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Choose the correct answer:

In a parallelogram ABCD, P and Q are the mid – points of BC and CD then the area of the triangle ΔAPQ will be how many times the area of parallelogram.

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12

14

18

38

answer is D.

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Detailed Solution

The correct answer is option 4.
On graphing,

To find the area use:

A (A B C D) = A (Δ A P Q) + A (Δ A B P) + A (Δ A D Q) + A (Δ C P Q)

We know,

A r e a o f t r i a n g l e = 12 (b a s e) (h e i g h t) A r e a o f p a r a l l e l o g r a m = (b a s e) (h e i g h t)

Let us assume that the base of parallelogram be 2b and height be h.
Using

A r e a o f p a r a l l e l o g r a m = (b a s e) (h e i g h t)

A (A B C D) = (2 b) (h) ⇒ A (A B C D) = 2 b h

Consider

Δ A B P

Point P is the mid – point of BC which lies between the parallel lines.

P F = D E 2 ⇒ P F = h 2

Also,

A r e a o f t r i a n g l e = 12 (b a s e) (h e i g h t)

A (Δ A B P) = 12 (A B) (P F) ⇒ A (Δ A B P) = 12 (2 b) (h 2) ⇒ A (Δ A B P) = b h 2

Consider

Δ A Q D

A (Δ A Q D) = 12 (D Q) (D E) ⇒ A (Δ A Q D) = 12 (b) (h) ⇒ A (Δ A Q D) = b h 2

Consider

Δ C P Q

Now,

P Q ∥ B C

Δ C P Q

and

Δ C D B

∠ P C Q = ∠ B C Q (∵ c o m m o n a n g l e) ⇒ ∠ C Q P = ∠ C D B (∵ P Q ∥ B C) ⇒ ∠ C P Q = ∠ C B D (∵ P Q ∥ B C)

The above two triangles are similar by AAA.
Now,

A Δ C P Q A Δ C B D = C Q C D 2

⇒ A (Δ C P Q) = (b 2 b) 2 (A (Δ C B D)) ⇒ A (Δ C P Q) = A (Δ C B D) 4

To get the area of

Δ C B D

A (Δ C B D) = 12 (2 b) (h) = b h

On substituting the values

A Δ C P Q = b h 4

A (A B C D) = A (Δ A P Q) + A (Δ A B P) + A (Δ A D Q) + A (Δ C P Q) ⇒ 2 b h = A (Δ A P Q) + b h 2 + b h 2 + b h 4 ⇒ A (Δ A P Q) = 2 b h − (b h 2 + b h 2 + b h 4)

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Choose the correct answer:In a parallelogram ABCD, P and Q are the mid – points of BC and CD then the area of the triangle ΔAPQ will be how many times the area of parallelogram.

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Choose the correct answer:

In a parallelogram ABCD, P and Q are the mid – points of BC and CD then the area of the triangle ΔAPQ will be how many times the area of parallelogram.