In geometry, a triangle is quite an elementary figure. It is a threesided bounded figure. A triangle has 3 interior angles. Triangles can be classified on the basis of number of sides as well as number of angles, such as  equilateral, isosceles, scalene and acuteangled, obtuseangled, rightangled triangle. There are various concepts related to triangles. Centroid is an important property of a triangle. The centroid of a triangle is the center of the triangle. It is defined as the point of intersection of all the three medians of a triangle; where, a median is a line segment joining the midpoint of a side and opposite vertex.
If we have an object, then we can say that the centroid of that object is its center. The centroid of a triangle is the center point of it where its center of gravity is located. The centroid of the triangle separates the medians in the ratio 2:1 ; i.e. the length of the portion of median from vertex to centroid is twice the length of portion of median from centroid to the midpoint. We come across several concepts and formulae related to centroid of a triangle. It is quite commonly used concept in geometry. In this page, we will learn more about centroid of a triangle and various applications related to it.
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The
centroid of triangle can be got by finding the average of the
xcoordinate's value and the average of the ycoordinate's value of all
the vertices of the triangle.
Centroid size: Centroid of any geometric figure is a point. And so, it looks like a dot.
In the figure shown below, the three vertices of the triangle are $A(x_1, y_1)$ , $B(x_2, y_2)$ and $C(x_3, y_3)$.
Centroid of points, A, B and C is $\frac{x_1 + x_2 + x_3}{3}$, $\frac{y_1 + y_2 + y_3}{3}$.
Shapes  Figure 
$\bar {x}$  $\bar{y}$  Area 

1  Triangle    $\frac{h}{3}$  $\frac{bh}{2}$  
2  Semicircle  0  $\frac{4R}{3\pi}$  $\frac{1}{2}$$\pi R^2$  
3  Quarter circle  $\frac{4R}{3\pi}$ 
$\frac{4R}{3\pi}$  $\frac{1}{4}$$\pi R^2$  
4  Sector of a circle

$\frac{2R}{3 a}$ sin a  0  a$R^2$ 

5  Parabola  0  $\frac{3h}{5}$  $\frac{4ah}{3}$ 

6  Semi Parabola  $\frac{3a}{8}$  $\frac{3h}{5}$ 
$\frac{2ah}{3}$ 

7  Arc of Circle  Points on the curve, r = $\rho $ from $\theta$ = $\alpha$ to $\theta$ = $\alpha$ 
$\frac{\rho sin \alpha}{\alpha}$  0 
2 $\alpha$$\rho$ 
8 
Right Triangular  $\frac{b}{3}$  $\frac{h}{3}$  $\frac{bh}{2}$  
9  Parabolic spandrel 
$\frac{3a}{4}$  $\frac{3h}{10}$  $\frac{ah}{3}$ 
Given that the vertices are A(4, 8), B(2, 6) and C(0, 10)
A(4, 8) is $A(x_1, y_1)$
C(0, 10) is $C(x_3, y_3)$
The formula to calculate the centroid of the triangle is $(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$.
= $(\frac{4 + 2 + 0}{3}, \frac{8 + 6 + 10}{3})$
= $(\frac{6}{3}, \frac{24}{3})$
= (2, 8)
The centroid coordinates of the triangle with vertices A(4, 8), B(2, 6) and C(0, 10) is (2, 8).
Given that the vertices are A(5, 4), B(6, 1) and C(9, 0).
A(5, 4) is $A(x_1, y_1)$
B(6, 1) is $B(x_2, y_2)$
C(9, 0) is $C(x_3, y_3)$
The formula to calculate the centroid of the triangle is $(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$.
= $(\frac{5 + 6 + 9}{3}, \frac{4 + 1 + 0}{3})$
= $(\frac{20}{3}, \frac{5}{3})$
= (6.67, 1.67)
The centroid coordinates of the triangle with vertices A(5, 4), B(6, 1) and C(9, 0)is (6.67, 1.67).
More topics in Centroid  
Centroid of a Triangle  
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