**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 Δ*AFC*= (Area of Δ*ABC*)/ 2. - Since
*CH*is a median of the triangle*CFG*, we have area of Δ*CHF*= (Area of Δ*CFG*)/ 2. - Since
*FH*is a median of the triangle*AFE*, we have area of Δ*AFH*= (Area of Δ*AFC*)/ 4.

Hence, area of Δ

CHF= area of Δ

AFC̶ area of ΔAHF= area of ΔAFC̶ (area of ΔAFC)/ 4= 3(area of Δ

AFC)/ 4.

Therefore, area of Δ*AFC* : area of Δ*CHF* = area of Δ*AFC* : 3(area of Δ*AFC*)/ 4 = 4 : 3.

Multiplying both sides by 2, it follows that area of Δ*ABC* : area of Δ*CGF* = 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 Δ*CGS*. We have *s* = (6 + 8 + 10)/2 = 12 cm. Area of Δ*CGS*

Area of triangle *CDG* = 1/2 × area of triangle *CGS* = 12 cm^{2}.

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 cm^{2}.

**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 2*s* = *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 *m*_{1} = 9 cm, *m*_{2} = 12 cm and *m*_{3} = 15 cm, say. Let 2*s* = *m*_{1} + *m*_{2} + *m*_{3} = 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

Excellent sir

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