Example 1. Consider the triangle formed by the medians of a given triangle. Prove that this triangle will have an area equal to three-fourths the area of the given triangle.
Example 2. The length of three medians of a triangle are 9 cm, 12 cm and 15 cm. Find the area of the triangle.
Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own. Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.
Solution 1. Let the given triangle be ABC, whose medians are AD, BE and CF.
Construction. Construct two lines FG (= BE) parallel to the median BE and CG (= AD) parallel to the median AD.
Proof. Note that EF = BD = CD and AD || CG. Therefore, AGEF is a parallelogram. Hence H is the mid-point of AE and FG.
- Since CF is a median of the triangle ABC, we have area of ΔAFC = (Area of ΔABC)/ 2.
- Since CH is a median of the triangle CFG, we have area of ΔCHF = (Area of ΔCFG)/ 2.
- Since FH is a median of the triangle AFE, we have area of ΔAFH = (Area of ΔAFC)/ 4.
Hence, area of ΔCHF
= area of ΔAFC ̶ area of ΔAHF = area of ΔAFC ̶ (area of ΔAFC)/ 4
= 3(area of ΔAFC)/ 4.
Therefore, area of ΔAFC : area of ΔCHF = area of ΔAFC : 3(area of ΔAFC)/ 4 = 4 : 3.
Multiplying both sides by 2, it follows that area of ΔABC : area of ΔCGF = 4 : 3, as required.
Alternatively, we can prove this result using vectors. Let ABC be the triangle and let AB = c, BC = a and CA = b. Vector area of △ABC is given by
Again, the three medians are given by
It follows that AD + BE + CF = 0. This means that they can also be treated as the sides of another triangle. Now, area of the triangle formed by the medians
Therefore, area of ∆ABC is equal to 4/3 times the area of the triangle formed by the medians.
Solution 2. Consider the triangle ABC, whose medians are AD, BE and CF. It is given that AD = 9 cm, BE = 12 cm and CF = 15 cm. Let the medians meet at the point G, centroid of the triangle.
We know that, centroid of a triangle divide the median in the ratio 2 : 1.
Since G is the point of trisection, we have
- GD =1/3 of AD =1/3 × 9 cm = 3 cm.
- GE =1/3 of BE = 1/3 ×12 cm = 4 cm.
- GF =1/3 of CF = 1/3 ×15 cm = 5 cm.
Now, produce AD up to S such DS = GD = 3 cm and join S to B and C.
In triangle CGS, we have
- GS = GD + DS = 3 + 3 = 6 cm.
- CG = 2/3 of CF = 2/3 × 15 cm = 10 cm.
- CS = BG = 2/3 of BE = 2/3 × 12 cm = 8 cm.
Hence, we can find the area of ΔCGS. We have s = (6 + 8 + 10)/2 = 12 cm. Area of ΔCGS
Area of triangle CDG = 1/2 × area of triangle CGS = 12 cm2.
Finally, we know that the three medians divide the triangle into six smaller triangles of equal area. Hence area of triangle ABC = 6 × area of triangle CDG = 72 cm2.
Alternatively, we can use the following result.
Result. If u, v and w are the lengths of the medians of the triangle, then the area of the triangle is given by
Proof. Here 2s = u + v + w, so that s = (u + v + w)/2. Required area of the triangle with sides u, v and w is
In view of Example 1, area of the original triangle is
Here, we can take u = 9 cm, v = 12 cm and w = 15 cm. Using this relation, area of the required triangle
Here, the length of three medians are m1 = 9 cm, m2 = 12 cm and m3 = 15 cm, say. Let 2s = m1 + m2 + m3 = 36 cm, so that s = 18 cm. If the three medians of any triangle is given, it’s area will be given by the formula