In this exercise you will use the principle above to determine the centroid of p

Question

In this exercise you will use the principle above to determine the centroid of pentagon which is shaped like a child's drawing of a house.
Let A, B, C, D and E be points in the plane such that the quadrilateral ABCD is a square with side length s and such that E lies outside this square and such that AEB is a triangle such that AEB is a right triangle; let P be the pentagon having the above vertices.

First determine the centroid of the square and the centroid of the triangle; then use the principle in the previous exercise to determine the centroid of P. (You willl need to choose coordinates for the vertices; try to make choices which simplify your calculations. You should prefer arguments which take advantage of the symmetry to arguments which use brute force calculations.)

Explanation / Answer

Easiest way would be to drop a bisector from two of the angles. The intersection will be the centroid. This is assuming a regular pentagon. If it is an irregular pentagon, you could tie a weight to a string. Hold the pentagon by a vertex while also holding the string. This line passes through the center of gravity. Do the same for another vertex. The lines will intersect at the centroid (aka center of gravity). This method would work for a regular pentagon also since the lines would be the angle bisectors.

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