Question : The medians CD and BE of a triangle ABC interest each other at O. The ratio $ar(\triangle ODE): ar(\triangle ABC)$ is equal to:
Option 1: $12:1$
Option 2: $4:3$
Option 3: $3:4$
Option 4: $1:12$
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Correct Answer: $1:12$
Solution :
In a triangle, the medians divide each other into segments in the ratio $2:1$, with the centroid (the point of intersection of the medians) closer to the vertex of the triangle.
In $\triangle ABC$, if $BE$ and $CD$ are medians intersecting at $O$, then $OB:OE = 2:1$
$\triangle ODE\sim\triangle ODE$
The ratio of their areas is the square of the ratio of their corresponding sides.
$\frac{ar(\triangle ODE)}{ar(\triangle BCO)}=\frac{OE^2}{OB^2}=\frac{1^2}{2^2}$
$⇒\frac{ar(\triangle ODE)}{ar(\triangle BCO)}=\frac{1 }{4}$
$⇒ar(\triangle BCO)=4×ar(\triangle ODE)$ $[\because ar(\triangle BCO)=\frac{1 }{3}×ar(\triangle ABC)]$
$⇒\frac{1 }{3}×ar(\triangle ABC)=4×ar(\triangle ODE)$
$\therefore ar(\triangle ABC)=12×ar(\triangle ODE)$
Hence, the correct answer is $1:12$.
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