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Question Number 4971 by Rasheed Soomro last updated on 28/Mar/16

Commented by Rasheed Soomro last updated on 28/Mar/16

$W r i t e p r o c e s s a l s o .$
Commented by Rasheed Soomro last updated on 29/Mar/16

$H o w ?$
Commented by Yozzii last updated on 29/Mar/16

$o n e - q u a r t e r t h e a r e a o f t h e s q u a r e$
Commented by Yozzii last updated on 29/Mar/16

$T h e l i n e s f r o m t h e m i d p o i n t s o n t h e s i d e s o f t h e s q u a r e$ $t r i s s e c t t h e h y p o t e n u s e o f t h e s m a l l e r t r i a n g l e . I w a s$ $t r y i n g t o u p l o a d m y a n s w e r a s p i c s$ $b u t i t f a i l e d t o u p l o a d .$
Commented by Yozzii last updated on 29/Mar/16

Commented by Yozzii last updated on 30/Mar/16

$T h e i n n e r t r i a n g l e s a r e r e f l e c t i o n s o f$ $e a c h o t h e r i n t h e d i a g o n a l . T h e d i a g o n a l$ $o f t h e s q u a r e i s a l s o t r i s s e c t e d . S o$ $S T o n e t h i r d A C . S T, A C, R Q a n d N J$ $a r e p a r a l l e l . S o, t h e a r e a y o u w a n t$ $i s m a d e u p o f a r e a s o f t w o c o n g r u e n t$ $t r a p e z i a . △ R O Q i s i s o s c e l e s s o$ $t h e l i n e f r o m O t h a t f a l l s o n R Q$ $n o r m a l l y b i s s e c t s t h e l i n e R Q .$ $U s i n g t h i s, t h e l e n g t h O M i s f o u n d$ $t o b e \frac{a \sqrt{2}}{4} . L i k e w i s e, B F = \frac{a \sqrt{2}}{4} s i n c e$ $△ R O Q ≊ △ B N J . S o, M F = O B - (B F + O M)$ $M F = a \sqrt{2} - \frac{2 a \sqrt{2}}{4} = \frac{a \sqrt{2}}{2} .$ $\frac{M F}{2} i s t h e^{'} {h e i g h t}^{'} o f e a c h t r a p e z i u m .$ $Y o u s e e i n m y d i a g r a m t h a t L K = \frac{1}{3} R Q = \frac{a \sqrt{2}}{6}$ $a n d S T = \frac{a \sqrt{2}}{3} .$ $∴ A_{t} = \frac{1}{2} (L K + S T) \frac{M F}{2} + \frac{1}{2} (U V + S T) \frac{M F}{2}$ $A_{t} = \frac{M F}{4} (L K + U V + 2 S T)$ $A_{t} = \frac{a \sqrt{2} / 2}{4} (2 L K + 2 S T)$ $A_{t} = \frac{a \sqrt{2}}{4} (\frac{a \sqrt{2}}{6} + \frac{a \sqrt{2}}{3})$ $A_{t} = \frac{2 a^{2}}{4} (\frac{9}{18}) = \frac{a^{2}}{4} .$
Commented by Yozzii last updated on 30/Mar/16

$T h e o r i g i n a l s o l u t i o n I w a n t e d t o s h a r e$ $i s a b i t l e n g t h y b e c a u s e w i t h o n l y a s s u m i n g$ $t h a t t h e f i g u r e i s a s q u a r e a n d t h a t$ $t h e l a r g e i n n e r t r i a n g l e s h a v e v e r t i c e s$ $a s 2 s i d e m i d - p o i n t s a n d 1 v e r t e x o f t h e$ $s q u a r e, I h a d t o p r o v e e v e r y t h i n g$ $e l s e, s u c h a s p a r a l l e l i s m a n d t r i s s e c t i o n .$ $I t {w a s n}^{'} t s p e c i f i e d i f t h e f i g u r e w a s a r e c t a n g l e$ $o r s q u a r e o r i f t h o s e v e r t i c e s a r e$ $r e a l l y m i d p o i n t s o f t h e s i d e s o f t h e$ $f i g u r e . O n e c o u l d a t t e m p t t h e$ $g e n e r a l c a s e w h e r e t h e v e r t i c e s o f$ $t h e t r i a n g l e c o m e f r o m o n e r e c t a n g l e$ $v e r t e x a n d t w o p o i n t s t h a t l i e a n y w h e r e$ $o n t w o s i d e s o f t h e r e c t a n g l e t h a t d o n o t c o n t a i n t h e$ $r e c t a n g l e v e r t e x .$
	Answered by Rasheed Soomro last updated on 30/Mar/16

	Commented by Rasheed Soomro last updated on 30/Mar/16

	$A l l s m a l l t r i a n g l e s h a v e e q u a l$ $a r e a . S o m e t r i a n g l e s a r e c o n g r u e n t$ $a l s o . T h e r e a r e s o m e t r i a n g l e s, w h i c h a r e$ $n o t c o n g r u e n t b u t d e s p i t e t h i s t h e y h a v e e q u a l$ $a r e a, b e c a u s e t h e y h a v e e q u a l b a s e s$ $a n d a l t i t u d e s . F o r e x a m p l e :$ $△ A I Q ≇ △ I Q M$ $B u t t h e s e t r i a n g l e s h a v e e q u a l b a s e s$ $A Q = Q M a n d c o m m o n a l t i t u d e I V$ $a n d h e n c e e q u a l a r e a .$ $N o w n u m b e r o f t r i a n g l e s i n w h o l e$ $i s 24,$ $N u m b e r o f t r i a n g l e s i n s h a d e d a r e a$ $i s 6 .$ $∴ T h e s h a d e d a r e a i s \frac{6}{24} = \frac{1}{4} p a r t o f t h e$ $w h o l e .$
	Commented by Yozzii last updated on 30/Mar/16

	$N i c e .$
	Commented by FilupSmith last updated on 31/Mar/16

	$A mazing and simple to understand!$
	Answered by Rasheed Soomro last updated on 31/Mar/16

	Commented by Rasheed Soomro last updated on 31/Mar/16

	$A n o t h e r a n s w e r :$ $H, I, J & K a r e m i d p o i n t s o f t h e$ $s i d e s o f t h e ◻ ABCD a n d i t c a n$ $b e s h o w n t h a t :$ $◼ HIJK = \frac{1}{2} o f ◼ ABCD (◼ f o r a r e a (I)$ $o f s q u a r e)$ $N o w,$ $◻ HIJK c o n t a i n s 24 s m a l l t r i a n g l e s$ $a n d$ $p o l y g o n PQRSTU c o n t a i n s 12 s m a l l$ $t r i a n g l e s .$ $S o,$ $A r e a PQRSTU = \frac{12}{24} = \frac{1}{2} o f ◼ HIJK (II)$ $F r o m (I) & (II)$ $A r e a PQRSTU = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} o f ◼ ABCD$
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