The points of concurrency is an important topic in the study of properties of triangles. A median of a triangle is the line segment that joins the midpoint of a side to its opposite vertex. The centroid is the point of concurrency of all the three medians of the triangle. In the case of a triangular plate, the centroid is equivalent to the center of gravity of the plate where the entire weight of the triangle is balanced. In triangle ABC, D, E and F are correspondingly the midpoints of AB, BC and CA. The three medians AE, BF and CD meet at a common point G, which is called the centroid of the triangle ABC.

Median Centroid

The Medians of a triangle intersect at a point called the centroid of the triangle, which is at a distance of two thirds of the lengths of the medians from the corresponding vertices. If G is the centroid of the triangle ABC, according to the centroid theorem we have,
AG = $\frac{2}{3}$ AE , BG = $\frac{2}{3}$ BF and CG = $\frac{2}{3}$ CD

In other words the centroid is a point of trisection of each of the medians and it divides the medians in the ratio 2:1.

Centroid Formula

The coordinates of the centroid of a triangle is given by,

$(\frac{x_{1}+x_{2}+x_{3}}{3}$, $\frac{y_{1}+y_{2}+y_{3}}{3})$
where (x1, y1), (x2, y2) and (x3, y3) are the vertices of the triangle.

Example:
Find the centroid of the triangle whose vertices are given by (2, -3), (0, 5) and (4, 7).
Solution:
Using the formula, the coordinates of the centroid is $(\frac{2+0+4}{3},\frac{-3+5+7}{3})$ = (2, 3).

Properties of Centroid

1. The centroid of a triangle divides each of the median in the ratio 2 : 1.
2. The centroid falls within the triangle for all types of triangles acute, obtuse and right triangle.
3. Any line drawn through the centroid divides two sides of the triangle proportionally. Let the line through centroid G cut the sides AB and AC of triangle ABC at P and Q. According to Property stated above
$\frac{AP}{PB}$ = $\frac{AQ}{QC}$

Solved Example

Question: Prove that the centroid of a parallelogram is the point of intersection of its diagonals. Solution:

Consider the triangles ABC and BCD with the diagonal BC as the base.
O is the midpoint of BC    (Diagonals bisect in a parallelogram)
Hence AO and CO are medians on BC for triangles ABC and BCD.

This means the centroids G1 and G1 of the two triangles should lie correspondingly on AO and CO.
The two triangles ABC and BCD together make up the parallelogram ABCD.
Hence the centroid of the parallelogram is the centroid of the line segment G1G2.

The centroid of the line segment G1G2 is the midpoint O.
Hence the centroid of a parallelogram is the point of intersection of its diagonals.

Equilateral Triangle Centroid

In an equilateral triangle the medians, altitudes, angle bisectors and perpendicular bisectors of the sides all coincide. Hence, the points of concurrency of these lines also coincide. That means the centroid coincides with the orthocenter, incenter and circumcenter of the triangle. Centroid Construction

As centroid is the point of concurrence of the medians of the triangle, it is sufficient we construct any two medians of a triangle. The point of intersection of these medians is the centroid of the triangle. 1. Bisect sides AB and BC at D and E.
2. Join the medians AE and CD.
3. The point of intersection G of AE and CD is the centroid.

Centroid Table

A List of centroids for some common plane shapes as shown below. The x and y coordinates of the centroid in tune with the assumption on the position of the shapes are given along with the corresponding areas.

 Shape Position Coordinates Area xc yc Line Segment $\frac{l}{2}$ - length = l units Right Triangle $\frac{b}{3}$ $\frac{h}{3}$ $\frac{1}{2}$ bh sq.units Triangle $\frac{a+b}{3}$ $\frac{h}{3}$ $\frac{1}{2}$bh sq.units Rectangle $\frac{b}{2}$ $\frac{h}{2}$ bh sq.units Semi Circle r $\frac{4r}{3\pi }$ $\frac{1}{2}$$\pi$r2 sq.units

These coordinates are relevant only to the positions and measurements as shown. When the shapes are placed differently on the coordinate plane, the coordinates of the centroid will also change.

Centroid Practice Problems

Practice Problems

Question 1: In triangle PQR, G is the centroid. Given PG = 14, GS = 6 and GU = 5. Find the lengths of GT, QG, RG, PT, RS and QU.
(Answer : GT = 7, QG = 10, RG = 12, PT = 21, RS = 18 and QU = 15).
Question 2: Find the coordinate of the centroid of the triangle XYZ whose vertices are, X (4, -5),  Y (0, -7)  and Z ( -3, 5).
(Answer: G is the point ($\frac{1}{3}$, -$\frac{7}{3}$)