Question : Three medians AD, BE, and CF of $\triangle ABC$ intersect at G. The area of $\triangle ABC$ is $36\text{ cm}^2$. Then the area of $\triangle CGE$ is:
Option 1: $12\text{ cm}^2$
Option 2: $6\text{ cm}^2$
Option 3: $9\text{ cm}^2$
Option 4: $18\text{ cm}^2$
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Correct Answer: $6\text{ cm}^2$
Solution :
Given: AD, BE, and CF are medians of $\triangle ABC$ intersect at G.
The area of $\triangle ABC$ = $36\text{ cm}^2$
The median of the triangle divides the triangle into equal areas.
$\therefore$ Area of $\triangle CGE$ = $ \frac{\text{Area of} \triangle ABC }{6}=\frac{36}{6}=6\text{ cm}^2$
Hence, the correct answer is $6\text{ cm}^2$.
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