In triangle ABC, medians AD, BE, and CF intersect at the centroid G. What is the ratio of the area of triangle GAB to the area of triangle ABC?

Correct option is B

​Given :

In triangle ABC, medians AD, BE and CF intersect at the centroid G.

Formula Used :
​$$\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$$​​

Solution :

The centroid G divides each median in the ratio 2 : 1 from the vertex.

Hence, the perpendicular distance (height) of G from base AB is one–third of the perpendicular distance of vertex C from base AB.

Since triangles GAB and ABC have the same base (AB), their areas are proportional to their heights.

​$$\frac{\text{Area of } \triangle GAB}{\text{Area of } \triangle ABC}$$​​

=$$\frac{\text{Height from } G}{\text{Height from } C}$$​​

= $$\frac{1}{3}$$​​
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