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Properties of Parallelograms

If A D is med...

Question

If AD is median of ΔABC and P is a point on AC such that area (ΔADP) : area (ΔABD) = 2 : 3, then ar(ΔPDC) : ar(ΔABC) is

3 : 5

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2 : 5

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1 : 5

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1 : 6

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Solution

The correct option is D
1 : 6

$Let us draw a line P Q passing through A parallel to B C . Since A D is a median, B D = D C ∴ Δ A B D and Δ A D C have equal bases and they are also between same parallels (P Q and B C) . ∴ A r e a (Δ A B D) = A r e a (Δ A D C) . Let the area be 3 x . Now, A r e a (Δ A B D) : A r e a (Δ A D P) = 3 : 2 \Rightarrow 3 x : A r e a (Δ A D P) = 3 : 2 \Rightarrow A r e a (Δ A D P) = 2 x \Rightarrow A r e a (Δ P D C) : A r e a (Δ A B C) = (A r e a (Δ A D C) - A r e a (Δ A D P)) : (A r e a (Δ A B D) + A r e a (Δ A D C)) = (3 x - 2 x) : (3 x + 3 x) = 1 : 6$

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Similar questions

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

A D

△ A B C

P

A C

Area (△ A D P) : Area (△ A B D) = 2 : 3

Area (△ P D C) : Area (△ A B C) = 1 : a

a

A D

△ A B C

P

A C

a r (A D P) : a r (A B D) = 2 : 3

a r (P D C) : a r (A B C)

If AD is median of ΔABC and P is a point on AC such that ar(ΔADP) : ar(ΔABD) = 2 : 3, then ar(ΔPDC) : ar(ΔABC) is ______ .

A D

△ A B C

P

A C

a r (△ A D P)

a r (△ A B D)

= 2 : 3.

a r (△ P D C)

a r (△ A B C) .

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