# 8.3: Medians and centroid - Mathematics

A median of a triangle is the segment joining a vertex to the midpoint of the opposing side.

Theorem (PageIndex{1})

The three medians of any nondegenerate triangle intersect in a single point. Moreover, the point of intersection divides each median in the ratio 2:1.

The point of intersection of medians is called the centroid of the tri- angle; it is usually denoted by M. In the proof we will apply Exercise 3.4.3 and Exercise7.3.1; their complete solutions are given in the hits.

Proof

Consider a nondegenerate triangle (ABC). Let ([AA']) and ([BB']) be its medians. According to Exercise 3.4.3, ([AA']) and ([BB']) have a point of intersection; denote it by (M). Draw a line (ell) thru (A') parallel to ((BB')). Applying Exercise7.3.1 for ( riangle BB'C) and (ell), we get that (ell) cross ([B'C]) at some point (X) and

(dfrac{CX}{CB'} = dfrac{CA'}{CB} = dfrac{1}{2};)

that is, (X) is the midpoint of ([CB']).

Since (B') is the midpoint of ([AC]) and (X) is the midpoint of ([B'C]), we get that

(dfrac{AB'}{AX} = dfrac{2}{3}.)

Applying Exercise 7.3.1 for ( riangle XA'A) and the line ((BB')), we get that

[dfrac{AM}{AA'} = dfrac{AB'}{AX} = dfrac{2}{3};]

that is, (M) divides ([AA']) in the ratio 2:1.

Note that 8.3.1 uniquely defines (M) on ([AA']). Repeating the same argument for medians ([AA']) and ([CC']), we get that they intersect at (M) as well, hence the result.

Exercise (PageIndex{1})

Let (square ABCD) be a nondegenerate quadrangle and (X, Y, V, W) be the midpoints of its sides ([AB], [BC], [CD]), and ([DA]). Show that (square XYVW) is a parallelogram.

Hint

Use the idea from the proof of Theorem (PageIndex{1}) to show that ((XY) parallel (AC) parallel (VW)) and ((XV) parallel (BD) parallel (YW)).

Every triangle has a single point within it that allows the triangle to balance perfectly, if the triangle is made of only one material. That point is called the centriod or the ‘center of gravity’ or ‘center of mass’ of the triangle.

Mathematically a centroid of a triangle is defined as the point where three medians of a triangle meet. It is one of the three points of concurrency in a triangle along with the incenter, circumcenter, and orthocenter. A centroid is represented typically by the symbol ‘G’.

Shown below is a ΔABC with centroid ‘G’.

## Apollonius’ Theorem In a triangle, it is true that the sum of the squares of two of its sides is equal to the sum of half the square of the third side and twice the square of the median corresponding to this third side. Where a, b, and c, are the legs and mb is the median corresponding to side b.

## Special Property of the Centroid

In the exercise below, the centroid (point C) has already been found for you. For this exercise, find the area of each portion of the overall triangle. Each portion (there are 6 total) is shaded a different color. First select the Area tool , which is found under the Angle menu . Then simply click on each of the individual portions. Answer the questions that follow. What do you notice about the area of each smaller triangle?

Recall that the area of a shape is the amount of space inside that shape. Based on this, what could we say the centroid is the center of?

Assume that the triangle portions all have a mass (similar to weight) that is distributed evenly across their entire area. Based on this, what could we now say the centroid is the center of? (*Hint, we could also call it the balancing point).

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Concepts covered in Mathematics 8th Standard Maharashtra State Board chapter 4 Altitudes and Medians of a triangle are Altitudes of a Triangle, Median of a Triangle, Constructing an Altitude of a Triangle, Draw Medians of a Triangle.

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## Altitudes And Medians Of A Triangle

ii. Draw a perpendicular from vertex P on the side QR using a set - square. Name the point where it meets side QR as M. Seg PM is an altitude on side QR.

iii. Considering side PR as a base, draw an altitude QX on side XZ. Seg QX is an altitude on side PR.

iv. Consider side PQ as a base, draw an altitude RN on seg PQ. Seg RN is an altitude on side PQ.

Seg PM, seg QO, seg RN are the altitudes of ∆PQR. The point of concurrence i.e., the orthocentre is denoted by the point O.

Draw an obtuse-angled ∆STV. Draw its medians and show the centroid.

To draw an obtuse-angled ∆STV.

i. Draw a base line of any length, mark it TV. At T draw an obtuse angle mark that line point S. Join S and V points. ΔSTV thus formed is an obtuse angled triangle.

ii. Find the mid-point A of side TV, by constructing the perpendicular bisector of the line segment TV. Draw AS.

iii. Find the mid-point B of side SV, by constructing the perpendicular bisector of the line segment SV. Draw seg BT.

iv. Find the mid-point C of side ST, by constructing the perpendicular bisector of the line segment ST. Draw seg CV.

Seg AS, seg BT and seg CV are medians of ∆STV.

Their point of concurrence is denoted by O.

Draw an obtuse-angled ∆LMN. Draw its altitudes and denote the orthocentre by ‘O’.

To draw an obtuse-angled ∆LMN.

i. Draw a base line of any length, mark it MN. At M draw an obtuse angle mark that line point L. Join L and N points. ΔLMN thus formed is an obtuse angled triangle.

ii. To draw an altitude from vertex L, extend side MN of the triangle from point M with a dashed line, as shown in the figure, and then draw the perpendicular lines from M.

iii. Considering side LN as a base, draw an altitude MP on side LN. Seg MP is an altitude on side LN.

iv. To draw altitude from vertex N, extend side LM of the triangle from point M with dashed line, as shown in the figure, and then draw the perpendicular line from vertex N.

v. Now for the orthocentre, as all the altitudes do not intersect we'll have to extend them so that they can meet giving us an orthocentre of the triangle.

vi. Hence, extend the altitude LQ, from point Q MP from point M, and NR from point R.

vii. The ortho centre of the Obtuse triangle lies outside the triangle.

viii. The point O denotes the orthocentre of the obtuse-angled ∆LMN.

Draw a right angled ∆XYZ. Draw its medians and show their point of concurrence by G.

To draw an right angled ∆XYZ.

i. Draw a base line of any length, mark it YZ. At Y draw a right angle mark that line point X. Join X and Z points. ΔXYZ thus formed is right angled triangle.

ii. Find the mid-point A of side YZ, by constructing the perpendicular bisector of the line segment YZ. Draw AX.

iii. Find the mid-point B of side XZ, by constructing the perpendicular bisector of the line segment XZ. Draw seg BY.

iv. Find the mid-point C of side XY, by constructing the perpendicular bisector of the line segment XY. Draw seg CZ.

Seg AX, seg BY and seg CZ are medians of ∆XYZ.

Their point of concurrence is denoted by G.

Draw an isosceles triangle. Draw all of its medians and altitudes. Write your observation about their points of concurrence.

i. Draw an isosceles triangle and name it as PQR.

An isosceles triangle is that triangle whose base is the side which is not equal to the other two sides or An isosceles triangle is a triangle which has two equal sides.

ii. Now, mark the mid-point i.e., A, B, C, of all the sides of the triangle and join it with the opposite vertex i.e., P, Q, R. The line segment i.e., PA, QB, RC hence found are the median of the triangle.

iii. Mark the point of concurrence as 'O'.

iv. Again, draw perpendicular line segment from each vertex.

v. Mark the point of concurrence X.

Here we see that both the point of concurrence of medians and altitudes coincides.

In the case of isosceles triangle, the two sides that are equal meet at a vertex, that lies directly above the midpoint of the base. Because of this, the altitude that runs from P to the base intersects the base at its midpoint, making it the median from P to the base as well, which is same for the other two sides also.

Therefore, in an isosceles triangle, the altitude and median are the same line segment, which is shown through the bold line in the above-given figure.

Point G is the centroid of ∆ABC.

(1) If l(RG) = 2.5 then l(GC) = .

(2) If l(BG) = 6 then l(BQ) = .

(3) If l(AP) = 6 then l(AG) = . and l(GP) = .

1) If then, as we know that the centroid divides each median in the ratio 2:1.

2) If then , as we know that the centroid divides each median in the ratio 2:1.

Since we have to find I(BQ), and from the figure it can be seen that,

Therefore, I(BQ) = 6 + 3

3) If then and l(GP) = 2, as we know that the centroid divides each median in the ratio 2:1 --------(i)

The Center of Gravity is the same as the centroid when the density is the same throughout.

Center of gravity, center of mass and centroid are all the same for simple solids.

They are often marked by a cross or dot and sometimes the letters CG or just G For a torus the centroid is at the very center
(even though there is no part of the torus there!) For a right solid cone the centroid is on
the center line and ¼ of the way from the base The center of gravity of a car can be very hard to figure out, as there is lots of empty space and materials of different density (such as the engine vs the seats).

When doing calculations we can often replace an object with its center of gravity.

### Example: You drop a hammer!

It may spin a little, but its center of gravity will fall straight down.

It also drops faster and faster due to gravity.

(The only complication is air resistance, which affect its motion more as it goes faster.)

A force that goes through the center of gravity won't cause any rotation. In fact you can balance an object by supporting it directly below its center of gravity.

## Make and Find a Centroid!

You can learn to find the centroid, and prove to yourself that it really is the center of gravity (CG) of the triangle, using a piece of sturdy cardboard (like poster board or chipboard), a ruler, pencil, and scissors.

Use the ruler to draw out any kind of triangle you want: acute, right, obtuse. In every triangle, the centroid is always inside the triangle!

Measure and locate the midpoint of each side of the triangle. Mark the midpoint clearly. Connect the three midpoints with their opposite vertices. Those lines are the medians.

Where the medians cross is the centroid. Cut out the triangle carefully. Hold it over your index finger, so the centroid is on the tip of your finger. Let go with the other hand. The triangle should balance perfectly!

Construct the centroid of  Δ ABC whose sides are AB = 6cm, BC = 7cm,  and AC = 5cm

Draw  ΔABC using the given measurements. Construct the perpendicular  bisectors of any two sides  (AC and BC) to find the  mid points D and E of AC  and BC respectively. Draw the medians AE and  BD and let them meet at G. The point G is the centroid of the given  ΔABC.

Construct  Δ ABC whose sides are AB = 6cm, BC = 4cm and AC = 5.5cm and locate  its orthocentre.

Draw  ΔABC using the given measurements. Construct altitudes from any two  vertices (A and C) to their opposite  sides (BC and AB respectively). The point of intersection of the altitudes H is the orthocentre of the given  Δ ABC.

Where is the orthocenter located in each type of triangle ?

Draw an example of each type of triangle and locate its orthocenter. Δ ABC is an acute triangle. The three altitudes intersect at G, a point inside the triangle

ΔKLM  is a right triangle. The two legs LM and KM, are also altitudes. They intersect at the triangle's right angle. This implies that the orthocenter is on the triangle at M, the vertex of the right angle of the triangle

Δ YPR is an obtuse triangle. The three lines that contain the altitudes intersect at W, a point that is outside the triangle.

Find the coordinates of the centroid of  ΔJKL shown below. We know that the centroid is two thirds of the distance from each vertex to the midpoint of the opposite side.

Choose the median KN. Find the coordinates of N, the midpoint of JL.

Find the distance from vertex K to midpoint N. The distance from K(5, 2) to N(5, 8) is 8 - 2, or 6 units.

Determine the coordinates of the centroid, which is 2/3 ⋅6  or  4 units up from vertex K along the median KN.

Hence, the coordinates of centroid P are Apart from the stuff given above, i f you need any other stuff in math, please use our google custom search here.

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## Euler Line

In any non-equilateral triangle the orthocenter (H), the centroid (G) and the circumcenter (O) are aligned. The line that contains these three points is called the Euler line. In an equilateral triangle all three centers are in the same place.

The relative distances between the triangle centers remain constant.

Distances between centers:

It is true that the distance from the orthocenter (H) to the centroid (G) is twice that of the centroid (G) to the circumcenter (O). Or put another way, the HG segment is twice the GO segment: When the triangle is equilateral, the barycenter, orthocenter, circumcenter, and incenter coincide in the same interior point, which is at the same distance from the three vertices.

This distance to the three vertices of an equilateral triangle is equal to from one side and, therefore, to the vertex, being h its altitude (or height).