Consider-quadrilateral-ABCD-same-as-in-Q-1378-with-same-conditions-restrictions-Pl-refer-the-Question-again-What-could-be-possible-minimum-and-maximum-area-of-the-quadrila

Question Number 1395 by Rasheed Soomro last updated on 28/Jul/15

C o n s i d e r \boldsymbol quadrilateral \boldsymbol ABCD s a m e a s i n Q 1378 w i t h

s a m e c o n d i t i o n s / r e s t r i c t i o n s (P l r e f e r t h e Q u e s t i o n a g a i n) .

∙ W h a t c o u l d b e p o s s i b l e \boldsymbol minimum a n d \boldsymbol maximum \boldsymbol area

o f t h e q u a d r i l a t e r a l ?

∙ W h e n q u s d r i l a t e r a l h a s \boldsymbol minimum \boldsymbol area w h a t i s t h e v a l u e / s

o f \boldsymbol m ∠ \boldsymbol A ? S i m i l a r l y w h a t i s v a l u e / s o f \boldsymbol m ∠ \boldsymbol A i n c a s e o f

\boldsymbol maximum \boldsymbol area ?

Commented by prakash jain last updated on 28/Jul/15

minimum area \to 0

maximum area \to {(\frac{a + b + c + d}{4})}^{2}

m ∠ A for \max area \to \frac{π}{2}

Commented by Rasheed Soomro last updated on 29/Jul/15

I n o r d e r t o l e a r n I p u t m y q u e s t i o n s (r a i s e d i n m y m i n d)

b e f o r e y o u . I w a n t t o b e s a t i s f i e d, s o \boldsymbol Please \boldsymbol reply .

m a x i m u m a r e a \to {(\frac{\boldsymbol a + \boldsymbol b + \boldsymbol c + \boldsymbol d}{4})}^{2}

∙ Y o u h a v e u s e d s y m b o l^{'} \to^{'} i n a b o v e . D o e s t h i s m e a n^{'} t e n d s t o^{'} ?

∙ C o u l d w e r e p l a c e i t w i t h^{'} =^{'} e q u a l l y ?

∙ D o e s t h e a b o v e e x p r e s s i o n c a n b e u s e d a s a f o r m u l a f o r

t h e a r e a a \boldsymbol quadrilateral \boldsymbol having \boldsymbol one \boldsymbol angle \boldsymbol of \frac{\boldsymbol π}{2} \boldsymbol rad . I f y e s

t h e n i t s h o u l d b e u s e d f o r a r e a o f r e c t a n g l e o f \boldsymbol x \boldsymbol by \boldsymbol y .

B u t a b o v e f o r m u l a g i v e s t h e a r e a (\frac{x + y}{2}) i n s t e a d o f x y!

I n o t h e r w o r d s :

B e i n g q u a d r i l a t e r a l m a x i m u m a r e a o f p a r a l l e l o g r a m i s

a r e a o f r e c t a n g l e . I f t h e d i m e n s i o n s o f ∥ g r a m a r e

x a n d y t h e n m a x i m u m a r e a s h o u l d b e x y . B u t t h e a b o v e

f o r m u l a g i v e s t h e r e s u l t \frac{x + y}{2}!

Commented by Rasheed Soomro last updated on 29/Jul/15

I t h i n k a s \boldsymbol a r e a \to 0, \boldsymbol m ∠ \boldsymbol A \to 0 o r \boldsymbol π

Commented by prakash jain last updated on 29/Jul/15

You can have area \to 0 with two opposite

side a and c, a \neq b near 0 .

Commented by Rasheed Ahmad last updated on 31/Jul/15

M a x i m u m a r e a o f p a r l l e l o g r a m

w i t h d i m e n s i o n s x a n d y i s t h e

a r e a o f r e c t a n g e w i t h s a m e

d i m e n s i o n s t h a t i s x y i f I a m r i g h t .

B u t y o u r f o r m u l a {(\frac{a + b + c + d}{4})}^{2}

g i v e s {(\frac{x + y}{2})}^{2} . P l e a s e e x p l a i n

a s I w a n t t o l e a r n f r o m y o u .