let the triangle be OAB and let O be the origin. The line joining O to the mid-point of AB has vector equation. [Equation goes here - download the original to see it.] The vector joining B to the mid-point of OA (which is X) has equation [Equation goes here - download the original to see it.] The line joining X to B has vector equation: [Equation goes here - download the original to see it.] At the centre of mass the two lines join, hence [Equation goes here - download the original to see it.] That is, [Equation goes here - download the original to see it.] Therefore, [Equation goes here - download the original to see it.] Uncoupling [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] That is G lies [Equation goes here - download the original to see it. ] from X or [Equation goes here - download the original to see it.] rom B and similarly for every median.
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Contents of
The Chi squared distribution
1 The chi squared Distribution
2 The Concept of a Centre of Mass
3 Centre of Mass of two-dimensional objects
4 Triangular lamina
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