Exploration 4

Centroids: Centers of Gravity

By: John Watson


Let’s start this exploration with a review of vocabulary related to triangles…

The medians of a triangle are the three lines that join each vertex with the midpoints of the opposite side. The centroid of a triangle is the point of concurrency of its three medians.

The centroid of a triangle is also referred to as its “center of gravity”.

The center of gravity of a 12” ruler would be located in the middle of the ruler at 6 inches. This makes sense because there are 6 inches of ruler on either side of this point.

Likewise, the centroid of a triangle is its center of gravity because each median divides the triangle in half, creating six smaller triangles of equal area shown below…

But how do we know that the 6 triangles have equal area?

First, let’s think of how we calculate the area of a triangle. The area of a triangle can be found using the formula…

A = b(h)   

b = the length of the base

h = the length of the height

If we know that two triangles have bases of the same length and heights of the same length, then we know that the triangles have the same area.


Let’s take a look at the median, BE, which divides ABC into two triangles,
BEA and BEC

(i) Point E is the midpoint of AC, so the segments EA and EC are of equal length. So, we can say that BEA and BEC have bases of the same length.

(ii) Notice that BEA and BEC also have the same height, which is equal to the length of BI.

(iii) BEA and BEC have equal bases and heights, so they must also have the same area. Let’s say the Area of BEA = Area of BEC= X

(iv) Since EA and EC are of equal length, we can say that DEA and DEC have bases of the same length. Notice that DEA and DEC also have the same height, which is the length of segment DH.

(v) DEAand DEC have equal bases and heights, so they must also have the same area. Let’s say the Area of DEA = Area of DEC= Y


Now let’s look at BDA and BDC

(vi)      Notice how…
Area of BDA = Area of BEA – Area of DEA
Area of BDA
= X – Y

Also,
Area of BDC = Area of BEC – Area of DEC
Area of BDC
= X – Y

Thus, by transitivity, we know that the Area of BDA = Area of BDC.


Now let’s take a look at another median GC that divides ABC into two triangles, CGA and CGB

(vii) Point G is the midpoint of AB, so the segments GA and GB are of equal length. So, we can say that CGA and CGB have bases of the same length. Notice that CGA and CGB also have the same height, which is equal to the length of CJ.

(viii) CGA and CGBhave equal bases and equal heights, so they must also have the same area. Let’s say the Area of CGA = Area of CGB = N

(ix) Since the segments GA and GB are of equal length, DGA and DGB have bases of the same length. Notice that DGA and DGB also have the same height, which is equal to the length of DK.

(x) DGA and DGB have equal bases and heights, so they must also have the same area. Let’s say the Area of DGA = Area of DGB = M


Now let’s look at CDA and BDC

(xi) Notice how…
Area of CDA = Area of CGA – Area of DGA
Area of CDA
= N – M

Also,
Area of BDC = Area of CGB – Area of DGB
Area of BDC
= N – M

Thus, by transitivity, the Area of CDA = Area of BDC.


(xii)

From (vi) we know that… Area of BDA = Area of BDC
From (xi) we know that… Area of CDA = Area of BDC

Then, by transitivity, Area of BDA = Area of BDC = Area of CDA !!!
Since all of these three triangles are of equal area, we know they each make up 1/3 of the area of ABC.

(xiii)  

From (v) we know that the Area of DEA = Area of DEC. Since they are of equal area, they both make up ½ of the Area of CDA.

From (x) we know that the Area of DGA = Area of DGB. Since they are of equal area, they both make up ½ of the Area of BDA.

If we use an argument similar to (iv), we can show that the Area of CDF = Area of BDF. Since they are of equal area, they both make up ½ of the Area of BDC.

 

(xiv)   

Now, from (xii) and (xiii), we know that each of the area of the 6 small triangles is equal to half of 1/3 of the area of ABC.

So, we can say that each of the 6 triangles above makes up 1/6 of the total area. This makes the centroid D the center of gravity!


Try this!
Find a large piece of cardboard and cut out a triangle. Try to balance your triangle on the end of a sharpened pencil. Assuming your cardboard is uniformly dense, your balancing point will be the centroid.

   

 


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