Exploring Analyic Geometry with Mathematica®

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Centroid of a Triangle

tricent.html

Exploration

"tricent_1.gif"

Graphics saved as "tri08.eps".

Show that the centroid of a triangle, as illustrated in the figure, is on a line at a distance "tricent_2.gif" from the base of the triangle.

Approach

Place the triangle in a convenient position as shown in the figure. Create equations for the moments of inertia of infinitesimal areas on either side of the centroid line. Use integration to find the ordinate of the line.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Create lines for two sides of the triangle whose vertices are (0,0), (d,0) and (a,b).

Clear[a,b,d];
L1=Line2D[{a,b},{0,0}];
L2=Line2D[{a,b},{d,0}];

Construct a horizontal line at a general coordinate, y, which is the height of the centroid.

Clear[y];
L=Line2D[Point2D[0,y],0];

The width of the triangle, W, is the difference between the abscissa of the intersection points of the sides and the horizontal line.

W=XCoordinate2D[Point2D[L,L2]]-
  XCoordinate2D[Point2D[L,L1]] //Simplify

"tricent_3.gif"

The moments of inertia on each side of the horizontal line must be equal.

Clear[yB];
Solve[Integrate[W*(yB-y),{y,0,yB}]==
      Integrate[W*(y-yB),{y,yB,b}],yB]

"tricent_4.gif"

Discussion

Notice that the centroid only depends on the height of the triangle, b. The centroid's height does not depend on the horizontal location of the apex point, a, nor on the width of the base, d.


Copyright © 1999-2007 Donald L. Vossler, Descarta2D Publishing
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