Centroid

In Geometry, the centroid is an important concept related to a triangle. A triangle is a three-sided bounded figure with three interior angles. Based on the sides and angles, a triangle can be classified into different types such as

  • Scalene triangle
  • Isosceles triangle
  • Equilateral triangle
  • Acute-angled triangle
  • Obtuse-angled triangle
  • Right-angled triangle

The centroid is an important property of a triangle. Let us discuss the definition of centroid, formula, properties and centroid for different geometric shapes in detail.

Centroid Definition

The centroid is the centre point of the object. The point in which the three medians of the triangle intersect is known as the centroid of a triangle. It is also defined as the point of intersection of all the three medians. The median is a line that joins the midpoint of a side and the opposite vertex of the triangle. The centroid of the triangle separates the median in the ratio of 2: 1. It can be found by taking the average of x- coordinate points and y-coordinate points of all the vertices of the triangle.

Properties of centroid

The properties of the centroid are as follows:

  • The centroid is the centre of the object
  • It is the centre of gravity
  • It should always lie inside the object
  • It is the point of concurrency of the medians

Centroid Formula

Centroid of a Triangle

Let’s consider a triangle. If the three vertices of the triangle are A(x1, y1), B(x2, y2), C(x3, y3), then the centroid of a triangle can be calculated by taking the average of X and Y coordinate points of all three vertices. Therefore, the centroid of a triangle can be written as

Centroid of a triangle = ((x1+x2+x3)/3, (y1+y2+y3)/3)

Centroid of Different Shapes

Here, the list of centroid formula is given for different geometrical shapes.

Shapes Figure ȳ Area
Triangular area Triangular area h/3 bh/2
Quarter-circular area Quarter-circular area 4r/3π 4r/3π πr2/4
Semicircular area Semicircular area 0 4r/3π πr2/2
Quarter-elliptical area Quarter-elliptical area 4a/3π 4b/3π πab/4
Semi elliptical area Semi elliptical area 0 4b/3π πab/2
Semiparabolic area Semiparabolic area 3a/8 3h/5 2ah/3
Parabolic area Parabolic area 0 3h/5 4ah/3
Parabolic spandrel Parabolic spandrel 3a/4 3h/10 ah/3

Solved Problem

Question:

Find the centroid of the triangle whose vertices are A(2, 6), B(4,8), and C(6,12).

Solution:

Given:

A(x1, y1) = A(2, 6)

B(x2, y2) = B(4,9)

C(x3, y3) = C(6,15)

We know that the formula to find the centroid of a triangle is = ((x1+x2+x3)/3, (y1+y2+y3)/3)

Now, substitute the given values in the formula

Centroid of a triangle = ((2+4+6)/3, (6+9+15)/3)

= (12/3, 30/3)

= (4, 10)

Therefore, the centroid of the triangle for the given vertices A(2, 6), B(4,8), and C(6,12) is (4, 10).

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