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Question Number 14797 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

Commented by mrW1 last updated on 05/Jun/17

$w e k n o w w h e n t h e t o p p o i n t o f a$ $t r i a n g l e i s m o v e d p a r a l l e l t o t h e b a s e$ $l i n e, t h e a r e a o f t h e t r i a n g l e r e m a i n s$ $u n c h a n g e d .$ $w e m o v e t h e p o i n t E i n t h i s w a y s u c h$ $t h a t i t l i e s o n t h e l i n e B D .$ $l e t M = i n t e r s e c t i o n p o i n t o f A C a n d B D .$ $A_{Δ A C E} = \frac{1}{2} \times A C \times M B$ $= \frac{1}{2} \times D B \times (\frac{B D}{2} - B E)$ $= \frac{1}{2} \times (D E + B E) \times (\frac{D E + B E}{2} - B E)$ $= \frac{1}{2} \times (10 + 4) \times (\frac{10 + 4}{2} - 4)$ $= 21$ $i n g e n e r a l, w i t h p = D E a n d q = B E$ $A_{Δ A E C} = \frac{(p + q) (p - q)}{4}$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

$A B C D, s q u a r e . D E = 10, E B = 4 .$ $. . . . . . . . . . . . . \sqrt{. . . . . . . . . . . . .} . . . . . . . . . . \sqrt{. . . . . . . . . . . . . .}$ $S_{A E C} = ?$
Commented by RasheedSoomro last updated on 04/Jun/17

$AB = ?$
Commented by ajfour last updated on 04/Jun/17

Commented by ajfour last updated on 04/Jun/17

$C a t t h e t o p c o r n e r, A a t t h e$ $b o t t o m . O r i g i n o f c o o r d i n a t e s$ $i s c e n t e r o f s q u a r e ., a x e s a l o n g$ $d i a g o n a l s o f s q u a r e .$ $F u r t h e r l e t s i d e o f s q u a r e = a \sqrt{2}$ $E q u a t i o n o f c i r c l e s :$ ${(x + a)}^{2} + y^{2} = 100$ ${(x - a)}^{2} + y^{2} = 16$ $s u b t r a c t i n g w e g e t x c o o r d i n a t e$ $o f p o i n t s o f i n t e r s e c t i o n :$ $4 a x = 84 \Rightarrow a x = 21$ $S_{△ A E C} = \frac{1}{2} (2 a) x = a x = 21 .$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

$h e l l o m r A j f o u r .$ $i t h i n k e q u a t i o n s o f c i r c l e s s h o u l d$ $b e a s b e l o w :$ ${(x + \frac{a \sqrt{2}}{2})}^{2} + y^{2} = 100$ ${(x - \frac{a \sqrt{2}}{2})}^{2} + y^{2} = 16 .$
Commented by ajfour last updated on 04/Jun/17

$y e s, i n s o l v i n g i a s s u m e d$ $d i a g o n a l o f s q u a r e a s 2 a .$ $a n s w e r s h o u l d b e 21 s t i l l . .$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

$w h e n y o u a s s u m e s q u a r e s i d e = a$ $t h e n d i a g o n a l l e n g t h = a \sqrt{2} . (n o t : 2 a)$
Commented by ajfour last updated on 04/Jun/17

$t h a n k s, b u t j u s t n o w i e d i t e d,$ $s i d e o f s q u a r e a s a \sqrt{2}, d i a g o n a l$ $l e n g t h i s o f c o u r s e 2 a t h e n .$
Commented by mrW1 last updated on 05/Jun/17

$A_{A E C} = \frac{1}{2} \times 14 \times (\frac{14}{2} - 4) = 7 \times 3 = 21$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$s o b e a u t i f u l a n d n i c e!$ $d e a r m r W 1 . p l e a s e s h o w u s : h o w ?$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$n i c e a n d b e a u t i f u l . t h a n k a l o t m y m a s t e r .$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Jun/17

	$s q u a r e s i d e s = a$
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