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Question Number 182676 by yaslm last updated on 12/Dec/22

	Answered by Acem last updated on 12/Dec/22

	$I f w e c o n s i d e r t h a t t h e s e q u a r e {i s n}^{'} t a s p e c i a l$ $c a s e o f t h e r e c t a n g l e, t h e n t h e r e q u i r e d r e c t a n g l e s$ $Missing \left or extra \right$
	Answered by TheSupreme last updated on 14/Dec/22
	$(a,b are the sided) ab=A 2(a+b)=p A=p ab=2(a+b) b(a−2)=2a b = ((2a)/(a−2)) {(a,b)∈R_+ ^2 : b=((2a)/(a−2))}$
	$(a, b a r e t h e s i d e d)$ $a b = A$ $2 (a + b) = p$ $A = p$ $a b = 2 (a + b)$ $b (a - 2) = 2 a$ $b = \frac{2 a}{a - 2}$ ${(a, b) \in R_{+}^{2} : b = \frac{2 a}{a - 2}}$
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