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Triangles between Same Parallels

ABCD is a par...

Question

$A B C D$ is a parallelogram. $X a n d Y$ are mid-points of $B C a n d C D$ respectively. Then area $(Δ A X Y) = \frac{3}{k}$ area(parallelogram $A B C D$ ).Find k

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Solution

$G i v e n : A B C D i s a p a r a l l e l o g r a m . X a n d Y a r e m i d - p o i n t s o f B C a n d C D r e s p e c t i v e l y . S i n c e X a n d Y a r e t h e m i d - p o i n t s o f s i d e s B C a n d C D r e s p e c t i v e l y i n △ B C D ∴ X Y ∥ B D a n d X Y = \frac{1}{2} B D . A r e a (△ C Y X) = \frac{1}{4} a r e a (△ D B C) = \frac{1}{8} a r e a (p a r a l l e l o g r a m A B C D) . . . . (1) [S i n c e a r (△ D B C) = \frac{1}{2} a r e a (p a r a l l e l o g r a m (A B C D)] P a r a l l e l o g r a m A B C D a n d △ A B X a r e b e t w e e n t h e s a m e p a r a l l e l s A D a n d B C a n d B X = \frac{1}{2} B C ∴ a r e a (△ A B X) = \frac{1}{4} a r e a (p a r a l l e l o g r a m A B C D) . . . . . (2) S i m i l a r l y, a r e a (△ A Y D) = \frac{1}{4} a r e a (p a r a l l e l o g r a m A B C D) N o w, a r (△ A X Y) = a r (p a r a l l e l o g r a m A B C D) - [a r (△ A B X) + a r (△ A Y D) + a r (△ C Y X)] \Rightarrow a r (△ A X Y) = a r (p a r a l l e l o g r a m A B C D) - (\frac{1}{4} + \frac{1}{4} + \frac{1}{8}) a r (A B C D) a r (A X Y) = (1 - \frac{5}{8}) a r (p a r a l l e l o g r a m A B C D) = \frac{3}{8} a r (p a r a l l e l o g r a m A B C D) .$

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