If G is the centroid of triangle ABC, Prove that Area of triangle GAB = 1/3 Area of triangle ABC.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Given `Delta ABC` with centroid G, prove that `"Area"Delta GAB=1/3"Area"Delta ABC` .

The centroid of a triangle is the intersection of the medians. A median is the segment drawn from a vertex of the triangle to the midpoint of the side opposite the vertex.

Let M be the midpoint of `bar(AC)` , N the midpoint of `bar(BC)` and O the midpoint of `bar(AB)` . Let P be the intersection of `bar(MN),bar(CO)` .

(1)The proof is easy if you know that the centroid divides each median into 2 segments in the ratio of 2:1 with the segment from the vertex longer than the segment to the side. Assuming you know this, then use the formula for the area of a triangle `"Area"=1/2bh` and let the base be `bar(AB)` . The the area of `Delta ABC=1/2(AB)(CO)` and the area of `Delta GAB=1/2(AB)(GO)` where `GO=1/3CO` . Then the proof is complete.

(2) If you do not know this, we can use similar triangles:

Recall that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half the length of the third side.

So `Delta CMN ` is similar to `Delta CAB` by AA similarity. Also `Delta GAB` is similar to `Delta GMN` by AA similarity. Both triangles have sides in ratio 2:1.

Then GM=2AN, GN=2BM, GO=2GP

GO+GP+CP=CO But `CP=1/2CO` so

GO+GP=1/2CO But GO=2GP so

3GP=1/2CO ==> GP=1/6CO ==>GO=1/3CO

Now `"Area" Delta CAB=1/2(AB)(CO)` and `"Area" Delta GAB=1/2(AB)(GO)=1/2(AB)(1/3CO)=1/3"Area"Delta CAB` as required.

Approved by eNotes Editorial
An illustration of the letter 'A' in a speech bubbles

Let's assume that vertices of triangle have following coordinates:

`A=(x_A,y_A)`, `B=(x_B,Y_B)`, `C=(x_C,y_C)`.

Since centroid is, in a way, a generalisation of arithmetic mean it's coordinates are:

`G=(1/3(x_A+x_B+x_C),1/3(y_A+y_B+y_C))`.

If we have coordinates of triangles vertices we can calculate its area by following formula:

`T_(ABC)=1/2|(x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y_A)|` that is area of triangle ABC.

Let's calculate area of triangle ABG:

`T_(ABG)=1/2|(x_A-1/3x_A-1/3x_B-1/3x_C)(y_B-y_A)-`

` ` `(x_A-x_B)(1/3y_A+1/3y_B+1/3y_C-y_A)|=`

`1/2|(2/3x_A-1/3x_B-1/3x_C)(y_B-y_A)-(x_A-x_B)(-2/3y_A+1/3y_B+1/3y_C)|`

`=1/2|2/3x_Ay_B-2/3x_Ay_A-1/3x_By_B+1/3x_By_A-1/3x_Cy_B+1/3x_Cy_A`

`+2/3x_Ay_A-1/3x_Ay_B-1/3x_Ay_C-2/3x_By_A+1/3x_By_B+1/3x_By_C|=`

`1/2|1/3x_Ay_B-1/3x_By_A-1/3x_Cy_B+1/3x_Cy_A-1/3x_Ay_C+1/3x_ByC|=`

` 1/3[1/2|(x_A-x_C)(y_B-y_A)-(x_A-x_B)(y_C-y_A)|] =1/3T_(ABC)`

If you look at the begining of this calculation and at the end you'll see that this proves that area of triangle ABG is 1/3 of the area of triangle ABC. 

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial