**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

Eis the midpoint ofAC, so the segmentsEAandECare of equal length. So, we can say thatBEAandBEChave bases of the same length.(ii) Notice that

BEAandBECalso have the same height, which is equal to the length ofBI.(iii)

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

EAandECare of equal length, we can say thatDEAandDEChave bases of the same length. Notice thatDEAandDECalso have the same height, which is the length ofsegment DH.(v)

DEAandDEChave equal bases and heights, so they must also have the same area. Let’s say theArea 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

Gis the midpoint ofAB, so the segmentsGAandGBare of equal length. So, we can say thatCGAandCGBhave bases of the same length. Notice thatCGAandCGBalso have the same height, which is equal to the length ofCJ.(viii)

CGAandCGBhave equal bases and equal heights, so they must also have the same area. Let’s say the Area ofCGA= Area ofCGB= N(ix) Since the segments

GAandGBare of equal length,DGAandDGBhave bases of the same length. Notice thatDGAandDGBalso have the same height, which is equal to the length ofDK.(x)

DGAandDGBhave equal bases and heights, so they must also have the same area. Let’s say the Area ofDGA= Area ofDGB= 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 ofDEC. Since they are of equal area, they both make up ½ of the Area ofCDA.From (x) we know that the Area of

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

CDF= Area ofBDF. Since they are of equal area, they both make up ½ of the Area ofBDC.

(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

Dthe 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.