 Students who are interested to learn how to find the centroid of a triangle can refer to this page. You can find different types of questions in the Worksheet on Finding the Centroid of a Triangle. Before finding the centroid of the triangle recall the formula for finding the centroid of a triangle. You can solve the problems easily only if you learn the formulas related to the centroid of a triangle. Gain immense practice using this Free Printable Centroid of a Triangle Worksheet with Answers and gain adequate skills regarding the concept.

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Centroid of a Triangle Worksheet PDF

Example 1.
Calculate the coordinates of the centroid of the triangle ABC is A(6,2) B(0,2) C(1,4)

Solution:

We know that
Centroid formula =( x1 + x2 + x3/3, y1 + y2 + y3/3)
(6 + 0 + 1 /3 , 2 + 2 + 4 /3)
(7/3, 8/3)

Example 2.
A(3,x), B(4,3) and C( y,2) are the vertices of the triangle ABC whose centroid is the origin and calculate the values of x and y.

Solution:

Given centroid of the triangle ABC is the origin
We know that the coordinates of the point dividing (x1, y1) and (x2,y2) in the ratio m:n is given by
(m1x1 + m2xx1/m1 + m2 , m1y2 + m2y1/m1 + m2)
(m1, m2) = (0,0)
(0,0) = (3 + 4 + y/3, x + 3 + 2 /3)
(0,0) = (7 + y/3, x + 5/3)
0 = 7 + y/3 and 0 = x + 5/3
y = -7 and x = -5

Example 3.
Find the centroid of the triangle whose vertices are (1,2) (3,4) and (5,6).

Solution:

Let the vertices of the triangle be A(1,2) (3,4) and (5,6).
The centroid of triangle (G) = (x1 + x2 + x3/3, y1 + y2 + y3 /3)
= (1 + 3 + 5/3 , 2 + 4 + 6/3)
= (9/3 ,2/3)
=( 3,4)

Example 4.
If the centroid of the triangle is at (4,2) and two of its vertices are (3,2) and (5,2) then find the third vertex of the triangle.

Solution:

Let the vertices of a triangle be A(3,2) B(5,2) and c(x3,y3)
The centroid of the triangle is (4,2)
Centroid formula = (x1 + x2 + x3/3 , y1 + y2 + y3/3)
(4,2) = (3 + 5 + x3/3 , 2 + 2 + y3/3)
(4,2) = (8 + x3/3, 4 + y3/3)
8 + x3/3 = 4
8 + x3 = 12
x3 = 12 – 8
x3 = 4.
4 + y3/3 = 2
4 + y3 = 6
y3 = 6-4
y3 = 2
The third vertex is (4,2)

Example 5.
Find the length of the median through A of a triangle whose vertices are A(1,3) B(1,1) and C(5,1)

Solution:

AD is the median of the ∆ABC.
D is the mid point of BC
Midpoint of a line = (x1 + x2/2, y1 + y2/2)
Midpoint of BC =( 1 + 5/2 , 1 + ½)
= (6/2, 2/2) = (3,1)
Length of the median AD = √(x2 – x1)² + (y3 – y1)²
= √(3 + 1)² + (0 + 3)²
= √4² + 3²
= √16 + 9
√25 = 5
The length of the median AD is 5 units.

Example 6.
The vertices of a triangle are (1,2) (h,3) and (4,k) if the centroid of the triangle is at the point (5,1) then find the value of √(h + k)² + (h + 3k)²

Solution:

Let the vertices A(1,2) B(h,3) and c(4,k).
Centroid of a ∆ABC = (x1 + x2 + x3/3, y1 + y2 + y3/3).
(5,1) = (1 + h + 4/3, 2 + 3 + k/3)
(5 + h/3, 5 + k/3)
5 + h/3 = 5
5 + h = 15
h = 15 – 5
h = 10
5 + k/3 = 1
5 + k = 3
k = 3 – 5
k = -2.
The value of √(h + k)² + (h + 3k)²
= √(10 – 2)² + (10+ 3(-2))²
= √8² + (10 – 6)²
= √8² + 4²
= √64 + 16
= √80
= 8.9

Example 7.
The A(h,6) B(2,3) and C(6,k) are the coordinates of vertices of a triangle whose centroid is G(1,5) find h and k.

Solution:

Given that
A(x1,y2) = A(h,6)
B(x2,y2) = B(2,3)
C(x3,y3) = C(6,k)
Centroid G(x,y) = G(1,5)
G is the centroid.
By centroid formula
x = x1 + x2 + x3/3
1 = h + 2 + 6/3
3 = h + 2+ 6
3 = h + 8
h = 3 – 8
h = -5.
y = y1 + y2 + y3/3
5 = 6 + 3 + k/3
15 = 9 + k
k = 15 – 9
k = 6
(h,k) = (-5, 6)

Example 8.
Find the coordinates of the point of trisection of the line segment AB with A(2,7) and B(4,8).

Solution:

Let P(x1, y1) and Q(x2, y2) divide the line AB into 3 equal parts.
AP = PQ = QB
AP/PB = AP/PQ + QB = AP/AP + AP = QB/2AP = ½
So, P divides AB in the ratio 1:2.
x1 = 1×4 + 2×2/2 + 1 = 0
y1 = 1×8 + 2×7/2 + 1 = 2
(x1,y1) = (0,2)
Also
AQ/QB = AP+PB/QB. QB + QB/QB = 2/1
x2 = 2×4 + 1×2/2 + 1 = 8 +⅔ = 10/3
y2 = 2×8 + 1×7/2 + 1 = 16 + 7/3 = 23/3
(x2, y2) = (10/3, 23/3)

Example 9.
The coordinates of the centroid of the triangle PQR are (2,5) if a = (6,5) and R = (11,8) calculate the coordinates of the vertex D.

Solution:

Let G be the centroid of DPQR
Whose coordinates are (2,5)
And let (x,y) be the coordinates of vertex P. Coordinates of G are
G(2,5) = G(x + 6 + 11/3, y + 5 + 8/3)
2 = x + 5/3
6 = x + 5
6 – 5 = x
x = 1
5 = y + 13/3
15 = y + 13
15 – 3 = y
y = 2
Coordinates of vertex P are (1,2).

Example 10.
A(5,x), B(4,3) and C(y,2) are the vertices of the triangle ABC whose centroid is the origin and calculate the values of x and y.

Solution:

Given the centroid of the triangle WBC is the origin.
(0,0) = (5+4+y/3, x+3+2/3)
(0,0) = (9 + y/3 , x + 5/3)
9 + y/3 = 0
9 + y = 0
y = -9.
x+5/3 = 0
x + 5 = 0
x = -5
(x,y) = (-9,-5)