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Question Number 55044 by behi83417@gmail.com last updated on 16/Feb/19

Commented by behi83417@gmail.com last updated on 16/Feb/19

$a s s h o w n i n p i c t u r e :$ $f i n d l o c u s o f c e n t r o i d o f t r i a n g l e :$ $1) w h e n : B a n d C, a r e f i x e d a n d : A m o v e s$ $o n t h e {c i r c l e}^{'} s p e r i m e t e r .$ $2) C, i s f i x e d a n d A a n d B a r e m o v i n g o n$ ${c i r c l e}^{'} s p e r i m e t e r, a n d p e r i m e t e r o f$ $t r i a n g l e i s c o n s t a n t a n d e q u a i l s t o : 5 .$
Commented by mr W last updated on 17/Feb/19

$p l e a s e c h e c k t h e q u e s t i o n 2) . {i t}^{'} s n o t$ $p o s s i b l e t o h a v e a t r i a n g l e w i t h$ $p e r i m e t e r 5 i n a n c i r c l e w i t h r a d i u s 1 .$
	Answered by mr W last updated on 17/Feb/19

	Commented by mr W last updated on 17/Feb/19

	$(1)$ $r = C G = \frac{2}{3} R \cos θ = \frac{2}{3} \cos θ$ $o r$ $x = \frac{2}{3} \cos^{2} θ = \frac{\cos 2 θ + 1}{3}$ $y = \frac{2}{3} \cos θ \sin θ = \frac{\sin 2 θ}{3}$ $\Rightarrow {(x - \frac{1}{3})}^{2} + y^{2} = {(\frac{1}{3})}^{2}$ $\Rightarrow t h e l o c u s i s a c i r c l e w i t h r a d i u s \frac{1}{3}$
	Commented by mr W last updated on 17/Feb/19

	$(2)$ $t h e l o c u s o f G i s a l s o a c i r c l e .$
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