The Centroid of the Triangle (Powerful geometry tricks to score 100%)

The point of intersection of all three Medians of the triangle is called the ‘Centroid’ of the triangle. The Median of a triangle is a line segment joining a vertex to the midpoint of its opposite side, thus bisecting the side.

The Median divides a triangle into two parts of equal areas. Here Area of ∆BAD =Area of ∆DAC.

All the Medians of a triangle are concurrent (intersect at a single point), the point of concurrence is called the Centroid of the triangle. The Centroid always lies inside the triangle. 

• Finding the centroid of a triangle: The centroid divides each median in the ratio of 2:1

Here AG:GP=BG:GQ=CG:GR=2:1 

• All the medians divide a triangle into 6 parts of equal areas.

The Centroid and the Center of Gravity

A homogeneous triangular sheet can be balanced on its Centroid. This means that the triangle is stabilized in a horizontal position when you place its Centroid on the tip of a Pencil.

In a homogeneous object, the density of matter is evenly distributed throughout its body.

The triangle made by Medians

Find the area of the triangle when the lengths of its medians are given.

Here, Area of ∆ABC: Area of ∆LMN=4:3

Step 1: Draw a line PD=GP, by extending the line AP.

Let AP=3x, BQ=3y & CR=3z

Note: The Centroid divides each median in the ratio of 2:1

∆BPD ⩭ ∆CPG (Congruent Triangles) 

∠BPD=∠CPG (Vertically Opposite)

BP=PC & GP=PD

As we know that all the medians divide a triangle into 6 parts of equal areas. 

If the area of ∆ABC=6 unit

Then the area of ∆DBG=2 unit

Thus, the area of ∆ABC=3×area of ∆DBG

Now, Since ∆DBG ~ ∆MLN (Similar) 

Note: Ratio of respective sides=2:3

So, the area of ∆DBG: area of ∆MLN=2²: 3²=4:9

The area of ∆DBG=4/9×area of ∆MLN

1/3×area of ∆ABC=4/9×area of ∆MLN

Area of ∆ABC=4/3×area of ∆MLN

Right angle property of Medians

In ∆ABC, G is the Centroid of the triangle. If AG=BC then find the value of ∠BGC.

Let, AP=3x 

Then AG=2x, GP=x (Property), and BC=2x (Given)

Here, ∆BPG (BP=PG) & ∆CPG (PC=PG) are the Isosceles triangles.

In ∆BGC, the sum of all internal angles =180°

∠CBG+∠BGC+∠GCB=180°

2(k+k’)=180°

k+k’=180°

∠BGC=90°

Coordinates of the Centroid of the triangle

Find the coordinates of the Centroid of the triangle if the coordinates of its vertices are given:

Coordinates of Centroid G=(x,y)

(x,y)=(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})

Stewart’s Theorem

To find the relation between the Cevian and the sides of a triangle, Stewart’s Theorem is used.

Let, In △ABC the lengths of the sides: BC=a, CA=b, and AB=c. Here we have to find the length of the Cevian AD.

Let ∠ADB=θ then ∠ADC=180°-θ (Linear Pair)

Here BD=m, DC=n, and BC=a

Let AD=d

In △ABD, Use Cosine Formula

c²=m²+d²-2md×Cos(θ)  ……….(i)

In △ADC, Use Cosine Formula

b²=n²+d²-2nd×Cos(180°-θ)

b²=n²+d²+2nd×Cos(θ)  ……….(ii)

Note: Cos(180°-θ)= -Cos(θ) 

Now we have to eliminate the Cosine term from the equations. So multiply equation (i) by n and equation (ii) by m, and after that add both equations.

n.c²=n.m²+n.d²-2mnd×Cos(θ) 

m.b²=m.n²+m.d²+2mnd×Cos(θ)

After addition:

m.b²+n.c²=m.n.(m+n)+d².(m+n)

m.b²+n.c²=(m+n)(d²+m.n)

m.b²+n.c²=a.(d²+m.n) ………(iii)

This equation (iii) is known as Stewart’s Theorem.

In case m=n or AD is the Median of the triangle ABC.

m.b²+n.c²=(m+n)(d²+m.n)

Put n=m

m.(b²+c²)=2m.(d²+m²)

(b²+c²)=2.(d²+m²) ………(iv)

Here m=a/2

This equation (iii) is known as Apollonius’s Theorem. (Centroid of the triangle)

The Length of the Median (Apollonius’s Theorem)

Let, the length of the median AP=d, BC=a, and BP=PC=m (BC=2m)

b²+c²=2(m²+d²)

Put m=BC/2=a/2

b²+c²=1/2*(a²+4d²)

The sum of the squares of the lengths of medians

Apply Apollonius Theorem:

AB²+AC²=1/2*(BC²+4AP²)

AC²+BC²=1/2*(AB²+4CR²)

BC²+AB²=1/2*(AC²+4BQ²)

Add all equations:

2(AB²+BC²+CA²)=1/2*[(AB²+BC²+CA²)+4(AP²+BQ²+CR²)]

3/2*(AB²+BC²+CA²)=2(AP²+BQ²+CR²)

(AB²+BC²+CA²)=4/3*(AP²+BQ²+CR²)

Triangle, made by joining the midpoints of the sides

In the given figure: 

RQ∥BC & RQ=½*BC

PQ∥AB & PQ=½*AB

PR∥AC & PR=½*AC

The area of ∆PQR=1/4×area of ∆ABC

Note: The centroid of an equilateral triangle also works as its incenter, circumcenter, and orthocenter. In other words, all are the same. 

Difference between the Centroid of the triangle and its Circumcenter

Que 1: In the given figure, E is the centroid of the triangle and  BCDE is a Parallelogram then find the coordinates of the point ‘D’.

Solution: The coordinates of Centroid (E)= (\frac{3-2+5}{3}, \frac{4+1+3}{3}) =(2, 8/3)

Let the Coordiantes of point D= (a,b)

Use Formula: The coordinates of mid-point between points M(x₁,y₁) and N(x₂,y₂)= (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Since, O is the mid-point of BD and EC.

So, The coordinates of the Mid-point of BD= The coordinates of the Mid-point of EC

(\frac{5+2}{2}, \frac{8/3+3}{2})=(\frac{-2+a}{2}, \frac{1+b}{2}).

\frac{-2+a}{2}=\frac{7}{2}.

a=9 ………..(i)

\frac{1+b}{2}=\frac{17/3}{2}.

b=14/3 …………(ii)

The coordinates of Point D= (9, 14/3)

Que 2: Find the coordinates of the centroid of a triangle whose sides are: 12x² -20xy+ 7y²=0 and 2x-3y+4=0.

Solution: 12x² -20xy+ 7y²=0, represents a pair of straight lines so first disassociate those two equations.

12x² -6xy -14xy+ 7y²=0

6x(2x-y) -7y(2x-y)=0

(6x-7y)(2x-y)=0

Now, 6x-7y=0 ……….. Side AB

2x-y=0 ………… Side BC

2x-3y= -4 ………. Side CA

Now, We have to find the coordinates of the vertices of △ABC by taking pair of straight lines and solving them by elimination method.

To find the coordinates of vertex A, we have to solve the equations of lines AB and AC.

6x- 7y=0   ……(i)

(2x- 3y= -4)×3 =6x -9y= -12 ……(ii)

Equation (i)-(ii): 2y=12

y=6

Put this value of y in equation (i): x=7

So, the coordinates of A=(7,6)

Similarly, the Coordinates of B=(0,0)

The coordinates of C=(1,2)

Now, the Coordinates of Centroid= (\frac{7+0+1}{3}, \frac{6+0+2}{3})= (\frac{8}{3}, \frac{8}{3})

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