Mathematics Education

EMAT 6680, Professor Wilson

Exploration Number 6, Triangle of Medians by Ursula Kirk

1.      Construct a triangle and its medians.

On Triangle ABC, we can see medians: k, l and m. These medians were be found by constructing a segment from the vertex of the triangle to the midpoint of the opposite side. The three medians of triangle ABC meet at point O, which is the centroid of the triangle.

2.      Construct a second triangle with the three sides having the lengths of the three medians from your first triangle.

Here, we have our original triangle ABC and a second triangle FDC made out the medians of triangle ABC.

3.       Find some relationship between the two triangles.

·                     Relationship Number 1

When triangle ABC is equilateral, the triangle of medians FDC is also equilateral.

·         Relationship Number 2

When triangle ABC is isosceles, the triangle of medians FDC is also isosceles.

·         Relationship Number 3

When triangle ABC is a right triangle, the triangle of medians FDC is not always a right triangle.

There is one case where the resulting triangle FDC is a right triangle.

4.      Show the area of the triangle of medians is 3/4 the area of the original triangle.

The area of triangle ABC is 21.51 and the area of triangle FDC is 16.14. We know that the area of triangle FDC has to be ¾ the area of triangle ABC. Therefore:

Area FDC =

5.      Proof the area of the triangle of medians is 3/4 the area of the original triangle.

We must show that the area of triangle CDF is 3/4 the area of triangle ABC. Since CDBG is a parallelogram, then triangle CDB and triangle ACD have congruent areas.

Area of Triangle CDO =

Area of Triangle CDB =

Since OF || DA and OF bisects AB then OF also bisects DB

Then BO = JF and therefore OC = . Next we substitute   for OC

The area of triangle CDO =  [  (BC) ] h

The area of triangle CDB =  (BC) h

Because of the definition of medians, we know that the area of triangle ABC is twice the area of triangle BDC

Area of triangle ABC = 2 [ (BC)] h

Area of triangle FCD is twice the area of triangle CDO.

Area of triangle CDO = 2 [ *  (BC) ] h

After cancelling out the fractions:

Area of triangle CDF =  [ BC * h ]

Area of triangle ABC = BC * h

If those 2 formulas are placed in a ratio the [BC * h] cancels out leaving us with the ratio of areas of triangle CDF and ABC to be