A (2 × 3) rectangle and a (3 × 4) rectangle are contain within a square without over laping at any inferior point , and the sides of the square are parallel to the sides of the two given rectangles. what is the smallest possible area of the square.


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Question Number 7519 by Tawakalitu. last updated on 01/Sep/16

$A (2 \times 3) r e c t a n g l e a n d a (3 \times 4) r e c t a n g l e a r e c o n t a i n w i t h i n$ $a s q u a r e w i t h o u t o v e r l a p i n g a t a n y i n f e r i o r p o i n t, a n d t h e$ $s i d e s o f t h e s q u a r e a r e p a r a l l e l t o t h e s i d e s o f t h e t w o g i v e n$ $r e c t a n g l e s . w h a t i s t h e s m a l l e s t p o s s i b l e a r e a o f t h e s q u a r e .$
Commented by Rasheed Soomro last updated on 02/Sep/16

$25$
Commented by Yozzia last updated on 02/Sep/16

$I t h i n k 5^{2} a l s o .$
Commented by Rasheed Soomro last updated on 02/Sep/16

$S u m o f s m a l l e r s i d e s o f t h e r e c t a n g l e s$ $s h o u l d b e m e a s u r e o f t h e s i d e o f r e q u i r e d s q u a r e$ $(i n c a s e m a x i m u m o f b i g g e r s i d e s i s l e s s o r e q u a l$ $t o t h e m e n t i o n e d s u m)$ $S o i n t h i s c a s e 2 + 3 = 5 i s t h e m e a s u r e o f t h e s i d e$ $o f t h e s q u a r e .$ $H e n c e a r e a o f t h e s q u a r e i s o f c o u r s e 5^{2} = 25 .$
Commented by Tawakalitu. last updated on 03/Sep/16

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