$A D$ is a median of $△ A B C$ and $P$ is a point on $A C$ such that $Area (△ A D P) : Area (△ A B D) = 2 : 3$ and $Area (△ P D C) : Area (△ A B C) = 1 : a$ . Then the value of $a$ is:

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Solution

Given, $A D$ is a median of $△ A B C$ and $P$ is a point on $A C$ such that $Area (△ A D P) : Area (△ A B D) = 2 : 3$
$\Rightarrow \frac{Area (△ A D P)}{Area (△ A B D)} = \frac{2}{3}$
$\Rightarrow Area (△ A D P) = 2 x$ and $Area (△ A B D) = 3 x$
Again $A D$ being the median divides the $△ A B C$ in two triangles of equal area.
$\Rightarrow Area (△ A B D) = Area (△ A D C) = 3 x$
Now, $Area (△ A D C) = Area (△ A D P) + Area (△ D P C)$
or, $3 x = 2 x + Area (△ P D C)$
or, $Area (△ P D C) = x$
Again, $Area (△ A B C) = Area (△ A B D) + Area (△ A D C) = 3 x + 3 x = 6 x$
Therefore, $Area (△ P D C) : Area (△ A B C) = x : 6 x = 1 : 6$

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