If AD is median of $△ A B C$ and P is a point on AC such that $a r (△ A D P) : a r (△ A B D) = 2 : 3$ , then $a r (△ P D C) : a r (△ A B C)$ is

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Solution

AD is the median of $△ A B C,$ P is a a point on AC such that $a r (△ A D P) : a r (△ A B D) = 2 : 3$
Let area of $△ A D P = 2 x^{2}$
Then area of $△ A B D = 3 x^{2}$

But area of $△ A B D = \frac{1}{2} a r e a △ A B C$
$∴ A r e a △ A B C = 2 \times a r e a o f △ A B D = 2 \times 3 x^{2} = 6 x^{2}$
and area of $△ P D C = a r (△ A D C) - a r (△ A D P) = a r (△ A B D) - a r (△ A D P) = 3 x^{2} - 2 x^{2} = x^{2} ∴ R a t i o = x^{2} : 6 x^{2} = 1 : 6$

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